Broadcast Routing(dynamic)

In some applications hosts need to send messages to many (or) all other hosts like weather reports, stock market updates (or) live radio programs.

 i.e, sending a packet to all destinations simultaneously is called Broadcasting.

 Different methods of Broadcasting:-

  • first method is to send a packet to all destinations. This is a method wasteful of Band width and source needs to know the complete list of all destinations.

so this is least desirable one.

  • flooding is another way to broadcast a packet, the problem with flooding is that it generates too many packets and also consumes too much of Band width.
  • Third way is to use multi destination routing

In this technique each packet contains a list of destinations (or) a bit map for those destinations.

when a packet arrives at a router,  the router checks all the output lines it requires. The router generates a new copy of the packet for each output line after sufficient number of hops each packet will carry only one destination.

i.e, multi destination routing is like separately addressed packets (to B,C,D,E & D) must follow the same route one of them pays full fare and rest are free.

  • The fourth type of method is to use sink tree (or) spanning tree.

A spanning tree is a subset of subnet that includes all the routers but contains no loops.

if each router knows which of it’s lines belong to spinning tree then it broadcasts packet to all the lines except the one it arrived on.

This is efficient method in terms of Band width usage but problem is to maintain the knowledge of all the nodes of spanning tree at a routes.

  • Last method is to use Reverse path forwarding to approximate behavior of spanning tree.

Consider a subnet and it’s sink tree for router I as root node and how reverse path algorithm works in figure (C) 

on the first hop I sends packets to F, H, J & N. on the second hop eight packets are generated among them 5 are given to preferred paths indicated as circles (A,D,G,O,M)

of the 6 packets generated in third hop only 3 are given to preferred paths (C,E & K) the others are duplicates.

in the fourth hop to B and L after this broadcasting terminates.

advantages of reverse path forwarding:-

  • it is easy to implement.
  • it does not require routers to known about spanning trees.
  • it does not require any special mechanism to stop the process (as like flooding).

The principle is : if a packet arrives on a line if it is preferred one to reach source it gets forwarded.

if it arrives on a line that is not preferred one that packet is discarded as a duplicate.


when a packet arrives at ‘L’ the preferred paths are N and P so it forward the packets to both N and P and if a packet arrives at ‘K’ , there the preferred path is M and N is not preferred so it forwards packet to M and discards to N.

This is reverse path forwarding.

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Asynchronous Transfer Model(ATM)


Asynchronous Transfer Model is another important connection oriented Network.

Why we call it asynchronous is most of  the transmission in telephone systems is synchronous (closed tied to a clock) but ATM is not such type.

ATM was designed in 1990’s, it was the cell ray protocol designed by the ATM forum and was adopted by ITU-T.

Design goals:-

  1.  A technology is required that provides large data rates with the high data rate transmission media available (Optical Fiber Communication) and this media requires less susceptible to noise.
  2. The system must interface with existing systems to provide wide-area inter connectivity.
  3. The cost for such a system should not be more.
  4. The new system must be connection-oriented type.
  5. The new system must be able to work with the existing tele-communication hierarchies like local loops, long distance carriers etc.

The problems associated with existing networks:-

The design goals come into picture for ATM, since there exists some problems that are associated with the existing systems.

Frame Networks:-

Before ATM we have data communications at DLL are based on frame switching and frame networks .

i.e, different protocols use frames of varying size (frame has data and header). If header size is more than that of actual data  there is a burden so some protocols have enlarged the size of data unit relative to the header.

if there is no data in such cases there is a wastage , so there is to provide variable frame sizes to the users.

Mixed N/w Traffic:-

If there exists variable frame sizes

  1. The switches Multiplexers and routers must require an elaborate Software to manage variable size frames.
  2. Internet working among different frame N/w ‘s become slow and expensive too.

suppose we have two networks generating frames of variable sizes that is N/W 1 is connected to line 1 and the frame is X. N/W 2is connected to line 2 and of having 3 frames of equal sizes A,B,C are connected to a TDM.

If X has arrived a bit earlier than A,B,C (having more priority than X) on the output line . The frames has to wait for a time to move on to the output line, this causes delay for line 2 N/W.

i.e, Audio and video frames are small so mixing them with conventional data traffic often creates unacceptable delays and makes shared frame links unusable for audio and video information.

but we need to send all kinds of traffic over the same links.

Cell Networks:-

so a solution to frame internet working is by adopting a concept called cell networking.

In a cell N/W we use a small data unit of fixed size called cell so all types of data are loaded into identical cells and are multiplexed with other cells and are routed through the cell N/W.

because each cell is small and of same size the problems associated with multiplexing different sized frames are avoided.

Asynchronous TDM:-

ATM  uses asynchronous TDM- hence the name Asynchronous Transfer Model.

i.e, it multiplexes data coming from different channels. it also uses fixed size slots called cells.

ATM Mux’rs fill a slot with a cell from any input channel that has a cell and slot is empty if there is no cell.

ATM architecture:-

ATM was going to solve all the world’s networking and tele-communications problems by merging voice, data, cable TV, telex,telegraph…… and everything else into a single integrated system that could do everything for everyone.

i.e, ATM was much successful than OSI and is now widely used in telephone system for moving IP packets.

ATM is a cell-switched N/W the user access devices are connected through a user-to- N/W interface (UNI) to the switches inside the N/W. The switches are connected through N/W-to-N/W interface (NNI) as shown in the following figure

Virtual Connection:-

two end points is accomplished through transmission paths (TP’s), Virtual Paths (VP’s) and Virtual Circuits (VC’s)

ATM Virtual Circuits:-

Since ATM N/w’s are connection-oriented, sending data requires a connection , first sending a packet to setup the connection.

  • as setup packet travels though the sub net all the routers on the path make an entry in their internal tables noting for existence and reserving the resources.
  • connections are often called virtual circuits and most ATM N/W’s support permanent virtual circuits.  i.e, for permanent connections b/w two hosts.
  • after establishing a connection either side can transmit data.
  • all information is in small, fixed size packets called cells.
  • cell routing is done in Hard ware at high speed.
  • fixed size cells makes the building of Hard ware routers easier with short, fixed length cells.
  • variable length IP packets have to be routed by Software which is a slower process.
  • ATM uses the Hardware that can setup to copy one incoming cell to multiple output lines (ex:-TV).
  • All cells follow the same route to the destination.
  • cell delivery is not guaranteed but their order is.
  • if cells lost along the way it is up to higher protocol levels to recover from lost cells but this also not guarantee.
  • ATM N/W’s are organized like traditional WAN’s with lines and switches.
  • the most common speeds for ATm are
  • 155 Mbps-used for high definition TV.
  • 155.52 Mbps-used for AT & T’s SONET transmission system
  • 622 Mbps-4 X 155 Mbps channels can be sent over it.



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Parallel Polarization

Parallel Polarizaton:-

Parallel polarization means \overrightarrow{E} field lies in the XZ-plane (y=0) that is the plane of incidence, the figure illustrates the case of parallel polarization

then the incident and reflected fields is given by

\overrightarrow{E_{is}} = E_{i}(-\sin \theta_{i }\overrightarrow{a_{z}}+ \cos \theta_{i }\overrightarrow{a_{x}} )e^{-j(\overrightarrow{k}.\overrightarrow{r})}

\overrightarrow{E_{is}} = E_{i}(-\sin \theta_{i }\overrightarrow{a_{z}}+ \cos \theta_{i }\overrightarrow{a_{x}} )e^{-j(x\overrightarrow{a_{x}}+y\overrightarrow{a_{y}}+z\overrightarrow{a_{z}}).(k_{i}\cos \theta _{i}\overrightarrow{a_{z}}+k_{i}\sin \theta _{i}\overrightarrow{a_{x}})}

\overrightarrow{E_{is}} = E_{i}(-\sin \theta_{i }\overrightarrow{a_{z}}+ \cos \theta_{i }\overrightarrow{a_{x}} )e^{-j(k_{i}\cos \theta _{i}z+xk_{i}\sin \theta _{i})}

\overrightarrow{E_{is}} = E_{i}(-\sin \theta_{i }\overrightarrow{a_{z}}+ \cos \theta_{i }\overrightarrow{a_{x}} )e^{-j\beta _{1}(z\cos \theta _{i}+x\sin \theta _{i})}

\overrightarrow{H_{is}} = \frac{E_{i}}{\eta _{1}}e^{-j\beta _{1}(z\cos \theta _{i}+x\sin \theta _{i})}\overrightarrow{a_{y}}

now the reflected wave is 

\overrightarrow{E_{rs}} = E_{r}(\sin \theta_{r }\overrightarrow{a_{z}}+ \cos \theta_{r }\overrightarrow{a_{x}} )e^{-j\beta _{1}(-z\cos \theta _{r}+x\sin \theta _{r})}

\overrightarrow{H_{rs}} =\frac{ H_{r}}{\eta _{1}}e^{-j\beta _{1}(-z\cos \theta _{r}+x\sin \theta _{r})}\overrightarrow{a_{y}}

since k_{i}=k_{r}=\beta _{1} =\omega \sqrt{\mu_{1} \epsilon_{1} }

let’s find out  \overrightarrow{k}  then by using the equation \overrightarrow{k}.\overrightarrow{E_{s}} =0\overrightarrow{H_{s}} = \frac{k}{\omega \mu }X \overrightarrow{E_{s}} = \overrightarrow{a_{k}}X \frac{E_{s}}{\eta } .

and the transmitted fields in the second  medium is  

\overrightarrow{E_{ts}} = E_{t}(-\sin \theta_{t }\overrightarrow{a_{z}}+ \cos \theta_{t}\overrightarrow{a_{x}} )e^{-j\beta _{2}(z\cos \theta _{t}+x\sin \theta _{t})}

\overrightarrow{H_{ts}} = \frac{H_{t}}{\eta _{2}}e^{-j\beta _{2}(z\cos \theta _{t}+x\sin \theta _{t})}\overrightarrow{a_{y}}

where \beta _{2} = \omega \sqrt{\mu _{2}\epsilon _{2}}.

Transmission coefficient:-

as \theta _{r} = \theta _{i} and also that the tangential components of \overrightarrow{E} and \overrightarrow{H} are continuous at the boundary z=0.


(E_{i}+E_{r}) \cos \theta _{i} = E_{t}\cos \theta _{t}

H_{tan1}-H_{tan2} = \overrightarrow{J_{s}} , H_{tan1} = H_{tan2}  since \overrightarrow{J_{s}} =0

\frac{E_{i}}{\eta _{1}}-\frac{E_{r}}{\eta _{1}}=\frac{E_{t}}{\eta _{2}}------EQN(1)

E_{i}+E_{r}=E_{t}\frac{\cos \theta _{t}}{\cos \theta _{i}}------EQN(2)

(1)+(2) implies 

2E_{i}=E_{t}(\frac{\eta _{1}}{\eta _{2}}+\frac{\cos \theta _{t}}{\cos \theta _{i}})

2E_{i}=E_{t}\frac{(\eta _{1}\cos \theta _{i}+\eta _{2}\cos \theta _{t})}{\eta _{2}\cos \theta _{i}}

\tau _{parallel} = \frac{E_{t}}{E_{i}} = \frac{2\eta _{2}\cos \theta _{i}}{(\eta _{1}\cos \theta _{i}+\eta _{2}\cos \theta _{t})}.

Reflection coefficient:-

from EQN (1)

\frac{E_{i}}{\eta _{1}}-\frac{E_{r}}{\eta _{1}}=\frac{E_{t}}{\eta _{2}}------EQN(1)

\frac{E_{i}}{\eta _{1}}-\frac{E_{r}}{\eta _{1}}= \frac{2E_{i}\eta _{2}\cos \theta _{i}}{\eta _{2}(\eta _{1}\cos \theta _{i}+\eta _{2}\cos \theta _{t})}

after simplification

E_{r}\frac{(\eta _{2}\cos \theta _{t}-\eta _{1}\cos \theta _{i})}{(\eta _{1}\cos \theta _{i}+\eta _{2}\cos \theta _{t})}=E_{i}

\rho _{parallel} = \frac{E_{r}}{E_{i}}=\frac{(\eta _{2}\cos \theta _{t}-\eta _{1}\cos \theta _{i})}{(\eta _{1}\cos \theta _{i}+\eta _{2}\cos \theta _{t})}

1+\rho _{parallel} = \tau _{parallel}(\frac{\cos \theta _{t}}{\cos \theta _{i}}).


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Surface impedance

At high frequencies, the current is almost confined to a very thin sheet at the surface of the conductor which is used in many applications.

The  surface impedance may be defined as the ratio of the tangential component of the electric field \overrightarrow{E_{tan}} at the surface of the conductor to the current density (linear) \overrightarrow{J_{s}} which flows due to this electric field.

given as Z_{s} (or) \eta _{s} = \frac{\overrightarrow{E_{tan}}}{\overrightarrow{J_{s}}}.

\overrightarrow{E_{tan}}   is the Electric field strength parallel to and at the surface of the conductor.

and \overrightarrow{J}  is the total linear current density which flows due to \overrightarrow{E_{tan}}.

The \overrightarrow{J_{s}} represents the total conduction per meter width flowing in this sheet.

Let us consider a conductor of the type plate, is placed at the surface y=0 and the current distribution in the y-direction is given by 


Assume that the depth of penetration (\delta) is very much less compared with the thickness of the conductor.

J_{s}= \int_{0}^{\infty } \overrightarrow{J}.\overrightarrow{dy}

J_{s}= \int_{0}^{\infty } J_{o}e^{-\gamma y}dy

J_{s}= J_{o}(e^{-\gamma y})_{0}^{\infty }

J_{s}= \frac{J_{o}}{\gamma }

from ohm’s law \overrightarrow{J_{o}} = \sigma \overrightarrow{E_{tan}}  

E = \frac{J_{o}}{\sigma } .

then  \eta _{s} = \frac{\gamma }{\sigma } .

Z_{s}  (or)  \eta _{s} = \frac{\gamma }{\sigma } .

we know that \gamma = \sqrt{j\omega \mu (\sigma +j\omega \epsilon )}  

for good conductors \sigma > > \omega \epsilon .

then \gamma \approx \sqrt{j\omega \mu \sigma } 

\eta _{s} = \frac{\gamma }{\sigma } = \sqrt{\frac{j\omega \mu }{\sigma }} .

therefore the surface impedance of a plane good conductor which is very much thicker than the skin depth is equal to the characteristic impedance of the conductor.

This impedance is also known s input impedance of the conductor when viewed as transmission line conducting energy into the interior of metal.

when the thickness of the plane conductor is not greater compared to the depth of penetration , reflection of wave occurs at the back surface of the conductor.




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oblique incidence

when a uniform plane wave  is incident obliquely (making an angle \theta _{i} other than 90^{o}) to the boundary between the two media then it is known as oblique incidence.

Now consider the situation that is more general case  that is the oblique incidence.

In this case the EM wave (incident wave) not strikes normally the boundary. i.e,  the incident wave is not propagating  along any standard axes (like x,y and z).

Therefore EM wave is moving in a random direction then the general form is    \overrightarrow{E} = E_{o} \cos (\omega t-\overrightarrow{k}.\overrightarrow{r})  

it is also in the form \overrightarrow{E} = E_{o} \cos (\overrightarrow{k}.\overrightarrow{r}-\omega t).

then \overrightarrow{k} = k_{x}\overrightarrow{a_{x}}+k_{y}\overrightarrow{a_{y}}+k_{z}\overrightarrow{a_{z}} is called the wave number vector (or) the propagation vector.

and \overrightarrow{r} = x\overrightarrow{a_{x}}+y\overrightarrow{a_{y}}+z\overrightarrow{a_{z}} is called the position vector (from origin to any point on the plane of incidence) , then the magnitude of \overrightarrow{k} is related to \omega according to the dispersion.

k^{2} = k_{x}^{2}+k_{y}^{2}+k_{z}^{2} = \omega ^{2}\mu \epsilon

\overrightarrow{k} X \overrightarrow{E} = \omega \mu \overrightarrow{H}.

\overrightarrow{k} X \overrightarrow{H} = -\omega \epsilon \overrightarrow{E}.

\overrightarrow{k} . \overrightarrow{H} =0.

\overrightarrow{k} . \overrightarrow{E} =0.

i.    \overrightarrow{E},\overrightarrow{H} and \overrightarrow{k} are mutually orthogonal.

ii.  \overrightarrow{E}  and  \overrightarrow{H} lie on the plane \overrightarrow{k} . \overrightarrow{r} = k_{x}x+k_{y}y+k_{z}z=constant.

then the \overrightarrow{H} field corresponding to \overrightarrow{E} field is \overrightarrow{H} = \frac{1}{\omega \mu } (\overrightarrow{k} X \overrightarrow{E}) = \frac{\overrightarrow{a_{k}} X \overrightarrow{E}}{\eta }.

Now choose oblique incidence of a uniform plane wave at a plane boundary.

the plane defined by the propagation vector \overrightarrow{k} and a unit normal vector \overrightarrow{a_{n}} to the boundary is called the plane of incidence.

the angle \theta _{i} between \overrightarrow{k} and \overrightarrow{a_{n}} is the angle of incidence.

both the incident and reflected waves are in medium 1 while the transmitted wave is in medium 2 .


\overrightarrow{E_{i}} =E_{i}\cos (k_{ix}x+k_{iy}y+k_{iz}z-\omega _{i}t)

\overrightarrow{E_{r}} =E_{i}\cos (k_{rx}x+k_{ry}y+k_{rz}z-\omega _{r}t)

\overrightarrow{E_{t}} =E_{t}\cos (k_{tx}x+k_{ty}y+k_{tz}z-\omega _{t}t).

the wave propagates 

  1.     \omega _{i}=\omega _{r}=\omega _{t}=\omega.
  2.     k_{ix} = k_{rx}=k_{tx}=k_{x}.
  3.    k_{iy} = k_{ry}=k_{ty}=k_{y}.

(1) indicates that all waves are propagating with same frequency. (2) and (3) shows that the tangential components of propagation vectors be continuous.

k_{i} \sin \theta _{i}=k_{r} \sin \theta _{r}  implies  k_{i} = k_{r} =\beta _{1} =\omega \sqrt{\mu _{1}\epsilon _{1}}    since \theta _{r}=\theta _{i}.

k_{i} \sin \theta _{i}=k_{t} \sin \theta _{t}   implies k_{t} =\beta _{2} =\omega \sqrt{\mu _{2}\epsilon _{2}}.

\frac{\sin \theta _{t}}{sin \theta _{i}}=\sqrt{\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}}}

now velocity u=\frac{\omega }{k}.

then from Snell’s law       r_{1}\sin \theta _{i} = r_{2}\sin \theta _{t},  where r_{1} and r_{2}  are the refractive indices of the two media.




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Poynting theorem


Poynting theorem is used to get an expression for propagation of energy in a medium.

It gives the relation between energy stored in a time-varying magnetic field and the energy stored in time-varying electric field and the instantaneous power flow out of a given region.

EM waves propagates through space from source to destination. In order to find out power in a uniform plane wave it is necessary to develop a power theorem for the EM field known as poynting theorem.

The direction of power flow is perpendicular to \vec{E} and \overrightarrow{H} in the direction of plane containing \overrightarrow{E} and \overrightarrow{H}.

i.e, it gives the direction of propagation .

\overrightarrow{P} = \overrightarrow{E}X\overrightarrow{H}  Watts/m2  (or)   VA/m2.


from Maxwell’s  equations \overrightarrow{\bigtriangledown } X \overrightarrow{H} = \overrightarrow{J}+\overrightarrow{J_{d}}

\overrightarrow{\bigtriangledown } X \overrightarrow{H} = \overrightarrow{J}+\frac{\partial \overrightarrow{D}}{\partial t}

\overrightarrow{J}=\overrightarrow{\bigtriangledown } X \overrightarrow{H}-\frac{\partial \overrightarrow{D}}{\partial t}

the above equation has units of the form current density Amp/m2. When it gets multiplied by \overrightarrow{E} V/m. The total units  will  have of the form power per unit volume.

\overrightarrow{J}\rightarrow Amp/m2\overrightarrow{E}\rightarrow Volts/m.

EJ\rightarrow Amp. Volt/m3 \rightarrow Watts/m3 \rightarrow Power/volume.

\overrightarrow{\bigtriangledown } X \overrightarrow{H} = \overrightarrow{J}+\frac{\partial \overrightarrow{D}}{\partial t}

\overrightarrow{E}.(\overrightarrow{\bigtriangledown } X \overrightarrow{H}) = \overrightarrow{E}.\overrightarrow{J}+\overrightarrow{E}.\frac{\partial \overrightarrow{D}}{\partial t}

\overrightarrow{E}.(\overrightarrow{\bigtriangledown } X \overrightarrow{H}) = \overrightarrow{E}.\overrightarrow{J}+\overrightarrow{E}.\frac{\partial (\epsilon \overrightarrow{E})}{\partial t}

by using the vector identity 

\overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H}) = \overrightarrow{H}.(\overrightarrow{\bigtriangledown }X \overrightarrow{E} )-\overrightarrow{E}.(\overrightarrow{\bigtriangledown }X \overrightarrow{H} )


\overrightarrow{H}.(\overrightarrow{\bigtriangledown }X \overrightarrow{E} )-\overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H}) = \overrightarrow{E}.\overrightarrow{J}+\overrightarrow{E}.\frac{\partial (\epsilon \overrightarrow{E})}{\partial t}.

from the equation of Maxwell’s \overrightarrow{\bigtriangledown } X \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t} = -\frac{\partial (\mu \overrightarrow{ H})}{\partial t}

\overrightarrow{H}.(-\frac{\partial (\mu \overrightarrow{ H})}{\partial t} )-\overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H}) = \overrightarrow{E}.\overrightarrow{J}+\overrightarrow{E}.\frac{\partial (\epsilon \overrightarrow{E})}{\partial t}

\overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H})=-\overrightarrow{E}.\overrightarrow{J}-\overrightarrow{E}.\frac{\partial (\epsilon \overrightarrow{E})}{\partial t}-\overrightarrow{H}.(\frac{\partial (\mu \overrightarrow{ H})}{\partial t} )------EQN(I)

by using the vector identity  \frac{\partial (\overrightarrow{A}.\overrightarrow{B})}{\partial t}=\overrightarrow{A}.\frac{\partial \overrightarrow{B}}{\partial t}+\overrightarrow{B}.\frac{\partial \overrightarrow{A}}{\partial t}

if \overrightarrow{A} = \overrightarrow{B}    \Rightarrow \frac{\partial (\overrightarrow{A}.\overrightarrow{A})}{\partial t}=2 \overrightarrow{A}.\frac{\partial \overrightarrow{A}}{\partial t}

\overrightarrow{A}.\frac{\partial \overrightarrow{A}}{\partial t}=\frac{1}{2}\frac{\partial A^{2}}{\partial t}

from EQN(I)   ,  \overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H})=-\overrightarrow{E}.\sigma\overrightarrow{E }-\frac{1}{2}\epsilon\frac{\partial E^{2}}{\partial t}-\frac{1}{2}\mu \frac{\partial H^{2}}{\partial t}

\overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H})=-\sigma E^{2}-\frac{1}{2}\frac{\partial }{\partial t}(\epsilon E^{2}+\mu H^{2})

by integrating the above equation by over  a volume 

\int_{v}\overrightarrow{\bigtriangledown }.(\overrightarrow{E} X \overrightarrow{H}) dv=-\int_{v}\sigma E^{2}dv-\int_{v}\frac{1}{2}\frac{\partial }{\partial t}(\epsilon E^{2}+\mu H^{2})dv

by converting the volume integral to surface integral

\oint_{s}(\overrightarrow{E} X \overrightarrow{H}). \overrightarrow{ds}=-\int_{v}\sigma E^{2}dv-\int_{v}\frac{1}{2}\frac{\partial }{\partial t}(\epsilon E^{2}+\mu H^{2})dv.

the above equation gives the statement of Poynting theorem.

Poynting theorem:-  

It states that the net power flowing out of a given volume is equal to the time rate of decrease in the energy stored with in that volume V and the ohmic losses.






Lag compensator

Lag compensator:-

A Lag compensator has a Transfer function of the form G(s) = \frac{s+z_{c}}{s+p_{c}}------EQN(I)

G(s) = \frac{s+\frac{1}{\tau }}{s+\frac{1}{\beta \tau }},    where \beta =\frac{z_{c}}{p_{c}}> 1     and \tau > 0

Pole-Zero Plot of Lag compensator:-

i.e, the pole is located to the right of the zero.

Realization of Lag compensator as Electrical Network:-

The lag compensator can be realized by an electrical Network.

Assume impedance of source is zero  [Z_{s} =0] and output load impedance to be infinite [Z_{R}=\infty ] .

The transfer function is \frac{E_{o}(s)}{E_{i}(s)} = \frac{(R_{2}+\frac{1}{Cs})}{R_{1}+(R_{2}+ \frac{1}{Cs})}

\frac{E_{o}(s)}{E_{i}(s)} = \frac{R_{2}Cs+1}{(R_{1}+R_{2})Cs+1}

after simplification

\frac{E_{o}(s)}{E_{i}(s)} =\frac{R_{2}}{(R_{1}+R_{2})}(\frac{s+\frac{1}{R_{2}C}}{s+\frac{1}{(R_{1}+R_{2})C}})

after comparing the above equation with the transfer function of lag compensator has a zero at  Z_{c} =\frac{1}{R_{2}C}  and has a pole at p_{c}=\frac{1}{(R_{1}+R_{2})C}=\frac{1}{\beta \tau } .

from the pole \beta =\frac{(R_{1}+R_{2})}{R_{2}} and  \tau =R_{2}C.

therefore  the transfer function  has a zero at -\frac{1}{\tau }   and a pole at -\frac{1}{\beta \tau }.

\frac{E_{o}(s)}{E_{i}(s)} =\frac{1}{\beta }\frac{s+\frac{1}{\tau }}{s+\frac{1}{\beta \tau }} = (\frac{\tau s+1}{\beta \tau s+1})--------EQN(II).

the values of the three parameters R_{1} , R_{2}  and C are determined from the two compensator parameters \tau  and \beta.

using the EQN(II)   

 \tau =R_{1}C> 0,    \beta =\frac{(R_{1}+R_{2})}{R_{2}}> 1.

there is an additional degree of freedom in the choice of the values of the network components which is used to set the impedance level of the N/w.

the gain is \left | G(j\omega ) \right |=\left | \frac{E_{o}(j\omega )}{E_{i}(j\omega )} \right | = \left | \beta(\frac{1+j\omega \tau }{1+\beta \tau j\omega }) \right |

D.C gain at \omega =0  is \beta  which is greater than 1.

Let the zero-frequency gain as unity, then the Transfer function is G(j\omega ) = (\frac{1+j\omega \tau }{1+\beta \tau j\omega }).

Frequency-response of Lag compensator:-

Note:-“lag” refers to the property that the compensator adds positive phase to the system over some appropriate frequency range.

G(j\omega ) = (\frac{1+j\omega \tau }{1+\beta \tau j\omega }),   let  \beta =1.

the frequency response of lag compensator is \left | G(j\omega ) \right |= \sqrt{\frac{1+\omega ^{2}\tau ^{2}}{1+\omega ^{2}\beta ^{2}\tau ^{2}}}

at \omega =\frac{1}{\tau } \Rightarrow\left | G(j\omega ) \right |= \sqrt{\frac{2}{1+\beta ^{2}}}.

(1+j\omega \tau )\rightarrow has a slope +20 dB/decade with corner frequency \frac{1}{\tau }.

(1+\beta \tau j\omega)\rightarrow slope is -20 dB/decade with corner frequency \frac{1}{\beta \tau }.

\Phi = \angle G(j\omega )=tan^{-1}\omega \tau -tan^{-1}\beta \omega \tau

to find at which frequency the phase is minimum , differentiate \Phi w.r to \omega and equate it to zero.

\Phi = tan^{-1}(\frac{\omega \tau-\beta\omega \tau}{1+\beta \omega^{2} \tau^{2}})

\frac{d\Phi }{d\omega }=0

\frac{1}{1+(\frac{\omega \tau-\beta \omega \tau}{1+\alpha \omega^{2} \tau^{2}})^{2}}(\frac{((1+\beta \omega^{2} \tau^{2})\tau (1-\beta ))-(\omega \tau (1-\beta )2\omega \beta \tau ^{2})}{(1+\beta \omega^{2} \tau^{2})^{2}})=0

{((1+\beta \omega^{2} \tau^{2})\tau (1-\beta ))-(\omega \tau (1-\beta )2\omega \beta \tau ^{2})}=0

\tau (1-\beta )(1+\beta \omega^{2} \tau^{2}-2\omega^{2} \beta\tau ^{2})=0

\tau (1-\beta)(1-\omega^{2} \beta \tau ^{2})=0

\because \tau \neq 0    implies (1-\beta)=0\Rightarrow \beta =1   , which is invalid because \beta > 1.

(1-\omega^{2} \beta \tau ^{2})=0\Rightarrow \omega^{2} =\frac{1}{\beta \tau ^{2}}.

\omega =\frac{1}{\sqrt{\beta} \tau }  , at this \omega  lead compensator has minimum phase given by 

\Phi _{m} = tan^{-1}(\frac{1-\beta }{2\sqrt{\beta}})

tan \Phi _{m} = \frac{1-\beta }{2\sqrt{\beta}} implies sin \Phi _{m} = \frac{1-\beta}{1+\beta }.

\beta =\frac{1-sin \Phi _{m}}{1+sin \Phi _{m}}.

at \omega =\omega _{m} ,    \left | G(j\omega ) \right | = \frac{1}{\sqrt{\beta }}.

Choice of \beta :-

Any phase lag at the gain cross over frequency of the compensated system is undesirable.

To prevent the effects of lag compensator , the corner frequency of the lag compensator must be located substantially lower than the \omega _{gc} of compensated system.

In the high frequency range , the lag compensator has an attenuation of 20 log(\beta ) dB, which is used to obtain required phase margin.

The addition of a lag compensator results in an improvement in the ratio of control signal to noise in the loop.

high frequency noise signals are attenuated by a factor \beta > 1, while low-frequency control signals under go unit amplification (0 dB gain).

atypical value of \beta =10.

Procedure for bode-plot of a lead compensator:-

Step 1:- Sketch the Bode-plot of the uncompensated system with the gain k. Set the value of k according to the steady-state error requirement.

Measure the gain cross over frequency and the phase margin of uncompensated system.

Step 2:-  find \omega _{gc}^{'} at which phase angle of uncompensated system is 

-180^{o} + given Phase Margin+ \epsilon.

\epsilon =5^{o}(or)15^{o}   is a good assumption for phase-lag contribution.

Step 3:- find gain of the uncompensated system at \omega _{gc}^{'} and equate it to 20 log (\beta)  and then find \beta.

Step 4:- choose the upper corner frequency of the compensator to one octave to one decade  below \omega _{gc}^{'} and find \tau value.

Step 5:- Calculate phase lag of compensator  at \omega _{gc}^{'}, if it is less than \epsilon go to next step.

Step 6:- Draw the Bode plot of compensated system  to meet the desired specifications.

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Lead Compensator

Lead compensator:-

A Lead compensator has a Transfer function of the form G(s) = \frac{s+z_{c}}{s+p_{c}}------EQN(I)

G(s) = \frac{s+\frac{1}{\tau }}{s+\frac{1}{\alpha \tau }},    where \alpha =\frac{z_{c}}{p_{c}}< 1     and \tau > 0

Pole-zero plot of Lead compensator:-

i.e, the pole is located to the left of the zero.

  • A lead compensator speeds up the transient response and increases margin of stability of a system.
  • It also helps  to increase the system error constant through a limited range.

Realization of Lead compensator as an Electrical Network:-

The lead compensator can be realized by an electrical Network.

Assume impedance of source is zero  [Z_{s} =0] and output load impedance to be infinite [Z_{R}=\infty ] .

The transfer function is \frac{E_{o}(s)}{E_{i}(s)} = \frac{R_{2}}{R_{2}+(R_{1}|| \frac{1}{Cs})}

\frac{E_{o}(s)}{E_{i}(s)} = \frac{R_{2}}{R_{2}+\frac{(R_{1}\frac{1}{Cs})}{(R_{1}+\frac{1}{Cs})}}

\frac{E_{o}(s)}{E_{i}(s)} = \frac{R_{2}(R_{1}+\frac{1}{Cs})}{R_{2}(R_{1}+\frac{1}{Cs})+R_{1}\frac{1}{Cs}}

after simplification

\frac{E_{o}(s)}{E_{i}(s)} =\frac{s+\frac{1}{R_{1}C}}{s+\frac{1}{R_{1}C}+\frac{1}{R_{2}C}}

\frac{E_{o}(s)}{E_{i}(s)} =\frac{s+\frac{1}{R_{1}C}}{s+\frac{1}{(\frac{R_{2}}{R_{1}+R_{2}})R_{1}C}}  by comparing this equation with the transfer function of lead compensator has a zero at  Z_{c} =\frac{1}{R_{1}C}  and the pole is p_{c}=\frac{1}{(\frac{R_{2}}{R_{1}+R_{2}})R_{1}C} .

from the pole \alpha =\frac{R_{2}}{R_{1}+R_{2}} and  \tau =R_{1}C.

therefore  the transfer function  has a zero at -\frac{1}{\tau }   and a pole at -\frac{1}{\alpha \tau }.

\frac{E_{o}(s)}{E_{i}(s)} =\frac{s+\frac{1}{\tau }}{s+\frac{1}{\alpha \tau }} = \alpha (\frac{\tau s+1}{\alpha \tau s+1})--------EQN(II).

the values of the three parameters R_{1} , R_{2}  and C are determined from the two compensator parameters \tau  and \alpha.

using the EQN(II)   

 \tau =R_{1}C> 0,    \alpha =\frac{R_{2}}{R_{1}+R_{2}}< 1.

there is an additional degree of freedom in the choice of the values of the network components which is used to set the impedance level of the N/w.

the gain is \left | G(j\omega ) \right |=\left | \frac{E_{o}(j\omega )}{E_{i}(j\omega )} \right | = \left | \alpha (\frac{1+j\omega \tau }{1+\alpha \tau j\omega }) \right |

D.C gain at \omega =0  is \alpha  which is less than 1.

attenuation \frac{1}{\alpha } is used to determine the steady state performance.

while using a lead N/w , it is important to increase the loop gain by an amount of \frac{1}{\alpha }.

A lead compensator is visualized as a combination of a N/w and an amplifier.

Note:-“lead” refers to the property that the compensator adds positive phase to the system over some appropriate frequency range.

Frequency-response of a lead compensator:-

G(j\omega ) = \alpha (\frac{1+j\omega \tau }{1+\alpha \tau j\omega }),   let  \alpha =1.

the frequency response of lead compensator is 

\Phi = \angle G(j\omega )=tan^{-1}\omega \tau -tan^{-1}\alpha \omega \tau

to find at which frequency the phase is maximum , differentiate \Phi w.r to \omega and equate it to zero.

\Phi = tan^{-1}(\frac{\omega \tau-\alpha \omega \tau}{1+\alpha \omega^{2} \tau^{2}})

\frac{d\Phi }{d\omega }=0

\frac{1}{1+(\frac{\omega \tau-\alpha \omega \tau}{1+\alpha \omega^{2} \tau^{2}})^{2}}(\frac{((1+\alpha \omega^{2} \tau^{2})\tau (1-\alpha ))-(\omega \tau (1-\alpha )2\omega \alpha \tau ^{2})}{(1+\alpha \omega^{2} \tau^{2})^{2}})=0

{((1+\alpha \omega^{2} \tau^{2})\tau (1-\alpha ))-(\omega \tau (1-\alpha )2\omega \alpha \tau ^{2})}=0

\tau (1-\alpha )(1+\alpha \omega^{2} \tau^{2}-2\omega^{2} \alpha \tau ^{2})=0

\tau (1-\alpha )(1-\omega^{2} \alpha \tau ^{2})=0

\because \tau \neq 0    implies (1-\alpha )=0\Rightarrow \alpha =1   , which is invalid because \alpha < 1.

(1-\omega^{2} \alpha \tau ^{2})=0\Rightarrow \omega^{2} =\frac{1}{\alpha \tau ^{2}}.

\omega =\frac{1}{\sqrt{\alpha} \tau }   , at this \omega  lead compensator has maximum phase given by 

\Phi _{m} = tan^{-1}(\frac{1-\alpha }{2\sqrt{\alpha }})

tan \Phi _{m} = \frac{1-\alpha }{2\sqrt{\alpha }}  implies sin \Phi _{m} = \frac{1-\alpha }{1+\alpha }.

({1+\alpha })sin \Phi _{m} = {1-\alpha }

sin \Phi _{m}+\alpha sin \Phi _{m} = {1-\alpha }

\alpha =\frac{1-sin \Phi _{m}}{1+sin \Phi _{m}}.

at \omega =\omega _{m} ,    \left | G(j\omega ) \right | = \frac{1}{\sqrt{\alpha }}.

when there is a need for phase leads of more than 60^{o}, two cascaded lead networks are used where each N/w provides half of the required phase.

for phase leads more than 60^{o}\alpha decreases sharply and if single N/w is used \alpha will be too low.

Choice of \alpha :-

In choosing parameters of compensator \tau depends on R_{1} and C . The \tau value may be anything but for \alpha there is a constraint. It depends on inherent noise in Control systems.

from the lead N/w , it’s been observed that the high frequency noise is amplified by \frac{1}{\alpha } while low frequencies by unity.

more (or) less \alpha should not be less than 0.07.

Procedure for bode-plot of a lead compensator:-

Step 1:- Sketch the Bode-plot of the uncompensated system with the gain k. Set the value of k according to the steady-state error requirement.

Measure the gain cross over frequency and the phase margin of uncompensated system.

Step 2:- using the relation 

Additional phase lead required = specified phase margin- Phase Margin of uncompensated system.

\epsilon  is a margin of safety required by the fact that the gain cross over frequency will increase due to compensation.

for example :-  \epsilon =5^{o} is a good assumption for -40 dB/decade.

\epsilon =15^{o}  (or) 20^{o} \Rightarrow -60 dB/decade.

Step 3:- Set the maximum phase of the lead compensator    \Phi _{m} = Additional phase lead required  and compute \alpha =\frac{1-sin \Phi _{m}}{1+sin \Phi _{m}}.

Step 4:- Find the frequency at which the uncompensated system has a gain of -20 log(\frac{1}{\sqrt{\alpha }}) dB, which gives the new gain cross over frequency.

with \omega _{gc} as the gain cross over frequency the system has a phase margin of \gamma _{1}

where as with \omega _{gc}^{'} as the gain cross over frequency the system has a phase margin of \gamma _{2}

Step 5:- Now \omega _{gc}^{'} = \omega _{m} = \frac{1}{\tau \sqrt{\alpha }}.    find the value of \tau and the transfer function of lead compensator  \frac{1+j\omega \tau }{1+\alpha j\omega \tau }.

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Introduction to Root Locus

The introduction of a feedback to a system causes some instability , therefore an unstable system can not perform the control task requires of it.

while in the analysis of  a given system, the very first investigation that needs to be made is whether the system is stable or not?

However, the determination of stability of a system is necessary but not sufficient.

A stable system with low damping is also unwanted.

a design problem in which the designer is required to achieve the desired performance for a system by adjusting the location of its close loop poles in the S-plane by varying one (or) more system parameters.

The Routh’s criterion obviously does not help much in such problems.

for determining the location of closed-loop poles one may resort to the classical techniques of factoring the characteristic equation and determining it’s roots.

when the degree is higher (or) repeated calculations are required as a system parameter is varied for adjustments.

a simple technique, known as the root locus technique, for finding the roots of the ch.eqn introduced by W.R.Evans.

This technique provides a graphical method of plotting the locus of the roots in the S-plane as a given system parameter is varied from complete range of values (may be from zero to infinity).

The roots corresponding to a particular value of the system parameter can then be located on the locus (or) value of the parameter for a desired root location can be determined from the locus.

Root Locus:-

  • In the analysis and design for stable systems and gives information about transient response of control systems.
  • It gives information about absolute stability and relative stability of a system.
  • It clearly shows the ranges of stability and instability.
  • used for higher order differential equations.
  • value of k for a particular root location can be determined.
  • and the roots for a particular k can be determined using Root Locus.


ch. equation is 1+G(s)H(s)=0

let G(s)H(s)=D(s)



To find the whether the roots are on the Root locus (or) not

They have to satisfy ‘2’ criteria known as

  1. Magnitude Criterion.
  2. Angle Criterion.

Magnitude criterion:-

\left | D(s) \right |=1

\left | G(s)H(s) \right |=1

the magnitude criterion states that  s=s_{a}  will be a point on root locus, if for that value of s 

i.e, \left | G(s)H(s) \right |=1

Angle criterion:-

\angle D(s) = \angle G(s)H(s)=\pm 180^{o}(2q+1)

where q=0,1,2……….

if \angle D(s) = \pm 180^{o}(2q+1) is odd multiple of 180^{o}, a point s on the root locus, if \angle D(s) is odd multiple of at s=s_{a} of 180^{o}, then that point is on the root locus.

Root Locus definition:-

The locus of roots of the Ch. eqn in the S-plane by the variation of system parameters (generally gain k) from 0 to \infty is known as Root locus.

It is a graphical method 

-\infty to 0       \rightarrow     Inverse Root Locus

0  to \infty    \rightarrow  Direct Root Locus

generally Root Locus means Direct Root Locus.

D(s) = G(s)H(s) = k \frac{(s+z_{1})(s+z_{2})(s+z_{3})....}{(s+p_{1})(s+p_{2})(s+p_{3})....}

\left | D(s) \right | = k \frac{\left | s+z_{1} \right |\left | s+z_{2} \right |\left | s+z_{3} \right |....}{\left | s+p_{1} \right |\left | s+p_{2} \right |\left | s+p_{3} \right |....}

\left | D(s) \right | = k \frac{\prod_{i=1}^{m}\left | s+z_{i} \right |}{\prod_{i=1}^{n}\left | s+p_{i} \right |}

m= no .of zeros

n= no.of poles

from magnitude criterion \left | D(s) \right | = 1

k =\frac{\prod_{i=1}^{n}\left | s+p_{i} \right |}{\prod_{i=1}^{m}\left | s+z_{i} \right |}

The open loop gain k corresponding to a point s=s_{a} on Root Locus can be calculated 

k= product of length of vectors from open loop poles to the point s=s_{a}/product of length of vectors from open loop zeros to the point s=s_{a}.

from the Angle criterion,

\angle D(s) = \angle (s+z_{1})+\angle (s+z_{2})+\angle (s+z_{3})..... -\angle (s+p_{1})+\angle (s+p_{2})+\angle (s+p_{3}).....

\angle D(s) = \sum_{i=1}^{m}\angle (s+z_{i}) -\sum_{i=1}^{n}\angle (s+p_{i})

\sum_{i=1}^{m}\angle (s+z_{i}) -\sum_{i=1}^{n}\angle (s+p_{i})=\pm 180^{o}(2q+1)

i.e,( sum of angles of vectors from Open Loop zeros to point s=s_{a})-(sum of angles of vectors from Open Loop poles to points=s_{a}=\pm 180^{o}(2q+1)

where q=0,1,2………


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Inductance of a Co-axial cable, Solenoid & Toroid

Inductance of a Co-axial cable:-

Consider a coaxial cable with inner conductor having radius a and outer conductor of radius b and the current is flowing in the cable along z-axis  in which the cable is placed such that the axis of rotation of the cable co-incides with z-axis.

the length of the co-axial cable be d meters.

we know that the \overrightarrow{H_{\phi }} between the region a< \rho < b  is \overrightarrow{H_{\phi }}=\frac{I}{2\pi \rho }\overrightarrow{a_{\phi }}

and \overrightarrow{B_{\phi }}=\frac{\mu _{o}I}{2\pi \rho }\overrightarrow{a_{\phi }}   since \overrightarrow{B_{\phi }}=\mu _{o}\overrightarrow{H_{\phi }}

as L = \frac{\lambda }{I}, here the flux linkage \lambda =1\phi

where \phi-Total flux coming out of the surface

\phi = \oint_{s} \overrightarrow{B}.\overrightarrow{ds}

Since the magnetic flux will be radial plane extending from \rho =a   to   \rho =b and z=0 to z=d.

\overrightarrow{ds_{\phi }} = d\rho dz \overrightarrow{a_{\phi }}

\phi = \oint_{s} \overrightarrow{B}.\overrightarrow{ds}

\phi = \oint_{s}\frac{\mu _{o}I}{2\pi \rho }\overrightarrow{a_{\phi }} .d\rho dz \overrightarrow{a_{\phi }}

\phi = \int_{\rho =a}^{b}\int_{z =0}^{d}\frac{\mu _{o}I}{2\pi \rho } d\rho dz

\phi = \frac{\mu _{o}I}{2\pi } ln[\frac{b}{a}] d

L=\frac{\phi }{I} =\frac{\mu _{o}d}{2\pi } ln[\frac{b}{a}]  Henries.


Inductance of a Solenoid:-

Consider a Solenoid of N turns and  let the current flowing inside it is  ‘I’ Amperes. The length of the solenoid is ‘l’ meters and ‘A’ is its cross sectional area.

\phi – Total flux coming out of solenoid.

flux linkage \lambda = N\phi      \Rightarrow \lambda = NBA----------EQN(I)        \because \frac{\phi }{A} = B

L = \frac{\lambda }{I}, from the definition

As B is the Magnetic flux density given B= \frac{flux}{unit area} =\frac{\phi }{A}

from EQN(I) ,

\lambda = N\mu _{o}HA-------EQN(II)  because B = \mu _{o}H

The field strength H of a solenoid is H = \frac{NI}{l} A/m

EQN (II) becomes  \lambda = N \mu _{o}\frac{NI}{l}A

\lambda = \frac{N^{2} \mu _{o}IA}{l}

from the inductance definition L = \frac{\lambda }{I}

L= \frac{N^{2} \mu _{o}A}{l}   Henries.

Inductance of a Toroid:-

Consider a toroidal ring with N-turns and carrying current I.

let the radius of the toroid be ‘R’ and the total flux emerging be \phi

then flux linkage  \lambda = N\phi

the magnetic flux density inside a toroid is given by B = \frac{\mu _{o}NI}{2\pi R}

\lambda = NBA

where A is the cross sectional area of the toroid then \lambda = N \frac{\mu_{o} NI}{2\pi R}A

\lambda = \frac{\mu_{o} N^{2}I}{2\pi R}A

L=\frac{\lambda }{I}

L=\frac{\mu_{o} N^{2}A}{2\pi R}  Henries.

if the toroid has a height ‘h’ , inner radius \rightarrow r_{1} and outer radius \rightarrow r_{2} then its Inductance is L=\frac{\mu_{o} N^{2}h}{2\pi }ln[\frac{r_{1}}{r_{2}}]   Henries.

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Classification (or) topologies of feedback Amplifiers

There are 4 different combinations possible with negative feedback in Amplifiers as given below

  1. Voltage-Series.
  2. Current-Series.
  3. Voltage-Shunt.
  4. Current-Shunt.

The first part represents the type of sampling at the output .

  • i.e ,  Voltage- Shunt connection.
  • Current-Series connection.

and the second part represents the type of Mixing at the input

  • Series- Voltage is applied at the input.
  • Shunt-Current is applied at the input.

For any Amplifier circuit we require 

  • High Gain
  • High Band Width
  • High Input Impedance
  • and Low Output Impedance.

Classification of feedback Amplifiers is also known as feedback Topologies.

Voltage-Series feedback Connection:-

at i/p side connection is Series and at o/p side connection used is Shunt  since o/p is collected is voltage.

Series connection increases i/p impedance and Voltage at the o/p indicates a decrease in o/p impedance.

i.e, Z_{if} = Z_{i}(1+A\beta )    and    Z_{of} = \frac{Z_{o}}{(1+A\beta )}.

Current-Series feedback Connection:-

Series connection increases i/p impedance and Current at the o/p indicates an increase in o/p impedance.

i.e, Z_{if} = Z_{i}(1+A\beta )    and    Z_{of} = Z_{o}(1+A\beta ).

Voltage-Shunt feedback Connection:-

In this connection, both i/p and o/p impedance decreases .

i.e, Z_{if} = \frac{Z_{i}}{(1+A\beta )}    and    Z_{of} = \frac{Z_{o}}{(1+A\beta )}.

Current-Shunt feedback Connection:-

Shunt connection decreases i/p impedance and Current at the o/p indicates an increase in o/p impedance.

i.e, Z_{if} = \frac{Z_{i}}{(1+A\beta )}    and    Z_{of} = Z_{o}(1+A\beta ).

Effect of negative feedback on different topologies:-

Type of f/b

Voltage gain


Band Width with f/b i/p impedance



Voltage-Series decreases increases increases decreases
Current-Series decreases increases increases increases
Voltage-Shunt decreases increases decreases decreases
Current-Shunt decreases increases decreases increases

Similarly negative feedback decreases noise and harmonic distortion for all the four topologies.

Note:-  for any of the characteristics in the above table, increase ‘s shown by multiplying the original value with (1+A\beta ) and decrease ‘s shown by dividing with (1+A\beta ).



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continuity equation

The continuity equation of the current is based on the principle of conservation of charge that is  the charge can neither be created not destroyed.

consider a closed surface S with a current density \overrightarrow{J}, then the total current I crossing the surface S is given by the volume V

The current coming out of the closed surface is I_{out} = \oint_{s} \overrightarrow{J}.\overrightarrow{ds}

since the direction of current is in the direction of positive charges, positive charges also move out of the surface because of the current I.

According to principle of conversation of charge, there must be decrease of an equal amount of positive charge inside the closed surface.

therefore the time rate of decrease of charge with in a given volume must be equal to the net outward current flow through the closed surface of the volume.

I_{out} = \frac{-dQ_{in}}{dt}= \oint_{s} \overrightarrow{J}.\overrightarrow{ds}

By Divergence theorem \oint_{s} \overrightarrow{J}.\overrightarrow{ds} = \oint_{v}\overrightarrow{\bigtriangledown }.\overrightarrow{J} dv

\frac{-dQ_{in}}{dt}= \oint_{v}\overrightarrow{\bigtriangledown }.\overrightarrow{J} dv

Q_{in}=\int_{v}\rho _{v}dv    implies  -\frac{dQ_{in}}{dt}=-\frac{d}{dt}\int_{v}\rho _{v}dv

for a constant surface the derivative becomes the partial derivative 

\oint_{v}\overrightarrow{\bigtriangledown }.\overrightarrow{J} dv =-\int_{v}\frac{\partial \rho _{v}}{\partial t}dv   -this is for the whole volume.

for a differential volume \overrightarrow{\bigtriangledown }.\overrightarrow{J} dv =-\frac{\partial \rho _{v}}{\partial t}dv

\overrightarrow{\bigtriangledown }.\overrightarrow{J} =-\frac{\partial \rho _{v}}{\partial t} , which is called as continuity of current equation (or) Point form (or) differential form of the continuity equation.

This equation is derived based on the principle of conservation charge states that there can be no accumulation of charge at any point.

for steady  (dc) currents     \overrightarrow{\bigtriangledown }.\overrightarrow{J} =0 \Rightarrow -\frac{\partial \rho _{v}}{\partial t}=0

from \overrightarrow{\bigtriangledown }.\overrightarrow{J} =0. The total charge leaving a volume is the same as total charge entering it. Kirchhoff’s law follows this equation.

This continuity equation states that the current (or) the charge per second, diverging from a small volume per unit volume is equal to the time rate of decrease of charge per unit volume at every point.


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Effect of negative feedback on Band width of an Amplifier

Let the Band Width of an amplifier without feedback is = BW. Band width of an amplifier with negative feed back is BW_{f}= BW (1+A\beta ). Negative feedback increases Band width.

Proof:-  Consider an amplifier with gain ‘A’

Now the frequency response of the amplifier is as shown in the figure. Frequency response curve means gain (dB) Vs frequency (Hz)

the frequency response of an amplifier consists of three regions

  1. Low frequency region (< f_{1} -lower cut off frequency).
  2. Mid frequency region ( between f_{1} and f_{2}).
  3. High frequency region ( the region > f_{2} -upper cutoff frequency)

Gain in low- frequency region is given asA_{vl} = \frac{A_{v}}{1-j\frac{f_{1}}{f}}---------EQN(I),

A_{v} -open loop gain,

f– frequency,

f_{1}– lower cut off frequency, where Gain in constant region is A_{v}.

Gain in High-frequency region is A_{vh} = \frac{A_{v}}{1+j\frac{f}{f_{2}}}  .

In low-frequency region:-

since open loop gain  in low-frequency region is A_{vl} and gain with feedback is A_{vlf} = \frac{A_{vl}}{1+A_{vl}\beta }

From EQN(I) A_{vl} = \frac{A_{v}}{1-j\frac{f_{1}}{f}}   after substituting A_{vl} in the above equation 

A_{vlf} = \frac{\frac{A_{v}}{1-j\frac{f_{1}}{f}}}{1+\frac{A_{v}}{1-j\frac{f_{1}}{f}}\beta }    

        =\frac{A_{v}}{1-j\frac{f_{1}}{f}+A_{v}\beta }

  = \frac{A_{v}}{1+A_{v}\beta-j\frac{f_{1}}{f} }

Now by dividing the whole expression with (1+A_{v}\beta )

A_{vlf}= \frac{\frac{A_{v}}{(1+A_{v}\beta )}}{\frac{1+A_{v}\beta-j\frac{f_{1}}{f}}{(1+A_{v}\beta )} }

A_{vlf}= \frac{\frac{A_{v}}{(1+A_{v}\beta )}}{1-j\frac{f_{1}}{f}\frac{1}{(1+A_{v}\beta )} }

A_{vlf} = \frac{A_{vf}}{1-j\frac{f_{1}^{'}}{f}}, where A_{vf} = \frac{A_{v}}{1+A_{v}\beta }  and f_{1}^{'} = \frac{f_{1}}{1+A_{v}\beta }

for example lower cut-off frequency f_{1} =20 Hz  implies f_{1}^{'} = \frac{20}{1+A_{v}\beta }  is decreasing with negative feedback.

In High-frequency region:-

Gain with out feed back in High frequency region is A_{vh} = \frac{A_{v}}{1+j\frac{f}{f_{2}}}

Now Gain with negative feed back is A_{vhf} = \frac{A_{vh}}{1+A_{vh}\beta }

Substituting A_{vh} in the above equation 

A_{vhf} = \frac{\frac{A_{v}}{1+j\frac{f}{f_{2}}}}{1+\frac{A_{v}}{1+j\frac{f}{f_{2}}}\beta }

A_{vhf}=\frac{A_{v}}{1+j\frac{f}{f_{2}}+A_{v}\beta }

A_{vhf}= \frac{A_{v}}{1+A_{v}\beta+j\frac{f}{f_{2}} }

Now by dividing the whole expression with (1+A_{v}\beta )

A_{vhf}= \frac{\frac{A_{v}}{(1+A_{v}\beta )}}{\frac{1+A_{v}\beta+j\frac{f}{f_{2}}}{(1+A_{v}\beta )} }

A_{vhf}= \frac{\frac{A_{v}}{(1+A_{v}\beta )}}{1+j\frac{f}{f_{2}}\frac{1}{(1+A_{v}\beta )} }

A_{vhf} = \frac{A_{vf}}{1+j\frac{f}{f_{2}^{'}}}, where A_{vf} = \frac{A_{v}}{1+A_{v}\beta }  and f_{2}^{'} = f_{2}(1+A_{v}\beta )

for example lower cut-off frequency f_{2} =20K Hz  implies f_{1}^{'} = 20 K (1+A_{v}\beta )  is increasing with negative feedback.

In Mid-frequency region:-

Gain with out feed back is A_{v}

and the gain with negative feed back is A_{vf} = \frac{A_{v}}{1+A_{v}\beta }

With out feedback With feedback
lower cut-off frequency  is f_{1} lower cut-off frequency f_{1}^{'}= \frac{f_{1}}{1+A_{v}\beta }, increases
upper cut-off frequency is f_{2} upper cut-off frequency is f_{2}^{'} = f_{2}(1+A_{v}\beta )
BW = f_{2}-f_{1} BW_{f} = f_{2}^{'}-f_{1}^{'} increases

Thus negative feedback decreases lower cut-off frequency and increases upper cut-off frequency.

\therefore Over all gain decreases with  negative feedback and Band Width increases.


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Full Wave Rectifier

Full Wave Rectifier (FWR) contains two diodes D_{1} and D_{2}.

FWR converts a.c voltage into pulsating DC in two-half cycles of the applied input signal.

Here  we use a  Transformer, whose secondary winding has been split equally into two half waves with a common center tapped connection ‘c’.

This configuration results in each diode conducting in turn when it’s anode terminal is positive with respect to Center point ‘c’ of  the Transformer.

Working of Full Wave Rectifier:-

During positive half cycle of applied i/p signal

  • point ‘P’ is more positive ‘c’.
  • point ‘Q’ is more negative ‘c’.

i.e, Diode D_{1} is Forward Biased and D_{2} is Reverse Biased , under this condition the equivalent circuit is as shown below

\therefore V_{o} \approx V_{i} =i_{L}R_{L}, when there is no diode resistance.

Similarly the conditions of diodes will be reversed for the negative half cycle of i/p signal.

  • point ‘P’ is negative ‘c’.
  • point ‘Q’ is positive ‘c’.

i.e, Diode D_{1} is Reverse Biased and D_{2} is Forward Biased , under this condition the equivalent circuit is  and output voltage is V_{o} \approx i_{L}R_{L}.

the i/p and o/p wave forms are as shown below

FWR is advantageous compared to HWR in terms of its efficiency and ripple factor.

Ripple Factor (\Gamma):-

\Gamma = \frac{V_({ac})rms}{V_{dc}} = \sqrt{\frac{(V_{rms})^2}{V_{dc}}-1}

to find out V_{rms} and V_{dc} of output signal

\therefore V_{rms} = \sqrt{\frac{1}{\pi }\int_{0}^{\pi }V_{m}^{2} (\sin ^{2}\omega t ) d\omega t}                     ,     \therefore V_{dc} = \frac{1}{\pi } \int_{0}^{\pi }V_{m} (\sin \omega t ) d\omega t

V_{rms} = \sqrt{\frac{1}{\pi }\int_{0}^{\pi }V_{m}^{2} (\frac{1-\cos 2\omega t}{2} ) d\omega t}          ,               V_{dc} = \frac{-V_{m}}{\pi } [-2 ]

V_{rms} = \sqrt{\frac{V_{m}^{2}}{\pi }\int_{0}^{\pi } }(\frac{1}{2}\pi )                                           ,              V_{dc} = \frac{2V_{m}}{\pi }

V_{rms} = \frac{V_{m}}{\sqrt{2}}.

I_{rms} = \frac{V_{rms}}{R_{L}}=\frac{V_{m}}{\sqrt{2}R_{L}} = \frac{I_{m}}{\sqrt{2}}      and  I_{dc} = \frac{2I_{m}}{\pi }.

now the ripple factor results to be  \Gamma = \sqrt{\frac{(\frac{V_{m}}{\sqrt{2}})^{2}}{(\frac{2V_{m}}{\pi })^{2}}-1}

                                                                        \Gamma = \sqrt{\frac{V_{m}^{2}\pi ^{2}}{8V_{m}^{2}}-1}

                                                                        \Gamma = \sqrt{\frac{\pi ^{2}}{8}-1}

                                                                        \Gamma = 0.482


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Time-domain representation of SSB-SC signal

Let the signal produced by SSB-SC modulator is S(t), a Band pass signal

S_{USB}(t)=S_{I}(t) cos \omega _{c}t-S_{Q}(t) sin \omega _{c}t, where S_{I}(t) is  In-Phase component of S(t) obtained by

i. Multiplying S(t) with \cos \omega _{c}t.

ii. and passing the product through a LPF with suitable cut-off frequency.

S_{I}(t) = S(t) \cos \omega _{c}t

By finding the Fourier Transform of in-phase component

S_{I}(f) = \frac{1}{2}(S(f-f_{c})+S(f+f_{c}))

after restricting the signal S_{I}(f) between   -B\leq f \leq B

S_{I}(f) =\left\{\begin{matrix} \frac{1}{2}(S(f-f_{c})+S(f+f_{c})),-B\leq f \leq B & \\0 , otherwise & \end{matrix}\right.

Similarly S_{Q}(t) is the quadrature phase component of s(t), obtained by multiplying S(t) with \sin \omega _{c}t  and by passing the resultant signal through a LPF .

S_{Q}(t) = S(t) \sin \omega _{c}t

By finding the Fourier Transform of Q-phase component

S_{Q}(f) = \frac{1}{2j}(S(f-f_{c})-S(f+f_{c}))

after restricting the signal S_{Q}(f) between   -B\leq f \leq B

S_{Q}(f) =\left\{\begin{matrix} \frac{1}{2j}(S(f-f_{c})-S(f+f_{c})),-B\leq f \leq B & \\0 , otherwise & \end{matrix}\right.

Now Let’s assume S(f) is the required frequency spectrum of SSB-SC signal when only USB has been transmitted.


from the above figure,

one can obtain S(f-f_{c}) ,  S(f+f_{c}) by shifting the signal S(f) towards right by f_{c}  and  left by  f_{c}

Now by adding  S(f-f_{c})  and  S(f+f_{c})

from the above figure, S_{I}(f) results to be

from the frequency spectrum of S_{I}(f) , the time-domain representation turns out to be S_{I}(t)=\frac{1}{2}A_{c}m(t)-----EQN(I)


The resultant signals S(f-f_{c})-S(f+f_{c}) and S_{Q}(f) 

from the frequency spectrum of S_{Q}(f) turns out to be 

S_{Q}(f) =\left\{\begin{matrix} \frac{1}{2j}A_{c}M(f),-B\leq f \leq 0 & \\ -\frac{1}{2j}A_{c}M(f),0\leq f \leq B & \end{matrix}\right.

since Signum function is 

Sign(f) =\left\{\begin{matrix} +1,f>0 & \\ -1 f<0 & \end{matrix}\right.

S_{Q}(f) when expressed in terms of Signum function s_{Q}(f) = \frac{j}{2}A_{c}M(f)(-sign(f))

s_{Q}(f) = (-jsign(f)M(f))\frac{A_{c}}{2}

By using Hilbert transform of m(t) , the time-domain representation turns out to be S_{Q}(t)=\frac{1}{2}A_{c}\widehat{m(t)}-----EQN(II)

From EQN’s (I) and (II) , the time-domain representation of SSB-SC signal results

S_{USB}(t) = \frac{A_{c}}{2}m(t)\cos \omega _{c}t-\frac{A_{c}}{2}\widehat{m(t)}\sin \omega _{c}t.

similarly, SSB signal when only LSB has been transmitted 

S_{LSB}(t) = \frac{A_{c}}{2}m(t)\cos \omega _{c}t+\frac{A_{c}}{2}\widehat{m(t)}\sin \omega _{c}t


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Analog ElectronicCircuits Lab

Expt-1-i PN-Junction Diode Characteristics

Expt-1-ii-Zener Diode Characteristics

Expt-2-Full Wave Rectifier with and without filters

Expt-3-Common Emitter Characteristics

Expt-4- Clippers


Expt-6-single stage CE Amplifier-frequency response

Expt-7-Inverting amplifier

Expt-8-Non-inverting Amplifier

Expt-9-Differentiator and Integrator using op-amp

Expt-10-square and triangular wave generators 


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Input and Output characteristics of transistor in Common Base Configuration

Input Characteristics:-

Input characteristics in Common Base configuration means input voltage Vs input current by keeping output voltage  as constant.

i.e, V_{EB} Vs I_{E} by keeping V_{CB} constant.

Therefore the curve between Emitter current I_{E} and Emitter to Base voltage V_{EB} for a given value of Collector to Base voltage V_{CB} represents input characteristic.

common base input and output characteristics

for a given output voltage  V_{CB}, the input circuit acts as a PN-junction diode under Forward Bias.

from the curves there exists a cut-in (or) offset (or) threshold voltage V_{EB} below which the emitter current is very small  and a  substantial amount of Emitter-current flows after cut-in voltage ( 0.7 V for Si and 0.3 V for Ge).

the emitter current I_{E} increases rapidly with the small increase in V_{EB}. with the low dynamic input resistance of a transistor.

i.e, r_{i} =\frac{\Delta V_{EB}}{\Delta I_{E}}|_{V_{CB}\approx Constant}

input resistance =\frac{change in input voltage}{change in emitter current}|V_{CB}{\approx Constant}

This is calculated by measuring the slope of the input characteristic.

i.e, input characteristic determines the input resistance r_{i}.

The value of r_{i} varies from point to point on the Non-linear portion of the characteristic and is about 100\Omega in the linear region.

Output Characteristics:-

Output Characteristics are in between output current Vs output voltage with input current as kept constant.

i.e, f(I_{c},V_{CE})_{I_{E} = Constant}

i.e, O/p characteristics are in between V_{CB} Vs I_{c} by keeping I_{E} as constant.

basically it has 4 regions of operation Active region, saturation region,cut-off region and reach-through region.

active region:-

from the active region of operation I_{c} is almost independent of I_{E} 

i.e, I_{c}\approx I_{E}

when V_{CB} increases, there is very small increase in I_{c} .

This is because the increase in V_{CB} expands the collector-base depletion region and shortens the distance between the two depletion regions.

with I_{E} kept constant the increase in I_{c} is so small. transistor operates in it’s normal operation mode in this region.

saturation region:-

here both junctions are Forward Biased.

Collector current I_{c} flows even when V_{CB}=0(left of origin)  and this current reaches to zero when V_{CB} is increased negatively.

cut-off region:-

the region below the curve I_{E}=0 ,transistor operates in this region  when  the two junctions are Reverse Biased. 

I_{c}\neq 0 even though I_{E}=0 mA.  this is because of collector leakage current (or) reverse-saturation current I_{CO} (or) I_{CBO}.

punch through/reach through region:-

I_{c} is practically independent of V_{CB} over certain transistor operating region of the transistor.

  • If V_{CB} is increased beyond a certain value, I_{c} eventually increases rapidly because of avalanche (or) zener effects (or) both this condition is known as punch through (or) reach through region.
  • If transistor is operated beyond the specified output voltage (V_{CB}) transistor breakdown occurs.
  • If V_{CB} is increased beyond certain limit, the depletion region(J_{c}) of o/p junction penetrates into the base until it makes contact with emitter-base depletion region. we call this condition as punch-through (or) reach-through effect.
  • In this region , the large collector current destroys the transistor.
  • To avoid this V_{CB} should be kept in safe limits specified by the manufacturer

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Early effect in Common Base Configuration

Early effect (or) Base-width modulation:-

In Common Base configuration in the Reverse Bias, As the voltage V_{CC} increases, the space-charge width between collector and base tends to increase,  with the result that the effective width of the base decreases. This dependency of Base-width on the Collector to emitter voltage is known as the early effect.

The early effect has three consequences:-

  1. There is less chance for recombination with in the base region. Hence \alpha increases with increasing \left | V_{CB} \right |.
  2. The charge gradient is increased with in the base and consequently, the current of minority carriers injected across the emitter junction increases.
  3. For extremely large voltages, the effective Base-width may be reduced to zero, causing voltage break-down in the transistor. This phenomenon is called the Punch-through.

For higher values of V_{CB}, due to early effect the value of \alpha increases, for example \alpha changes say from 0.98 to 0.985. Hence there is a very small positive slope in the CB output characteristics and hence the output resistance is not zero.

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Hall effect

When a transverse magnetic field ‘B’is applied to a specimen (of metal (or) Semi conductor) carrying a current Ian Electric field E is induced perpendicular to both I and B. This phenomenon is known as Hall effect.

The figure shows the  experimental arrangement to observe Hall effect  Now 

I \rightarrow Current flowing in the semi conductor (x-direction)

B\rightarrow Applied Magnetic field (z-direction)

E\rightarrow Induced Electric field is along y-direction perpendicular to both I and B.

Now charge carrier electron is moving under the influence of two fields both electric field(E) and Magnetic field(B). 

i.e, electron is under the influence of both E and B, E applies some force on electron similarly B.

under equilibrium F_{E} = F_{B}

qE = Bqv_{d}------EQN(I), where v_{d} is the drift velocity

Electric field Intensity due to Hall effect is E=\frac{V_{H}}{d}--------------EQN(II)

V_{H} is the Hall voltage between plates 1 and 2.

and d- is the distance between the two plates.

In an N-type Semi conductor, the current is due to electrons , plate 1 is negatively charged compared to plate 2.

The current density J related to charge density \rho is J = \rho v_{d}------------EQN(III)

J = \frac{Current}{Area}=\frac{I}{A}=\frac{I}{Wd}

W- width of the specimen, d- height of the specimen.

From EQN(I) E=Bv_{d} and From EQN(II) V_{H}=Ed

up on multiplying with ‘d’ on both sides E d = Bd v_{d}

V_{H} = Bd v_{d}

V_{H} = B d \frac{J}{\rho }    from EQN(III)

V_{H} = B\frac{I}{Wd\rho }d

V_{H} = \frac{BI}{\rho W}

V_{H} = \frac{1}{\rho } . \frac{BI}{W}, let Hall coefficient R_{H} = \frac{1}{\rho }

V_{H} = R_{H}. \frac{BI}{W} .

Uses of Hall effect (or) Applications of Hall effect:-

  • Hall effect specifies the type of semi conductor that is P-type (or) N-type.when R_{H} is positive it’s a P-type semi conductor and  R_{H}  negative means  it’s  N-type semi conductor.
  • It is used to find out carrier concentrations ‘n’ and ‘p’ , by using either \rho = nq  or \rho =pq.
  • To find out mobilities \mu _{n} and \mu _{p} using the equation \mu =\sigma R_{H}.
  • Some other applications of Hall effect are measurement of velocity, sorting,limit sensing etc.
  • used to measure a.c power and the strength of Magnetic field and also finds the angular position of static magnetic fields in a magnetic field meter.
  • used in Hall effect multiplier, which gives the output proportional to product of two input signals.


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E due to infinite line charge distribution

Consider an infinitely long straight line carrying uniform line charge with density \rho _{L} C/m and lies on Z-axis from -\infty to +\infty.

Consider a point P at which Electric field intensity has to be determined which is produced by the line charge distribution.

from the figure let the co-ordinates of P are (0,\rho ,0) ( a point on y-axis) and assume dQ is a small differential charge confirmed to a point  M (0,0,Z) as co-ordinates.

\therefore dQ produces a differential field \overrightarrow{dE} 

\overrightarrow{dE}=\frac{dQ}{4\pi \epsilon _{o}R^{2}}\widehat{a_{r}}

the position vector \overrightarrow{R}=-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }} and the corresponding unit vector \widehat{a_{r}} =\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{\sqrt{\rho ^{2}+Z^{2}}}

\therefore \overrightarrow{dE} =\frac{dQ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})

therefore \overrightarrow{dE} =\frac{\rho _{L}dZ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})

then the Electric field strength \overrightarrow{E} produced by the infinite line charge distribution \rho _{L} is 

\overrightarrow{E} = \int \overrightarrow{dE}

\overrightarrow{E} = \int_{z=-\infty }^{\infty }\frac{\rho _{L}dZ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})

to solve this integral  let Z= \rho \tan \theta \Rightarrow dZ=\rho \sec ^{2}\theta d\theta

as Z\rightarrow -\infty \Rightarrow \theta \rightarrow \frac{-\pi }{2}

Z\rightarrow \infty \Rightarrow \theta \rightarrow \frac{\pi }{2}

\therefore \overrightarrow{E} = \int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} \frac{\rho _{L}}{4\pi\epsilon _{o}}(\frac{-\rho ^{2}\\sec ^{2}\theta \tan \theta d\theta \overrightarrow{a_{z}}+\rho ^{2}\sec ^{2}\theta d\theta \overrightarrow{a_{\rho }}}{(\rho ^{2}+\rho ^{2}\tan ^{2}\theta )^{\frac{3}{2}}})

\overrightarrow{E} = \int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} \frac{\rho _{L}}{4\pi\epsilon _{o}}(\frac{-\rho ^{2}\\sec ^{2}\theta \tan \theta d\theta \overrightarrow{a_{z}}+\rho ^{2}\sec ^{2}\theta d\theta \overrightarrow{a_{\rho }}}{\rho ^{3}\sec ^{3}\theta })

\overrightarrow{E} = \frac{\rho _{L}}{4\pi\epsilon _{o}\rho }[\int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} -\sin \theta \overrightarrow{a_{z} }d\theta +\int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} \cos\theta d\theta \overrightarrow{a_{\rho }}]

\overrightarrow{E}= \frac{\rho _{L}}{4\pi\epsilon _{o}\rho }[0+2\overrightarrow{a_{\rho }}]

\therefore \overrightarrow{E}= \frac{\rho _{L}}{2\pi\epsilon _{o}\rho }\overrightarrow{a_{\rho }} Newtons/Coulomb.

\overrightarrow{E} is a function of \rho   only, there is no \overrightarrow{a_{z}} component and \rho is the perpendicular distance from the point P to line charge distribution \rho _{L}.

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Energy Density in Electrostatic Fields

To determine the Energy present in an assembly of charges (or) group of charges one must first determine the amount of work necessary to assemble them.

It is seen that , when a unit positive charge is moved from infinity to a point in a field, the work is done by the external source and energy is expended.

If the external source is removed then the unit positive charge will be subjected to a force exerted by the field and will be moved in the direction of force.

Thus to hold the charge at a point in an electrostatic field, an external source has to do work , this energy gets stored in the form of Potential Energy when the test charge is hold at a point in a field.

when external source is removed , the Potential Energy gets converted to a Kinetic Energy.

In order to derive the expression for energy stored in electrostatics (i.e, the expression of such a Potential Energy)

Consider an empty space where there is no electric field at all, the Charge Q_{1} is moved from infinity to a point in the space ,let us say the point as P_{1}, this requires no work to be done to place a charge Q_{1} from infinity to a point P_{1} in empty space.

i.e, work done = 0 for placing a charge Q_{1} from infinity to a point P_{1} in empty space.

now another charge Q_{2} has to be placed from infinity to another point P_{2} . Now there has to do some work to place Q_{2} at P_{2} because there is an electric field , which is produced by the charge Q_{1} and Q_{2} is required to move against the field of Q_{1}.

Hence the work required to be done is  Potential=\frac{work done}{unit charge}

i.e, V = \frac{W}{Q} \Rightarrow W = V X Q .

\therefore Work done to position Q_{2} at P_{2} = V_{21} X Q_{2}.

Now the charge Q_{3} to be moved from infinity to P_{3} , there are electric fields due to Q_{1} and Q_{2}, Hence total work done is due to potential at P_{3} due to charge at P_{1} and Potential at P_{3} due to charge at P_{2}.

\therefore Work done to position Q_{3} at P_{3} = V_{31}Q_{3}+V_{32}Q_{3}.

Similarly , to place a  charge Q_{n} at P_{n} in a field created by (n-1) charges is ,work done to position Q_{n} at P_{n}=V_{n1}Q_{n}+V_{n2}Q_{n}+V_{n3}Q_{n}+.......

\thereforeTotal Work done W_{E} =Q_{2}V_{21}+Q_{3}V_{31}+Q_{3}V_{32}+Q_{4}V_{41}+Q_{4}V_{42}+Q_{4}V_{43}+.... EQN(I)

The total work done is nothing but the Potential energy in the system of charges hence denoted as W_{E},

if charges are placed in reverse order (i.e, first Q_{4} and then Q_{3} and then  Q_{2}  and finally Q_{1} is placed)

work done to place Q_{3} \Rightarrow V_{34}Q_{3}

work done to place Q_{2} \Rightarrow V_{24}Q_{2}+V_{23}Q_{2}

work done to place Q_{1} \Rightarrow V_{14}Q_{1}+V_{13}Q_{1}++V_{12}Q_{1}

Total work done W_{E} =Q_{3}V_{34}+Q_{2}V_{24}+Q_{2}V_{23}+Q_{1}V_{14}+Q_{1}V_{13}+Q_{1}V_{12}+.... EQN(II)

EQN (I)+EQN(II) gives

2W_{E} =Q_{1}(V_{12}+V_{13}+V_{14}+....+V_{1n}) +Q_{2}(V_{21}+V_{23}+V_{24}+....+V_{2n})+Q_{3}(V_{31}+V_{32}+V_{34}+....+V_{3n})+.....

let V_{1}=(V_{12}+V_{13}+V_{14}+....+V_{1n}), V_{2}=(V_{21}+V_{23}+V_{24}+....+V_{2n}) and V_{n}=(V_{n1}+V_{n2}+V_{n3}+....+V_{nn-1}) are the resultant Potentials due to all the charges except that charge.

i.e, V_{1} is the resultant potential due to all the charges except Q_{1}.

2W_{E} = Q_{1}V_{1}+Q_{2}V_{2}+Q_{3}V_{3}+......+Q_{n}V_{n}

W_{E} =\frac{1}{2} \sum_{m=1}^{n}Q_{m}V_{m} Joules.

The above expression represents the Potential Energy stored in the system of n point charges.


W_{E} = \frac{1}{2}\int_{l}\rho _{l}dl. V  Joules

W_{E} = \frac{1}{2}\int_{s}\rho _{s}ds. V Joules

W_{E} = \frac{1}{2}\int_{v}\rho _{v}dv. V Joules  for different types of charge distributions.

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Switches-Circuit Switching

Switches are used in Circuit-Switched and Packet-Switched Networks.  The switches are used are different depending up on the structure and usage.

Circuit Switches (or) Structure of Circuit Switch:-

The Switches used in Circuit Switching are called Circuit-Switches

Space-Division Switch:-

  • The paths are separated spatially from one switch to other.
  • These were originally designed for analog circuits but currently used for both analog and digital Networks.

Cross-bar Switch:-

In this type of Switch we connect n inputs and m outputs using micro switches (Transistors) at each cross point to form a cross-bar switch of size n X m.

The number of cross points required = n X m.

As n and m increases, cross points required also increases, for example n=1000 and m=1000  requires n X m= 1000 X 1000 cross points. A cross-bar with these many number of cross points is impractical and statics show that 25% of the cross points are in use at any given time.

Multi stage Switch:-

The solution to Cross-bar Switch is Multi stage switching. Multi stage switching is preferred over cross-bar switches to reduce the number of cross points.  Here number of cross-bar switches are combined in several stages.

Suppose an N X N cross-bar Switch can be made into 3 stage Multi bar switch as follows.

  1.  N is divided into groups , that is N/n Cross-bars with n-input lines and k-output lines forms n X k cross points.
  2. The second stage consists of k Cross-bar switches with each cross-bar switch size as (N/n) X (N/n).
  3. The third stage consists of N/n cross-bar switches with each switch size as k X n.

The total number of cross points = 2kN + k (\frac{N}{n})^{2}, so the number of cross points required are less than single-stage cross-bar Switch = N^{2}.

for example k=2 and n=3 and N=9 then a Multi-stage switch looks like as follows.

The problem in Multi-stage switching is Blocking during periods of heavy traffic, the idea behind Multi stage switch is to share intermediate cross-bars. Blocking means times when one input line can not be connected to an output because there is no path available (all possible switches are occupied). Blocking generally occurs in tele phone systems and this blocking is due to intermediate switches.

Clos criteria gives a condition for a non-blocking Multi stage switch 

n = \sqrt{\frac{N}{2}}k\geq (2n-1)  and  Total no.of Cross points \geq 4N(\sqrt{2N}-1).

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Solved Example problems in Electro Magnetic Theory

  1. Convert Points P(1,3,5)  from Cartesian to Cylindrical and Spherical Co-ordinate system.

Ans. Given P(1,3,5) \fn_cm \Rightarrow x=1,y=3, z=5

Cylindrical :- \fn_cm \phi =tan^{-1}(\frac{y}{x})

                              \fn_cm =tan^{-1}(\frac{3}{1})

                                \fn_cm =75^{o}

Similarly \fn_cm \rho = \sqrt{x^{2}+ y^{2}} \Rightarrow \sqrt{1^{2}+ 3^{2}} = \sqrt{10}=3.16

\fn_cm P(\rho ,\phi ,z)= P(3.16,71.5^{o},5)

Spherical :- \fn_cm r= \sqrt{x^{2}+y^{2}+z^{2}}

                        \fn_cm r= \sqrt{1^{2}+3^{2}+5^{2}}

                      \fn_cm r= \sqrt{35}

                    \fn_cm r=5.91

\fn_cm \theta =tan^{-1}(\frac{\sqrt{x^{2}+y^{2}}}{z})

\fn_cm \theta =tan^{-1}(\frac{\sqrt{1^{2}+3^{2}}}{5})

\fn_cm \theta =32.31^{o}

\fn_cm \phi =tan^{-1}(\frac{y}{x})

\fn_cm \phi =tan^{-1}(\frac{3}{1})

\fn_cm \phi = 75^{o}

\fn_cm P(r,\theta ,\phi ) = P(5.91,32.31^{o},71.5^{o})


Phase Locked Loop (PLL)

Demodulation of an FM signal using PLL:-

Let the input to PLL is an FM signal S(t) = A_{c} \sin (2 \pi f_{c}t+2\pi k_{f} \int_{0}^{t}m(t)dt)

let  \Phi _{1} (t) = 2\pi k_{f} \int_{0}^{t}m(t)dt ------Equation (I)

 Now the signal at the output of VCO is FM signal (another FM signal, which is different from input FM signal) Since Voltage Controlled Oscillator is an FM generator.

\therefore b(t) = A_{v} \cos (2 \pi f_{c}t+2\pi k_{v} \int_{0}^{t}v(t)dt)

the corresponding phase    \Phi _{2} (t) = 2\pi k_{v} \int_{0}^{t}v(t)dt ------Equation (II)

It is observed that S(t) and b(t) are out of phase by 90^{o}. Now these signals are applied to a phase detector , which is basically a multiplier

\therefore the error signal e(t) =S(t) .b(t)

e(t) =A_{c} \sin (2 \pi f_{c}t+2\pi k_{f} \int_{0}^{t}m(t)dt). A_{v} \cos (2 \pi f_{c}t+2\pi k_{v} \int_{0}^{t}v(t)dt)

e(t) =A_{c}A_{v} \sin (2 \pi f_{c}t+\phi _{1}(t)). \cos (2 \pi f_{c}t+\phi _{2}(t))

on further simplification , the product yields a higher frequency term (Sum) and a lower frequency term (difference)

e(t) =A_{c}A_{v}k_{m} \sin (4 \pi f_{c}t+\phi _{1}(t)+\phi _{2}(t))- A_{c}A_{v}k_{m}\sin (\phi _{1}(t)-\phi _{2}(t))

e(t) =A_{c}A_{v}k_{m} \sin (2 \omega _{c}t+\phi _{1}(t)+\phi _{2}(t))- A_{c}A_{v}k_{m}\sin (\phi _{1}(t)-\phi _{2}(t))

This product e(t) is given to a loop filter , Since the loop filter is a LPF it allows the difference and term and rejects the higher frequency term.

the over all output of a loop filter is 


Frequency domain representation of a Wide Band FM

To obtain the frequency-domain representation of Wide Band FM signal for the condition \beta > > 1 one must express the FM signal in complex representation (or) Phasor Notation (or) in the exponential form

i.e, Single-tone FM signal is S_{FM}(t)=A_{c}cos(2\pi f_{c}t+\beta sin 2\pi f_{m}t).

Now by expressing the above signal in terms of  Phasor notation (\because \beta > > 1 , None of the terms can be neglected)

S_{FM}(t) \simeq Re(A_{c}e^{j(2\pi f_{c}t+\beta sin 2\pi f_{m}t)})

S_{FM}(t) \simeq Re(A_{c}e^{j2\pi f_{c}t}e^{j\beta sin 2\pi f_{m}t})

S_{FM}(t) \simeq Re(e^{j2\pi f_{c}t} A_{c}e^{j\beta sin 2\pi f_{m}t})-------Equation(I)

Let    \widetilde{s(t)} =A_{c}e^{j\beta sin 2\pi f_{m}t}      is the complex envelope of FM signal.

\widetilde{s(t)} is a periodic function with period \frac{1}{f_{m}} . This \widetilde{s(t)} can be expressed in it’s Complex Fourier Series expansion.

i.e, \widetilde{S(t)} = \sum_{n=-\infty }^{\infty }C_{n} e^{jn\omega _{m}t}  this approximation is valid over [-\frac{1}{2f_{m}},\frac{1}{2f_{m}}] . Now the Fourier Coefficient  C_{n} = \frac{1}{T} \int_{\frac{-T}{2}}^{\frac{T}{2}} \widetilde{S(t)} e^{-jn2\pi f_{m}t}dt

T= \frac{1}{f_{m}}

C_{n} = \frac{1}{\frac{1}{f_{m}}} \int_{\frac{-1}{2f_{m}}}^{\frac{1}{2f_{m}}} \widetilde{S(t)} e^{-jn2\pi f_{m}t}dt

C_{n} = f_{m} \int_{\frac{-1}{2f_{m}}}^{\frac{1}{2f_{m}}} A_{c}e^{j\beta sin 2\pi f_{m}t} e^{-jn2\pi f_{m}t}dt

C_{n} = f_{m} \int_{\frac{-1}{2f_{m}}}^{\frac{1}{2f_{m}}} A_{c}e^{{j\beta sin 2\pi f_{m}t-jn2\pi f_{m}t}}dt

C_{n} = f_{m} \int_{\frac{-1}{2f_{m}}}^{\frac{1}{2f_{m}}} A_{c}e^{j({\beta sin 2\pi f_{m}t-n2\pi f_{m}t})}dt

let x=2\pi f_{m}t       implies   dx=2\pi f_{m}dt

as x\rightarrow \frac{-1}{2f_{m}} \Rightarrow t\rightarrow -\pi     and    x\rightarrow \frac{1}{2f_{m}} \Rightarrow t\rightarrow \pi

C_{n} = \frac{A_{c}}{2\pi } \int_{-\pi }^{\pi } e^{j({\beta sin x-nx})}dx

let J_{n}(\beta ) = \frac{1}{2\pi } \int_{-\pi }^{\pi } e^{j({\beta sin x-nx})}dx   as    n^{th}  order Bessel Function of first kind then   C_{n} = A_{c} J_{n}(\beta ).

Continuous Fourier Series  expansion of  

\widetilde{S(t)} = \sum_{n=-\infty }^{\infty }C_{n} e^{jn\omega _{m}t}

\widetilde{S(t)} = \sum_{n=-\infty }^{\infty }A_{c} J_{n} (\beta )e^{jn\omega _{m}t} 

Now substituting this in the Equation (I)

S_{WBFM}(t) \simeq Re(e^{j2\pi f_{c}t} \sum_{n=-\infty }^{\infty }A_{c} J_{n} (\beta )e^{jn\omega _{m}t})

S_{WBFM}(t) \simeq A_{c} Re( \sum_{n=-\infty }^{\infty }J_{n} (\beta ) e^{j2\pi f_{c}t} e^{jn\omega _{m}t})

S_{WBFM}(t) \simeq A_{c} Re( \sum_{n=-\infty }^{\infty }J_{n} (\beta ) e^{j2\pi (f_{c}+nf _{m}t)})

\therefore S_{WBFM}(t) \simeq A_{c} \sum_{n=-\infty }^{\infty }J_{n} (\beta ) cos 2\pi (f_{c}+nf _{m}t)

The  Frequency spectrum  can be obtained by taking Fourier Transform 

S_{WBFM}(f) = \frac{A_{c}}{2}\sum_{n=-\infty }^{\infty }J_{n}(\beta ) [\delta (f-(f_{c}+nf_{m}))+\delta (f+(f_{c}-nf_{m}))]

n value wide Band FM signal
0 S_{WBFM}(f) = \frac{A_{c}}{2}\sum_{n=-\infty }^{\infty }J_{0}(\beta ) [\delta (f-f_{c})+\delta (f+f_{c})]
1 S_{WBFM}(f) = \frac{A_{c}}{2}\sum_{n=-\infty }^{\infty }J_{1}(\beta ) [\delta (f-(f_{c}+f_{m}))+\delta (f+(f_{c}+f_{m}))]
-1 S_{WBFM}(f) = \frac{A_{c}}{2}\sum_{n=-\infty }^{\infty }J_{-1}(\beta ) [\delta (f-(f_{c}-f_{m}))+\delta (f+(f_{c}-f_{m}))]

From the above Equation it is clear that 

  • FM signal has infinite number of side bands at frequencies (f_{c}\pm nf_{m})for n values changing from -\infty to  \infty.
  • The relative amplitudes of all the side bands depends on the value of  J_{n}(\beta ).
  • The number of significant side bands depends on the modulation index \beta.
  • The average power of FM wave is P=\frac{A_{c}^{2}}{2} Watts.






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Matched Filter, impulse response h(t)

Matched Filter can be considered as a special case of Optimum Filter. Optimum Filter can be treated as Matched Filter when the noise at the input of the receiver is White Gaussian Noise.

Transfer Function of Matched Filter:-

Transfer Function of Optimum filter is H(f)=\frac{k X^{*}(f)e^{-j2\pi fT}}{S_{ni}(f)}

if input noise is white noise , its Power spectral density (Psd) is S_{ni}(f)=\frac{N_{o}}{2}.

then H(f) becomes H(f)=\frac{k X^{*}(f)e^{-j2\pi fT}}{\frac{N_{o}}{2}}

H(f)=\frac{2k}{N_{o}}X^{*}(f)e^{-j2\pi fT}-----Equation(I)

From the properties of Fourier Transforms , by Conjugate Symmetry property  X^{*}(f) = X(-f)

Equation (I) becomes

H(f)=\frac{2k}{N_{o}}X(-f)e^{-j2\pi fT}------Equation(II)

From Time-shifting property of Fourier Transforms

x(t)\leftrightarrow X(f)

From Time-Reversal Property  x(-t)\leftrightarrow X(-f)

By Shifting the signal x(-t) by T Seconds in positive direction(time) ,the Fourier Transform is given by  x(T-t)\leftrightarrow X(-f)e^{-j2\pi ft}

Now the inverse Fourier Transform of the signal from the Equation(II) is

F^{-1}[H(f)]=F^{-1}[\frac{2k}{N_{o}}X(-f)e^{-j2\pi fT}]


Let the constant \frac{2k}{N_{o}} is set to 1, then the impulse response of Matched Filter will become h(t) = x(T-t).

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Mutual Information I(X ; Y) Properties

Property 1:- Mutual Information is Non-Negative

Mutual Information is given by equation I(X ; Y) =\sum_{i=1}^{m}\sum_{j=1}^{n}P(x_{i}, y_{j})\log _{2} \frac{P(\frac{x_{i}}{y_{j}})}{P(x_{i})}---------Equation(I)

we know that P(\frac{x_{i}}{y_{j}})=\frac{P(x_{i}, y_{j})}{P(y_{j})}-------Equation(II)

Substitute Equation (II) in Equation (I)

I(X ; Y) =\sum_{i=1}^{m}\sum_{j=1}^{n}P(x_{i}, y_{j})\log _{2}\frac{P(x_{i}, y_{j})}{P(x_{i})P(y_{j})}

The above Equation can be written as

I(X ; Y) =-\sum_{i=1}^{m}\sum_{j=1}^{n}P(x_{i}, y_{j})\log _{2}\frac{P(x_{i})P(y_{j})}{P(x_{i}, y_{j})}

-I(X ; Y) =\sum_{i=1}^{m}\sum_{j=1}^{n}P(x_{i}, y_{j})\log _{2}\frac{P(x_{i})P(y_{j})}{P(x_{i}, y_{j})}------Equation(III)

we knew that \sum_{k=1}^{m} p_{k}\log _{2}(\frac{q_{k}}{p_{k}})\leq 0---Equation(IV)

This result can be applied to Mutual Information I(X ; Y) , If p_{k} = P(x_{i}, y_{j}) and q_{k} be P(x_{i}) P( y_{j}), Both p_{k} and q_{k} are two probability distributions on same alphabet , then Equation (III) becomes

-I(X ; Y) \leq 0

i.e, I(X ; Y) \geq 0  , Which implies that Mutual Information is always Non-negative (Positive).

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Example Problems in Electro Magnetic Theory Wave propagation

  1. A medium like Copper conductor which is characterized by the parameters \bg_black \sigma = 5.8 X 10^{7} Mho's/meter and \epsilon _{r}=1,\mu _{r}=1 uniform plane wave of frequency 50 Hz. Find \alpha ,\beta ,v,\eta  and \lambda.

Ans.  Given \bg_black \bg_black \sigma = 5.8 X 10^{7} Mho's/meter  ,     \bg_black \epsilon _{r}=1,\mu _{r}=1    and \bg_white f= 50 Hz

\bg_white \alpha =? ,\beta =? ,v = ?,\eta =? and \bg_white \lambda =?

Find the Loss tangent \bg_white \frac{\sigma }{\omega \epsilon } = \frac{5.8X 10^{7}}{2 \pi X50X\epsilon _{o}\epsilon _{r}}

                                              \bg_white \bg_white \frac{\sigma }{\omega \epsilon } = \frac{5.8X 10^{7}}{100\pi X\epsilon _{o}}

                                            \bg_white \bg_white \frac{\sigma }{\omega \epsilon } = 2.08 X 10 ^{16}> > 1

So given medium is a Conductor (Copper)

then \bg_white \alpha (or) \beta =\sqrt{\frac{\omega \mu \sigma }{2}}

                         \bg_white =\sqrt{\frac{5.8X10^{7}X2\pi X 50X\mu _{o}}{2}}

                      \bg_white \alpha = 106.99  , \bg_white \beta =106.99.

\bg_white v_{p}=\frac{\omega }{\beta }  \bg_white =\frac{2\pi X50}{106.99}\bg_white =2.936 meters/Sec.

\bg_white \lambda =\frac{2\pi }{\beta }=\frac{2\pi }{106.99}=0.0587 meters.

\bg_white \eta =\sqrt{\frac{j\omega \mu }{(\sigma +j\omega \epsilon )}}

    \bg_white =\sqrt{\frac{jX2\pi X50X\mu _{o}}{(5.8X10^{7}+j2\pi X50X\epsilon _{o})}}

    \bg_white = \sqrt{\frac{j 3.947 X10^{-4}}{(5.8X10^{7}+j 2.78 X10^{-9})}}

    \bg_white = \sqrt{\frac{ 3.947 X10^{-4}\angle 90^{o}}{(5.8X10^{7}\angle -2.74 X 10^{-15})}}

    \bg_white = \sqrt{0.68 X10 ^{-11}}\angle \frac{90-(2.74 X 10^{-5})}{2}

\eta = 2.6 X 10^{-6}\angle 45^{o}.

2. If \bg_white \epsilon _{r}=9,\mu =\mu _{o} for a medium in which a wave with a frequency of \bg_white f= 0.3 GHz is propagating . Determine the propagation constant and intrinsic impedance of the medium when \bg_white \sigma =0.

Ans: Given \bg_white \epsilon _{r}=9,  \bg_white \mu =\mu _{o} , \bg_white f=0.3GHz and \bg_white \sigma =0.

\bg_white \gamma =?,\eta =?

Since \bg_white \sigma =0, the given medium is a lossless Di-electric.

which implies \bg_white \alpha = \frac{\sigma }{2}\sqrt{\frac{\mu }{\epsilon }} =0.

\bg_white \beta = \omega \sqrt{\mu \epsilon }

    \bg_white =2\pi X o.3X10^{9}\sqrt{\mu _{o}X9\epsilon _{o}}

    \bg_white = 18.86.

\bg_white \eta = \sqrt{\frac{\mu }{\epsilon }}

\bg_white \eta = \sqrt{\frac{\mu_{o}\mu _{r} }{\epsilon_{o}\epsilon _{r} }}

\bg_white \eta = \sqrt{\frac{\mu_{o} }{9\epsilon_{o} }}

\bg_white \eta = \frac{120\pi }{3}

\bg_white \eta = 40 Ω.


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Delta modulation and Demodulation

DM is done on an over sampled message signal in its basic form, DM provides a stair case approximated signal to the over sampled version of message signal.

i.e, Delta Modulation (DM) is a Modulation scheme in which an incoming  message signal is over sampled (i.e, at a rate much higher than the Nyquist rate f_{s}> 2f_{m}) to purposely increase the correlation between adjacent samples of the signal. Over sampling is done to permit the use of a sample Quantizing strategy for constructing the encoded signal.

Signaling rate and Transmission Band Width are quite large in PCM. DM is used to overcome these problems in PCM . 

DM transmits one bit per sample. 

The process of approximation in Delta Modulation is as follows:-

The difference between the input (x[nT_{s}]) and the approximation (x[(n-1)T_{s}]) is quantized into only two levels \pm \Delta corresponding to Positive and negative differences.

i.e, If the approximation  (x[(n-1)T_{s}]) falls below the signal (x[nT_{s}])at any sampling epoch(the beginning of a period)output signal level is increased by \Delta.

On the other hand the approximation   (x[(n-1)T_{s}]) lies above the signal (x[nT_{s}]) , output signal level is diminished by \Delta provided that the input signal does not change too rapidly from sample to sample.

it is observed that the  change in stair case approximation lies with in  \pm \Delta .

This process can be illustrated in the following figure

Delta Modulated System:- The DM system consists of Delta Modulator and Delta Demodulator.

Delta Modulator:- 

Mathematical equations involved in DM Transmitter are 

error signal: e[nT_{s}]=x[nT_{s}]-x_{q}[(n-1)T_{s}]

Present sample of the (input) sampled signal: x[nT_{s}]

last sample approximation of stair case signal: x_{q}[(n-1)T_{s}]

Quantized  error signal( output of one-bit Quantizer): e_{q}[nT_{s}]

if         x[nT_{s}]\geq x_{q}[(n-1)T_{s}] \Rightarrow e_{q}[nT_{s}] = \Delta.

and  x[nT_{s}]< x_{q}[(n-1)T_{s}] \Rightarrow e_{q}[nT_{s}] = -\Delta.

encoding has to be done on the after Quantization that is when the output level is increased by \Delta from its previous quantized level, bit ‘1’ is transmitted .

similarly when output is diminished by \Delta from the previous level  a ‘0’ is transmitted.

from the accumulator x_{q}[nT_{s}]=x_{q}[(n-1)T_{s}]+ e_{q}[nT_{s}]

                                               e_{q}[nT_{s}]=e[nT_{s}]+ q[nT_{s}]

where q[nT_{s}] is the Quantization error.


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Fourier Series and it’s applications

The starting point of Fourier Series is the development of representation of signals as linear combination (sum of) of a set of basic signals.

f(t)\approx C_{1}x_{1}(t)+C_{2}x_{2}(t)+.......+C_{n}x_{n}(t)+....

The alternative representation if  a set of complex exponentials are used,

f(t)\approx C_{1}e^{j\omega _{o}t}+C_{2}e^{2j\omega _{o}t}+.......+C_{n}e^{jn\omega _{o}t}+....

The resulting representations are known as Fourier Series in Continuous-Time [Fourier Transform in the case of Non-Periodic signal]. Here we focus on representation of Continuous-Time and Discrete-Time periodic signals in terms of basic signals as Fourier Series and extend the analysis to the Fourier Transform representation of broad classes of aperiodic, finite energy signals.

These Fourier Series & Fourier Transform representations are most powerful tools used

  1. In the analyzation of signals and LTI systems.
  2. Designing of Signals & Systems.
  3. Gives insight to S&S.

The development of Fourier series analysis has a long history involving a great many individuals and the investigation of many different physical phenomena.

The concept of using “Trigonometric Sums”, that is sum of harmonically related sines and cosines (or) periodic complex exponentials are used to predict astronomical events.

Similarly, if we consider the vertical deflection f(t,x) of the string at time t and at a distance x along the string then for any fixed instant of time, the normal modes are harmonically related sinusoidal functions of x.

The scientist Fourier’s work, which motivated him physically was the phenomenon of heat propagation and diffusion. So he found that the temperature distribution through a body can be represented by using harmonically related sinusoidal signals.

In addition to this he said that any periodic signal could be represented by such a series.

Fourier obtained a representation for aperiodic (or) non-periodic signals not as weighted sum of harmonically related sinusoidals but as weighted integrals of Sinusoids that are not harmonically related, which is known as Fourier Integral (or) Fourier Transform.

In mathematics, we use the analysis of Fourier Series and Integrals in 

  1. The theory of Integration.
  2. Point-set topology.
  3. and in the eigen function expansions.

In addition to the original studies of vibration and heat diffusion, there are numerous other problems in science and Engineering in which sinusoidal signals arise naturally, and therefore Fourier Series and Fourier T/F’s plays an important role.

for example, Sine signals arise naturally in describing the motion of the planets and the periodic behavior of the earth’s climate.

A.C current sources generate sinusoidal signals as voltages and currents. As we will see the tools of Fourier analysis enable us to analyze the response of an LTI system such as a circuit to such Sine inputs.

Waves in the ocean consists of the linear combination of sine waves with different spatial periods (or) wave lengths.

Signals transmitted by radio and T.V stations are sinusoidal in nature as well.

The problems of mathematical physics focus on phenomena in Continuous Time, the tools of Fourier analysis for DT signals and systems have their own distinct historical roots and equally rich set of applications.

In particular, DT concepts and methods are fundamental to the discipline of numerical analysis , formulas for the processing of discrete sets of data points to produce numerical approximations for interpolation and differentiation were being investigated.

FFT known as Fast Fourier Transform algorithm was developed, which suited perfectly for efficient digital implementation and it reduced the time required to compute transform by orders of magnitude (which utilizes the DTFS and DTFT practically).

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Block Diagram of Digital Communication System/Elements of DCS

A General Communication System can be viewed as a Transmitting unit and a Receiving Unit connected by a medium(Channel). Obviously Transmitter and Receiver consists of various sub systems (or) blocks.

Our basic aim is to understand the various modules and sub systems in the system. If we are trying to understand the design and various features of DCS , it is plus imperative that we have to understand how we should design a transmitter and we must understand how to design a very good quality Receiver. Therefore one must know the features of the channel to design a good Transmitter as well as receiver that is the channel and it’s contribution will come repeatedly in digital Communications. 

Source:- the primary block (or) the starting point of a DCS is an information source, it may be an analog/digital source , for example the signal considered is analog in nature, then the signal generated by the source is some kind of electrical signal which is random in nature. if the signal is a speech signal (not an electrical signal) that has to be converted into electrical signal by means of a Transducer, which can be considered as a part of source itself.

Sampling & Quantization:- the secondary block involves the conversion of analog to discrete signal 

this involves the following steps

Sampling:- it is the process that involves in the conversion of Continuous Amplitude Continuous Time (CACT) signal into Continuous Amplitude Discrete Time (CADT) signal.

Quantization:- it is the process that involves in the conversion of Continuous Amplitude Discrete Time (CADT) signal into Discrete Amplitude Discrete Time (DADT) signal.

Source Encoder:-  An important problem in  Digital Communications is the efficient representation of data generated by a Discrete Source, this is accomplished by source encoder.

” The process of representation of incoming data  from a Discrete source into a more suitable form required for Transmission is known as source encoding”

 Note:-The blocks Sampler, Quantizer followed by an Encoder constructs ADC (Analog to Digital Converter).

∴ the output of Source encoder is a Digital Signal, the advantages of Source coding are

  • It reduces the Redundancy.
  • Minimizes the average bit rate.

Channel encoder:-Channel coding is also known as error control coding. Channel coding is a technique which reduces the probability of error P_{e} by reducing Signal to Noise Ratio at the expense of Transmission Band Width.The device that performs the channel coding is known as Channel encoder.

Channel encoding increases the redundancy of incoming data , this also involves error detection and error correction  along with the channel decoder at the receiver.

Spreading Techniques:- Spread Spectrum techniques are the methods by which a signal generated with a particular Band Width is deliberately spread in the frequency domain, resulting in a signal with a wider Band width.

There are two types of spreading techniques available

  1. Direct Sequence Spread Spectrum Techniques.
  2. Frequency Hopping Spread Spectrum Techniques.

The output of a spreaded signal is very much larger than incoming sequence. Spreading increases the BW required for transmission, which is a disadvantage even though spreading is done for high security of data.

SS techniques are used in Military applications.

Modulator:- Spreaded sequence is modulated by using digital modulation schemes like ASK, PSK, FSK etc depending up on the requirement, now the transmitting antenna transmits the modulated data into the channel.

Receiver:- Once you understood the process involved in transmitter Block. One should perform reverse operations in the receiver block. 

i.e the input of the demodulator is demodulated after that de-spreaded and then the channel decoder removes the redundancy added by the channel encoder ,the output of channel decoder is then source decoded and is given to Destination.

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Why Digital Communication is preferred over Analog Communication?


Communication is the process of establishing Connection (or) link between two points (which are separated by some distance) and transporting information between those two points. The electronic equipment used for communication purpose is called Communication equipment. The equipment when assembled together forms a communication system.

Examples of different types of communications are

  • Line Telephony & Telegraphy.
  • Radio Broadcasting.
  • Point-to-Point Communication.
  • Mobile Communication.
  • TV Broadcasting.
  • Radar and Satellite Communications etc.

Why Digital?

A General Communication system has two devices and a medium (channel) connecting those two devices. This can be understood that a Transmitter and Receiver are separated by a medium called as Communication channel. To transport an information-bearing signal from one point to another point over a communication channel either Analog or digital modulation techniques are used.

Now Coming to the point, Why Digital communication is preferred over analog Communication?

Why are communication systems, military and commercial alike, going digital?

  1. There are many reasons; the primary advantage is the ease with which digital signals compared with analog signals are generated. That is the generation of digital signals is much easier compared to analog signals.
  2. Propagation of Digital pulse through a Transmission line:-

When an ideal binary digital pulse propagating along a Transmission line. The shape of the waveform is affected by two mechanisms

  • Distortion caused on the ideal pulse because all Transmission lines and Circuits have some Non-ideal frequency Transfer function.
  • Unwanted electrical noise (or) other interference further distorts the pulse wave form.

Both of these mechanisms cause the pulse shape to degrade as a function of line length. During the time that the transmitted pulse can still be reliably identified (i.e. before it is degraded to an ambiguous state). The pulse is amplified by a digital amplifier that recovers its original ideal shape. The pulse is thus “re-born” (or) regenerated.

Circuits that perform this function at regular intervals along Transmission system are called “regenerative repeaters’. This is one of the reasons why Digital is preferred over Analog.

3.Digital Circuits Vs Analog Circuits:-

Digital Circuits are less subject to distortion and Interference than are analog circuits because binary digital circuits operate in one of two states FULLY ON (or) FULLY OFF to be meaningful, a disturbance must be large enough to change the circuit operating point from one state to another. Such two state operation facilitates signal representation and thus prevents noise and other disturbances from accumulating in transmission.

However, analog signals are not two-state signals, they can take an infinite variety of shapes with analog circuits and even a small disturbance can render the reproduced wave form unacceptably distorted. Once the analog signal is distorted, the distortion cannot be removed by amplification because accumulated noise is irrecoverably bound to analog signals, they cannot be perfectly generated.

 4. With digital techniques, extremely low error rates and high signal fidelity is possible through error detection and correction but similar procedures are not available with analog techniques.

5.  Digital circuits are more reliable and can be produced at a lower cost than analog circuits also; digital hardware lends itself to more flexible implementation than analog hardware.

Ex: – Microprocessors, Digital switching and large scale Integrated circuits.

6. The combining of Digital signals using Time Division Multiplexing (TDM) is simpler than the combining of analog signals using Frequency Division Multiplexing (FDM).

7. Digital techniques lend themselves naturally to signal processing functions that protect against interference and jamming (or) that provide encryption and privacy and also much data communication is from computer to computer (or) from digital instruments (or) terminal to computer, such digital terminations are normally best served by Digital Communication links.

8. Digital systems tend to be very signal-processing intensive compared with analog systems.

Apart from pros there exists a con in Digital Communications that is non-graceful degradation when the SNR drops below a certain threshold, the quality of service can change suddenly from very good to very poor. In contrast most analog Communication Systems degrade more gracefully.


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Maxwell’s Equations in Point (or Differential form) and Integral form

Maxwell’s Equations for time-varying fields in point and Integral form are:
  1. \overrightarrow{\bigtriangledown }X\overrightarrow{H}=\overrightarrow{J}+\frac{\partial \overrightarrow{D}}{\partial t}      \Rightarrow \oint_{l}\overrightarrow{H}.\overrightarrow{dl}=\oint_{s}\overline{J}.\overrightarrow{ds}+\int_{s}\frac{\partial \overrightarrow{D}}{\partial t}.\overrightarrow{ds}.
  2. \overrightarrow{\bigtriangledown }X\overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}       \Rightarrow \oint_{l}\overrightarrow{E}.\overrightarrow{dl}=-\int_{s}\frac{\partial \overrightarrow{B}}{\partial t}.\overrightarrow{ds} . 
  3. \overline{\bigtriangledown }.\overrightarrow{D}=\rho _{v}            \Rightarrow \oint_{s}\overrightarrow{D}.\overrightarrow{ds}=\int_{v}\rho _{v}dv.
  4. \overrightarrow{\bigtriangledown }.\overrightarrow{B}=0      \Rightarrow \oint_{s}\overrightarrow{B}.\overrightarrow{ds}=0.

The 4 Equations above are known as Maxwell’s Equations. Since Maxwell contributed to their development and establishes them as a self-consistent set.  Each differential Equation has its integral part. One form may be derived from the other with the help of Stoke’s theorem (or) Divergence theorem.

word statements of the field Equations:-

A word statement of the field Equations is readily obtained from their mathematical statement in the integral form.

1.\overrightarrow{\bigtriangledown }X\overrightarrow{H}=\overrightarrow{J}+\frac{\partial \overrightarrow{D}}{\partial t} \Rightarrow \oint_{l}\overrightarrow{H}.\overrightarrow{dl}=\oint_{s}\overline{J}.\overrightarrow{ds}+\int_{s}\frac{\partial \overrightarrow{D}}{\partial t}.\overrightarrow{ds}.

i.e, The magneto motive force (\because \oint_{l}\overrightarrow{H}.\overrightarrow{dl}\rightarrow is m.m.f)around a closed path is equal to the conduction current plus the time derivative of the electric displacement through any surface bounded by the path.

 2. \overrightarrow{\bigtriangledown }X\overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}\Rightarrow \oint_{l}\overrightarrow{E}.\overrightarrow{dl}=-\int_{s}\frac{\partial \overrightarrow{B}}{\partial t}.\overrightarrow{ds}.

The electro motive force (\because \oint_{l}\overrightarrow{E}.\overrightarrow{dl}\rightarrow is e.m.f)around a closed path is equal to the time derivative of the magnetic displacement through any surface bounded by the path.

3.\overrightarrow{\bigtriangledown }.\overrightarrow{D}=\rho _{v}  \Rightarrow \oint_{s}\overrightarrow{D}.\overrightarrow{ds}=\int_{v}\rho _{v}dv.

The total electric displacement through the surface enclosing a volume is equal to the total charge within the volume.

4.    \overrightarrow{\bigtriangledown }.\overrightarrow{B}=0  \Rightarrow \oint_{s}\overrightarrow{B}.\overrightarrow{ds}=0.

The net magnetic flux emerging through any close surface is zero.

the time-derivative of electric displacement is called displacement current. The term electric current is then to include both conduction current and displacement current. If the time-derivative of electric displacement is called an electric current, similarly \frac{\partial \overrightarrow{B}}{\partial t} is known as magnetic current, e.m.f as electric voltage and m.m.f as magnetic voltage.

the first two Maxwell’s Equations can be stated as 

  1. The magnetic voltage around a closed path is equal to the electric current through the path.
  2. The electric voltage around a closed path is equal to the magnetic current through the path.
Maxwell’s Equations for static fields in point and Integral form are:

Maxwell’s Equations of static-fields in differential form and integral form are:

  1. \overrightarrow{\bigtriangledown } X\overrightarrow{H}=\overrightarrow{J}        \Rightarrow \oint_{l}\overrightarrow{H}.\overrightarrow{dl}=\oint_{s}\overline{J}.\overrightarrow{ds}.
  2. \overline{\bigtriangledown } X\overrightarrow{E}=0           \Rightarrow \oint_{l}\overrightarrow{E}.\overrightarrow{dl}=0.
  3. \overline{\bigtriangledown }.\overrightarrow{D} = \rho _{v}            \Rightarrow \oint_{s}\overrightarrow{D}.\overrightarrow{ds}=\int_{v}\rho _{v}dv.
  4. \overline{\bigtriangledown }.\overrightarrow{B} = 0             \Rightarrow \oint_{s}\overrightarrow{B}.\overrightarrow{ds}=0.

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Inconsistensy in Ampere’s law (or) Displacement Current density

Faraday’s experimental law has been used to obtain one of Maxwell’s equations in differential form \overrightarrow{\bigtriangledown }X \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t} , which shows that a time-varying Magnetic field produces an Electric field.

From Ampere’s  Circuital law which is applicable to Steady Magnetic fields

\overrightarrow{\bigtriangledown } X \overrightarrow{H}=\overrightarrow{J}

By taking divergence of Ampere’s law the Ampere’s law is not consistent with time-varying fields

\overrightarrow{\bigtriangledown }.(\overrightarrow{\bigtriangledown } X \overrightarrow{H})=\overrightarrow{\bigtriangledown }.\overrightarrow{J}

\overrightarrow{\bigtriangledown } . \overrightarrow{J}=0 ,since \overrightarrow{\bigtriangledown }.(\overrightarrow{\bigtriangledown } X \overrightarrow{H})=0---------Equation(1)

the divergence of the curl is identically zero which implies \overrightarrow{\bigtriangledown }.\overrightarrow{J}=0------Equation(2), but from the continuity equation \overrightarrow{\bigtriangledown }.\overrightarrow{J} = -\frac{\partial \rho _{v}}{\partial t}-------Equation(3) which is not equal to zero, as \frac{\partial \rho _{v}}{\partial t}\neq 0 is an unrealistic limitation(i.e we can not assume \frac{\partial \rho _{v}}{\partial t} as zero) .

\therefore to make a compromise between the above two situations we must add an unknown term \overrightarrow{G} to Ampere’s Circuital law

i.e, \overrightarrow{\bigtriangledown }X\overrightarrow{H} = \overrightarrow{J}+\overrightarrow{G}

then by taking the Divergence of the above equation

\overrightarrow{\bigtriangledown }.(\overrightarrow{\bigtriangledown }X\overrightarrow{H}) = \overrightarrow{\bigtriangledown }.\overrightarrow{J}+\overrightarrow{\bigtriangledown }.\overrightarrow{G}------Equation(4)

from Equation(1),Equation(4) becomes     \overrightarrow{\bigtriangledown }.\overrightarrow{J}+\overrightarrow{\bigtriangledown }.\overrightarrow{G}=0

\overrightarrow{\bigtriangledown }.\overrightarrow{G}=-\overrightarrow{\bigtriangledown }.\overrightarrow{J}

thus \overrightarrow{\bigtriangledown }.\overrightarrow{G} = \frac{\partial \rho _{v}}{\partial t}---------Equation(5)

from Maxwell’s first Equation \overrightarrow{\bigtriangledown }.\overrightarrow{D}=\rho _{v} 

then Equation (5) becomes \overrightarrow{\bigtriangledown }.\overrightarrow{J} = \frac{\partial }{\partial t} (\overrightarrow{\bigtriangledown }.\overrightarrow{D})

\overrightarrow{\bigtriangledown }.\overrightarrow{G} =\overrightarrow{\bigtriangledown }. \frac{\partial \overrightarrow{D}}{\partial t}

then   \overrightarrow{G} = \frac{\partial \overrightarrow{D}}{\partial t}

\overrightarrow{\bigtriangledown }X\overrightarrow{H} = \overrightarrow{J}+\frac{\partial \overrightarrow{D}}{\partial t}

This is the equation obtained which does not disagree with the continuity equation. It is also consistent with all other results. This is a second Maxwell’s Equation is time-varying fields so the term \frac{\partial \overrightarrow{D}}{\partial t} has the dimensions of current density Amperes/Square-meter. Since it results from a time-varying electric flux density (\overrightarrow{D} ) , Maxwell termed it as displacement current density \overrightarrow{J_{D}}.

\overrightarrow{\bigtriangledown }X\overrightarrow{H} = \overrightarrow{J}+\frac{\partial \overrightarrow{D}}{\partial t}

\overrightarrow{\bigtriangledown }X\overrightarrow{H} = \overrightarrow{J}+\overrightarrow{J_{D}}

\overrightarrow{J_{D}}=\frac{\partial \overrightarrow{D}}{\partial t}

up to this point three current densities are there \overrightarrow{J}=\sigma \overrightarrow{E} , \overrightarrow{J}=\rho _{v} \overrightarrow{v} and \overrightarrow{J_{D}}= \frac{\partial\overrightarrow{D} }{\partial t}.

when the medium is Non-conducting medium \overrightarrow{\bigtriangledown }X \overrightarrow{H}=\frac{\partial\overrightarrow{D} }{\partial t}

the total displacement current crossing any given surface is expressed by the surface integral I_{d} = \oint_{s} \overrightarrow{J_{D}}.\overrightarrow{ds}

I_{d} = \oint_{s} \frac{\partial \overrightarrow{D}}{\partial t}.\overrightarrow{ds}

from Ampere’s law \oint_{s}(\overrightarrow{\bigtriangledown }X \overrightarrow{H}).\overrightarrow{ds}=\int_{s} \overrightarrow{J}.\overrightarrow{ds} +\oint_{s} \frac{\partial \overrightarrow{D}}{\partial t}.\overrightarrow{ds}

\oint_{l}\overrightarrow{H}.\overrightarrow{dl}=\int_{s} \overrightarrow{J}.\overrightarrow{ds} +\oint_{s} \frac{\partial \overrightarrow{D}}{\partial t}.\overrightarrow{ds}

\oint_{l}\overrightarrow{H}.\overrightarrow{dl}=I +\oint_{s} \frac{\partial \overrightarrow{D}}{\partial t}.\overrightarrow{ds}

\oint_{l}\overrightarrow{H}.\overrightarrow{dl}=I +I_{d}

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Correlation Receiver(Special case of Optimum Receiver)

we know that one form of Optimum filter is Matched filter, we will now derive another form of Optimum filter that is different from Matched filter Let the input to the Optimum filter is v(t) which is a noisy input that is v(t)=x(t)+n(t)

from the figure output of the filter after sampling at t=T_{b} seconds is v_{o}(T_{b})


v_{o}(t) = \int_{-\infty }^{T_{b}}v(\tau ) h(t-\tau )d\tau

at t= T_{b} output becomes  v_{o}(T_{b}) = \int_{-\infty }^{T_{b}}v(\tau ) h(T_{b}-\tau )d\tau -----Equation(1)

Now by substituting h(\tau )=x_{2}(T_{b}-\tau )-x_{1}(T_{b}-\tau )

h(T_{b}-\tau )=x_{2}(T_{b}-T_{b}+\tau )-x_{1}(T_{b}-T_{b}+\tau )

h(T_{b}-\tau )=x_{2}(\tau )-x_{1}(\tau )------Equation(2)

by substituting the  Equation(2)  in Equation(1) over the limits [0,T_{b}]

v_{o}(T_{b}) = \int_{0 }^{T_{b}}v(\tau ) (x_{2}(\tau )-x_{1}(\tau ))d\tau

v_{o}(T_{b}) = \int_{0 }^{T_{b}}v(\tau ) x_{2}(\tau ) d\tau-\int_{0}^{T_{b}}v(\tau )x_{1}(\tau )d\tau

Now by replacing \tau with t the above equation becomes 

v_{o}(T_{b}) = \int_{0 }^{T_{b}}v(t ) x_{2}(t ) dt-\int_{0}^{T_{b}}v(t )x_{1}(t )dt-----Equation(3)

The equation (3) suggests that the Optimum Receiver can be implemented as shown in the figure, this form of the Receiver is called  as correlation Receiver. This receiver requires the integration operation be ideal with zero initial conditions. Correlation Receiver performs coherent-detection.

in general Correlation Receiver can be approximated with Integrate and dump filter.

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Electric Potential (V)

Electric field intensity \overrightarrow{E} can be calculated by using either Coulomb’s law/Gauss’s law . when the charge distribution is symmetric another way of obtaining \overrightarrow{E} is from the electric scalar potential V

Assume a test charge Q_{t} at A in an Electric field, let points A and B are located at r_{A}and r_{B} units from the origin O,from Coulomb’s law the force acting on a test charge Q_{t} is \overrightarrow{F}= Q_{t}\overrightarrow{E}

The work done in moving a point charge Q_{t} along a differential length \overrightarrow{dl} is dW is given by dW = -\overrightarrow{F}.\overrightarrow{dl}

dW = -Q_{t}\overrightarrow{E}.\overrightarrow{dl}

so the total work done in moving a point charge Q_{t} from A to B is W=-Q_{t}\int_{A}^{B}\overrightarrow{E}.\overrightarrow{dl}

the direction of work done is always opposite to the direction of displacement.

where A is the initial point and B is the final point. Dividing the work done by the charge Q_{t} gives the potential energy per unit charge denoted by V_{AB},this is also known as potential difference between  the two points A and B.

Thus V_{AB} = \frac{W}{Q_{t}}= -\int_{A}^{B}\overrightarrow{E}.\overrightarrow{dl}

if we take B as initial point and A as final point , then V_{BA} = \frac{W}{Q_{t}}= -\int_{B}^{A}\overrightarrow{E}.\overrightarrow{dl}----Equation(1)

To derive the expression for V in terms of charge Q and distance r , we can use the concept of Electric field intensity \overrightarrow{E} produced by a charge Q, which is placed at a distance r

i.e, \overrightarrow{E} = \frac{Q}{4\pi \epsilon _{o}r^{2}}\overrightarrow{a_{r}}

from Equation(1) V_{BA} = -\int_{B}^{A}\overrightarrow{E}.\overrightarrow{dl}

V_{AB}= -\int_{r_{A}}^{r_{B}}\frac{Q}{4\pi \epsilon _{o}r^{2}}\overrightarrow{a_{r}}.dr \overrightarrow{a_{r}}  since \overrightarrow{dl}=dr.\overrightarrow{a_{r}}

V_{AB}= -\int_{r_{A}}^{r_{B}}\frac{Q}{4\pi \epsilon _{o}r^{2}}dr

V_{AB}= -\frac{Q}{4\pi \epsilon _{o}}\left [ \frac{-1}{r} \right ]_{r_{A}}^{r_{B}}

V_{AB}= -\frac{Q}{4\pi \epsilon _{o}}[\frac{1}{r_{B}}-\frac{1}{r_{A}}]


similarly, V_{BA}=V_{A}-V_{B} 

where V_{A} and V_{B} are the scalar potentials at the points A and B respectively. If A is  located at \infty with respect to origin ,with zero potential V_{A} =0 and B is located at a distance r with respect to origin. then the work done in moving a charge from  A (infinity) to B is given by 

V_{AB} = \frac{Q}{4\pi \epsilon _{o}r_{B}}  here r_{B} = r 

\therefore V = \frac{Q}{4\pi \epsilon _{o}r} volts.

hence the potential at any point is the potential difference between that point and a chosen point at which the potential is zero. In other words assuming Zero potential at infinity .

The potential at a distance r from a point charge is the work done per unit charge by an external agent in transferring a test charge from infinity to that point.

i.e, V = -\int_{\infty }^{r}\overrightarrow{E}.\overrightarrow{dl}

So a point charge Q_{1} located at a point P with position vector \overline{r_{1}} then the potential at another point Q with a position vector \overline{r} is 

V_{at \overline{r}} = \frac{Q_{1}}{4\pi \epsilon _{o}\left | \overline{r} -\overline{r_{1}}\right |}

As like \overrightarrow{E} superposition principle is applicable to V also that is for n point charges Q_{1},Q_{2},Q_{3},Q_{4},........Q_{n} located at points with position vectors  \overline{r_{1}},\overline{r_{2}},\overline{r_{3}},.......\overline{r_{n}}

then the potential at \overline{r} is 

V_{at \overline{r}} = \frac{Q_{1}}{4\pi \epsilon _{o}\left | \overline{r} -\overline{r_{1}}\right |}+\frac{Q_{2}}{4\pi \epsilon _{o}\left | \overline{r} -\overline{r_{2}}\right |}+\frac{Q_{3}}{4\pi \epsilon _{o}\left | \overline{r} -\overline{r_{3}}\right |}+........+\frac{Q_{n}}{4\pi \epsilon _{o}\left | \overline{r} -\overline{r_{n}}\right |}


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few problems on Auto correlatioon Function(ACF) and Energy Spectral Density(ESD)

  1. Find the Auto correlation function of x(t) = \frac{1}{\sqrt{2\pi }}\exp ^{\frac{-t^{2}}{2}}.

Ans. We know that  Auto correlation function forms fourier transform pair with Energy Spectral Density function

ACF\leftrightarrow ESD

R_{xx}(t)\leftrightarrow S(f)

the Fourier Transform of  e^{-ct^{2}}\leftrightarrow \frac{\sqrt{\pi }}{c}e^{-\pi ^{2}f^{2}}

\frac{1}{\sqrt{2\pi }}e^{\frac{-t^{2}}{2}}\leftrightarrow \frac{1}{\sqrt{2\pi }}.\sqrt{\frac{\pi }{(1/2)}}e^{\frac{-\pi ^{2}f^{2}}{(1/2)}} here c = \frac{1}{2}

\frac{1}{\sqrt{2\pi }}e^{\frac{-t^{2}}{2}}\leftrightarrow \frac{1}{\sqrt{2\pi }}.\sqrt{2\pi }e^{-2 \pi ^{2}f^{2}}

\frac{1}{\sqrt{2\pi }}e^{\frac{-t^{2}}{2}}\leftrightarrow e^{-2 \pi ^{2}f^{2}}

x(t)\leftrightarrow X(f)

\therefore the Fourier Transform of x(t) is X(f) and is X(f) = e^{-2\pi ^{2}f^{2}} and the Energy Spectral Density S(f) = \left | X(f) \right |^{2}

S(f) = e^{-4\pi ^{2}f^{2}}

By finding the inverse Fourier Transform of S(f) gives the Auto Correlation Function

S(f) = e^{\frac{-\pi ^{2}f^{2}}{(1/4)}}

e^{\frac{-\pi ^{2}f^{2}}{(1/4)}}\leftrightarrow \frac{e^{\frac{-t^{2}}{4}}}{\sqrt{4\pi }}

e^{\frac{-\pi ^{2}f^{2}}{(1/4)}}\leftrightarrow \frac{e^{\frac{-t^{2}}{4}}}{2\sqrt{\pi }}

\therefore the ACF of the given signal is inverse Fourier Transform of S(f) which is R_{xx}(t) = \frac{e^{\frac{-t^{2}}{4}}}{2\sqrt{\pi }}.


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Figure of merit of FM

The block diagram of FM Receiver in the presence of noise is as follows

The incoming signal at the front end of the receiver is an FM signal S_{FM}(t) = A_{c} cos(2\pi f_{c}t+2\pi k_{f}\int m(t )dt )--------Equation(1) got interfered by Additive noise n(t), since the FM signal has a transmission band width B_{T},the Band Pass filter characteristics are also considered over the band of interest i.e from\frac{-B_{T}}{2} to \frac{B_{T}}{2}.

The output of Band Pass Filter is x(t) = S_{FM}(t)+n_{o}(t)------------------Equation(2) is passed through a Discriminator for simplicity simple slope detector (discriminator followed by envelope detector) is used, the output of discriminator is v(t) this signal is considered over a band of (-W,W) by using a LPF .

The input noise to the BPF is n(t),  the  resultant output noise is band pass noise n_{o}(t)

n_{o}(t) = n_{I}(t)cos \omega _{c}t-n_{Q}(t)sin \omega _{c}t-----------Equation(3)

phasor representation of Band pass noise is n_{o}(t)=r(t)cos (\omega _{c}t +\Psi (t)) where r(t)=\sqrt{n_{I}^{2}(t)+n_{Q}^{2}(t)} and \Psi (t) = \tan ^{-1}(\frac{n_{Q}(t)}{n_{I}(t)}).

n_{I}(t),n_{Q}(t) are orthogonal, independent and are Gaussian.

r(t)– follows a Rayleigh’s distribution and \Psi (t) is uniformly distributed over (0,2\pi )r(t) and \Psi (t) are separate random processes.

substituting Equations (1), (3) in (2)

x(t) = S_{FM}(t)+n_{o}(t)

x(t)= A_{c} cos(2\pi f_{c}t+2\pi k_{f}\int m(t )dt )+n_{I}(t)cos \omega _{c}t-n_{Q}(t)sin \omega _{c}t

x(t)= A_{c} cos(2\pi f_{c}t+2\pi k_{f}\int m(t )dt )+r(t)cos (\omega _{c}t +\Psi (t))

x(t)= A_{c} cos(2\pi f_{c}t+\Phi (t) )+r(t)cos (\omega _{c}t +\Psi (t))-----Equation(4) where \Phi (t)=2\pi k_{f}\int m(t )dt.

now the analysis is being done from it’s phasor diagram/Noise triangle as follows

x(t) is the resultant of two phasors A_{c} cos(2\pi f_{c}t+\Phi (t) ) and r(t)cos (\omega _{c}t +\Psi (t)).

\theta (t)-\Phi (t) = \tan ^{-1}(\frac{r(t)\sin (\Psi (t)-\Phi (t))}{A_{c}+r(t)\cos (\Psi (t)-\Phi (t))})

\theta (t)-\Phi (t) = \tan ^{-1}(\frac{r(t)\sin (\Psi (t)-\Phi (t))}{A_{c}}) since r(t)< <A _{c}

\theta (t)=\Phi (t) +\tan ^{-1}(\frac{r(t)\sin (\Psi (t)-\Phi (t))}{A_{c}})

\theta (t)=\Phi (t) +\frac{r(t)\sin (\Psi (t)-\Phi (t))}{A_{c}} because \frac{r(t)}{A_{c}}< < 1\Rightarrow \tan ^{-1}\theta =\theta.

\theta (t) is the phase of the resultant signal x(t) and when this signal is given to a discriminator results  an outputv(t).

i.e, v(t)=\frac{1}{2\pi }\frac{d\theta (t)}{dt}

i.e, v(t)= \frac{1}{2\pi }\frac{d}{dt}(\Phi (t) +\frac{r(t)\sin (\Psi (t)-\Phi (t))}{A_{c}})---------Equation(5)

As \Phi (t) =2\pi k_{f}\int m(t)dt

\frac{d\Phi (t)}{dt}=2\pi k_{f}m(t)

the second term in the Equation n_{d}(t)=\frac{1}{2\pi }\frac{d}{dt}(\frac{r(t)\sin (\Psi (t)-\Phi (t))}{A_{c}}) where n_{d}(t) – denotes noise after demodulation.

this can be approximated to n_{d}(t)=\frac{1}{2\pi A_{c} }\frac{d}{dt}(r(t)\sin \Psi (t))-------Equation(6), which is a valid approximation. In this approximation r(t)\sin \Psi (t) is Quadrature-phase noise with power spectral density S_{NQ}(f) over (\frac{-B_{T}}{2},\frac{B_{T}}{2})

the power spectral density of n_{d}(t) will be obtained from Equation (6) using Fourier transform property \frac{d}{dt}\leftrightarrow j2\pi f

S_{Nd}(f)=\frac{1}{(2\pi A_{c})^{2}}(2\pi f)^{2}S_{NQ}(f)

S_{Nd}(f)=(\frac{f}{A_{c}})^{2}S_{NQ}(f)  , \left | f \right |\leq \frac{B_{T}}{2}

S_{Nd}(f)=0    elsewhere.

the power spectral density functions are drawn in the following figure

\therefore v(t) = k_{f}m(t)+n_{d}(t)-------Equation(7)) , from Carson’s rule \frac{B_{T}}{2}\geq W

the band width of v(t) has been restricted by passing it through a LPF.

Now, S_{Nd}(f)=(\frac{f}{A_{c}})^{2}S_{NQ}(f),\left | f \right |\leq W

S_{Nd}(f)=0 elsewhere.

To calculate Figure of Merit FOM = \frac{(SNR)_{output}}{(SNR)_{input}}

Calculation of (SNR)_{output}:-

output Noise power P_{no} = \int_{-W}^{W}(\frac{f}{A_{c}})^{2} N_{o}df

P_{no} = \frac{N_{o}}{A_{c}^{2}} \left ( \frac{f^{3}}{3} \right )^{W}_{-W}

P_{no} = \frac{N_{o}}{A_{c}^{2}} \left ( \frac{2W^{3}}{3} \right )------Equation(I)

The output signal power is calculated from k_{f}m(t) tha is  P_{so} = k_{f}^{2}P--------Equation(II)

(SNR)_{output} = \frac{P_{so}}{P_{no}}

From Equations(I) and (II)

(SNR)_{output} =\frac{\frac{N_{o}}{A_{c}^{2}} \left ( \frac{2W^{3}}{3} \right )}{k_{f}^{2}P}

(SNR)_{output} =\frac{3}{2}\frac{k_{f}^{2}PA_{c}^{2}}{N_{o}W^{3}}-------Equation(8)

Calculation of (SNR)_{input}:-

(SNR)_{input} = \frac{P_{si}}{P_{ni}}

input signal power P_{si}= \frac{A_{c}^{2}}{2}---------------Equation(III)

noise signal power  P_{ni}=N_{o}W--------------Equation(IV)

from Equations (III) and (IV)

(SNR)_{input} = \frac{A_{c}^{2}}{2WN_{o}}-------------------Equation(9)

Now the Figure of Merit of FM is FOM = \frac{(SNR)_{output}}{(SNR)_{input}}

FOM = \frac{\frac{3}{2}\frac{k_{f}^{2}PA_{c}^{2}}{N_{o}W^{3}}}{\frac{A_{c}^{2}}{2WN_{o}}} 

FOM_{FM} = \frac{3k_{f}^{2}P}{W^{2}}--------Equation(10)

to match this with AM tone(single-tone) modulation is used i.e, m(t) = cos \omega _{m}t then the signal power P = \frac{1}{2}  and W = f_{m}

FOM_{FM} = \frac{3k_{f}^{2}}{f_{m}^{2}}\frac{1}{2}

FOM_{FM} = \frac{3}{2} (\frac{k_{f}}{f_{m}})^{2}

FOM_{FM} = \frac{3}{2}\beta ^{2} 

since for tone(single-tone) modulation \beta = \frac{k_{f}}{f_{m}}.

when you compare single-tone FM with AM FOM_{FM (single-tone)} =FOM_{AM(single-tone)}

\frac{3}{2}\beta ^{2} > \frac{1}{3}

\beta > \frac{\sqrt{2}}{3}

\beta > 0.471.

the modulation index \beta >0.471. will be beneficial in terms of noise cancellation, this is one of the reasons why we prefer WBFM over NBFM.


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Capture effect in Frequency Modulation

The Amplitude Modulation schemes like AM,DSB-SC and SSB-SC systems can not handle inherent Non-linearities in a really good manner where as FM can handle it very well.

Let us suppose un Modulated FM carrier S(t) = A_{c}cos\omega _{c}(t)

S(t) = A_{c}cos(\omega _{c}(t)+\phi (t))

By considering un modulated FM carrier in terms of frequency(by neglecting phase) i.e S(t) = A_{c}cos (\omega _{c}t) has been interfered by a near by interference located at a frequency (\omega _{c}+\omega ) where \omega is a small deviation from \omega _{c}.

the nearby inerference is I cos(\omega _{c} + \omega )t 

when the original signal got interfered by this near by interference , the received signal is r(t)= A_{c}cos \omega _{c}t + I cos(\omega _{c}+\omega )t 

r(t)= (A+ I cos \omega t)cos \omega _{c}t -I sin \omega t sin\omega _{c}t   Let A_{c}=A

r(t) = E_{r}(t) cos (\omega _{c}t+\Psi _{d}(t))

now the phase of the signal is \Psi _{d}(t) = tan^{-1} (\frac{I sin \omega t}{A+I cos \omega t })

as A> > I implies \frac{I}{A}< < 1

\Psi _{d}(t) = tan^{-1} (\frac{I sin \omega t}{A})

since \frac{I}{A}< < 1 , \tan ^{-1}\theta = \theta

\Psi _{d}(t) \approx \frac{I sin \omega t}{A}

As the demodulated signal is the output of a discriminator y _{d}(t) =\frac{d}{dt} (\frac{I sin \omega t}{A})

y _{d}(t) =\frac{I\omega }{A} ({cos \omega t}) , which is the detected at the output of the demodulator.

the detected output at the demodulator is y_{d}(t) in the absence of message signal  i.e, m(t)=0.

i.e, when message signal is not being transmitted at the transmitter but detected some output y_{d}(t) which is nothing but the interference. 

As ‘A’ is higher the interference is less at t=0 the interference is \frac{I\omega }{A} and is a linear function of \omega, when \omega is small interference is less. That is \omega is closer to \omega _{c} interference is less in FM. 

Advantage of FM :- is Noise cancellation property , any interference that comes closer with the carrier signal (in the band of FM) more it will be cancelled. Not only that it overridden by the carrier strength A_{c} but also exerts more power in the demodulated signal.

This is known as ‘Capture effect’ in FM which is a very good property of FM. Over years it has seen that a near by interference is 35 dB less in AM where as the near by interference in FM is 6 dB less this is a big advantage.

Two more advantages of FM over AM are: 

  1. Non-linearity in the Channel ,FM cancels it very nicely due to it’s inherent modulation and demodulation technique.
  2. Capture effect( a near by interference) FM overrides this by A_{c}.
  3. Noise cancellation.



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Propagation of plane EM wave in conducting medium (or) lossy dielectrics

A lossy dielectric medium is one which an EM wave as it propagates losses power owing to imperfect dielectric,that is a lossy dielectric is an imperfect conductor that is a partially conducting medium (\sigma \neq 0) . 

where as a lossless dielectric is a  (\sigma =0) perfect dielectric,then wave equations for conductors are also holds good here 

i.e, \bigtriangledown ^{2}\overrightarrow{E}=\mu \sigma \frac{\partial \overrightarrow{E}}{\partial t} +\mu \epsilon \frac{\partial ^{2}\overrightarrow{E}}{\partial t^{2}}

\frac{\partial }{\partial t}= jw

then \bigtriangledown ^{2}\overrightarrow{E}= j\omega \mu \sigma \overrightarrow{E}+\mu \epsilon(j\omega ) ^{2}\overrightarrow{E}

\bigtriangledown ^{2}\overrightarrow{E}= (\sigma + j \omega \epsilon )j\omega \mu \overrightarrow{E}

\bigtriangledown ^{2}\overrightarrow{E}= \gamma ^{2} \overrightarrow{E}

\bigtriangledown ^{2}\overrightarrow{E}- \gamma ^{2} \overrightarrow{E}=0---------Equation (1)

Equation (1) is called helm holtz equation and \gamma is  called propagation constant.

\gamma ^{2} =j\omega \mu (\sigma +j\omega \epsilon )

\gamma ^{2}= j\omega \mu \sigma -\omega ^{2}\mu \epsilon

Since \gamma is a complex quantity it can be expressed as \gamma = \alpha +j\beta

\alpha– is attenuation constant measured in Nepers/meter.

\beta-is phase constant measured in radians/meter.

(\alpha +j\beta ) ^{2}= j\omega \mu \sigma -\omega ^{2}\mu \epsilon

\alpha^{2} +2j\alpha \beta-\beta ^{2} = j\omega \mu \sigma -\omega ^{2}\mu \epsilon

by equating real and imaginary parts separately \alpha ^{2}-\beta ^{2}= -\omega ^{2}\mu \epsilon------Equation(2)

 and 2\alpha \beta =\omega \mu \sigma

\alpha =\frac{\omega \mu \sigma}{2\beta }

 by substituting  \alpha value in the equation (2)   \frac{\omega ^{2}\mu ^{2}\sigma ^{2}}{4\beta ^{2}}-\beta ^{2}=-\omega ^{2}\mu \epsilon

{\omega ^{2}\mu ^{2}\sigma ^{2}}-4\beta ^{4}=-4\omega ^{2}\beta ^{2}\mu \epsilon

4\beta ^{4}-4\omega ^{2}\beta ^{2}\mu \epsilon -{\omega ^{2}\mu ^{2}\sigma ^{2}}=0

let \beta ^{2}=t 

4t^{2}-4\omega ^{2}t\mu \epsilon -{\omega ^{2}\mu ^{2}\sigma ^{2}}=0

t^{2}-\omega ^{2}t\mu \epsilon -\frac{\omega ^{2}\mu ^{2}\sigma ^{2}}{4}=0

the roots of the above quadratic expression are

t=\frac{\omega ^{2}\mu \epsilon \pm \sqrt{\omega ^{4}\mu ^{2}\epsilon ^{2}-4(-\frac{\omega ^{2}\mu ^{2}\sigma ^{2}}{4})}}{2}

t=\frac{\omega ^{2}\mu \epsilon \pm \sqrt{\omega ^{4}\mu ^{2}\epsilon ^{2}(1+\frac{\sigma }{\omega \epsilon })^{2}}}{2}

\beta ^{2}=\frac{\omega ^{2}\mu \epsilon \pm \sqrt{\omega ^{4}\mu ^{2}\epsilon ^{2}(1+\frac{\sigma }{\omega \epsilon })^{2}}}{2}

\beta =\sqrt{\frac{\omega ^{2}\mu \epsilon \pm \sqrt{\omega ^{4}\mu ^{2}\epsilon ^{2}(1+\frac{\sigma }{\omega \epsilon })^{2}}}{2}}

\beta =\sqrt{\frac{\omega ^{2}\mu \epsilon (1+ \sqrt{(1+\frac{\sigma }{\omega \epsilon })^{2}})}{2}}




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Drift and Diffusion currents

The flow of charge (or) current through a semi conductor material is of two types. Similarly the net current that flows through a PN diode is also of two types (i) Drift current and  (ii) Diffusion current.

Drift current:-

When an Electric field is applied across the semi conductor, the charge carriers attains certain velocity known as drift velocity v_{d}= \mu E with this velocity electrons move towards positive terminal and holes move towards negative terminal of the battery. This movement of charge carriers constitutes a current known as ‘Drift current’.

Drift current is defined as the flow of electric current due to the motion of the charge carriers under the influence of an external field.

Drift current density due to free electrons J_{n} = qn\mu _{n}E atoms/Cm2  and the Drift current density due to free holes J_{p} = qp\mu _{p}E atoms/Cm2.

The current densities are perpendicular to the direction of current flow.

Diffusion Current:-

It is possible for an electric current to flow in a semi conductor even in the absence of the applied Electric field (or) voltage provided there exists a concentration gradient.

concentration gradient exists if the number of electrons (or) holes is greater in one region than other region in a semi conductors.

Now the charge carriers move from higher concentration to that lower concentration of same type charged regions.

The  resulting current is known as diffusion current.

Diffusion current density (J_{P}) due to holes is  J_{p} = -q D_{p}\frac{dp}{dx}    A/Cm2 .

Diffusion current density (J_{P}) due to holes is J_{n} = q D_{n}\frac{dn}{dx}    A/Cm2 .

\therefore Total current in a semi-conductor is the sum of drift and diffusion currents

In P-type total current density is J= qp\mu _{p} E-qD_{p}\frac{dp}{dx} .

In N-type total current density is J= qn\mu _{n} E+qD_{n}\frac{dn}{dx}.



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Conductivity of a Semi conductor

In a pure Semi conductor number of electrons = number of holes. Thermal agitation (increase in temperature) produces new electron-hole pairs and these electron-hole pair combines produces new charge particles.

one particle is of negative charge which is known as free electron with mobility \mu _{n} another in with positive charge known as free hole with mobility \mu _{p}.

two particles moves in opposite direction in an electric field \overrightarrow{E} and constitutes a current.

The total current density (J) with in the semi conductor.

\overrightarrow{J} = \overrightarrow{J_{n}} + \overrightarrow{J_{p}}

Total conduction current density = conduction current density due to electrons + conduction current density due to holes.

J_{n}= nq\mu _{n}E.

J_{p}= pq\mu _{p}E.

n- number of electrons/Unit-Volume.

p-number of holes/Unit-Volume.

E- applied Electric field strength V/m.

q-charge of electron/hole \approx 1.6X10^{-19}C.

J = nq\mu _{n}E +pq\mu _{p}E.

J = (n\mu _n +p\mu _{p})qE.

J=\sigma E.

where \sigma = (n\mu _{n}+p\mu _{p})q is the conductivity of semi conductor.

Intrinsic Semi conductor:-

In an  intrinsic semi conductor n=p=n_{i}

\therefore conductivity \sigma _{i}= (n_{i}\mu _{n}+ n_{i}\mu _{p})q

\sigma _{i}= n_{i}(\mu _{n}+ \mu _{p})q

where J_{i} is the current density in an intrinsic semi conductor J_{i} = \sigma _{i} E

Conductivity in N-type semi conductor:-

In N-type n> > p 

number of electrons > > number of holes

\therefore \sigma _{N}\simeq n\mu _{n}q

J _{N}= n\mu _{n}q.

Conductivity in P-type semi conductor:-

In P-type p> > n

number of holes > > number of electrons

\therefore \sigma _{p} \approx p\mu _{p}q.

J_{P}= p\mu _{p}q E.



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Current components of a PNP Transistor

The various Current components which flow across a PNP Transistor are as shown in the figure.

For Normal operation 

  • Emitter Junction J_{E} is Forward Biased.
  • collector Junction J_{C} is Reverse Biased.

The current flows into Emitter is Emitter current I_{E},  I_{E} = I_{hE}+I_{eE}.

This current consists of two components

  • I_{hE} or I_{pE}– Current due to majority carriers(holes).
  • I_{eE}  or I_{nE}– Current due to minority carriers(electrons).

since I_{eE} is very small I_{E} \simeq I_{hE}-----------Equation(1)

All the holes crossing the Emitter junction J_{E} do not reach the Collector junction because some of them combine with the electrons in the N-type Base.

I_{hC} – is the hole current in the Collector.

∴ Base current = Total hole current in Emitter – hole current in Collector.

i.e, I_{B} = I_{hE}-I_{hC}----------------Equation(2).

If emitter were open circuited I_{E} = 0 Amperes which implies  I_{E} = I_{hE} from Equation(1) I_{hE}\approx 0 Amperes.

Under these conditions, Base-Collector junction acts as Reverse-Biased Diode and gives rise to a small reverse-Saturation current known as I_{CO}.

when I_{E} \neq 0  , Total Collector current  I_{C} is the sum of current due to holes in the Collector and Reverse Saturation current I_{CO}.

i.e, I_{C} = I_{hC}+I_{CO}.

i.e, In a PNP Transistor I_{CO} consists of holes moving across J_{C} (from Base to Collector) that is I_{hCO} and electrons crossing the junction J_{C} (from Collector to Base) constitutes I_{eCO}.

I_{CO} = I_{hCO}+I_{eCO}

i.e, I_{E} = 0  \Rightarrow I_{C} = I_{CO} only

when I_{E} \neq 0 \Rightarrow I_{C} = I_{hC}+I_{CO}.

\therefore Total current in the transistor is given by  I_{E} = I_{B}+I_{C}.

\therefore The general expression for Collector current is I_{C} = \alpha I_{E}+I_{CO}

I_{C} =\frac{\alpha }{(1-\alpha )} I_{B}+\frac{1}{(1-\alpha )}I_{CO}.



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Electro Magnetic Wave Equation

Assume a Uniform, Homogeneous,linear,isotropic and Stationary medium with Non-zero current i.e, \overrightarrow J_{c}(\sigma \overrightarrow{E}) \neq 0.

When an EM wave is travelling in a conducting medium in which  \overrightarrow J\neq 0. The wave is rapidly attenuated in a conducting medium and in a good conductors, the attenuation is so high at Radio frequencies. The wave penetrates the conductor only to a small depth.

choose the equation \overrightarrow{\bigtriangledown }X \overrightarrow{H} = \overrightarrow{J_{C}}+\overrightarrow{J_{D}} the time-domain representation of it is    \overrightarrow{\bigtriangledown }X \overrightarrow{H} = \overrightarrow{J_{C}}+\frac{\partial \overrightarrow{D}}{\partial t}   

since  \frac{\partial }{\partial t} =j\omega in phasor-notation \overrightarrow{\bigtriangledown }X \overrightarrow{H} = \sigma \overrightarrow{E}+j\omega \epsilon \overrightarrow{E} , \because \overrightarrow{J_{C}}=\sigma \overrightarrow{E} and \overrightarrow{D}=\epsilon \overrightarrow{E} . 

\overrightarrow{\bigtriangledown }X \overrightarrow{H} = \sigma \overrightarrow{E}+\frac{\epsilon\partial \overrightarrow{E}}{\partial t}--------------Equation (1)

 By differentiating the above equation with respect to time\frac{\partial(\overrightarrow{\bigtriangledown } X \overrightarrow{H})}{\partial t} =\sigma \frac{\partial \overrightarrow{E}}{\partial t} + \epsilon \frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}}--------------Equation (2)

From the Maxwell Equation \overrightarrow{\bigtriangledown } X \overrightarrow{E} =-\frac{\partial \overrightarrow{B}}{\partial t}

By taking curl on both sides \overrightarrow{\bigtriangledown } X \overrightarrow{\bigtriangledown } X \overrightarrow{E} =- \overrightarrow{\bigtriangledown } X \frac{\partial \overrightarrow{B}}{\partial t}

\overrightarrow{\bigtriangledown } X \overrightarrow{\bigtriangledown } X \overrightarrow{E} =- \overrightarrow{\bigtriangledown } X \frac{\partial \overrightarrow{(\mu H)}}{\partial t}  since \overrightarrow{B} = \mu \overrightarrow{H}

\overrightarrow{\bigtriangledown } X \overrightarrow{\bigtriangledown } X \overrightarrow{E} =-\mu (\overrightarrow{\bigtriangledown } X \frac{\partial \overrightarrow{ H}}{\partial t}) 

By using the vector identity 

\bigtriangledown ^{2}\overrightarrow{E} = \overrightarrow{\bigtriangledown }(\overrightarrow{\bigtriangledown }. \overrightarrow{E}) -\overrightarrow{\bigtriangledown } X \overrightarrow{\bigtriangledown } X \overrightarrow{E}

\bigtriangledown ^{2}\overrightarrow{E} = \overrightarrow{\bigtriangledown }(\overrightarrow{\bigtriangledown }. \overrightarrow{E}) -(-\mu (\overrightarrow{\bigtriangledown } X \frac{\partial \overrightarrow{ H}}{\partial t}))

\bigtriangledown ^{2}\overrightarrow{E} = \overrightarrow{\bigtriangledown }(\overrightarrow{\bigtriangledown }. \overrightarrow{E}) +\mu (\overrightarrow{\bigtriangledown } X \frac{\partial \overrightarrow{ H}}{\partial t})

From Equation (2) 

\mu (\overrightarrow{\bigtriangledown } X \frac{\partial \overrightarrow{ H}}{\partial t})=\mu \sigma \frac{\partial \overrightarrow{E}}{\partial t}+\mu \epsilon \frac{\partial ^{2}\overrightarrow{E}}{\partial t^{2}}

\bigtriangledown ^{2}\overrightarrow{E}-\overrightarrow{\bigtriangledown }(\overrightarrow{\bigtriangledown }.\overrightarrow{E})=\mu \sigma \frac{\partial \overrightarrow{E}}{\partial t}+\mu \epsilon \frac{\partial ^{2}\overrightarrow{E}}{\partial t^{2}}

\bigtriangledown ^{2}\overrightarrow{E}=\overrightarrow{\bigtriangledown }(\overrightarrow{\bigtriangledown }.\overrightarrow{E})+\mu \sigma \frac{\partial \overrightarrow{E}}{\partial t}+\mu \epsilon \frac{\partial ^{2}\overrightarrow{E}}{\partial t^{2}}

from Maxwell’s equation \overrightarrow{\bigtriangledown }.\overrightarrow{D} = \rho _{v}

\overrightarrow{\bigtriangledown }.\overrightarrow{E} = \frac{\rho _{v}}{\epsilon }

\bigtriangledown ^{2}\overrightarrow{E}=\overrightarrow{\bigtriangledown }(\frac{\rho _{v}}{\epsilon })+\mu \sigma \frac{\partial \overrightarrow{E}}{\partial t}+\mu \epsilon \frac{\partial ^{2}\overrightarrow{E}}{\partial t^{2}}---------EquationI

This is called Wave equation (Electric field)for a general medium when {\rho _{v}}\neq 0.

wave equation for Magnetic field of a general  medium is 

\bigtriangledown ^{2}\overrightarrow{H}=\overrightarrow{\bigtriangledown }(\frac{\rho _{v}}{\epsilon })+\mu \sigma \frac{\partial \overrightarrow{H}}{\partial t}+\mu \epsilon \frac{\partial ^{2}\overrightarrow{H}}{\partial t^{2}}---------EquationII.

Case 1:-  wave equation for a conducting medium

for a conductor the net charge inside an isolated conductor is \rho _{v}=0, then the wave equation for a conducting medium (\sigma \neq 0) is

\bigtriangledown ^{2}\overrightarrow{E}=\mu \sigma \frac{\partial \overrightarrow{E}}{\partial t}+\mu \epsilon \frac{\partial ^{2}\overrightarrow{E}}{\partial t^{2}}---------EquationI

\bigtriangledown ^{2}\overrightarrow{H}=\mu \sigma \frac{\partial \overrightarrow{H}}{\partial t}+\mu \epsilon \frac{\partial ^{2}\overrightarrow{H}}{\partial t^{2}}---------EquationII

The above equations are known as wave equations for conducting medium and are involving first and second order time derivatives, which are well known equations for damped(or) attenuated waves in absorbing medium of homogeneous, isotropic such as metallic conductor.

Case 2:-  Wave equation for free space/ Non-conducting medium/loss-less medium/Perfect Di-electric medium

The conditions of free space are \rho =0,\sigma =0,\overline{J} =0,\mu =\mu _{o} and \epsilon =\epsilon _{o}

By substituting the above equations in general wave equations, the resulting wave equations for non-conducting medium are

\bigtriangledown ^{2}\overrightarrow{E}=\mu_{o} \epsilon_{o} \frac{\partial ^{2}\overrightarrow{E}}{\partial t^{2}}---------EquationI

\bigtriangledown ^{2}\overrightarrow{H}=\mu_{o} \epsilon_{o} \frac{\partial ^{2}\overrightarrow{H}}{\partial t^{2}}---------EquationII.

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Automatic Gain Control (AGC)

Why AGC is required in a Radio Receiver?

Let us discuss about the facts why we need AGC in a Radio Receiver , as we all know that the voltage gain available at the Receiver from antenna to demodulator in several stages of amplification is very high, so that it can amplify a very weak signal But what if the signal is much stronger at the front end of the receiver ?

If same gain (gain maintained for an incoming weak signal) is maintained by different stages of the Receiver for a stonger  incoming signal, the signal is further amplified by these stages and the received signal strength is far beyond the expectations which can be avoided. so we need to have a mechanism which will measure the stength of the input signal and accordingly adjust the gain. AGC does precisely this job and improves the dynamic range of the antenna to (60-100)dB by adjusting the gain of the Intermediate Frequency and sometimes the Radio Frequency stages.

It is generally observed that as a result of fading, the amplitude of the IF carrier signal at the detecor input may vary  as much as 30 (or) 40 dB this results in the corresponding variation in general level of reproduced signal at the receiver output.

At IF carrier minimum loud speaker output becomes inaudible and mixed up with noise.

At IF  carrier maximum loud speaker output becomes intolerably large.

Therefore a properly designed AGC reduces the amplitude variation due to fading from a high value of (30-40)dB to (3-4)dB.

 Basic need of AGC or AVC:-

AGC is a sub system by means of which the overall gain of a receiver is varied automatically with the variations in the stregth of the received signal to keep the output substantially constant. 

i.e, the overall requirement of an AGC circuit in a receiver is to maintain a constant output level.

Some of the factors that explain why AGC is needed:-

  • When a Receiver without AGC/AVC is tuned to a strong station, the received signal may overload the subsequent IF and AF stages this overloading causes carrier distortion in the incoming signal this can be prevented by using manual gain control on first RF stage but now a days AGC circuits are used for this purpose.
  • When the Receiver is tuned from one station to another, difference in signal strengths of the two stations causes an unpleasant loud output if signal is moving from a weak station to a strong station unless we initially keep the volume control very low before changing the tuning from one station to another . Changing the volume control every time before attempting to re-tunethe receiver is howeve cumbersome. Therefore AGC/AVC enables the user to listen to a station without constantly monitoring the volume control.
  • AGC is particularly important for mobile Receivers.
  • AGC helps to smooth out the rapid fading which may occur with long distance short-wave reception.


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Energy Spectal Density of rectangular pulse and solved problems

As we all know the Rectangular pulse is defined as x(t) = rect(\frac{t}{T}), exists for a duration of T sec symmetrical with respect to y-axis as shown 

Fourier Transform is X(f) 

X(f) = \int_{\frac{-T}{2}}^{\frac{T}{2}}x(t)e^{-j2\pi ft}dt

X(f) = \int_{\frac{-T}{2}}^{\frac{T}{2}} 1.e^{-j2\pi ft}dt

X(f) = \left [ \frac{e^{-j2\pi ft}}{j2\pi f} \right ] ^{\frac{T}{2}}_{\frac{-T}{2}}

X(f) = \frac{1}{\pi f}\left [ \frac{e^{j\pi ft}-e^{-j\pi ft}}{2j} \right ]

X(f) = \frac{T \sin \pi fT}{\pi fT}

X(f) = X(f) = sinc (\pi fT) 

\therefore The Energy Spectral Density of the given signal x(t) will be S(f) = \left | X(f) \right |^{2}

s(f)= T^{2}sinc^{2}(\pi fT).

Pb2. Let x(t) = sinc(T\pi t) and  y(t) = \frac{dx(t)}{dt} find the Energy Spectral Density S(f) of y(t).

Ans:-  The Energy Spectral Density of y(t)is S(f)

i.e,  S(f) = \left | Y(f) \right |^{2}

y(t) \leftrightarrow \frac{dx(t)}{dt}, then Fourier Tansform of  y(t)is

Y(f) \leftrightarrow j2\pi f X(f)

as x(t) = sinc(\pi Tt)\leftrightarrow X(f)= \frac{1}{T}rect(\frac{f}{T})

then Y(f) \leftrightarrow j2\pi f X(f)

\left | Y(f) \right |^{2} \leftrightarrow \left | j2\pi f \frac{1}{T}rect(\frac{f}{T}) \right | ^{2}

\left | Y(f) \right |^{2}\leftrightarrow \frac{4\pi ^{2}f^{2}}{T^{2}}rect^{2}(\frac{f}{T})

\therefore S(f) = \frac{4\pi ^{2}f^{2}}{T^{2}}rect^{2}(\frac{f}{T})

Pb3. The Energy contained with in the band [0, f], f>0 E(f) = \frac{1}{\sigma ^{3}}( {2-(2+2\sigma f+\sigma ^{2}f^{2})e^{-\sigma f}}) then find the ESD S(f), for any f>0 Energy of a signal can be defined \int_{-\infty }^{\infty }\left | S(f) \right | df

Ans:- Given E(f) = \int_{0}^{f}\left | S(f) \right | df

S(f ) = \frac{dE(f)}{df}

E(f) = \frac{1}{\sigma ^{3}}( {2-(2+2\sigma f+\sigma ^{2}f^{2})e^{-\sigma f}})

\frac{dE(f)}{df} = 0- \frac{1}{\sigma ^{3}}[2(-\sigma )e^{-\sigma f}+2\sigma f(-\sigma )e^{-\sigma f}+2\sigma e^{-\sigma f}+\sigma ^{2}f^{2}(-\sigma )e^{-\sigma f}+\sigma ^{2}(2f )e^{-\sigma f}]

\frac{dE(f)}{df} = - \frac{1}{\sigma ^{3}}[-2\sigma e^{-\sigma f}-2\sigma^{2} fe^{-\sigma f}+2\sigma e^{-\sigma f}+\sigma ^{3}f^{2}e^{-\sigma f}+2\sigma ^{2}fe^{-\sigma f}]

\frac{dE(f)}{df} = - \frac{1}{\sigma ^{3}}\sigma ^{3}f^{2}e^{-\sigma f}

\therefore S(f)=f^{2}e^{-\sigma f}

Pb:-find the Auto correlation function of  x(t) = \frac{1}{\sqrt{2\pi }}e^{\frac{-t^{2}}{2}}.

Ans:-  Auto Correlation Function \leftrightarrowEnergy spectral Density

R_{xx} (t) \leftrightarrow S(f)

we know that Fourier Transform of e^{-ct^{2}}\leftrightarrow \sqrt{\frac{\pi }{c}}e^{\frac{-\pi ^{2}f^{2}}{c}}

By using the above rule

\frac{1}{\sqrt{2\pi }}e^{\frac{-t^{2}}{2}}\leftrightarrow \frac{1}{\sqrt{2\pi }} \sqrt{\frac{\pi }{(1/2)}}e^{\frac{-\pi ^{2}f^{2}}{(1/2)}}, here  c=1/2

\frac{1}{\sqrt{2\pi }}e^{\frac{-t^{2}}{2}}\leftrightarrow \frac{1}{\sqrt{2\pi }} \sqrt{{2\pi }}e^{-2\pi ^{2}f^{2}}

\frac{1}{\sqrt{2\pi }}e^{\frac{-t^{2}}{2}}\leftrightarrow e^{-2\pi ^{2}f^{2}}

\therefore X(f) =e^{-2\pi ^{2}f^{2}}

Now the Energy Spectral Density S(f) = \left | X(f) \right |^{2}

S(f) =e^{-4\pi ^{2}f^{2}}

by finding the inverse Fourier Transform of S(f)

e^{-4\pi ^{2}f^{2}}\Rightarrow e^{\frac{-\pi ^{2}f^{2}}{(1/4)}}

e^{\frac{-\pi ^{2}f^{2}}{(1/4)}}\leftrightarrow \frac{1}{\sqrt{4\pi }}e^{\frac{-t^{2}}{4}}, c=1/4

Auto Correlation Function = \frac{1}{2\sqrt{\pi }}e^{\frac{-t^{2}}{4}}

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Electric field due to infinite line charge distribution

Consider an infinitely long straight line carrying uniform line charge with density \rho _{L} C/m and lies on Z-axis from -\infty to +\infty.

Consider a point P at which Electric field intensity has to be determined which is produced by the line charge distribution.

from the figure let the co-ordinates of P are (0,\rho ,0) ( a point on y-axis) and assume dQ is a small differential charge confirmed to a point  M (0,0,Z) as co-ordinates.

\therefore dQ produces a differential field \overrightarrow{dE} 

\overrightarrow{dE}=\frac{dQ}{4\pi \epsilon _{o}R^{2}}\widehat{a_{r}}

the position vector \overrightarrow{R}=-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }} and the corresponding unit vector \widehat{a_{r}} =\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{\sqrt{\rho ^{2}+Z^{2}}}

\therefore \overrightarrow{dE} =\frac{dQ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})

therefore \overrightarrow{dE} =\frac{\rho _{L}dZ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})

then the Electric field strength \overrightarrow{E} produced by the infinite line charge distribution \rho _{L} is 

\overrightarrow{E} = \int \overrightarrow{dE}

\overrightarrow{E} = \int_{z=-\infty }^{\infty }\frac{\rho _{L}dZ}{4\pi \epsilon _{o}}({\frac{-Z\overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{(\rho ^{2}+Z^{2})^{\frac{3}{2}}}})

to solve this integral  let Z= \rho \tan \theta \Rightarrow dZ=\rho \sec ^{2}\theta d\theta

as Z\rightarrow -\infty \Rightarrow \theta \rightarrow \frac{-\pi }{2}

Z\rightarrow \infty \Rightarrow \theta \rightarrow \frac{\pi }{2}

\therefore \overrightarrow{E} = \int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} \frac{\rho _{L}}{4\pi\epsilon _{o}}(\frac{-\rho ^{2}\\sec ^{2}\theta \tan \theta d\theta \overrightarrow{a_{z}}+\rho ^{2}\sec ^{2}\theta d\theta \overrightarrow{a_{\rho }}}{(\rho ^{2}+\rho ^{2}\tan ^{2}\theta )^{\frac{3}{2}}})  

\overrightarrow{E} = \int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} \frac{\rho _{L}}{4\pi\epsilon _{o}}(\frac{-\rho ^{2}\\sec ^{2}\theta \tan \theta d\theta \overrightarrow{a_{z}}+\rho ^{2}\sec ^{2}\theta d\theta \overrightarrow{a_{\rho }}}{\rho ^{3}\sec ^{3}\theta })  

\overrightarrow{E} = \frac{\rho _{L}}{4\pi\epsilon _{o}\rho }(\int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} -\sin \theta \overrightarrow{a_{z} }d\theta +\int_{\theta = \frac{-\pi }{2}}^{ \frac{\pi }{2}} \cos\theta d\theta \overrightarrow{a_{\rho }})

\overrightarrow{E}= \frac{\rho _{L}}{4\pi\epsilon _{o}\rho }(0+2\overrightarrow{a_{\rho }})

\therefore \overrightarrow{E}= \frac{\rho _{L}}{2\pi\epsilon _{o}\rho }\overrightarrow{a_{\rho }} Newtons/Coulomb.

\overrightarrow{E} is a function of \rho only, there is no \overrightarrow{a_{z}} component and \rho is the perpendicular distance from the point P to line charge distribution \rho _{L}.



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Compensators are corrective sub systems to compensate the deficiency in the performance of the plant or system, so given a plant and a set of specifications suitable compensators are to be designed so that the overall system will meet given specifications. Proper selection of performance specifications is the most important step in the design of compensators.

i.e, All the control systems are designed to achieve specific objectives that is the requirements are defined for the control system. A good control system has less error, good accuracy, good speed of response, good relative stability, good damping which will not cause unusual Overshoots etc.

For stationary performance of the system, gain is adjusted first but gain adjustment alone can not provide satisfactory results. When gain increases, Steady-state behavior of system improves but results into poor Transient response (or) even instability.

The desired behavior of a system is specified in terms of 

  • Transient response.
  • Steady-state error(e_{ss}).

e_{ss} →is usually specified in terms of constants k_{a},k_{v} and k_{p} for u(t),r(t) and p(t) as inputs.

Transient response → it measures relative stability and speed of response which are specified in time or frequency domain.

In time domain the measure of relative stability is in terms of \xi or M_{p}, while the speed response is measured in terms of rise time t_{r} , settling time t_{s} or natural frequency \omega _{n}. Where as in frequency-domain the measure of relative stability is given by Resonant peak M_{r} or Phase margin \varphi _{pm} and the speed response is measured by Resonant frequency \omega _{r} or band width \omega _{b}.

Once a set of performance specifications have been selected, the next step is to chose the appropriate compensator. There exists Electrical,hydraulic,pneumatic and mechanical compensators and in this context we prefer Electrical compenators.

An external device which is used to alter the behavior of the system so as to achieve given specifications is called as compensator [ compensators are added to the original system].

Compensators can be added to the system in series or in parallel or in combination of both.

Series Compensation:-

The flow of signal in series scheme is from lower energy level towards higher energy level. This requires additional amplifiers to increase the gain and also provide necessary isolation. The number of components required in series scheme is more than in parallel scheme.


Parallel Compensation:-

In this compensation technique energy flow is from higher energy level to lower energy level.  As there is no need of any amplifiers additional components required are less.

Series-Parallel Compensation:-

This is a compensation technique which utilizes the advantages of both series and parallel compensation techniques.