Probability of error of ASK

Probability of error of ASK using coherent Detection:-

The equation of ASK from it’s basic definition

S_{ASK}(t)=\left\{\begin{matrix} \sqrt{2P_{s}}\cos 2\pi f_{c}t\ \rightarrow \ symbol\ '1'\\ 0\ \rightarrow \ symbol\ '0' \end{matrix}\right. .

The inputs to the optimum filter are x_{1}(t) \ and \ x_{2}(t)  when the symbols 1 and 0 are being transmitted at the transmitter.

\therefore x_{1}(t) = \sqrt{2P_{s}}\cos 2\pi f_{c}t .

x_{2}(t)=0.

we know that P_{e} of an optimum filter is P_{e} = \frac{1}{2}\ erfc(\frac{x_{o1}(t)-x_{o2}(t)}{2\sqrt{2}\sigma })

now chose the ratio \rho _{max}^{2} =(\frac{x_{o1}(t)-x_{o2}(t)}{\sigma })^{2}

from the Matched filter concept (\frac{x_{o1}(t)-x_{o2}(t)}{\sigma })^{2}=\frac{2}{N_{o}}\int_{-\infty }^{\infty }\left | X(f) \right |^{2}df-----EQN(1)

from Parsevel’s relation  \int_{-\infty }^{\infty }\left | X(f) \right |^{2}df =\int_{-\infty }^{\infty }\left | x(t) \right |^{2}dt----EQN(2)

from equations (1) and (2)

\rho _{max}^{2}=\int_{-\infty }^{\infty }\left | x(t) \right |^{2}dt

Here x(t) is the signal x(t) =x_{1}(t)-x_{2}(t) .

x(t) =\sqrt{2P_{s}}\cos 2\pi f_{c}t .

over a duration of [0, T_{b}] the symbols are transmitted

\int_{0}^{T_{b}}\left | x(t) \right |^{2}dt=2P_{s}\int_{0}^{T_{b}}\cos^{2} 2\pi f_{c}t \ dt .

                          =P_{s}T_{b}.

from equation(1)      \rho _{max}^{2} = \frac{2}{N_{o}} P_{s}T_{b}.

\rho _{max} = \sqrt{\frac{2P_{s}T_{b}}{N_{o}}} .

\therefore P_{e} = \frac{1}{2} \ erfc(\sqrt{\frac{2P_{s}T_{b}}{N_{o}}}).\frac{1}{2\sqrt{2}})

\therefore P_{e} = \frac{1}{2} \ erfc(\sqrt{\frac{P_{s}T_{b}}{4N_{o}}}) .

we know that carrier signal power is   P_{s} = \frac{A^{2}}{2}.

A=\sqrt{2P_{s}} .

\therefore P_{e} = \frac{1}{2} \ erfc(\sqrt{\frac{A^{2}T_{b}}{8N_{o}}}) .

\therefore P_{e}= \frac{1}{2} \ erfc(\sqrt{\frac{E_{b}}{4N_{o}}})  . since A^{2}T_{b} = P_{s}T_{b} =E_{b}.

probability of error  of coherent ASK is  P_{e}= \frac{1}{2} \ erfc(\sqrt{\frac{E_{b}}{4N_{o}}})   (or)  in terms of Q function as P_{e}= Q(\sqrt{\frac{E_{b}}{2N_{o}}}) .

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SNR in PCM (or) Signal-to-Quantization Noise Ratio in PCM system

we know that signal-to-Noise Ratio is defined as

SNR=\frac{S}{N}=\frac{Normalized\ signal\ power}{Normalized\ Noise \ power}.

let x(t) is a given signal with which is an arbitrary signal with Normalized signal power P watts.

and this x(t)  is some arbitrary signal oscillating between +x_{max}  and -x_{max} .

then the step size used in the Quantization process is \Delta =\frac{2 . x_{max}}{L}.

and L=2^{n} .

where L- is the number of Quantization levels.

             n- no.of bits required to encode each Quantization level.

\therefore \Delta =\frac{2 . x_{max}}{2^{n}} ----EQN(1).

Quantization Noise Power \bg_black N \ (or ) \ N_{Q} :-

if uniform (or) linear Quantization is used in PCM system, during the approximation process of x(t) with x_{q}(t) \ (or) \ \widehat{x}(t) there exists some error between these two signals . This error is called as Quantization error (or) noise.

x(t) \approx x_{q}(t)   

In discrete time domain e(nT_{S}) = x_{q}(nT_{S})-x(nT_{S}) .

Quantization error = Quantized signal- original signal.

we know that step size  is \Delta =\frac{2 . x_{max}}{L}.

Now to find out Quantization noise , assume it is uniformly distributed random variable .

now the Probability density function of this uniformly distributed random variable is f_{\epsilon }(\epsilon )

f_{\epsilon }(\epsilon ) = \left\{\begin{matrix} \frac{1}{\Delta } , \frac{-\Delta }{2}\leq0\leq \frac{\Delta }{2}.\\ 0,otherwise. \end{matrix}\right.

Mean square value of this random variable is with zero mean

E(\epsilon ^{2}) = \int_{-\infty }^{\infty } \epsilon ^{2}f_{\epsilon }(\epsilon )d\epsilon.

E(\epsilon ^{2}) = \int_{\frac{-\Delta }{2} }^{\frac{\Delta }{2} } \frac{1}{\Delta }d\epsilon .

simplification gives E(\epsilon ^{2}) = \frac{\Delta ^{2}}{12}.

Mean Square value= Quantization Noise Power.

\therefore N_{q} = \frac{\Delta ^{2}}{12}----EQN(2).

by substituting \Delta in equation (2) ,

N_{q} = \frac{(\frac{2 . x_{max}}{2^{n}})^{2}}{12} .

N_{q} = \frac{x_{max}^{2}}{3X2^{2n}} .

\therefore SQR \ in \ PCM system=\frac{Signal \ power}{Noise \ power} .

SQR = \frac{P}{\frac{x_{max}^{2}}{3(2^{2n})}} .

\frac{S}{N} = \frac{S}{N_q} = SQR=\frac{3P2^{2n}}{x_{max}^{2}} .

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SNR in DM (or) Signal-to-Quantization Noise Ratio in Delta Modulation system

we know that signal-to-Noise Ratio is defined as

SNR=\frac{S}{N}=\frac{Normalized\ signal\ power}{Normalized\ Noise \ power}.

let x(t) is a given signal which is a single tone signal x(t) =A_{m} \cos 2\pi f_{m}t \ where \ \omega _{m} = 2\pi f_{m}.

The maximum value of RMS signal power is P_{rms} = \frac{A_{m}^{2}}{2R}.

Normalized signal power P=\frac{A_{m}^{2}}{2}    with R=1.

we know that slope overload distortion can be eliminated if and only if A_{m}\leq \frac{\Delta}{2\pi f_{m}T_{s}}.

let A_{m}= \frac{\Delta}{2\pi f_{m}T_{s}}

By substituting A_{m} value in P  then the power results to be P= \frac{(\frac{\Delta}{2\pi f_{m}T_{s}})^{2}}{2} .

P= \frac{\Delta^{2}}{8\pi^{2} f_{m}^{2}T_{s}^{2}}----EQN(1)

Quantization Noise Power \bg_black N \ (or ) \ N_{Q} :-

if uniform (or) linear Quantization is used in DM system, during the approximation process of x(t) with x_{q}(t) \ (or) \ \widehat{x}(t) there exists some error between these two signals . This error is called as Quantization error (or) noise.

x(t) \approx x_{q}(t)   (approximation process)

In discrete time domain e(nT_{S}) = x_{q}(nT_{S})-x(nT_{S}) .

Quantization error = Quantized signal- original signal.

Now to find out Quantization noise , assume it is uniformly distributed random variable (+\Delta ,-\Delta )

now the Probability density function of this uniformly distributed random variable is f_{\epsilon }(\epsilon )

f_{\epsilon }(\epsilon ) = \left\{\begin{matrix} \frac{1}{2\Delta } , -\Delta \leq0\leq \Delta .\\ 0,otherwise. \end{matrix}\right.

Mean square value of this random variable is with zero mean

E(\epsilon ^{2}) = \int_{-\infty }^{\infty } \epsilon ^{2}f_{\epsilon }(\epsilon )d\epsilon.

E(\epsilon ^{2}) = \int_{\Delta }^{\Delta }\epsilon ^{2} \frac{1}{2\Delta }d\epsilon .

simplification gives E(\epsilon ^{2}) = \frac{\Delta ^{2}}{3}.

Mean Square value= Quantization Noise Power.

\therefore N_{q} = \frac{\Delta ^{2}}{3}----EQN(2).

The M signal is passed through a reconstruction Low pass Filter at the output of a DM Receiver . The Band width of this filter is f_{M}  in such a way that f_{M}\geq f_{m} \ and \ f_{M}< < f_{s}.

where f_{s}  is the sampling frequency of the signal.

now assume that Quantization noise is distributed over a frequency band up to f_{s}  and is given by \frac{\Delta ^{2}}{3} .

then the noise power N_{q}^{'}  distributed over f_{M}   will be

N_{q}^{'} = \frac{f_{M}}{f_{s}}\frac{\Delta ^{2}}{3}---EQN(3) .

\therefore SQR \ in \ DM system=\frac{Signal \ power}{Noise \ power} .

\frac{S}{N} = \frac{S}{N_q^{'}} = SQR=\frac{\frac{\Delta^{2}}{8\pi^{2} f_{m}^{2}T_{s}^{2}}}{\frac{f_{M}}{f_{s}}\frac{\Delta ^{2}}{3}} .

SQR_{DM} = \frac{3}{8\pi ^{2}f_{m}^{2}T_{s}^{3}f_{M}} \ where \ T_{s}=\frac{1}{f_{s}}.

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Optimum filter

The function of a receiver in a binary Communication system is to distinguish between two transmitted signals x_{1}(t)\ and \ x_{2}(t)  (or) (s_{1}(t)\ and \ s_{2}(t)) in the presence of noise.

The performance of Receiver is usually measured in terms of the probability of error Pe and the receiver is said to be optimum if it yields the minimum probability of error.

i.e, optimum receiver is the one with minimum probability of error Pe .

optimum receiver takes the form of Matched filter when the noise at the receiver input is white noise.

optimum receiver (or) optimum filter: –

The block diagram of optimum receiver is as shown in the figure below

the decision boundary is set to \frac{x_{o1}(T)+x_{o2}(T)}{2} .

Probability of error of optimum filter:-

The probability of error can be obtained as similar to Integrate and dump receiver. Here we will consider noise as Gaussian Noise.

The output of optimum filter is  y(t) = x_{o1}(t)+n_{o}(t) .

The output of sampler is  y(T) = \left\{\begin{matrix} x_{o1}(T)+n_{o}(T) \ for \ binary \ i/p \ '1'\\ x_{o2}(T)+n_{o}(T) \ for \ binary \ i/p \ '0' \end{matrix}\right.

suppose if Binary ‘1’ is transmitted then the input is x(t) = x_{1}(t) , to find the probability of error this transmitted ‘1’ should be received as ‘0’.

this is possible  when the condition  \left | y(T) \right | <\frac{x_{o1}(T)+x_{o2}(T)}{2} is true.

1 will be received as 0    \Rightarrow x_{o1}(T)+n_{o}(T) <\frac{x_{o1}(T)+x_{o2}(T)}{2} .

n_{o}(T) <\frac{x_{o2}(T)-x_{o1}(T)}{2} .

similarly a Binary ‘0’ will be  received as ‘1’ if and only if 

 is true.

1 will be received as 0     .

 .

the conditions are  summarized in the table

Noe the Probability Distribution Function of Gaussian noise with zero mean and standard deviation \sigma  is given by

f(n_{o}(T)) = \frac{1}{\sigma \sqrt{2\pi }} e^{-\frac{n_{o}^{2}(T)}{2}} .

Probability of error= probability ‘1’ will be received as ‘0’ =probability ‘0’ will be received as ‘1’.

\therefore P_{e} =  area under the curve 

   (or) area under the curve n_{o}(T) <\frac{x_{o2}(T)-x_{o1}(T)}{2} .

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Internetworking

until now, we assumed to a Network means homogeneous, which consists of number of machines all are using same protocol in each layer.

But in real a number of networks like LANs, WANs with numerous protocols in each layer are connected together to form internet.

The main reasons for internetworking are:

  • Some computers use TCP/IP, but some may use IBM’s SNA and a substantial number of telephone companies uses ATM N/W’s and some are still using Novell NCP/IPX (or) Apple talk.

i.e, The installed base of different N/W’s is large.

  • The cost of computers and N/W’s cheaper there is a possibility of having more N/W’s and protocols.

Let’s see an example, How Networks are connected? 

In this figure we have wide area ATM network ,  FDDI ring, wireless LAN and SNA’s main frame Network all are connected together.

The main reason for interconnecting all these Network’s is to 

  1. Allow users in one network will communicate with users in other networks.
  2. and also, to access data (from one network to other).

This happens when we exchange packets between number of networks (which is not a simple task).

How Networks differ?

Networks may differ in different ways, 

i.e, at physical layer- The modulation techniques differ at DLL – frame formats.

we just see here the differences in network layer.

when a sender in one network need to send packets to different Networks at the interfaces between Networks many problems may occur.

when packets from a CO-N/W to CL-N/W flows, we may require protocol conversion.

In some cases, Address conversions also be needed (i.e, a kind of directory system is required). Passing Multicast packets to a Network, which may not support Multicasting is a problem.

Packet size in different N/W’s is a big issue, a N/W supports a packet of size 8000 bytes when given to a N/W packet size as 1500 bytes.

i.e, The problems we may face when N/W’s differ are listed below. 

 

aliasing effect in Sampling

Effect of under sampling (aliasing effect):-

When a CT band limited signal is sampled at  f_{s} < 2f_{m} , then the successive cycles of the spectrum of the sampled signal overlap with each other as shown below

Some aliasing is produced in the signal this is due to under sampling.

aliasing is the phenomenon in which a high frequency component in the frequency spectrum of the signal takes as a low frequency component in the spectrum of the sampled signal.

Because of aliasing it is not possible to reconstruct x(t) from g(t) by low pass filtering.

The spectral components in the overlapping regions and hence the signal is distorted.

Since any information signal contains a large no.of frequencies so the decision of sampling frequency is always become a problem.

A signal is first passed through LPF  before sampling.

i.e, it is band limited by this LPF which is known as pre-alias filter.

To avoid aliasing

  1. Pre-alias filter must be used to limit the band width of the signal to f_{m}  Hz.
  2. Sampling frequency must be

Pre-alias filter means before sampling is passed through a LPF to make a perfect band limited signal.

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Capacitance of a Co-axial cable

A Co-axial cable is a Transmission line, in which two conductors are placed co-axially and are separated by some dielectric material with dielectric constant (or) permittivity  (\epsilon ).

a conductor is in the form of a cylinder with some radius, let the radius of inner conductor is ‘a’ meters and that of outer conductor be ‘b’ meters.

Now connect this co-axial conductor to a supply of ‘V’ volts , after applying ‘V’ assume positive charges are distributed on M_{2} and negative charges on M_{1} .

Now, a field is induced \overrightarrow{E} between M_{2}  and M_{1} because of flux lines, to find out \overrightarrow{E} at any point P  between these two conductors

location of P is out of the conductor M_{2} an inside the conductor M_{1}.

\therefore \overrightarrow{E}_{at P} = \overrightarrow{E}_{\ due \ to \ inner \ conductor \ M_{1}} .

assume a cylindrical co-ordinate system \rho ,\ \phi , \ z  and axis of cable coincides with z-axis this is similar to a line charge distribution \rho_{L} placed along the z-axis.

\rho _{L}=\frac{Q}{L} .

\therefore \overrightarrow{E}_{at P} = \frac{\rho_ {L}}{2\pi \epsilon _{o}\rho }\overrightarrow{a}_{\rho } .

\therefore V = -\int_{1}^{2}\overrightarrow{E}.\overrightarrow{dl} .

V = -\int_{1}^{2} \frac{\rho_ {L}}{2\pi \epsilon _{o}\rho }\overrightarrow{a}_{\rho }.(d\rho \overrightarrow{a}_{\rho }+d\phi \overrightarrow{a}_{\phi }+dz \overrightarrow{a}_{z }) .

V = -\int_{b}^{a} \frac{\rho_ {L}}{2\pi \epsilon _{o}\rho }\overrightarrow{a}_{\rho }.(d\rho \overrightarrow{a}_{\rho }) .

V = - \frac{\rho_ {L}}{2\pi \epsilon _{o} }(\ln a-\ln b) .

V = \frac{\rho_ {L}}{2\pi \epsilon _{o} }(\ln b-\ln a) .

V = \frac{\rho_ {L}}{2\pi \epsilon _{o} }\ln (\frac{b}{a}) .

V = \frac{Q}{2\pi \epsilon _{o}L }\ln (\frac{b}{a}) \ \because \ \rho _{L} = \frac{Q}{L} .

\therefore C_{co-axial} =\frac{Q}{V} = \frac{2\pi \epsilon_{o}L}{\ln (\frac{b}{a})} .

L- length of the conductors.

b-radius of the outer conductor.

a- radius of the inner conductor.

\epsilon – permittivity of the medium.

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Capacitance of a spherical conductor

choose two spherical conductor systems as shown in the figure with the inner conductor – M_{2}  -radius-a   and outer conductor – M_{1}  -radius-b .

Now induced field is directed from M_{2}   to  M_{1} .  then the potential difference between the two conductors is 

V= -\int_{1}^{2}\overrightarrow{E} .\overrightarrow{dl} .

by assuming a point P between the two conductors such that P is out of inner spherical conductor ( M_{2} ) an inside the outer conductor ( M_{1} ).

\therefore \overrightarrow{E}_{at P} = \frac{Q}{4\pi \epsilon _{o}r^{2} }\overrightarrow{a}_{r} .

\therefore V = -\int_{1}^{2}\overrightarrow{E}.\overrightarrow{dl} .

V = -\int_{1}^{2} \frac{Q}{4\pi \epsilon _{o}r^{2} }\overrightarrow{a}_{r}.(dr \overrightarrow{a}_{r }+d\phi \overrightarrow{a}_{\phi }+d\theta \overrightarrow{a}_{\theta }) .

V = -\int_{r=b}^{a} \frac{Q}{4\pi \epsilon _{o}r^{2} }\overrightarrow{a}_{r}.(dr \overrightarrow{a}_{r }+d\phi \overrightarrow{a}_{\phi }+d\theta \overrightarrow{a}_{\theta }) .

V = - \frac{Q}{4\pi \epsilon _{o} }(\frac{1}{a}-\frac{1}{b}) .

V = \frac{Q}{4\pi \epsilon _{o} }(\frac{1}{b}-\frac{1}{a}) .

\therefore C_{spherical} =\frac{Q}{V} = \frac{4\pi \epsilon _{o}}{(\frac{1}{b}-\frac{1}{a}) } .

b-radius of the outer conductor.

a- radius of the inner conductor.

\epsilon – permittivity of the medium.

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Capacitance (C)- Farads

When two conductors are embedded in a homogeneous dielectric medium as shown in the figure

M_{2} – conductor 2 carries a positive charge (+Q) and M_{1} -conductor 1 carries a negative charge (-Q). Assume the two charges are equal.

\therefore The resultant charge is zero.

we know that charge is carried on the surface as a surface charge density \rho _{s} and \overrightarrow{E}  is normal to the conducting surface.

since M_{2}  carries positive charge the flux is directed from M_{2}  to  M_{1} . flux starts from positive charge and ends at negative charge.

i.e, flux lines leaving one conductor must terminate at the surface of the other conductor.

\therefore  a field is induced between the two conductors  M_{2}  and  M_{1}  and the medium between these two conductors is a dielectric material.

Now, work must be done to carry a positive charge from M_{1} to  M_{2} , that is opposite to the induced field.

V_{21}= -\int_{2}^{1}\overrightarrow{E_{induced}}.\overrightarrow{dl}

V_{21}=V_{1}-V_{2} .

V= -\int_{2}^{1}\overrightarrow{E_{induced}}.\overrightarrow{dl}.

Now, a system as shown in the above figure will have a special physical property called capacitance.

Definition of Capacitance of a conductor:-

The capacitance of a conductor is defined as the physical property of the conductor which is the ability to store Electrical Energy

i.e, C=\frac{Q}{V}  Farads

(or)

Capacitance is the ration of charge of the conductor (either m M_{1} (or)  M_{2} ) to the voltage applied between the two conductors.

Procedure to find Capacitance:-

  • Assume Q  and determine V in terms of Q , or else assume V and determine Q in terms of V.
  • choose a suitable Co-ordinate system.
  • assume the two conducting plates are carrying charges +Q and -Q.
  • Determine  using Gauss’s law (or) by using Coloumb’s law  and find V= - \int \overrightarrow{E}.\overrightarrow{dl} .
  • Finally calculate  C=\frac{Q}{V} .

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Capacitance of Parallel Plate Capacitor

Choose two parallel conducting plates with charge densities separated by a distance ‘d’ meters as shown in the figure.

Assume charges are distributed uniformly on the plates.

Now apply a voltage source ‘V’ to these plates, then all positive charges are accumulated on conductor M_{2} similarly negative charges are accumulated on conductor  M_{1} .

\therefore the charges give rise to a field \overrightarrow{E}  in between them called as induced electric field.

To find out capacitance, choose a co-ordinate system x, y and z as shown in the figure

\therefore V =-\int \overrightarrow{E}.\overrightarrow{dl}.

Noe the potential difference is V=V_{2}-V_{1}.

V_{12}=V_{2}-V_{1}.

\therefore V =-\int_{1}^{2} \overrightarrow{E_{induced}}.\overrightarrow{dl}.

\overrightarrow{E}  is directed from 2 to 1 work has to be done in opposite direction from 1 to 2.

Now \rho _{s}=\frac{Q}{S} . assume two conducting plates has equal surface area S

\therefore  to find out \overrightarrow{E} at any point P in between the ‘2’ plates,  use the concept of infinite sheet of charge distributions with densities  \rho _{s}  and   -\rho _{s} .

E_{\rho _{s}} = \frac{\rho _{s}}{2\epsilon _{o}}(-\overrightarrow{a_{z}}) — from the positive charge distribution.

E_{-\rho _{s}} = \frac{-\rho _{s}}{2\epsilon _{o}}(\overrightarrow{a_{z}}) –with the negative charge distribution.

\therefore  Electric field intensity at P is the sum of electric field intensities due to two infinite charge distributions

\overrightarrow{E_{total}} = \overrightarrow{E_{\rho _{s}}}+\overrightarrow{E_{-\rho _{s}}} .

\overrightarrow{E_{total}} = \frac{\rho _{s}}{2\epsilon _{o}}(-\overrightarrow{a_{z}})+ \frac{-\rho _{s}}{2\epsilon _{o}}(\overrightarrow{a_{z}}) .

\overrightarrow{E_{total} }= \frac{\rho _{s}}{\epsilon _{o}}(-\overrightarrow{a_{z}}) .

Now the potential difference between the two conductors is \therefore V =-\int_{1}^{2} \overrightarrow{E_{induced}}.\overrightarrow{dl} .

\therefore V =-\int_{1}^{2} \frac{\rho _{s}}{\epsilon _{o}}(-\overrightarrow{a_{z}}).\overrightarrow{dl} .

we know that \overrightarrow{dl} = dx\ \overrightarrow{a_{x}}+ dy\ \overrightarrow{a_{y}}+ dz\ \overrightarrow{a_{z}} .

\therefore V =\int_{1}^{2} \frac{\rho _{s}}{\epsilon _{o}}(\overrightarrow{a_{z}}).\ dz \ \overrightarrow{a_{z}} .

V =\int_{0}^{d} \frac{\rho _{s}}{\epsilon _{o}}.\ dz .

V = \frac{\rho _{s}d }{\epsilon _{o}} .

V = \frac{Q\ d }{S\ \epsilon _{o}} \ \because \rho _{s}=\frac{Q}{S} .

\therefore C= \frac{Q}{V}=\frac{S\ \epsilon _{o} }{d} .

If the medium between two parallel plates is air  (or) free space (or) Vaccum  use \epsilon _{o}  or else use \epsilon .

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Infinite Line equivalent circuit

Consider the basic form of Transmission line with some impedance Z_{o} at the Load end.

an infinite line can be approximated by an equivalent finite line with load impedance Z_{o} as shown in the above figure, then the input impedance can be calculated from the voltage and current equations.

now at x= l , V=V_{R} \ and \ I=I_{R} .

V_{R}= V_{s}\cos h\gamma l-I_{s}Z_{o}\sin h\gamma l-----EQN(1).

I_{R}= I_{s}\cos h\gamma l-\frac{V_{s}}{Z_{o}}\sin h\gamma l----EQN(2).

The load voltage is given by the equation V_{R}=I_{R}Z_{o}.

V_{s}\cos h\gamma l-I_{s}Z_{o}\sin h\gamma l = Z_{o}(I_{s}\cos h\gamma l-\frac{V_{s}}{Z_{o}}\sin h\gamma l)

V_{s}Z_{O}\cos h\gamma l-I_{s}Z_{o}^{2}\sin h\gamma l = Z_{o}(I_{s}Z_{o}\cos h\gamma l-V_{S}\sin h\gamma l)

V_{s}(Z_{O}\cos h\gamma l+Z_{R}\sin h\gamma l) = I_{s}Z_{o}(Z_{o}\sin h\gamma l+Z_{R}\cos h\gamma l)

Z_{S}=\frac{V_{S}}{I_{S}}=Z_{o} \frac{(Z_{o}\cos h\gamma l+Z_{o}\sin h\gamma l)}{(Z_{o}\cos h\gamma l+Z_{o}\sin h\gamma l)}

Z_{S} \ (or) \ Z_{in}=Z_{o}.

represents the source (or) input impedance of an infinite Transmission Line.

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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 }

\overrightarrow{E}= \frac{\rho _{L}}{4\pi\epsilon _{o}\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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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  .

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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Application of Ampere’s circuit law-infinite sheet of current

consider an infinite sheet in the z=0 plane, which has uniform current density  \overrightarrow{k} = k_{y}\overrightarrow{a_{y}}    A/m .

Let us suppose the current is flowing in the positive y direction.

the sheet of current is assumed to be in rectangular co-ordinate system

\overrightarrow{H} = H_{x}\overrightarrow{a_{x}}+H_{y}\overrightarrow{a_{y}}+H_{z}\overrightarrow{a_{z}}

Let us suppose the conductor is carrying a current I , by right hand thumb rule magnetic field is produced around the conductor is right angles to the direction of I.

In this case of infinite sheet , the current is in the y-direction there is no component of H along the direction of y  and also the z components cancel each other because of opposite direction of the fields produced so only x components of H exists.

\overrightarrow{H} = \left\{\begin{matrix} H_{o}\ \overrightarrow {a_{x}} \ for \ z>0\\ -H_{o}\ \overrightarrow {a_{x}} \ for \ z<0 \end{matrix}\right.

from Ampere’s Circuit law  \oint \overrightarrow{H}. \overrightarrow{dl} = \int \left (\overrightarrow{H}_{1-2}. \overrightarrow{dl}_{1-2} +\overrightarrow{H}_{2-3}. \overrightarrow{dl}_{2-3}+\overrightarrow{H}_{3-4}. \overrightarrow{dl}_{3-4}+\overrightarrow{H}_{4-1}. \overrightarrow{dl}_{4-1} \right ) .

the component \oint \overrightarrow{H}. \overrightarrow{dl} = \int \left (\overrightarrow{H}_{1-2}. \overrightarrow{dl}_{1-2} +\overrightarrow{H}_{2-3}. \overrightarrow{dl}_{2-3}+\overrightarrow{H}_{3-4}. \overrightarrow{dl}_{3-4}+\overrightarrow{H}_{4-1}. \overrightarrow{dl}_{4-1} \right )

\overrightarrow{H}_{1-2}. \overrightarrow{dl}_{1-2}= \overrightarrow{H}_{1-0}. \overrightarrow{dl}_{1-0}+\overrightarrow{H}_{0-2}. \overrightarrow{dl}_{0-2}

\overrightarrow{H}_{1-2}. \overrightarrow{dl}_{1-2}= H_{o} \overrightarrow{a_{x}}.(\frac{a}{2})(-\overrightarrow{a_{z}})- H_{o} \overrightarrow{a_{x}}.(\frac{a}{2})(-\overrightarrow{a_{z}}) .

\overrightarrow{H}_{1-2}. \overrightarrow{dl}_{1-2}=0 .

similarly \overrightarrow{H}_{3-4}. \overrightarrow{dl}_{3-4} =0 .

\therefore \oint \overrightarrow{H}. \overrightarrow{dl} = \int \left (\overrightarrow{H}_{2-3}. \overrightarrow{dl}_{2-3}+\overrightarrow{H}_{4-1}. \overrightarrow{dl}_{4-1} \right) .

\oint \overrightarrow{H}. \overrightarrow{dl} = -(H_{o}\ \overrightarrow{a_{x}}).(b \ -\overrightarrow{a_{x}})+(H_{o}\ \overrightarrow{a_{x}}).(b \ \overrightarrow{a_{x}}) .

\oint \overrightarrow{H}. \overrightarrow{dl} = 2H_{o}\ b .

I= 2H_{o} \ b .

k_{y}\ b = 2H_{o} \ b .

H_{o} =\frac{1}{2} k_{y} .

 as   \overrightarrow{H} = \left\{\begin{matrix} H_{o}\ \overrightarrow {a_{x}} \ for \ z>0 \\ -H_{o}\ \overrightarrow {a_{x}} \ for \ z<0 \end{matrix}\right.

this will be changed to \overrightarrow{H} = \left\{\begin{matrix} \frac{1}{2} k_{y}\ \overrightarrow {a_{x}} \ for \ z>0 \\ -\frac{1}{2} k_{y}\ \overrightarrow {a_{x}} \ for \ z<0 \end{matrix}\right.

In general for a finite sheet of current density  \overrightarrow{k}  A/m  Magnetic field is generalised as     \overrightarrow{H} = \frac{1}{2} (\overrightarrow{k} X \overrightarrow{a_{n}}) .

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Application of Ampere’s circuit law-infinite sheet of current

consider an infinite sheet in the z=0 plane, which has uniform current density  \overrightarrow{k} = k_{y}\overrightarrow{a_{y}}    A/m .

Let us suppose the current is flowing in the positive y direction.

the sheet of current is assumed to be in rectangular co-ordinate system

\overrightarrow{H} = H_{x}\overrightarrow{a_{x}}+H_{y}\overrightarrow{a_{y}}+H_{z}\overrightarrow{a_{z}}

Let us suppose the conductor is carrying a current I , by right hand thumb rule magnetic field is produced around the conductor is right angles to the direction of I.

In this case of infinite sheet , the current is in the y-direction there is no component of H along the direction of y  and also the z components cancel each other because of opposite direction of the fields produced so only x components of H exists.

\overrightarrow{H} = \left\{\begin{matrix} H_{o}\ \overrightarrow {a_{x}} \ for \ z>0\\ -H_{o}\ \overrightarrow {a_{x}} \ for \ z<0 \end{matrix}\right.

from Ampere’s Circuit law  \oint \overrightarrow{H}. \overrightarrow{dl} = \int \left (\overrightarrow{H}_{1-2}. \overrightarrow{dl}_{1-2} +\overrightarrow{H}_{2-3}. \overrightarrow{dl}_{2-3}+\overrightarrow{H}_{3-4}. \overrightarrow{dl}_{3-4}+\overrightarrow{H}_{4-1}. \overrightarrow{dl}_{4-1} \right ) .

the component \oint \overrightarrow{H}. \overrightarrow{dl} = \int \left (\overrightarrow{H}_{1-2}. \overrightarrow{dl}_{1-2} +\overrightarrow{H}_{2-3}. \overrightarrow{dl}_{2-3}+\overrightarrow{H}_{3-4}. \overrightarrow{dl}_{3-4}+\overrightarrow{H}_{4-1}. \overrightarrow{dl}_{4-1} \right )

\overrightarrow{H}_{1-2}. \overrightarrow{dl}_{1-2}= \overrightarrow{H}_{1-0}. \overrightarrow{dl}_{1-0}+\overrightarrow{H}_{0-2}. \overrightarrow{dl}_{0-2}

\overrightarrow{H}_{1-2}. \overrightarrow{dl}_{1-2}= H_{o} \overrightarrow{a_{x}}.(\frac{a}{2})(-\overrightarrow{a_{z}})- H_{o} \overrightarrow{a_{x}}.(\frac{a}{2})(-\overrightarrow{a_{z}}) .

\overrightarrow{H}_{1-2}. \overrightarrow{dl}_{1-2}=0 .

similarly \overrightarrow{H}_{3-4}. \overrightarrow{dl}_{3-4} =0 .

\therefore \oint \overrightarrow{H}. \overrightarrow{dl} = \int \left (\overrightarrow{H}_{2-3}. \overrightarrow{dl}_{2-3}+\overrightarrow{H}_{4-1}. \overrightarrow{dl}_{4-1} \right) .

\oint \overrightarrow{H}. \overrightarrow{dl} = -(H_{o}\ \overrightarrow{a_{x}}).(b \ -\overrightarrow{a_{x}})+(H_{o}\ \overrightarrow{a_{x}}).(b \ \overrightarrow{a_{x}}) .

\oint \overrightarrow{H}. \overrightarrow{dl} = 2H_{o}\ b .

I= 2H_{o} \ b .

k_{y}\ b = 2H_{o} \ b .

H_{o} =\frac{1}{2} k_{y} .

 as   \overrightarrow{H} = \left\{\begin{matrix} H_{o}\ \overrightarrow {a_{x}} \ for \ z>0 \\ -H_{o}\ \overrightarrow {a_{x}} \ for \ z<0 \end{matrix}\right.

this will be changed to \overrightarrow{H} = \left\{\begin{matrix} \frac{1}{2} k_{y}\ \overrightarrow {a_{x}} \ for \ z>0 \\ -\frac{1}{2} k_{y}\ \overrightarrow {a_{x}} \ for \ z<0 \end{matrix}\right.

In general for a finite sheet of current density  \overrightarrow{k}  A/m  Magnetic field is generalised as     \overrightarrow{H} = \frac{1}{2} (\overrightarrow{k} X \overrightarrow{a_{n}}) .

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Ampere’s Circuit law

Ampere’s Circuit law states that the line integral of the tangential component of \overrightarrow{H} around a closed path is the same as the net current (Ienc) enclosed by the path.

 i.e, \oint \overrightarrow{H} .\overrightarrow{dl} = I_{enclosed} .

This is similar to Gauss’s law and can be applied to determine \overrightarrow{H} when the current distribution is symmetrical it’s a special case of Biot-savart’s law.

Proof:-

Consider a circular loop which encloses a current element . Let the current be in upward direction then the field is in anti- clock wise .

The current which is enclosed by the circular loop is of infinite length then  \overrightarrow{H}at  any point A is given by

\overrightarrow{H} = \frac{I_{enc}}{2\pi R} \overrightarrow{a}_{\phi } .

                          \overrightarrow{H}.\overrightarrow{dl} = \frac{I_{enc}}{2\pi R} \overrightarrow{a}_{\phi }.dl \overrightarrow{a}_{\phi } .

                        \overrightarrow{H}.\overrightarrow{dl} = \frac{I_{enc}}{2\pi R}.dl.

                      \overrightarrow{H}.\overrightarrow{dl} = \frac{I_{enc}}{2\pi R}.R d\phi .

                     \overrightarrow{H}.\overrightarrow{dl} = \frac{I_{enc}}{2\pi }d\phi .

                   \oint \overrightarrow{H} . \overrightarrow{dl} = \int_{\phi =0}^{2\pi } \frac{I_{enc}}{2\pi }d\phi .

                  \oint \overrightarrow{H} .\overrightarrow{dl} = I_{enc} .

which is known as the integral form of Ampere’s circuit law.

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Amplitude Shift Keying(ASK)

Amplitude Shift Keying (ASK) (or) On Off Keying (OOK) is the simplest Digital Modulation technique.

In this method, carrier amplitude is switched between two voltages ON and OFF levels depending up on the input binary sequence.

The carrier signal is a continuous wave (or) sinusoidal wave form

S(t)=A \cos 2\pi f_{c}t .

The normalized power is P=\frac{A^{2}}{2}

A=\sqrt{2P_{s}} .

The carrier signal can be expresses in terms of power as S(t)=\sqrt{2P_{s}} \cos 2\pi f_{c}t.

if energy per bit is E_{b} and the bit interval as T_{b} then the carrier signal is S(t)=\sqrt{\frac{2E_{b}}{T_{b}}} \cos 2\pi f_{c}t.

Now according to ASK Binary ‘1’ is represented with carrier voltage and Binary ‘0’ is represented with zero voltage.

\left\{\begin{matrix} S_{ASK}(t)=\sqrt{2P_{s}} \cos 2\pi f_{c}t\rightarrow \ Binary\ '1' \\ =0 \ \rightarrow \ Binary\ '0' \end{matrix}\right.

in terms of Energy and bit duration ASK signal can be written as 

\left\{\begin{matrix} S_{ASK}(t)=\sqrt{\frac{2E_{b}}{T_{b}}} \cos 2\pi f_{c}t\rightarrow \ Binary\ '1' \\ =0 \ \rightarrow \ Binary\ '0' \end{matrix}\right..

ASK Transmitter:-

The figure shows the ASK generator (or) ASK Transmitter

It is a simple product Modulator, which modulates the incoming binary sequence (in the form of a signal) with the carrier signal S(t)

i.e, S_{ASK}(t)=b(t).S(t)

b(t) represents the binary sequence in the form of a signal.

when the  input bit (or) symbol is Binary ‘1’ product Modulator passes the carrier signal and for Binary’0′, A zero output is given which blocks the carrier signal.

\left\{\begin{matrix} S_{ASK}(t)=\sqrt{2P_{s}} \cos 2\pi f_{c}t\rightarrow \ Binary\ '1' \\ =0 \ \rightarrow \ Binary\ '0' \end{matrix}\right..

Coherent ASK Detector:-

The figure shows the Block Diagram of coherent ASK/BASK Detector. The ASK signal S_{ASK}(t) is applied to the correlator ( The Block product Modulator followed up by the Integrator).

S_{ASK}(t) is multiplied by local carrier C(t) this carrier C(t) is phase locked with that of the carrier used in the Transmitter. As this is coherent reception.

The product S_{ASK}(t).C(t) is applied to the Integrator. The Integrator integrates the input over one bit interval T_{b} and the output is given to a threshold device. If the threshold voltage is set to 0 V.

the output of threshold device v(t) (or) v is either ‘1’ (or) ‘0’ based on the following condition.

                                                                     v> 0\rightarrow \ a \ symbol \ '0' \ is \ detected.

v\leq 0\rightarrow \ a \ symbol \ '1' \ is \ detected.

Note:- The input to demodulator is not S_{ASK}(t) always most of the times it is interfered with noise n(t) in the channel.

in coherent detection input to the demodulator is simply S_{ASK}(t) signal where as in Non-coherent detection the input is noisy ASK signal.

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Working of PN-junction Diode under Forward and Reverse Bias Conditions

In order to consider the working of a diode, we shall consider the effect of forward and Reverse Bias across PN-junction.

Forward Bias:-

Forward Bias means the Positive terminal of the Battery has been  connected to P-type and negative terminal to N-type in a PN-junction diode that is when an external voltage is applied to PN-junction in such a way that it cancels the barrier potential and permits the current flow such a bias  is called as Forward-Bais.

Under No Bias voltage condition, Near the junction the holes moves towards the junction and electrons as well forms a region known as Depletion region, the region depleted with immobile ions .

when the applied voltage V establishes an electric field opposite to the potential barrier , as a result the width of the potential barrier is reduced as it is very small

0.3 Volts in Ge diode and 

0.7 Volts in Si diode.

∴ a small voltage (V) is sufficient to completely eliminate the barrier that is the barrier is completely eliminated and the resistance at the junction becomes zero and the current flow across the diode can be explained as follows.

Now holes move towards junction simultaneously electrons since holes and electrons were repelled by the opposite terminals of the Battery, As the Battery voltage is sufficiently greater than barrier voltage electrons and holes gets sufficient energy to cross the barrier easily.

The continuous current in external circuit is due to electrons, the current in N-type material is due to movement of free electrons, when these electrons reaches the junction they combine with the holes at the junction and releases a new electron.Similarly, in the P-type region current is due to holes.

i.e, when an electron-hole combination takes place near the junction ,   A co-valent bond near positive terminal of the battery breaks down and it liberates an electron which moves towards positive terminal of the Battery as electron movement is  towards positive terminal of the Battery this can be treated as hole movement in opposite direction.

therefore the constant movement of electrons and holes towards opposite terminals creates a high forward current in the external circuit.

PN-juction Diode in Reverse-Bias:-

When an External voltage V is applied to a PN-junction in such a way(direction) that it increases the Potential barrier is called as Reverse Bias that is Positive terminal of the Battery connected to N-type and negative terminal to P-type. 

The applied voltage V acts in the Same direction to that of Potential Barrier.

that is when the PN-junction is Reverse Biased

  • The junction Potential Barrier width increases.
  • The junction offers higher resistance.
  • electrons and holes move away from the junction and a very small current flows through the junction because of  minority carriers known as Reverse saturation current.

 

Working /Operation of NPN and PNP Transistor

NPN Transistor Working:-

For Normal operation of NPN Transistor Emitter junction JE is Forward Biased and Collector junction JC is Reverse Biased.

The applied Forward Biased at Emitter-Base junction injects a large number of electrons into the N-region and these electrons have enough energy to overcome the JE junction and enter into the very thin lightly doped Base region.

Since Base is very lightly doped very few electrons recombine with the holes in the P-type Base region and constitutes a small Base current IB in μA.

The electrons in the Emitter region are more when compared to electrons in the Collector. Only 5% (or) 1% of injected electrons combines with the holes in Base to produce IB and remaining 95% (or0 99% of electrons diffuse into Collector region due to extremely small thickness of Base.

Since Collector junction is Reverse-Biased a strong Electro-static field develops between Base and Collector. The field immediately collects the diffused electrons which enters Collector junction and are collected by the Collector(Positive electrode).

Thus injected electrons from Emitter reaches Collector constituting a current known as I_{E}=I_{B}+I_{C} Thus Emitter current is sum of Base current and Collector current. I_{B} is very small in the Base region.

Current directions are  always from negative to positive and Majority carriers are electrons in NPN Transistor.

NPN Transistor is preferred over PNP since the mobility of electron is more than that of hole that is electron moves faster than holes.

PNP Transistor Working:-

For Normal operation of PNP Transistor Emitter junction JE is forward Biased and Collector junction JC is reverse biased.

The applied FB at Emitter-Base junction injects a large number of holes in the P-type emitter region and these holes have enough energy to enter into very thin lightly doped Base region. Base is very lightly doped N-type region. Therefore very few holes combines with the Base region and constitutes a small Base current IB (in Micro Amperes).

The holes in the Emitter region are more when compared to holes in the collector region.Only 5% or 1% of injected holes from Emitter combines with the electrons in the Base to produce IB and remaining 95% (or) 99% of holes diffuse into Collector region  due to extremely small thickness of Base.

Since Collector junction is Reverse-Biased a strong Electro-static field develops between Base and Collector. The field immediately collects the diffused holes which enters Collector junction and are collected by the Collector(negative electrode).

Thus injected holes from Emitter reaches Collector constituting a current known as I_{E}=I_{B}+I_{C}, I_{B} is very small in the Base region.

Majority carriers are holes in PNP Transistor.

 

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.

 

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 w.r.to ‘c’.
  • point ‘Q’ is more negative w.r.to ‘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 w.r.to ‘c’.
  • point ‘Q’ is positive w.r.to ‘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

 

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.

 

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 o/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.

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p style=”text-align: justify;”>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 ).

Colpitts oscillator

Colpitt’s  Oscillator is an excellent circuit and is widely used in commercial signal generators upto 100MHz.

It consists of a single-stage inverting amplifier and an LC phase shift Network.

The two capacitors C_{1} and C_{2} provides potential divider used for providing V_{f}. C_{1} is the feedback element and which provides positive feedback required for sustained Oscillations.

The amplifier circuit is a self-Bias Circuit with R_{1} , R_{2} and parallel combination of R_{E} with C_{E}.

V_{CC} is applied through a resistor  R_{C} (or) RFC choke some times. This RFC choke offers very high impedance to high frequency currents.

R_{C} value has chosen in such a way that it offers high impedance. Two coupling Capacitors C_{C1} and C_{C2} are used to block d.c currents, that means they do not permit d.c currents into tank circuit.

These capacitors C_{C1} and C_{C2} provides a path from Collector to Base through LC Network.

when V_{CC} is switched on , a transient current is produced in the tank circuit an consequently damped oscillations are setup in the circuit.

The oscillatory current in the tank circuit produces a.c voltages across C_{1} and C_{2} . If terminal 1 is more positive w.r. to 2 , then voltages across C_{1} and C_{2} are opposite thus providing a phase shift of 180^{o} between 1 and 2. 

as the transistor is operating in CE mode , it provides a phase shift of 180^{o}.

Therefore the over all phase shift provided by the circuit results 360^{o} which is an essential condition for developing oscillations.

If the feedback is adjusted so that the loop gain A\beta =1 then then the  circuit acts as an Oscillator.

The frequency of oscillation depends on the tank circuit and is varied by gang (or) group tuning of C_{1} and C_{2} means C_{1}=C_{2}.

working:-

The capacitors C_{1} and C_{2} are charged by V_{CC} and are discharged through the coil L setting up of oscillations with frequency 

f_{o}=\frac{1}{2\pi }\sqrt{\frac{1}{L}(\frac{1}{C_{1}}+\frac{1}{C_{2}})}.

these oscillations across C_{1} are applied to the Base-Emitter junction  and the amplified version of output is collected across Collector (the frequency of amplifier output is same as that of input of the amplifier) .

This amplified energy is given back to tank circuit to compensate losses.

therefore un damped oscillations results in the circuit.

Derivation for frequency of oscillations:-

chose \left | A\beta \right |\geq 1 for sustained oscillations.

Analysis(Qualitative):-

if Z_{1} , Z_{2}  and Z_{3}  are pure reactive elements  such that Z_{1}=\frac{1}{j\omega C_{1}} =\frac{-j}{\omega C_{1}} ,  Z_{2}=\frac{1}{j\omega C_{2}} =\frac{-j}{\omega C_{2}}   and  Z_{3}=j\omega L.

from the general condition for an Oscillator 

\left | A\beta \right | =1  \Rightarrow h_{ie}(Z_{1}+Z_{2}+Z_{3})+Z_{1}Z_{2}(1+h_{fe})+Z_{1}Z_{3}=0.

h_{ie}(-\frac{j}{\omega C_{1}}-\frac{j}{\omega C_{2}}+j\omega L)+\frac{j^{2}}{\omega ^{2}C_{1}C_{2}}(1+h_{fe})-\frac{j}{\omega C_{1}}.j\omega L=0

find the real and imaginary parts,

-j(\frac{1}{\omega C_{1}}+\frac{1}{\omega C_{2}}-\omega L)h_{ie}-\frac{1}{\omega ^{2}C_{1}C_{2}}(1+h_{fe})+\frac{L}{C_{1}}=0

equating imaginary part to zero  (\frac{1}{\omega C_{1}}+\frac{1}{\omega C_{2}}-\omega L)=0  ,  since h_{ie}\neq 0 .

\frac{\omega C_{1}+\omega C_{2}}{\omega^{2} C_{1}C_{2}}=\omega L.

after simplification 

\omega ^{2}=\sqrt{\frac{1}{L}(\frac{1}{C_{1}}+\frac{1}{C_{2}})}.

by substituting \omega =2\pi f    results f_{o}=\frac{1}{2\pi }\sqrt{\frac{1}{L}(\frac{1}{C_{1}}+\frac{1}{C_{2}})}.

substituting the value of \omega ^{2}  in the real part gives h_{fe}=\frac{C_{2}}{C_{1}}  . this is the condition for sustained oscillations.

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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 , then the total current I crossing the surface S is given by the volume V

The current coming out of the closed surface is 

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.

By Divergence theorem 

    implies  

for a constant surface the derivative becomes the partial derivative 

   -this is for the whole volume.

for a differential volume 

 , 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      

from . 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.

 

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 . 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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p style=”text-align: justify;”> 

Relation between Laplace and Fourier Transform

The Fourier transform  of a signal x(t) is given as 

X(j\omega ) = \int_{-\infty }^{\infty } x(t) e^{-j\omega t}dt----EQN(I)

Fourier Transform exists only if \int_{-\infty }^{\infty } \left | x(t) \right |dt< \infty  

we know that s=\sigma + j\omega 

X(S) = \int_{-\infty }^{\infty } x(t) e^{-s t}dt

X(S) = \int_{-\infty }^{\infty } \left | x(t)e^{-\sigma t} \right | e^{-j\omega t}dt----EQN(II)

if we compare Equations (I) and (II) both are equal when  \sigma =0.

i.e, X(S) =X(j\omega)| \right |_{s=j\omega } .

This means that Laplace Transform is same as Fourier transform when s=j\omega.

Fourier Transform is nothing but the special case of Laplace transform where  s=j\omega indicates the imaginary axis in complex-s-plane.

Thus Laplace transform is basically Fourier Transform on imaginary axis in the s-plane.

 

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Region of Convergence (ROC)

The range of values of the complex variable s for which Laplace Transform X(S)=\int_{-\infty }^{\infty }x(t)e^{-st}\ dt converges is called the Region of Convergence (ROC).

i.e, The region of Convergence (or) existence of signal’s Laplace transform X(S) is the set of values of s for which the integral defining the direct L T/F X(S) converges.

The ROC is required for evaluating the inverse L T/F of x(t) from X(S).

i.e, the operation of finding the inverse T/F requires an integration in the complex plane.

i.e, x(t)= \frac{1}{2\pi j}\int_{\sigma -j\infty }^{\sigma +j\infty }X(S) e^{St} \ ds .

The path of integration is along S-plane S = \sigma +j\omega that is along \sigma +j\omega with \omega varying from -\infty \ to \ \infty  and moreover , the path of integration must lie in the ROC for X(S).

for example the signal e^{-at}u(t) , this is possible if \sigma >-a  so the path of integration is shown in the figure

Thus to obtain x(t) = e^{-at}u(t)   from X(S) = \frac{1}{s+a}   , the integration is performed through this path for the function  \frac{1}{s+a} . such integration in the complex plane requires a back ground in the theory of functions of complex variables.

so we can avoid this integration by compiling a Table of L T/F’s . so for inverse L T/F’s we use this table instead of performing complex integration.

specific constraints on the ROC are closely associated with time-domain properties of x(t).

Properties of ROC/ constraints (or) Limitations:-

1.The ROC of  X(S) consists of strips parallel to the j\omega axis in the S-plane.

i.e, The ROC of X(S) consists of the values of s for which Fourier T/F of x(t)e^{-\sigma t} converges this is possible if x(t)e^{-\sigma t} is fully integrable thus the condition depends only on \sigma . Hence ROC is the strips (bands) which is only in terms of values of \sigma.

2. 

3. For Rational Laplace T/F’s , the ROC does not contain any poles. This is because X(S) is finite at poles and the integral can not be converge at this point.

4. If x(t) is of finite duration and absolutely integrable, then the ROC is the entire S-plane.

5. If x(t) is right-sided and if the line Re\left \{ s \right \} =\sigma _{o} is in the ROC, then all values of s for which Re\left \{ s \right \} > \sigma _{o} will also be in the ROC.

i.e, if the signal is x(t) = e^{-at}u(t)  right-sided [0 \ to \ \infty ]  then X(S) = \frac{1}{s+a}  for ROC : Re\left \{ s \right \} > -a .

6. If x(t) is left-sided and if the line Re\left \{ s \right \} =\sigma _{o} is in the ROC, then all values of s for which Re\left \{ s \right \} < \sigma _{o} will also be in the ROC.

7. If x(t) is two-sided and if the line Re\left \{ s \right \} =\sigma _{o} is in the ROC, then the ROC consists of a strip  in the s-plane that includes the line Re\left \{ s \right \} = \sigma _{o} .

for the both sided signal , the ROC lies in the region \sigma _{1} < Re\left \{ s \right \}<\sigma _{2} . This ROC is the strip parallel to j\omega  axis in the s-plane.

8. If the L T/F X(S) of x(t) is rational, then it’s ROC is bounded by poles (or) extends to infinity in addition no poles of X(s) are contained in the ROC.

If the function has two poles , then ROC will be area  between these two poles for two sided signal, if for single sided signal the area extends from one pole to infinity.

But is does not include any pole.

9. If the L T/F X(S) of x(t) is rational, then if x(t) is right-sided. The ROC is the region in the s-plane to the right of the right most pole and if x(t) is left-sided, the ROC is the region in the s-plane to the left of the left most pole.

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Analogy between Vectors and Signals

we have already defined the signal as any ordinary function of time. To understand more about signal we consider it as a problem. A problem is better understood (or) better remembered if it can be associated with some familiar phenomenon.

we always search for analogies while studying a new problem.

i.e, In the study of abstract problems analogies are very helpful. Particularly if the problem can be shown to be analogous to some concrete phenomenon.

It is then easier to gain some insight into the new problem from the knowledge of the analogous phenomenon.

There is a perfect analogy that exists between vectors and signals which leads to a better understanding of signal analysis. we shall now briefly review the properties of vectors.

Vectors:-

A vector is specified by magnitude and direction \overrightarrow{A}.

Let us consider two vectors \overrightarrow{V_{1}}  and \overrightarrow{V_{2}} . It is possible to find out the component of one vector along the other vector.

In order to find out the component of vector \overrightarrow{V_{1}} along  \overrightarrow{V_{2}} . Let us assume it as C_{12}V_{2} ,  which is only the magnitude.

how do we represent physically the component of one vector \overrightarrow{V_{1}} along  \overrightarrow{V_{2}} ? This is possible by finding the projection of one vector on to the other.

i.e, by drawing a perpendicular from \overrightarrow{V_{1}}   to   \overrightarrow{V_{2}}

\overrightarrow{V_{1}} = C_{12}\overrightarrow{V_{2}} + \overrightarrow{V_{e}} .

There exists two other possibilities.

but these are not suitable. \because  the error vectors are more in these cases.

\overrightarrow{V_{1}}.\overrightarrow{V_{2}}=V_{1}V_{2}\cos \theta .

If \theta is the angle between two vectors \overrightarrow{V_{1}}  and \overrightarrow{V_{2}} , the component of \overrightarrow{V_{1}}  along \overrightarrow{V_{2}} is

\frac{\overrightarrow{V_{1}}.\overrightarrow{V_{2}}}{\left | V_{2} \right |}=V_{1}\cos \theta-----EQN(1).

The component  of \overrightarrow{V_{1}}  along \overrightarrow{V_{2}} is C_{12}V_{2}----EQN(2).

\therefore (1) = (2) .

\frac{\overrightarrow{V_{1}}.\overrightarrow{V_{2}}}{\left | V_{2} \right |} = C_{12}V_{2} .

C_{12}=\frac{\overrightarrow{V_{1}}.\overrightarrow{V_{2}}}{V_{2}^{2}} .

If two vectors are orthogonal  \overrightarrow{V_{1}}.\overrightarrow{V_{2}} =0 .

i.e, C_{12} =0.

Signals:-

The concept of vector comparison & orthogonality can be extended to signals.

i.e, a signal is nothing but a single-valued function of independent variable. Assume two signals  f_{1}(t)   and f_{2}(t), now to approximate  f_{1}(t)  in terms of f_{2}(t)  over  t_{1}< t<t_{2} .

f_{1}(t) \approx C_{12}f_{2}(t) .

\therefore f_{1}(t) \approx C_{12}f_{2}(t)+f_{e}(t) .

f_{e}(t) = f_{1}(t)-C_{12}f_{2}(t) .

Now, we choose in order to achieve the best approximation.

 i.e, which keeps the error as minimum as possible.

One possible way for minimizing error  f_{e}(t) is to choose minimize the average value of f_{e}(t) .

i.e, as    \frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}dt.

But the process of averaging gives a false indication.

i.e, for example while approximating a function  \sin t  with a null function  f(t)=0  is

  f_{1}(t) =C_{12}f_{2}(t).

\sin t =0. \ \ 0\leq t\leq 2\pi.

indicates that  \sin t =0  during    0   to  2\pi   without any error 

i.e,  f_{e}(t) = f_{1}(t)-C_{12}f_{2}(t) .

f_{e}(t) = \sin t .

Average value of error is = \frac{1}{2 \pi } \int_{0}^{2\pi } \sin t \ dt =0 .

This seems to be error is zero but actually there exists some error.

To avoid this false indication, we choose to minimize the average of the square of the error

i.e, Mean Square Error \epsilon = \frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}f_{e}^{2}(t) \ dt .

\epsilon = \frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}(f_{1}(t)-C_{12}f_{2}(t))^{2} \ dt.

To find value which keeps error minimum  \frac{d\epsilon }{dC_{12}}=0 .

C_{12} = \frac{\int_{t_{1}}^{t_{2}}f_{1}(t)f_{2}(t) \ dt}{\int_{t_{1}}^{t_{2}}f_{2}^{2}(t) \ dt} .

C_{12}  Which is similar to C_{12}=\frac{\overrightarrow{V_{1}}.\overrightarrow{V_{2}}}{V_{2}^{2}} where \int_{t_{1}}^{t_{2}}f_{1}(t)f_{2}(t) \ dt   denotes the inner product between two  Real signals

\therefore  For the orthogonality of two signals C_{12} =0 

\Rightarrow \ \int_{t_{1}}^{t_{2}}f_{1}(t)f_{2}(t) \ dt =0.

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Properties of Laplace Transforms(Bi-lateral)

1.Linearity Property:-

x_{1}(t)\leftrightarrow X_{1}(S) \ \ \ ROC : R_{1}

x_{2}(t)\leftrightarrow X_{2}(S) \ \ \ ROC : R_{2}

a\ x_{1}(t)+b\ x_{2}(t)\leftrightarrow \ \ ?.

we know that  Laplace Transform of a signal  x(t)  is  X(S) = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ dt .

L\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \int_{-\infty }^{\infty }\ \left \{ a\ x_{1}(t)+b\ x_{2}(t) \right \} \ e^{-St}\ dt.

L\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \int_{-\infty }^{\infty }\ \left \{ a\ x_{1}(t)\ e^{-St}\ dt+b\ x_{2}(t) \ e^{-St}\ dt\right \}

L\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \ \ a \int_{-\infty }^{\infty }\ x_{1}(t)\ e^{-St}\ dt+b \int_{-\infty }^{\infty }\ x_{2}(t) \ e^{-St}\ dt.

L\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \ \ a X_{1}(S) +b X_{2}(S) .          ROC: R_{1} \cap R_{2}.

2.Time-shifting Property:-

x(t)\leftrightarrow X(S) \ \ \ ROC : R

x(t-t_{o})\leftrightarrow \ ?.

we know that  X(S) = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ dt .

L\left \{ x(t-t_{o}) \right \} = \int_{-\infty }^{\infty }\ x(t-t_{o}) \ e^{-St}\ dt.    Lett-t_{o}=\lambda \Rightarrow dt= d\lambda

t \ limits \ : \ -\infty \ to \ \infty, \ \ \ \lambda \ \ limits \ : \ \infty \ to \ -\infty

L\left \{ x(t-t_{o}) \right \} = \int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{-S(\lambda +t_{o})}\ d\lambda .

L\left \{ x(t-t_{o}) \right \} =e^{-St_{o}} \int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{-S\lambda }\ d\lambda.

x(t-t_{o})\leftrightarrow \ e^{-St_{o}}\ X(S) \ , \ \ ROC:R.

from the above equation x(t-t_{o})  forms a Laplace Transform pair with e^{-St_{o}} \ X(S).

3.Frequency-shifting Property:-

x(t)\leftrightarrow X(S) \ \ \ ROC : R

?\ \leftrightarrow \ X(S-S_{o}).

we know that  X(S) = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ dt .

L\left \{ e^{S_{o}t}\ x(t) \right \} = \int_{t=-\infty }^{\infty }\ e^{S_{o}t}\ x(t) \ e^{-St}\ dt.

L\left \{ e^{S_{o}t}\ x(t) \right \} = \int_{t=-\infty }^{\infty }\ \ x(t) \ e^{-(S-S_{o})t}\ dt .

e^{S_{o}t}x(t)\leftrightarrow \ X(S-S_{o}) \ , \ \ ROC:R.

from the above equation e^{S_{o}t}x(t)  forms a Laplace Transform pair with X(S-S_{o}).

4. Differentiation in time-domain:-

x(t)\leftrightarrow X(S) \ \ \ ROC : R

\frac{dx(t)}{dt}\leftrightarrow \ ?.

we know that  inverse Laplace Transform  x(t) =\frac{1}{2\pi \ j} \int_{\sigma - j\infty }^{\sigma +j\infty }\ X(S) \ e^{St} \ dS .

\frac{dx(t)}{dt} =\frac{1}{2\pi \ j} \int_{\sigma - j\infty }^{\sigma +j\infty }\ X(S) \ \frac{d(e^{St})}{dt} \ dS.

\frac{dx(t)}{dt} =\frac{1}{2\pi \ j} \int_{\sigma - j\infty }^{\sigma +j\infty }\ X(S) \ S \ e^{St} \ dS .

\frac{dx(t)}{dt} =\frac{1}{2\pi \ j} \int_{\sigma - j\infty }^{\sigma +j\infty } (\ S\ X(S)) \ e^{St} \ dS.

\frac{dx(t)}{dt}\leftrightarrow \ S\ X(S).

from the above equation \frac{dx(t)}{dt}  forms a Laplace Transform pair with S\ X(S)

Similarly  \frac{d^{n}x(t)}{dt^{n}}\leftrightarrow \ S^{n}\ X(S).

5.Differentiation in S-domain:-

x(t)\leftrightarrow X(S) \ \ \ ROC : R

?\leftrightarrow \frac{dX(S)}{dS}.

we know that  X(S) = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ dt .

\frac{dX(S)}{dS} = \int_{-\infty }^{\infty }\ x(t) \ \frac{de^{-St} }{dS}\ dt.

\frac{dX(S)}{dS} = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ (-t)\ dt .

\frac{dX(S)}{dS} = \int_{-\infty }^{\infty }\ (-t \ x(t)) \ e^{-St} \ dt.

\frac{dX(S)}{dS}\leftrightarrow \ -t\ x(t) \ \ \ ROC:R.

from the above equation \frac{dX(S)}{dS}  forms a Laplace Transform pair with -t\ x(t).

6. Time-reversal property:-

x(t)\leftrightarrow X(S) \ \ \ ROC : R

x(-t)\leftrightarrow \ \ ?.

we know that  X(S) = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ dt .

L\left \{ x(-t) \right \} = \int_{-\infty }^{\infty }\ x(-t) \ e^{-St} \ dt.          Let  -t=\ \lambda ,      -dt=\ d\lambda,  t \ limits \ : \ -\infty \ to \ \infty, \ \ \ \lambda \ \ limits \ : \ \infty \ to \ -\infty.

L\left \{ x(-t) \right \} = \int_{\lambda =\infty }^{-\infty }\ x(\lambda ) \ e^{S\lambda } \ (-d\lambda ) .

L\left \{ x(-t) \right \} = \int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{-(-S) \lambda } \ d\lambda.

x(-t)\leftrightarrow \ X(-S).

from the above equation x(-t)  forms a Laplace Transform pair with X(-S).

7. Time-Scaling property:-

x(t)\leftrightarrow X(S) \ \ \ ROC : R

x(at)\leftrightarrow \ \ ?.

we know that  X(S) = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ dt .

L\left \{ x(at) \right \} = \int_{-\infty }^{\infty }\ x(at) \ e^{-St} \ dt.          Let  at=\ \lambda ,      dt=\ \frac{d\lambda}{a},  t \ limits \ : \ -\infty \ to \ \infty, \ \ \ \lambda \ \ limits \ : \ -\infty \ to \ \infty.

L\left \{ x(at) \right \} = \frac{1}{a}\int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{(\frac{-S}{a})\lambda } \ d\lambda .

x(at)\leftrightarrow \frac{1}{a} \ X(\frac{S}{a}), \ \ \ if \ a>0.

x(-at)\leftrightarrow \frac{1}{a} \ X(\frac{-S}{a}), \ \ \ if \ a<0 \ and \ (a\neq -1).

8. Convolution in Time-domain:-

x_{1}(t)\leftrightarrow X_{1}(S) \ \ \ ROC : R_{1}

x_{2}(t)\leftrightarrow X_{2}(S) \ \ \ ROC : R_{2}

x_{1}(t) * x_{2}(t)\leftrightarrow \ \ ?.

we know that  X(S) = \int_{t=-\infty }^{\infty }\ x(t) \ e^{-St} \ dt .

L\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{t=-\infty }^{\infty }\ \left \{ x_{1}(t) * x_{2}(t) \right \}\ e^{-St} \ dt.

L\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{t=-\infty }^{\infty }\ \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau )\left \{ x_{2}(t-\tau ) \right \}\ e^{-St} \ dt \ d\tau.

L\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau )\left \{ \int_{t=-\infty }^{\infty } x_{2}(t-\tau )\ e^{-St} \ dt \right \} \ d\tau.

L\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau )\left \{ e^{-S\tau } X_{2}(S) \right \} \ d\tau.

L\left \{ x_{1}(t) * x_{2}(t) \right \} = \left \{ \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau ) e^{-S\tau } \ d\tau \right \} \ X_{2}(S).

L\left \{ x_{1}(t) * x_{2}(t) \right \} = \ X_{1}(S) \ X_{2}(S)

x_{1}(t) * x_{2}(t)\leftrightarrow \ X_{1}(S) \ X_{2}(S), ROC : R_{1} \cap R_{2}.

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Initial-value & Final -value Theorems:-

Initial-value Theorem:-

Use:- to find out the initial value of a signal x(t) without using inverse Laplace Transform.

x(0^{-})= \lim_{t\rightarrow 0^{-}}x(t)=\lim_{S\rightarrow \infty }s\ X(S).

Proof:-

we know that  L\left \{ \frac{dx(t)}{dt} \right \}\leftrightarrow S\ X(S)-x(0^{-}).

L\left \{ \frac{dx(t)}{dt} \right \}=\int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt.

i.e,       \int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt=S\ X(S)-x(0^{-}) .

\lim_{s\rightarrow \infty }\ \int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt= \lim_{s\rightarrow \infty }\ S\ X(S)- \lim_{s\rightarrow \infty }\ x(0^{-}) .

0= \lim_{s\rightarrow \infty }\ S\ X(S)- \lim_{s\rightarrow \infty }\ x(0^{-}) .

\lim_{s\rightarrow \infty }\ S\ X(S)= \lim_{s\rightarrow \infty }\ x(0^{-}) .

\ x(0^{-}) = \lim_{s\rightarrow \infty }\ S\ X(S) .

Hence proved.

final-value Theorem:-

Use:- to find out the final value of a signal x(t) without using inverse Laplace Transform.

x(\infty )= \lim_{t\rightarrow \infty }x(t)=\lim_{S\rightarrow 0 }s\ X(S).

Proof:-

we know that  L\left \{ \frac{dx(t)}{dt} \right \}\leftrightarrow S\ X(S)-x(0^{-}).

L\left \{ \frac{dx(t)}{dt} \right \}=\int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt.

i.e,       \int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt=S\ X(S)-x(0^{-}) .

\lim_{s\rightarrow 0 }\ \int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt= \lim_{s\rightarrow 0 }\ S\ X(S)- \lim_{s\rightarrow 0 }\ x(0^{-}) .

x(\infty )-x(0^{-})= \lim_{s\rightarrow 0 }\ S\ X(S)- \lim_{s\rightarrow 0 }\ x(0^{-}) .

x(\infty )-x(0^{-})= \lim_{s\rightarrow 0 }\ S\ X(S)- x(0^{-})

x(\infty )= \lim_{s\rightarrow 0 }\ S\ X(S) .

Hence proved.

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Derivation of Z-Transform/ inverse Z- Transform

Derivation of Z-Transform:-

If x[n]  is the given sequence then it’s Discrete Time Fourier Transform  is  X(e^{j\omega })  .

i.e,    x[n]\leftrightarrow X(e^{j\omega })

x[n]\ r^{-n}\leftrightarrow X(r\ e^{j\omega }).

DTFT of x[n] = \sum_{n=-\infty }^{\infty } x[n] e^{-j\omega n} .

DTFT of  x[n]\ r^{-n} = \sum_{n=-\infty }^{\infty } x[n] \ r^{-n}e^{-j\omega n} .

X(r\ e^{j\omega }) = \sum_{n=-\infty }^{\infty } x[n] \ \left ( r\ e^{j\omega } \right )^{-n} .

Let    Z=r\ e^{j\omega }   a complex-variable.

X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z ^{-n}   . is the Z-Transform of a sequence  x[n] .

Inverse Z-Transform:-

The inverse DTFT of  X(e^{j\omega })  is    x[n] = \frac{1}{2\pi } \int X(e^{j\omega }) \ e^{j\omega n} \ d\omega.

x[n] \ r^{-n} = \frac{1}{2\pi } \int X(r\ e^{j\omega }) \ e^{j\omega n} \ d\omega .

x[n] = \frac{1}{2\pi } \int X(r\ e^{j\omega }) \ (r\ e^{j\omega })^{n} \ d\omega .

x[n] = \frac{1}{2\pi\ j Z } \int X(Z) \ Z^{n} \ dz .    Let  r\ e^{j\omega } = Z \Rightarrow \j \ r \ e^{j\omega }\ d\omega = dZ     and  \ j \ Z \ d\omega = dZ \Rightarrow d\omega =\frac{dz}{j \ Z} .

x[n] = \frac{1}{2\pi\ j } \int X(Z) \ Z^{n-1} \ dz – represents the inverse Z-Transform.

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properties of Fourier Transforms

1.Linearity Property:-

x_{1}(t)\leftrightarrow X_{1}(j\omega )

x_{2}(t)\leftrightarrow X_{2}(j\omega)

a\ x_{1}(t)+b\ x_{2}(t)\leftrightarrow \ \ ?.

we know that  Fourier Transform of a signal  x(t)  is X(j\omega) = \int_{-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt .

F\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \int_{-\infty }^{\infty }\ \left \{ a\ x_{1}(t)+b\ x_{2}(t) \right \} \ e^{-j\omega t}\ dt.

F\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \int_{-\infty }^{\infty }\ \left \{ a\ x_{1}(t)\ e^{-j\omega t}\ dt+b\ x_{2}(t) \ e^{-j\omega t}\ dt\right \}

F\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \ \ a \int_{-\infty }^{\infty }\ x_{1}(t)\ e^{-j\omega t}\ dt+b \int_{-\infty }^{\infty }\ x_{2}(t) \ e^{-j\omega t}\ dt.

F\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \ a \ X_{1}(j\omega) +b \ X_{2}(j\omega) .

2.Time-shifting Property:-

x(t)\leftrightarrow X(j\omega )

x(t-t_{o})\leftrightarrow \ ?.

we know that  X(j\omega) = \int_{-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt .

L\left \{ x(t-t_{o}) \right \} = \int_{-\infty }^{\infty }\ x(t-t_{o}) \ e^{- j\omega t}\ dt.    Lett-t_{o}=\lambda \Rightarrow dt= d\lambda

t \ limits \ : \ -\infty \ to \ \infty, \ \ \ \lambda \ \ limits \ : \ \infty \ to \ -\infty

L\left \{ x(t-t_{o}) \right \} = \int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{- j\omega (\lambda +t_{o})}\ d\lambda .

L\left \{ x(t-t_{o}) \right \} =e^{-j\omega t_{o}} \int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{-j\omega \lambda }\ d\lambda.

x(t-t_{o})\leftrightarrow \ e^{- j\omega t_{o}}\ X(j\omega).

from the above equation x(t-t_{o})  forms  Fourier Transform pair with e^{- j\omega t_{o}} \ X(j\omega).

3.Frequency-shifting Property:-

x(t)\leftrightarrow X(\omega )

?\ \leftrightarrow \ X(\omega -\omega _{o}).

we know that  X(\omega ) \ or X(j\omega ) = \int_{-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt .

L\left \{ e^{j\omega _{o}t}\ x(t) \right \} = \int_{t=-\infty }^{\infty }\ e^{j \omega _{o}t}\ x(t) \ e^{-j\omega t}\ dt.

L\left \{ e^{j\omega _{o}t}\ x(t) \right \} = \int_{t=-\infty }^{\infty }\ \ x(t) \ e^{-(\omega -\omega _{o})t}\ dt .

e^{j\omega _{o}t}x(t)\leftrightarrow \ X(\omega -\omega _{o}).

from the above equation e^{j\omega _{o}t}\ x(t)  forms  Fourier Transform pair with X(\omega -\omega _{o}).

4. Differentiation in time-domain:-

x(t)\leftrightarrow X(\omega )

\frac{dx(t)}{dt}\leftrightarrow \ ?.

we know that  inverse Fourier  Transform  x(t) =\frac{1}{2\pi } \int_{\omega =\infty }^{\infty }\ X( \omega ) \ e^{j\omega t} \ d\omega .

\frac{dx(t)}{dt} =\frac{1}{2\pi } \int_{\omega =\infty }^{\infty }\ X(\omega ) \ \frac{d(e^{j\omega t})}{dt} \ d\omega.

\frac{dx(t)}{dt} =\frac{1}{2\pi } \int_{\omega =\infty }^{\infty }\ X(\omega ) \ j \omega \ e^{j\omega t} \ d\omega .

\frac{dx(t)}{dt} =\frac{1}{2\pi } \int_{\omega =\infty }^{\infty }\ (\ j \omega \ X(\omega )) \ e^{j\omega t} \ d\omega.

\frac{dx(t)}{dt}\leftrightarrow \ j \omega \ X(\omega ).

from the above equation \frac{dx(t)}{dt}  forms Fourier Transform pair with \ j \omega \ X(\omega )

Similarly  \frac{d^{n}x(t)}{dt^{n}}\leftrightarrow \ \ (j \omega) ^{n}\ X(\omega ).

5.Differentiation in w-domain:-

x(t)\leftrightarrow X(\omega )

?\leftrightarrow \frac{dX(\omega )}{d\omega }.

we know that  X(\omega ) = \int_{t =-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt .

\frac{dX(\omega )}{d\omega } = \int_{t=-\infty }^{\infty }\ x(t) \ \frac{d(e^{-j\omega t}) }{d\omega } \ dt.

\frac{dX(\omega )}{d\omega } = \int_{t=-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ (-j t)\ dt .

\frac{dX(\omega )}{d\omega } = \int_{t=-\infty }^{\infty }\ (-jt \ x(t)) \ e^{-j\omega t} \ dt.

\frac{dX(\omega )}{d\omega }\leftrightarrow \ -jt\ x(t).

from the above equation \frac{dX(\omega )}{d\omega }  forms Fourier Transform pair with -jt\ x(t).

6. Conjugation property:-

x(t)\leftrightarrow X(\omega )

x^{*}(t)\leftrightarrow \ \ ?.

we know that  X(\omega ) = \int_{t=-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt .

F\left \{ x^{*}(t) \right \} = \int_{-\infty }^{\infty }\ x^{*}(t) \ e^{-j\omega t} \ dt.

F\left \{ x^{*}(t) \right \} = \int_{t =-\infty }^{\infty }( x(t ) \ e^{j \omega t} \ dt )^{*} .

x^{*}(t)\leftrightarrow \ X^{*}(-\omega ).

from the above equation x^{*}(t)  forms Fourier Transform pair with X^{*}(-\omega ).

7. Time-Scaling property:-

x(t)\leftrightarrow X(\omega )

x(at)\leftrightarrow \ \ ?.

we know that  X(\omega ) = \int_{-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt .

F\left \{ x(at) \right \} = \int_{-\infty }^{\infty }\ x(at) \ e^{-j\omega t} \ dt.          Let  at=\ \lambda ,      dt=\ \frac{d\lambda}{a},  t \ limits \ : \ -\infty \ to \ \infty, \ \ \ \lambda \ \ limits \ : \ -\infty \ to \ \infty.

F\left \{ x(at) \right \} = \frac{1}{a}\int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{-j(\frac{\omega }{a})\lambda } \ d\lambda .

x(at)\leftrightarrow \frac{1}{a} \ X(\frac{\omega }{a}), \ \ \ if \ a>0.

x(-at)\leftrightarrow \frac{1}{a} \ X(\frac{-\omega }{a}), \ \ \ if \ a<0 \ and \ (a\neq -1).

8. Convolution in Time-domain:-

x_{1}(t)\leftrightarrow X_{1}(\omega )

x_{2}(t)\leftrightarrow X_{2}(\omega )

x_{1}(t) * x_{2}(t)\leftrightarrow \ \ ?.

we know that  X(\omega ) = \int_{t=-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt .

F\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{t=-\infty }^{\infty }\ \left \{ x_{1}(t) * x_{2}(t) \right \}\ e^{-j\omega t} \ dt.

F\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{t=-\infty }^{\infty }\ \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau )\left \{ x_{2}(t-\tau ) \right \}\ e^{-j\omega t} \ dt \ d\tau.

F\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau )\left \{ \int_{t=-\infty }^{\infty } x_{2}(t-\tau )\ e^{-j\omega t} \ dt \right \} \ d\tau.

F\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau )\left \{ e^{-j\omega \tau } X_{2}(\omega ) \right \} \ d\tau.

F\left \{ x_{1}(t) * x_{2}(t) \right \} = \left \{ \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau ) e^{-j\omega \tau } \ d\tau \right \} \ X_{2}(\omega ).

F\left \{ x_{1}(t) * x_{2}(t) \right \} = \ X_{1}(\omega ) \ X_{2}(\omega )

x_{1}(t) * x_{2}(t)\leftrightarrow \ X_{1}(\omega ) \ X_{2}(\omega ).

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Properties of Z-transforms

1.Linearity Property:-

x_{1}[n]\leftrightarrow X_{1}(Z) \ \ \ ROC :a_{1}< \left | Z \right | <b_{1}-R_{1}

x_{2}[n]\leftrightarrow X_{2}(Z) \ \ \ ROC :a_{2}< \left | Z \right | <b_{2}-R_{2}

a\ x_{1}[n]+b\ x_{2}[n]\leftrightarrow \ \ ?.

we know that  Bi-lateral Z- Transform of a signal  x[n]  is X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n} .

Z\left \{a\ x_{1}[n]+b\ x_{2}[n]\right \} = \sum_{n=-\infty }^{\infty } \ \ \left \{ a\ x_{1}[n]+b\ x_{2}[n] \right \} Z^{-n}.

Z\left \{a\ x_{1}[n]+b\ x_{2}[n]\right \} = \sum_{n=-\infty }^{\infty } \ \ \left \{ a\ x_{1}[n]\ Z^{-n}+b\ x_{2}[n] \ Z^{-n}\right \}

Z\left \{a\ x_{1}[n]+b\ x_{2}[n]\right \} = \ a\ \sum_{n=-\infty }^{\infty } x_{1}[n]\ Z^{-n}+b \ \sum_{n=-\infty }^{\infty } x_{2}[n] \ Z^{-n}.

Z\left \{a\ x_{1}[n]+b\ x_{2}[n]\right \} = \ a \ X_{1}(Z)+b \ X_{2}(Z) .          ROC: R_{1} \cap R_{2}.

2.Time-shifting Property:-

x[n]\leftrightarrow X(Z) \ \ \ ROC : R

x[n-k]\leftrightarrow \ ?.

we know that  X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n} .

Z\left \{ x[n-k] \right \} = \sum_{n=-\infty }^{\infty } x[n-k] \ Z^{-n}.    Let     n-k=m \Rightarrow m= n+k

Here n is a variable and k is a constant.

Z\left \{ x[n-k] \right \} = \sum_{m\ =-\infty }^{\infty } x[m] \ Z^{-(m+k)} .

Z\left \{ x[n-k] \right \} = Z^{-k} \sum_{m\ =-\infty }^{\infty } x[m] \ Z^{-m}.

Z\left \{ x[n-k] \right \} = Z^{-k} X(Z)

x[n-k]\leftrightarrow \ Z^{-k} X(Z)\ , \ \ ROC:R.

from the above equation x[n-k]  forms Z Transform pair with Z^{-k} X(Z).

3. Scaling  in-Z-domain property:-

x[n]\leftrightarrow X(Z) \ \ \ ROC :r_{1}< \left | Z \right | <r_{2}-R_{1}

a^{n}x[n]\leftrightarrow \ \ \ ? ,

we know that  X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n} .

Z\left \{ a^{n}\ x[n] \right \} = \sum_{n=-\infty }^{\infty }a^{n}\ x[n] \ Z^{-n}.

Z\left \{ a^{n}\ x[n] \right \} = \sum_{n=-\infty }^{\infty }\ x[n] \ (a^{-1}Z)^{-n} .

Z\left \{ a^{n}\ x[n] \right \} = X(a^{-1}Z)

a^{n}x[n]\leftrightarrow \ X(\frac{Z}{a}), \ \ \ if \ a>0.   \ \ ROC \ of \ a^{n}x[n] \ is :\left | a \right |\ r_{1}< \left | Z \right | <\left | a \right |\ r_{2} .

4. Time-reversal property:-

x[n]\leftrightarrow X(Z) \ \ \ ROC :r_{1}< \left | Z \right | <r_{2}-R_{1}

x[-n]\leftrightarrow \ \ ?.

we know that  X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n} .

Z\left \{ x[-n] \right \} = \sum_{n=-\infty }^{\infty } x[-n] \ Z^{-n} .  Let  -n=\ m ,

Z\left \{ x[-n] \right \} = \sum_{m=-\infty }^{\infty } x[m] \ (Z^{-1})^{-m}\ \ \ ROC :r_{1}< \left | Z^{-1} \right | <r_{2}.

x[-n]\leftrightarrow \ X(Z^{-1})\ \ \ ROC :\frac{1}{\left | r_{2} \right |}< \left | Z \right | <\frac{1}{\left | r_{1} \right |}.

from the above equation x[-n]  forms Z Transform pair with X(\frac{1}{Z}).

5. Differentiation in Z-domain:-

x[n]\leftrightarrow X(Z) \ \ \ ROC :r_{1}< \left | Z \right | <r_{2}-R_{1}

n\ x[n]\leftrightarrow \ \ ?.

we know that  X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n} .

\frac{d(X(Z))}{dZ} = \sum_{n=-\infty }^{\infty } x[n] \frac{d (Z^{-n})}{dZ} .

 

\frac{d(X(Z))}{dZ} = \sum_{n=-\infty }^{\infty } \ -n \ x[n] Z^{(-n-1)} .

\frac{d(X(Z))}{dZ} = -\left \ Z^{-n}\right ] Z^{-1} .

\frac{d(X(Z))}{dZ} = - Z^{-1} \ Z\left \{ n\ x[n] \right \} .

from the above equation \frac{d(X(Z))}{dZ}  forms Z-Transform pair with n\ x[n]  and the ROC is same as that of the original sequence x[n].

6. Convolution in Time-domain:-

x_{1}[n]\leftrightarrow X_{1}(Z) \ \ \ ROC :a_{1}< \left | Z \right | <b_{1}-R_{1}

x_{2}[n]\leftrightarrow X_{2}(Z) \ \ \ ROC :a_{2}< \left | Z \right | <b_{2}-R_{2}

x_{1}[n]* x_{2}[n]\leftrightarrow \ \ ?.

we know that  X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n} .

Z\left \{ x_{1}[n] *x_{2}[n]\right \} = \sum_{n=-\infty }^{\infty }(x_{1}[n] *x_{2}[n]) \ Z^{-n} .

Z\left \{ x_{1}[n] *x_{2}[n]\right \} = \sum_{k=-\infty }^{\infty }(x_{1}[k] x_{2}[n-k]) \sum_{n=-\infty }^{\infty }\ Z^{-n} .

Z\left \{ x_{1}[n] *x_{2}[n]\right \} = \sum_{k=-\infty }^{\infty }x_{1}[k] \sum_{n=-\infty }^{\infty }x_{2}[n-k] \ Z^{-n} .

Z\left \{ x_{1}[n] *x_{2}[n]\right \} = \sum_{k=-\infty }^{\infty }x_{1}[k] \ Z^{-k}\ X_{2}(Z) .

Z\left \{ x_{1}[n] *x_{2}[n]\right \} = \left \ Z^{-k} \right ]\ X_{2}(Z) .

Z\left \{ x_{1}[n] *x_{2}[n]\right \} = X_{1}(Z) \ X_{2}(Z) .

 

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Complex Convolution Theorem :-

Complex Convolution Theorem :-

Let two signals x_{1}[n] , \ x_{2}[n]  which are complex signals then the product of these two signals be x_{3}[n] = x_{1}[n]\ x_{2}[n] .

x_{1}[n]\leftrightarrow X_{1}(Z) \ \ \ ROC :a_{1}< \left | Z \right | <b_{1}-R_{1}

x_{2}[n]\leftrightarrow X_{2}(Z) \ \ \ ROC :a_{2}< \left | Z \right | <b_{2}-R_{2}

Z\left \{ x_{1}[n]\ x_{2}[n] \right \}\leftrightarrow \ \ ?.

we know that  X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n} .

Z\left \{ x_{3}[n]\right \} = \sum_{n=-\infty }^{\infty }(x_{1}[n]\ x_{2}[n]) \ Z^{-n} .

Z\left \{ x_{3}[n]\right \} = \sum_{n=-\infty }^{\infty }(\ x_{2}[n]\ Z^{-n}) \ \frac{1}{2\pi j} \oint_{c}X_{1}(v)\ v^{n-1}\ dv .

Z\left \{ x_{3}[n]\right \} = \sum_{n=-\infty }^{\infty }(\ x_{2}[n]\ (\frac{Z}{v} )^{-n} \ \frac{1}{2\pi j} \oint_{c}X_{1}(v)\ v^{-1}\ dv .

Z\left \{ x_{3}[n]\right \} = \ \frac{1}{2\pi j} \oint_{c}X_{2}(\frac{Z}{v} ) X_{1}(v)\ v^{-1}\ dv .

Parseval’s Relation :-

The two complex valued signals x_{1}[n] , \ x_{2}[n]   then the Parseval’s relation states that

\sum_{n=-\infty }^{\infty } x_{1}[n] \ x_{2}^{*}[n] = \ \frac{1}{2\pi j} \oint_{c} X_{1}(v)\ X_{2}^{*}(\frac{1}{v^{*}} ) v^{-1}\ dv .

Proof:-

By using complex convolution theorem

\sum_{n=-\infty }^{\infty } x_{1}[n] \ x_{2}^{*}[n]\ Z ^{-n} = \ \frac{1}{2\pi j} \oint_{c} X_{1}(v)\ X_{2}^{*}(\frac{Z^{*}}{v^{*}} ) v^{-1}\ dv .

in the above equation substitute Z=1 , then

\sum_{n=-\infty }^{\infty } x_{1}[n] \ x_{2}^{*}[n] = \ \frac{1}{2\pi j} \oint_{c} X_{1}(v)\ X_{2}^{*}(\frac{1}{v^{*}} ) v^{-1}\ dv .

Hence Parseval’s relation is proved.

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Initial value & Final value Theorems in Z-Transforms:-

Initial  value Theorem (Z-Transforms):-

If uni-lateral Z-Transform of x[n] is X_{+}(Z)  then  x(0) = \lim_{z\rightarrow \infty } X_{+}(Z) .

X_{+}(Z) = \sum_{n=0}^{\infty } x[n] \ Z^{-n} .

X_{+}(Z) = x(0) +x(1)\ Z^{-1}+ x(2)\ Z^{-2}+x(3)\ Z^{-3}+......

\lim_{Z\rightarrow \infty }X_{+}(Z) = \lim_{Z\rightarrow \infty }\left

\lim_{Z\rightarrow \infty }X_{+}(Z) = x(0) .

Final  value Theorem (Z-Transforms):-

If uni-lateral Z-Transform of x[n] is X_{+}(Z)  then  x(\infty ) = \lim_{z\rightarrow 1}(Z-1) \ X_{+}(Z) .

Z\left \{ x(n+1) \right \}-Z\left \{ x(n) \right \} = \lim_{k\rightarrow \infty }\sum_{n=0}^{k} (x[n+1]-x[n]) \ Z^{-n} .

Z\ X_{+}(Z) -x(0)-X_{+}(Z)= \lim_{k\rightarrow \infty }\sum_{n=0}^{k} (x[n+1]-x[n]) \ Z^{-n}.

\lim_{z\rightarrow 1}\left =\lim_{z\rightarrow 1} \lim_{k\rightarrow \infty }\sum_{n=0}^{k} \left

\lim_{z\rightarrow 1}\left = \lim_{z\rightarrow 1}\left \{ x(\infty )-x(0) \right \}.

\lim_{z\rightarrow 1}(Z-1)\ X_{+}(Z) = x(\infty ).

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Methods to find out inverse Z-transforms

Long division (or) Power Series Expansion Method:-

X(Z) = \sum_{n=-\infty }^{\infty }x[n]\ Z^{-n}

X(Z)can be expressed either in positive powers of Z or negative powers of Z.

if the sequence is causal X(Z) has negative powers of Z similarly the Non-causal sequence   X(Z)  negative powers of Z.

Let X(Z)=\frac{N(Z)}{D(Z)} =\frac{b_{0}+b_{1}Z^{-1}+b_{2}Z^{-2}+...........+b_{M}Z^{-M}}{a_{0}+a_{1}Z^{-1}+a_{2}Z^{-2}+...........+a_{N}Z^{-N}} .

X(Z) is causal it has ROC  \left | Z \right |> \left | r \right |  then X(Z) can be expressed as

X(Z) = x(0)+x(1)Z^{-1}+x(2)Z^{-2}+x(3)Z^{-3}+........

X(Z) is non-causal it has ROC \left | Z \right |<\left | r \right | then X(Z) can be expressed as

X(Z)=x(0)+x(-1)Z^{1}+x(-2)Z^{2}+x(-3)Z^{3}+...........

Partial fraction Method:-

Let X(Z)=\frac{N(Z)}{D(Z)} =\frac{b_{0}+b_{1}Z^{-1}+b_{2}Z^{-2}+...........+b_{M}Z^{-M}}{a_{0}+a_{1}Z^{-1}+a_{2}Z^{-2}+...........+a_{N}Z^{-N}}     and    a_{o}=1

if     M<N  ,  X(Z)   is a proper function .

if   M\geq N   ,  X(Z)   is improper function  so convert the improper function to proper function as

X(Z)=c_{o}+c_{1}Z^{-1}+c_{2}Z^{-2}+.........+c_{M-N}Z^{-(M-N)}+ \frac{N_{1}(Z)}{D(Z)}.

X(Z) = polynomial + rational \ proper \ function .

express    X(Z )  into powers of Z  as follows

X(Z ) = \frac{b_{o}Z^{N}+b_{1}Z^{N-1}+b_{2}Z^{N-2}+.......+b_{M}Z^{N-M}}{Z^{N}+a_{1}Z^{N-1}+a_{2}Z^{N-2}+.......+a_{N}}

then divide   X(Z)  by   Z

\frac{X(Z)}{Z} = \frac{b_{o}Z^{N-1}+b_{1}Z^{N-2}+b_{2}Z^{N-3}+.......+b_{M}Z^{N-M-1}}{Z^{N}+a_{1}Z^{N-1}+a_{2}Z^{N-2}+.......+a_{N}}  .

Now  express   \frac{X(Z)}{Z}  into partial fractions using different cases and find out the inverse Z-transform  for the function

X(Z ) = Z.(partial fraction \ expansion ) .

Convolution Method:-

express X(Z)   as a product of two functions X_{1}(Z)   and X_{2}(Z)  as follows X(Z) =X_{1}(Z) . X_{2}(Z)

then find the inverse Z- transforms of individual functions

x_{1}[n]\leftrightarrow X_{1}(Z)

x_{2}[n]\leftrightarrow X_{2}(Z)

by using convolution method find convolution of x_{1}[n]  and x_{2}[n]

i.e, x[n] = x_{1}[n] * x_{2}[n]

now x[n]   is the inverse Z-transform of X(Z) .

Cauchy Residue Theorem:-

f(Z) a function in Z if the derivative \frac{df(Z)}{dZ}   exists on and inside contour C and f(Z) has no poles at Z=Z_{o}   then.

\frac{1}{2\pi j}\oint_{c} \frac{f(Z)dZ}{Z-Z_{o}}=\left\{\begin{matrix} f(Z_{o}) \ if \ Z_{o} \ is\ inside \ C\\ 0 \ if \ Z_{o}\ is\ outside \ C \end{matrix}\right. .

if   (k+1)^{th} the derivative  of   f(Z)  exists on and has no poles at Z=Z_{o}   then.

\frac{1}{2\pi j}\oint_{c} \frac{f(Z)}{(Z-Z_{o})^k}dZ=\left\{\begin{matrix}\frac{1}{(k-1)!} \frac{d^{k-1}f(Z)} {dZ^{k-1}}\ if \ Z_{o} \ is\ inside \ C\\ 0 \ if \ Z_{o}\ is\ outside \ C \end{matrix}\right. .

the values on the right hand side are called Residue’s of the pole Z=Z_{o} .

if there are n no of poles inside C\frac{1}{2\pi j}\oint_{c} \frac{f(Z)dZ}{(Z-Z_{1})(Z-Z_{2})(Z-Z_{3})..(Z-Z_{n})}=\sum_{i=1}^{n}\lim_{Z\rightarrow Z_{i}} \left  .

 

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Relation between Laplace and Fourier Transform

The Fourier transform  of a signal x(t) is given as 

X(j\omega ) = \int_{-\infty }^{\infty } x(t) e^{-j\omega t}dt----EQN(I)

Fourier Transform exists only if \int_{-\infty }^{\infty } \left | x(t) \right |dt< \infty 

we know that s=\sigma + j\omega 

X(S) = \int_{-\infty }^{\infty } x(t) e^{-s t}dt

X(S) = \int_{-\infty }^{\infty } \left | x(t)e^{-\sigma t} \right | e^{-j\omega t}dt----EQN(II)

if we compare Equations (I) and (II) both are equal when  \sigma =0.

i.e, X(S) =X(j\omega)| \right |_{s=j\omega }.

This means that Laplace Transform is same as Fourier transform when s=j\omega.

Fourier Transform is nothing but the special case of Laplace transform where  s=j\omegaindicates the imaginary axis in complex-s-plane.

Thus Laplace transform is basically Fourier Transform on imaginary axis in the s-plane.

Normal incidence on a perfect conductor

Normal incidence on a perfect conductor

whenever an EM Wave travelling in one medium impinges second medium the wave gets partially transmitted and partially reflected depending up on the type of the second medium.

Assume the first case in Normal incidence that is Normal incidence on a Perfect conductor.

i.e an EM wave propagating in free space strikes suddenly a conducting Boundary which means the other medium is a conductor.

The figure shows a plane Wave which is incident normally upon a boundary between free space and a perfect conductor.

assume the wave is propagating in positive z-axis and the boundary is z=0 plane.

The transmitted wave  since the electric field intensity inside a perfect conductor is zero.

The incident   and reflected   waves are in the medium 1  that is free space.

The energy transmitted is zero so the energy absorbed by the conductor is zero and entire wave is reflected to the same medium

now incident wave is 

  in free space  for medium 1

 ( a wave propagating in positive z-direction) and the reflected wave is  ( a wave propagating in positive z-direction).

 .

by using tangential components  .

The resultant wave is   .

 .

the above equation is in phasor notation , converting the above equation into time-harmonic (or) sinusoidal variations

 .

This is the wave equation which represents standing wave , which is the contribution of incident and reflected waves. as this wave is stationary it does not progress.

it has maximum amplitude at odd multiples of   and minimum amplitude at multiples of  .

Similarly The resultant Magnetic field is

The resultant wave is   .

 .

the above equation is in phasor notation , converting the above equation into time-harmonic (or) sinusoidal variations

 .

this wave is  a stationary wave  it has minimum amplitude at odd multiples of   and maximum amplitude at multiples of  .

 

Nature of Magnetic materials

In order to find out the various types of materials in magnetic fields and their behaviour we use the knowledge of the action of magnetic field on a current loop with a simple model of an atom

Magnetic materials are classified on the basis of presence of magnetic dipole moments in the materials.

a charged particle with angular momentum always contributes to the permanent contributions to the angular moment of an atom
1. orbital magnetic dipole moment.
2. electron spin moment.
3. Nuclear spin magnetic moment.

Orbital Magnetic dipole Moment:-

The simple atomic model is one which assumes that there is a central positive nucleus surrounded by electrons in various circular orbits.

an electron in an orbit is analogous to a small current loop and as such experiences a torque in an external magnetic field, the torque tending to align the magnetic field produced by the orbiting electron with the external magnetic field.

Thus the resulting magnetic field at any point in the material would be greater than it would be at that point when the other moments were not considered.

so there are Quantum numbers which describes the orbital state of notion of electron in an atom there are n,l and ml
n-principal Quantum number, which determines the energy of an electron.
l-Orbital Quantum number which determines the angular momentum of orbit.
ml-magnetic Quantum number which determines the component of magnetic moment along the direction of an electric field.

electron spin Magnetic Moment:-

The angular momentum of an electron is called spin of the electron. as electron is a charged particle the spin of the electron produces magnetic dipole moment because electron is spinning about it’s own axis and thus generates a magnetic dipole moment.

\pm 9X 10^{-24} A-m^{2} is the value of electron spin when we consider an atom those electrons which are in shells which are not completely filled with contribute to a magnetic moment for the atom.

Nuclear spin Magnetic Moment:-

a third contribution of the moment of an atom is caused by nuclear spin this provides a negligible effect on the overall magnetic properties of material

That is the mass of the nucleus is much larger than an electron thus the dipole moments due to nuclear spin are very small.

so the total magnetic dipole moment of an atom is nothing but the summation of all the above mentioned .

Compression laws (A-law and u-law)

The laws used in compressor of a non-uniform Quantizer are known as compression laws.

two laws are available

  1. \mu– law.
  2. A-law.

\mu-law:-  A particular form of compression law that is used in practice is the so called  \mu– law which is defined by 

\left | v \right |= sgn(x){\frac{\ln (1+\mu \left | x \right |)}{(1+\mu )}}, for\ 0\leq \left | x \right |\leq 1.

where \left | v \right |  –  Normalized compressed output voltage.

             \left | x \right |   – Normalized input voltage to the compressor.

and  \mu  is a positive constant and its  value decides the curvature of \left | v \right |.

\mu =0   corresponds to no compression, which is the case of uniform quanization, the curve is almost linear as the value of \muincreases signal compression increases.

\mu =255  is the north american standard for PCM voice telephony.

for a given value of \mu, the reciprocal slope of the compression curve, which defines the quantum steps is given by the derivative of  \left | x \right |  with respect to \left | v \right | 

\frac{dx}{dv} = \frac{\ln (1+\mu )}{\mu } (1+\mu \left | x \right |) 

we see that therefore \mulaw is neither strictly linear nor strictly logarithmic.

i.e,

It is approximately linear at low levels of input corresponding to \mu \left | x \right |<<1. and  is approximately logarithmic at high input levels corresponding to \mu \left | x \right |>>1.

typical value of \mu =255 

A-law :- another compression law that is used in practice is the so called A- law defined by

\left | v \right | =\left\{\begin{matrix} sgn(x) (\frac{A\left | x \right |}{1+\ln A}), 0\leq \left | x \right |\leq \frac{1}{A}.\\ sgn(x) (\frac{1+ \ln A\left | x \right |}{1+\ln A}),\frac{1}{A} \leq \left | x \right |\leq 1. \end{matrix}\right.

A=1 corresponds to the case of uniform Quantization. A-law has been plotted for three different values of A. A=1, A=2, A=87.56.

typical value of A is 87.56 in European Commercial PCM standard which is being followed in India.

for both the \mu-law and A-law, the dynamic range capability of the compander improves with increasing \muand A respectively. The SNR for low-level signals increases at the expense of the SNR for high level signals.

to accommodate these two conflicting requirements (i.e, a reasonable SNR for  both low and high-level signals), a compromise is usually made in choosing the value of parameter \mufor the \mu-law and parameter A for the A-law. The typical values used in practice are \mu=255and A=87.56.

It is also interested to note that in actual PCM systems, the companding circuitry does not produce an exact replica of the non-linear compression curves shown in the compression law characteristics.

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Basic block diagram of analog communication system

Introduction:-

Communications refer to sending, receiving and processing of information by electrical means, that is it means exchanging information between transmitter and receiver.

In early 1840’s the type of communication used was Wire telegraphy later on the forms are as telephony, Radio communication (possible with the invention of triode tube, Satellite communications and fibre optics(with the invention of transistors and IC’s and semi-conductor devices), that means communications become more advanced with increasing emphasis on computer and other data communications.

A modern communication system is concerned with

before transmission:- 

  • sorting:- sorting for the right message.
  • Processing:- processing is to make that message more suitable for transmission.
  • storing:- storing that message before transmission.

then the actual transmission of that message takes place (processing and filtering  noise)

at the receiver:-

  • decoding:-decoding the original message.
  • storage:-storing a copy of that message.
  • interpretation:-and analyzing for the correctness of that message.

the different forms of modern communication systems includes Mobile communications,Computer communications, Radio telemetry etc.

to become familiar with communication systems one needs to know about amplifiers and oscillators that means fundamentals of electronic circuits must be known, with these concepts as a background the every day communication concepts like noise, modulation and information theory as well as various types of systems may be studied.

The most general form of Communication system ( one or two blocks may differ) is shown in the figure basic terminology used in Communication systems is message signal /information/data,channel,noise,modulation, encoding and decoding. Communication system is meant for communicating messages between Transmitter and Receiver (or) source & destination.

source:-

source or information source is the primary block in communication system which generates original message / actual message. 

i.e, selecting one message (actual message) from a group of messages itself is called as sorting data (or) information. Source generates message which may be in any form like words, code , symbols, sound signal, images, videos etc.among these the desired message has been selected and conveyed.

A transducer is one which converts one form of energy into electrical energy because the message from information source may not be always in electrical form, a transducer is used in between source and transmitter as a separate block sometimes (or) may be a part of Tx r.

Transmitter:-

Txr is meant for the following tasks

  • restriction of range of audio frequencies (i.e, limiting the bandwidth of the message signal).
  • Amplification.
  • Modulation. 

In general modulation is said to be the main function of the transmitter.

Channel:-

The medium that exists between transmitter and receiver is called as channel. The function of channel is to provide connection between transmitter  and receiver, two types of channels are  there wired/point to point  and wireless/broadcasting channels.

Point to point channels are generally wired channels(i.e, a physical medium exists) like Microwave links, optical fibre links etc. 

Microwave links:- these links are used in telephone transmission.In these type of links guided EM waves are used to transmit from Txr to Rxr.

optical fibre links:- used in low-loss high speed data transmission and uses optical fibers as the medium .

Broadcast channels:- the medium or channel is wireless here, in broadcasting a single transmitter can send information to many receivers simultaneously, satellite broadcasting system is one such system.

during the process of transmission and reception, the signal gets distorted due to noise in the channel, noise may interfere with the signal at any point but noise in the channel has greatest effect on the signal.

Receiver:-

The main function of the receiver is to reproduce the message signal in electrical form from the distorted received signal. This reproduction process is called demodulation (or) detection , in general this demodulation may be assumed as the reverse process of modulation carried out in transmission. 

there are a great variety of receivers in communication systems, the type of receiver chosen depends on type of modulation, operating frequency ,its range  and type of destination required. Most common receiver is superheterodyne receiver .

                            crystal receiver with head phones
                                  Radio receiver

so many types of receivers are available from a very simple crystal receiver with headphones to radar receiver etc.

Destination:- It is the final stage of any communication system. it would be a loud speaker / a display device/simply a load etc depending up on the requirement of the system.

flooding (static)

This is another type of static algorithm.

the main concept of flooding is to sent every incoming packet on a line to every other outgoing line except the line it arrived on.

flooding generates a large no.of duplicate packets, sometimes infinite unless we may take certain measures.

the measures are as follows:-

  • one measure is use of hop count in the header of each packet and decrement this count at each hop when count reaches to zero discard the packet.
  • How to take this hop count is another problem. Generally it is set to the length of path from source to destination and in worst cases the full diameter of the subnet.

  • another way is avoid sending a packet more than once through a router this is possible by using sequence no.
  • i.e, a source router (which generates packets) can put a sequence no. to each packet and each router will maintain a list of sequence nos. and if sees a packet with same sequence no in the list that packet is discarded (not flooded).

another way of flooding is of use selective flooding.

i.e, with this the router wouldn’t send every incoming packet on every line instead the router will send packets in a particular direction only.

i.e, east bound packets are sent on east side routers and similarly on  west side by west side routers.

even flooding is cumbersome, it has some uses

i.e,

  1. used in military applications.
  2. used in distributive data base applications in which to update all data bases concurrently.
  3. used in broadcast Routing.

flooding is used rather than any other algorithm since flooding chooses shorter path between two nodes where other algorithms may not.

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Quadrature Phase Shift Keying (QPSK) Transmitter and Receiver

Quadrature Phase Shift keying:-

The designing of digital communication system requires two important goals to achieve
1. To achieve low probability of error Pe.
2. To utilize Channel Band width efficiently.
QPSK is A Band width conserving modulation scheme, which is an example of Quadrature Carrier Multiplexing.
The modulation schemes such as ASK, PSK & FSK does not meet the Band width requirements of data Communication systems since the Bit rate and Baud rate are same in these schemes. Since the channel band width depends up on the bit rate (or) signalling rate of the modulation scheme. If two (or) more bits are combined into a symbol, then the signalling rate is reduced. Therefore the frequency of the carrier is also reduced, this reduces the transmission channel band width. Thus grouping of bits into symbols reduces Channel Band width.

Meaning of QPSK:-

In Quadri Phase Shift Keying as with Binary PSK information carried by the transmitted signal is contained in the phase of the carrier. The phase of the carrier Φc takes on one of four equally spaced values such as π/4, 3π/4, 5π/4 and 7π/4 that is in QPSK two successive bits are combined into a di-bit or symbol and each possible value of the phase corresponds to a unique di-bit.
for example the foregoing set of phase values are chosen to represent the gray encoded set of di-bits 10, 00, 01 and 11 , where only a single bit is changed from one di-bit to the next.

Generation of QPSK/ QPSK transmitter:-

Consider the generation and detection of QPSK signals. The figure shows a Block diagram of a typical QPSK Transmitter.The incoming binary sequence is first transmitted into polar form by a Non-Return to zero level encoder. Thus symbols 1 and 0 are represented by
√ Es   and –√ Es
This binary wave is next divided by means of a de-multiplexer into two separate binary waves. Consisting of the odd and even numbered input bits {be(t)} and {bo(t)} represents those two binary waves.
The two bit streams be(t) and bo(t) are modulated by two ortho-normal basis functions Φ1(t) and Φ2(t).finally, the two binary PSK signals are added to produce the desired QPSK signal.
i.e, SQPSK(t) = Se(t) + So(t).
SoPSK(t)= bo(t)* √(2/Ts)* cos 2πfc t
SePSK(t)= be(t)* √(2/Ts)* sin 2πfc t

SQPSK(t)= bo(t)* √(2/Ts)* cos 2πfc t + be(t)* √(2/Ts)* sin 2πfc t.

QPSK Receiver:-

The QPSK Receiver consists of a pair of correlators  called as In-phase channel and Quadrature phase channel with a common input.  The input x(t) is supplied with a pair of coherent reference signals Φ1(t) and Φ2(t).  The two correlators produces two signals x1(t) and x2(t) in response to the received signal x(t). these signals x1(t) and x2(t) are compared with threshold voltage 0V by the decision devices in the two channels.

If x1 >0, a decision has been made in favor of symbol ‘1’ for the in-phase channel output. but if x1<0 a decision has been made in favor of ‘0’. similarly for the Q-phase channel,

x2>0—-> a symbol ‘1’ is decided.

x2<0—-> a symbol ‘0’ is decided.

finally, these two binary sequences at the I-phase and Q-phase channel outputs are combined in a multiplexer to reproduce the original binary sequence at the Receiver output with the minimum probability of symbol error in the AWGN channel.

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Reconstruction filter(Low Pass Filter)

Reconstruction filter (Low Pass Filter) Procedure to reconstruct actual signal from sampled signal:-

Low Pass Filter is used to recover original signal from it’s samples. This is also known as interpolation filter.

An LPF is that type of filter which passes only low frequencies up to cut-off frequency and rejects all other frequencies above cut-off frequency.

For an ideal LPF, there is a sharp change in the response at cut-off frequency as shown in the figure.

i.e, Amplitude response becomes suddenly zero at cut-off frequency which is not possible practically that means an ideal LPF is not physically realizable.

i.e, in place of an  ideal LPF a practical filter is used.

In case of a practical filter, the amplitude response decreases slowly to zero (this is one of the reason why we choose  f_{s}>2f_{m})

This means that there exists a transition band in case of practical Low Pass Filter in the reconstruction of original signal from its samples.

Signal Reconstruction (Interpolation function):-

The process of reconstructing a Continuous Time signal x(t) from it’s samples is known as interpolation.

Interpolation gives either approximate (or) exact reconstruction (or) recovery of CT signal.

One of the simplest interpolation procedures is known as zero-order hold.

Another procedure is linear interpolation. In linear interpolation the adjacent samples (or) sample points are connected by straight lines.

We may also use higher order interpolation formula for reconstructing the CT signal from its sample values.

If we use the above process (Higher order interpolation) the sample points are connected by higher order polynomials (or) other mathematical functions.

For a Band limited signal, if the sampling instants are sufficiently large then the signal may be reconstructed exactly by using a LPF.

In this case an exact interpolation can be carried out between sample points.

Mathematical analysis:-

A Band limited signal x(t) can be reconstructed completely from its samples, which has higher frequency component fm Hz.

If we pass the sampled signal through a LPF having cut-off frequency of  fm  Hz.

From sampling theorem  

g(t) = x(t).\delta _{T_{s}}(t).

g(t)=\frac{1}{T_{s}}\left \{ 1+2\cos \omega _{s}t+2\cos 2\omega _{s}t+2\cos 3\omega _{s}t+..... \right \}.

g(t)     has a multiplication factor  \frac{1}{T_{s}}. To reconstruct  x(t)  (or)  X(f) , the sampled signal must be passed through an ideal LPF of Band Width of  f_{m}  Hz and gain  T_{s} .

\left | H(\omega ) \right |=T_{s} \ for \ -\omega _{m}\leq \omega \leq \omega _{m}.

h(t) = \frac{1}{2\pi } \int_{-\omega _{m}}^{\omega _{m}}T_{s}e^{j\omega t}\ d\omega.

h(t) = 2f_{m}T_{s} \ sinc(2\pi f_{m}t).

If sampling is done at Nyquist rate , then Nyquist interval is  T_{s} = \frac{1}{2f_{m}}.

 therefore  h(t) = \ sinc(2\pi f_{m}t).

h(t) = 0.      at all Nyquist instants  t= \pm \frac{n}{2f_{m}}  , when    g(t)    is applied at the input to this filter the output will be  x(t)  .

Each sample in g(t)  results a sinc pulse having amplitude equal to the strength of sample. If we add all these sinc pulses that gives the original signal  x(t) .

g(t) = x(kT_{s})\delta (t-kT_{s}).

x(t) =\sum_{k} x(kT_{s})\ h (t-kT_{s}) .

x(t) =\sum_{k} x(kT_{s})\ sinc(2\pi f_{m} (t-kT_{s})).

x(t) =\sum_{k} x(kT_{s})\ sinc(2\pi f_{m}t-k\pi ) .

This is known as interpolation formula

It is assumed that the signal  x(t) is strictly band limited but in general an information signal may contain a wide range of frequencies and can not be strictly band limited this means that the maximum frequency in the signal can not be predictable.

then it is not possible to select suitable sampling frequency  fs  .

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