Categories of Networks

Another alternating criterion for classifying N/w’s is their scale

i.e, the classification of multiple processor systems by their physical size.

Personal Area Network (PAN):-

PAN – is meant for one person. A wireless N/w connecting a computer with it’s mouse, keyboard and printer is a PAN also a PDA that controls the user’s hearing aid.

The next category is longer-range N/w’s  that is LAN, WAN, MAN -finally Inter network.

Those are categorized based on physical size, owner ship, the distance it covers and it’s physical architecture.

Local Area Network (LAN):-

  • LANs which are generally called as LANs are privately owned Networks within a single building (or) campus.
  • These are up to a few Km in size (10 m to 1 Km).
  • These are used to connect Personal computers (or) work stations in company office (or) factories to share resources (ex: Printers) and exchange information.
  • LANs are distinguished from other kinds of Networks by ‘3’ characteristics

i. their size.    ii. Transmission technology.      iii. topology.

  • LANs size is restricted that is worst case transmission time is bounded and is well known before in hand makes it possible to use certain kinds of designs that would not otherwise be possible. which also simplifies N/w management.
  • LANs may use a transmission technology consisting of a cable to which all the machines are attached.

Ex:-Telephone lines in rural areas.

  • LANs (traditional) may run at speeds of 10 Mbps to 100 Mbps and newer ones up to 10 Gbps.

(1 Mbps  \rightarrow  1000000 bits per second), (1 Gbps  \rightarrow  1000000000 bits per second).

  • The general possible topologies for LANs are bus, ring and star.

Bus Topology:-

In the Bus N/w (linear) at any instant at most one machine is the master and is allowed to transmit all other are required to refrain from sending.

An arbitration mechanism is needed to resolve conflicts when ‘2’ (or) more machines want to transmit simultaneously.

The arbitration mechanism may be centralized (or) de-centralized.

ex:- IEEE 802.3 ETHERNET a Bus based broad cast N/w is operating at 10 Mbps to 10 Gbps.

In Ethernet computers can transmit whenever they want to , if ‘2’ (or) more packets collide each computer just waits a random time and tries again later.

Ring topology:-

a second type broad casting system is the ring. In a ring each bit propagates on it’s own not waiting for the rest of the packet to which it belongs.

It also requires some arbitrating mechanism is required to the ring. IEEE 802.5 is a ring based LAN, which operates at 4 and 16 Mbps.

Ex:- FDDI.

LANs can be as simple as 2 pc’s and a printer (or) as long as with in a building.

LAN is used for sharing H/w (or) S/w (or) data.



Design issues of Network Layer


The N/w layer is concerned with getting packets from source and makes many hops at intermediate routers along the way to the destination.

i.e, N/w layer is the lowest layer that deals with end-to-end transmission.

To achieve goals of N/w Layer (NL), NL must know the topology of communication subnet and to choose appropriate paths through it.

Network Layer Design issues:-

The design issues of NL are

Services provided to the Transport Layer at the N/w -Transport layer interface while providing services we must keep the following factors in mind.

the services should be independent of the router technology .

the Transport layer should be shielded from the number, type and topology of the routers present.

The N/w addresses made available to the Transport layer should use a uniform numbering plan even across LAN’s and WAN’s.

depending on above goals , the designers of N/w layer have more freedom and has assuming that the N/w layer should provide ‘2’ types of services.

  1. Connection-oriented Service(COS)
  2. Connection-Less Service(CLS).

Implementation of Connection-Less N/w layer service:-

In this type of N/w layer service packets are injected into subnet individually and are routed independently.

Packets are called datagrams and subnet as datagram subnet. Here no advanced setup is required.

Suppose in Transport Layer, a Process   in Host wants to send a long message to process in Host it then adds a TL header to the message and handed over to N/w layer.

Now in the N/w layer, the packet size is small compared to this message so it breaks the message into 4 packets 1, 2, 3 and 4 and gives the packets to router A.

Each router in the datagram subnet will maintain a Table (Routing Table), which gives the information about where to send the packets.

So Router A has a table, each table consists of ‘2’ columns Destination (where to go) and line (for outgoing) for that Destination.

i.e. from A to reach Destination A, No line is used, from A to reach Destination D,  B Router is used.

Now packets 1, 2, 3 and 4  (from ) Host are given to Router A then Router A stores all these packets and forwards them depending on the Routing Table of ‘A’.

Initially packets 1, 2, and 3 are forwarded using C to later on it uses a new Router B to forward 4 to depending on traffic.

Implementation of Connection-Oriented N/w layer service:-

In this service the subnet is called Virtual Circuit subnet.

A path from Source Router to Destination Router is established before sending any packets. That Router is use for all traffic flowing over that connection.

Ex: – a telephone system.

When the connection is released, the Virtual Circuit is also terminated.

Let us see an example.

In this Host wants to send packets to Host 1,2,3 and 4.

Now will establishes a connection 1 with Host.

Whenever packet comes from Host , we use connection 1 with identifier as 1.

i.e, from Routing Table of A, a packet is coming from and its identifier is ‘1’ and uses the outgoing Router C with identifier ‘1’

Similarly for   use identifier as ‘2’. This is called label switching. The comparison between these Connection-Oriented and Connection Less Services (or) Virtual Circuit (or) Datagram sub nets is given in the table.

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Circuit Switched Networks

A Circuit Switched N/w consists of a set of switches connected by physical links.

A connection b/w ‘2’ stations is dedicated path made of one (or) more links. Each connection uses only one dedicated channel on each link.

i.e, each link is divided into n channels either by using TDM (or) FDM.

This circuit consists of 4 switches I, II, III and IV and Multiplexers with n=’3′ channels and one link.

In some circuits Multiplexing can be implicitly included in the switch fabric it self. In this circuit the end systems are connected to a switch for simplicity consider ‘2’ end systems A and M, connected to the switches I and III.

when A needs to communicate with M . A needs to request to a connection to M, which must be accepted by all switches and by M it self- which is called setup phase.

a channel circuit is reserved on each link and the combination of circuits forms a dedicated path. After establishing path data transfer can take place. The next phase is tear down.

i.e, after all data have been transferred. Generally circuit-switching takes place at the physical layer.

Before Communication (starting), the stations must make reservation for the resources like channels, switch buffers switch i/o ports switch processing time and are dedicated during the entire duration of data transfer until the tear down phase.

Data transferred is not packatized that is data is send as a continuous flow b/w source and destination.

there is end-to-end addressing in setup phase.

The 3 phases involved are:-

Circuit switched N/w’s requires ‘3’  setup phases

  1. Connection-setup.
  2. Data transfer.
  3. Tear down.

Setup Phase:-

A dedicated circuit is established before the ‘2’ communicating parties talk to each other.

i.e, creating  a dedicated channels b/w switches. To communicate A with M . initially a requesting process as follows

A to I, I to IV and IV to III, III to M and an acknowledgement in the reverse order after the reception of ‘ack’ a connection is established.

Data Transfer Phase:-

In this phase data transfer occurs b/w the ‘2’ devices.

Tear down phase:-

To disconnect , a signal is sent to each switch to release the resources by any one of station.

Efficiency of Circuit Switched Network:-

These are less efficient in terms of allocated resources. Since all the resources are allocated during the entire duration of the connection  and these resources are un available to other connections.

Delay in this type of N/w’s is due to establishment of connection , data transfer and disconnecting the circuit.

Switching at the physical layer in the traditional telephone N/w uses the circuit switching approach.

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Pseudo Noise Sequences (PN Sequences)

Pseudo Noise Sequences (PN Sequences):-

A PN Sequence is a periodic binary sequence with a noise like wave form (generally of high frequency) .

PN Sequences are commonly used to generate noise that is approximately “white”.

It has applications in scrambling, cryptography and spread spectrum communications.

It is commonly referred to as Pseudo Random Binary Sequence (PRBS) and they are widely used in communication standards these days.

The qualifier “Pseudo” implies that the sequence is not truly random. Actually, it is periodic with a large period and exhibits some characteristics of a random white sequence with in that period.

Generation of PN Sequences:-

PN Sequence is a periodic noise like wave form that is usually generated by means of a feedback shift Register, a general Block Diagram of which is shown below.

  1. A feedback Shift register consists of an ordinary shift register made up of ‘m’ no. of Flip-flops.(two state memory stages) 1,2,3….m. The data of one FF is shifted to the next FF whenever a clock pulse is applied and a logic circuit that are inter connected to form a multi loop feedback circuit.
  2. The FF’s in the shift register are regulated by a single timing clock. at each pulse (clock) of the clock, the state of each FF is shifted to the next one down the line.
  3. With each clock pulse the logic circuit computes a Boolean function of the states of the FF’s the result is then feedback as the i/p to the first FF, thereby preventing the shift register from emptying.

The PN Sequence so generated is determined by the length ‘m’ of the shift registers its initial state and the feedback logic.

Let denote the state of the FF after   clock pulse. This state may be represented by ‘0’ (or) ‘1’. The state of the shift register after the   clock pulse is then defined by the set {  } where

For the initial state, k is zero.

From the definition of a shift register.


Where   is the  i/p applied to the first Flip-Flop after clock pulse. According to the configuration described.     is a Boolean function of the individual states

For a specified length m this Boolean function uniquely determines the subsequence sequence of states and therefore the PN Sequence produced at the o/p of the final Flip-Flop in the shift register.

With ‘m’ no . of FF’s no. of possible states of the shift register is at most .

Therefore the PN Sequence generated by a feedback shift register must eventually become periodic with a period of at most .

A feedback shift register is said to be linear when the feedback logic consists of entirely modulo-2 (EX-OR gates) adders.

A zero state is not permitted because (i.e, a state with all FF’s set to zero) the shift register remains in the same state forever.

The period of a PN Sequence produced by a linear feedback shift register with ‘m’ Flip Flops can not exceed    .

When the period is exactly    , the PN Sequence is called a Maximal length Sequence  (or) m-Sequence.




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Hierarchial Routing(dynamic)

As the size of the N/w increases the entries in the routers routing table increases, this increase causes ‘2’ things to increase

  1. Memory consumed by Routing Tables.
  2. CPU processing time required to scan the entries in Routing Tables also increases.

and also the Band width needed also increases.

at particular point it is not possible almost all for a router to maintain routing tables still the size increases.

So the possible solutions for this is to use Hierarchial routing.

In this Hierarchial routing there are regions. The regions consists of no.of routers and he routers in a region are aware of how to route packets in it’s own regional routers but nothing about internal structure of other regions.

Hierarchial routing may be ‘2’ level Hierarchy as shown in the given figure.

Initially assume we don’t have Hierarchial routing that there exists 17 routers 1A,  1B, 2A, 2B, 2C, 2D……………….5E and the routing table for all these 17 routers by choosing no.of hops and destination line as parameters is given in the figure.

from router 1A to reach router 1B, it uses line 1B itself and the no.of hops are ‘1’. Similarly, from route 1A to router 4B it uses line 1C and no.of hops are ‘4’  as 1A to 1C to 3B to 4A to 4B.

if we use Hierarchial routing the no.of entries previously 17 are reduced to ‘7’. only.

The 17 outers are divided into 5 regions with region having some no.of routers.

If we observe the table , the table consists of 7 entries , the destination as region 2,3,4,5 but not routers 2A, 2B,…..etc.

but for it’s own region it is aware of other routers 1A, 1B, 1C  from 1A to reach region 4 it uses line 1C and no.of hops are 3……..

even there is a problem with Hierarchial routing is choosing the best path based on path lengths.

the best from 1A to 5C is Via region 2 rather than Via region 3.  If N/w size increases we go for other levels of Hierarchy that is a 3 level Hierarchy.


Kamour and Kleinrock (1979) discovered that the no.of levels for a N router subnet is  ln N entries for a router =e .ln (N).


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Burst and Random error correcting codes

Burst and Random error   correcting codes:-

To correct specific, well-defined classes of error patterns a variety of codes are designed but the problems of correcting random errors and burst errors are treated separately.

But in practical systems errors occur neither independently nor in well-defined bursts. Errors may occur in a fashion, that as a mixture of random and burst errors.

therefore either random error correcting codes (or) single-burst error correcting codes are insufficient and inefficient to correct errors which involves as a mixture of both random and burst errors.

So for channels in which both types of errors occur, there is a need for special types of codes, which will correct both random and burst errors simultaneously.

The most effective method uses the interlacing technique.

Interlace code:-

for an (n, k) cyclic code a (\lambda n,\lambda k)   interlaced code can be constructed by simply arranging \lambda code vectors of the original code into  rows of a rectangular array and transmitting them by column by column the resulting code is called interlaced code with an interlacing degree  \lambda.

In an interlaced code, a burst of length  \lambda (or) less will affect not more than one digit in each row [transmission is done on a column by column basis].

If the original code can correct single errors, then the interlaced code can correct single bursts of length  \lambda  (or) less.

If the original code can correct‘t’ errors (t>1) then the interlaced code can correct any combination of t bursts of length \lambda   (or) less.

consider for example a (15,7) BCH code is generated by g(x)=x^{8}+x^{4}+x^{2}+x+1 which will have d_{min}=5 (minimum distance) it is able to correct \leq \left [ \frac{d_{min}-1}{2} \right ]\Rightarrow \leq 2.

it may correct 2 errors. so for this code we can construct a (75, 35) interlaced code with  as (75, 35) with a burst error correcting capability of 10.

Now the message block length is 35 bits, this 35 bit message block is divided into 5 ,’7’ bit blocks as

and each 7 bit message block is converted into a 15 bit code word by using g(x).

These code words are arranged as five rows of a 5 X 15 matrix. The column of the matrix is transmitted in the order indicated as a 75 bit long code vector.

Error correcting capabilities:-

To illustrate the burst and random error correcting capabilities of this code.

assume that errors havee occures in bit positions 5,37 through 43 and 69.

at the decoder, the decoder operates on the rows of the table that each row has a maximum of 2 errors and from (15,7) BCCH code , we know that the code is able to correct up to ‘2’ errors per row.

therefore the error pattern occurred in the table can be corrected. The errors in bit positions 5 and 69 as random errors and from 37 to 43 as burst error.

while operating on the rows of the code array may be an obvious way to encode and decode an interlaced code which is not the simplest implementation.

The simplest implementation results from the property that if the original code is cyclic, then the interlaced code is also cyclic.

The polynomial in interlaced code is g(x^{\lambda }) if the original polynomial is g(x).

Thus encoding and decoding can be accomplished by using shift registers. The decoder for the interlaced code can be derived from the decoder for original code by replacing each shift register stage of the original decoder by \lambda stages without changing the other connections.

each shift register stage by \lambda stages without changing other connections. This allows the decoder to look at successive rows of the code array on successive decoder cycles.

[as first cycle–> first row, in second cycle–> second row…]

If the decoder for the original code was simple then the decoder for interlaced code will also be simple.

Therefore The interlacing technique is an effective tool for deriving long powerful codes from short optimal codes.

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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


  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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Multicast Routing(dynamic)

Some applications require widely-separated processes to work together as groups.

i.e, for example a distributed  data base system.

so there is a need to send a message to well defined groups normally large in size but small compared as a whole (system).

sending a message to such group is called Multi casting and the routing algorithm used is called Multicast Routing.

therefore some mechanism is required to create and destroy groups and allow processes to leave and join a group.

i.e, Routers learn about the hosts belong to which group, this is possible by ‘2’ ways

  1.  Either hosts must inform their routers about changes in groups.
  2. (or) routers must query their hosts periodically.

Now let us see how  to route messages in  Multicast routing 

Consider a N/W with ‘2’ groups 1 and 2 and some are members of both 1,2. a spanning tree for the left most router A is given in the figure.

when a process sends a multicast packet to a group the first router examines it’s spanning tree and prunes(cuts) it without having the members of the other group.

Then using this pruned trees, the router can send messages to the specific group only either to group 1 using Fig(a) and to groups using Fig(b).

while pruning we use Link State Routing (or) Distance Vector Routing.

if a router B is not a member of either 1 (or) 2 and it receives a multicast message if it doesn’t want to receive messages . It sends a PRUNE message saying  don’t send any multicast messages to it.

disadvantage of this algorithm is it scales poorly to large N/w’s  hence another alternative design is core-based tree.

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Design issues of Data Link Layer

Data Link Layer:-

The study of design principles of Data Link Layer deals with the algorithms for achieving reliable, efficient communication between ‘2’ adjacent machines at DLL.

adjacent means those two machines are connected by a wire— a co-axial cable, telephone line (or) point-to point wireless channel.

The essential property of a channel –wire like— means sending the bits in the same order as they are sent .

Let us suppose we have a system A and machine B and these two are connected by a wire and assume that there is no means of any software in any machine then the picture seems to be as follows 

as A just puts the bits on the wire and B just takes off the bits.

i.e, they have only a finite data rate and there is a non-zero propagation delay between the time a bit is sent and the time it is received and the communication devices makes errors occasionally.

these are the limitations that have been taken care for the efficiency of the data transfer.

The protocols used for communications must take all these factors into consideration.

i.e design issues  and also the nature of errors , their causes and how they can be detected and corrected are all taken care in DLL.

Design issues of DLL:-

The DLL has specific functions to carryout those are

  1. Providing a well-defined service interface to the N/w layer.
  2. Dealing with the transmission errors.
  3. Regulating the flow of data so that slow receivers are not swamped by fast senders.

i.e, DLL takes the packets coming from N/W layer and are encapsulated into frames for Transmission.

each frame contains a frame header and a trailer and a payload field for holding the packet.

the heart of DLL does the frame formation.

Services provided to the N/W Layer:-

The function of DLL is to provide services to N/W layer 


Optical Communication System

Optical Communication System:-

An Optical fiber transmission link comprises the elements shown in the given figure.

The key sections are

  1. A transmitter consisting of a light source and its associated drive circuitry.
  2. A cable offering mechanical and environmental protection to the optical fibers contained inside it.
  3. A receiver consisting of a Photo detector plus amplification and signal restoring circuitry.

Additional components include optical connectors, splicers, couplers (or) beam splitters and repeaters.


optical fiber is one of the most important elements in an optical fiber link. The cable may contain copper wires for powering repeaters which are needed for periodically amplifying and reshaping the signal in long distance communication.

The cable generally contains several cylindrical hair-thin glass fibers, each of which is an independent communication channel.

Similar to copper cables, the installation of optical fiber cables can be either aerial, in ducts, under sea (or) buried directly in the ground.

As a result of installation and (or) manufacturing limitations, individual cable lengths will range from several hundred meters to several kilo meters for long distance applications.

The real size and cable weight determines the actual length of a single cable section.

Cable in ducts—- shorter length

Aerial/ buried applications—– longer lengths.

The complete long-distance transmission line is formed by splicing (or) connecting together these cable sections.

In optical fibers attenuation is a function of wave length .

In early stages of technology, optical fibers were used in

First window: (800nm-900nm) wave length.

Later on optical fibers are used in the long-wave length region.

Long-wave length region (1100-1600) nm

Second window-centered around 1300nm.

Third window- centered around 1550nm.


Once the fiber cable is installed a light source (which is dynamically compatible with the fiber cores) is used to launch power into the fiber.

The electric i/p signal is either analog (or) digital form the Transmitter circuit converts this electric signal to an optical signal.

Optical source is a square-law device. In (800-900) nm region the light source is made up of  

Ga Al As and in long distance region (1100nm-1600nm) In Ga AsP is the alloy used.

after an optical signal (light) has been launched into the fiber, it will be attenuated and distorted with increasing distance because of scattering, absorption and dispersion mechanisms in the wave guide.


The attenuated and distorted , modulated optical power emerging from the fiber end will be detected by photo diode (or) photo detector.

Photo detector converts optical power into electrical signal (it also uses a square-law).

photo detectors are PIN diodes, Avalanche photo diodes and the type of material it is made up of is In Ga As.

further the electrical signal will be amplified and restored.

therefore the design of the receiver is more complex than that of transmitter.


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Colpitt’s 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}.


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.


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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Brewster’s angle

In parallel polarization the incident angle at which there is no reflection is called Brewster’s angle.

\rho _{parallel} = 0.

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

\frac{E_{r}}{E_{i}}=\frac{(\eta _{2}\cos \theta _{t}-\eta _{1}\cos \theta _{i})}{(\eta _{1}\cos \theta _{i}+\eta _{2}\cos \theta _{t})}=0.

(\eta _{2}\cos \theta _{t}-\eta _{1}\cos \theta _{i})=0.

\eta _{2}\cos \theta _{t}=\eta _{1}\cos \theta _{i}.

by squaring on both sides  \eta^{2} _{2}\cos^{2} \theta _{t}=\eta^{2} _{1}\cos^{2} \theta _{i}.

\eta^{2}_{2}(1-\sin^{2} \theta _{t})=\eta^{2} _{1}(1-\sin^{2} \theta _{i})....EQN(I)

By using Snell’s law  \frac{\sin \theta _{i}}{\sin \theta _{t}} =\sqrt{\frac{\mu _{2}\epsilon _{2}}{\mu _{1}\epsilon _{1}}} =\frac{r_{2}}{r_{1}}.

using the above equation  \sin^{2} \theta _{t} =\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}} \sin^{2} \theta _{i}. substituting this in EQN (I).

\eta^{2}_{2}(1-\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}} \sin^{2} \theta _{i})=\eta^{2} _{1}(1-\sin^{2} \theta _{i}).

(\eta^{2}_{2}-\eta^{2}_{2}\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}} \sin^{2} \theta _{i})=(\eta^{2} _{1}- \eta^{2} _{1}\sin^{2} \theta _{i}).

after simplification    \sin^{2} \theta _{i}(\eta^{2} _{1}-\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}}\eta^{2} _{2} )=(\eta^{2} _{1}-\eta^{2} _{2 }).

as \eta _{1} = \sqrt{\frac{\mu _{1}}{\epsilon _{1}}}  and  \eta _{2} = \sqrt{\frac{\mu _{2}}{\epsilon _{2}}}.

\sin^{2} \theta _{i}(\frac{\mu _{1}}{\epsilon _{1}}-\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}}\frac{\mu _{2}}{\epsilon _{2}} )=(\frac{\mu _{1}}{\epsilon _{1}}-\frac{\mu _{2}}{\epsilon _{2}})..

by simplification  \sin^{2} \theta _{i}= \frac{(1-\frac{\mu _{2}\epsilon _{1}}{\mu _{1}\epsilon _{2}} )}{(1-\frac{\epsilon^{2} _{1}}{\epsilon^{2} _{2}})}.

here \theta _{i} is called as Brewster’s angle.

Let us assume two mediums are lossless dielectrics and are non-magnetic  then

\mu _{1} =\mu _{2}=\mu _{0}.

\sin^{2} \theta _{Brewster}= \frac{1}{(1+\frac{\epsilon _{1}}{\epsilon _{2}})}.

\sin \theta _{Brewster}= \sqrt{\frac{\epsilon _{2}}{\epsilon _{2}+\epsilon _{1}}}.

\tan \theta _{Brewster}= \sqrt{\frac{\epsilon _{2}}{\epsilon _{1}}} = \frac{r_{2}}{r_{1}}.

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OSI reference model

The OSI model is based on ISO and is introduced in the year 1983 and was revised in 1995 .

This is also known as ISO-OSI model(International Standards Organization-Open System Interconnection model)

and is used to connect open systems(open- they are ready for communication)

The OSI model has 7 layers. These layers are formed by considering the following things

  1.  A layer should be created where a different abstraction is required.
  2. Each layer should perform a well defined function.
  3. A layer  boundaries should be chosen to minimize the information flow across the interface.
  4. The function of each layer chosen by keeping an eye toward defining international standardized protocols.
  5. The no.of layers chosen  that same function is not performed in the each layer and the function performed is not so small.

Now the model looks like this

Physical layer:-

Physical layer is connected with 

  • Transmitting raw bits over a communication channel.
  • i.e, the design issue makes sure that sending ‘1’  must be received as ‘1’ itself  but not as ‘0’.
  • How many volts is required to represent 1 and 0?
  • How many Nano Seconds a bit lasts?
  • whether transmission may proceed in both the directions (or) uni-directional and how the initial connection is established?
  • whether to terminate the connection (or) not?
  • i.e, most of the design issues deal with mechanical, electrical and timing interfaces & the physical transmission medium.

 At physical layer  the data rate, synchronization of bits, line configuration(point-to-point,Broadcasting) and the topology used and Transmission mode( simplex/duplex) are specified.

Data Link Layer:-

It gets services from physical layer and offers services to the N/w layer.

The DLL makes the raw transmission as reliable and is responsible for node to node delivery . It makes physical layer appears error free to the upper layer.

the main functions of DLL are:

Framing:- The DLL divides the stream of bits received from N/w layer into manageable data units called frames.

Physical addressing:- If frames are distributed to different systems on the N/w , DLL adds header to the frame to define physical address of sender and receiver of the frame.

Flow control:- DLL also keep a fast Transmitter from drowning a slow receiver in data. therefore it requires a mechanism for controlling the flow to prevent overwhelming of the receiver.

Error control:-

DLL provides a mechanism to 

  • detect damaged (or) lost frames and to re transmit this damaged (or) lost frames.
  • and needs a mechanism to prevent duplication of frames. error control is normally achieved through a trailer added at the end of the frame.
  • it accepts data from N/w layer and break up that data into data frames and transmit the frames sequentially. If transmission is reliable it is observed by acknowledgement frame.
  • MAC layer in DLL takes care of how to share channel and control the access in case of broad casting used for end-to-end (or) node-to-node delivery.

Network Layer:-

It controls the operation of subnet. The processes involved in N/w layer are


Link state Routing(dynamic)

Distance Vector routing was used in ARPANET (1979) till it is replaced by Link State Routing.

The two problems in Distance Vector Routing are

  1. DVR doesn’t take line Band width into account since the design metric is delay.(initially all lines are 56 Kbps so line Band width is not an issue but when lines are upgraded with 230 Kbps, 1.54 Kbps then the problem arises if we will not consider Band width).
  2. one more problem that occurs in DVR is count-to infinity problem.

for these reasons it was replaced by new algorithm known as  Link State  Routing algorithm (LSR) algorithm.

the main idea of LSR is as follows(for a Router)

Step 1:- Discover it’s neighbors and learn their N/W addresses.

Step 2:-Measure the delay (or) cost to each of it’s neighbors.

Step 3:- Construct a packet with the information a Router has learned.

Step 4:- Send this packet to all the Routers.

Step 5:-Compute the shortest path to every other Router.

for example,

S 1:-Learning about the neighbors

first of all, when a Router is booted to learn about it’s neighbors it will send a packet called ‘HELLO’ on each point-to- point line.

the Router on the other hand is expected to send back a reply saying who it is?

when two (or) more Routers are connected by a LAN, the situation is more complicated and the Routers are named uniquely to avoid any conflicts.

In the LAN A, C and F are connected to LAN, when a distance Router hears that 3 Routers are all connected to F, it is essential to know whether all 3 means same F (or) not?

To avoid this we can treat LAN as an additional node N as below

N in the above figure is an artificial node, the path from A to C is represented as ANC.

S2:- each Router in LSR requires to know an estimate of delay to each of it’s neighbors.

one way to measure delay send an ECHO packet and get reply immediately and then calculate the round-trip-delay t/2 .

for better results perform this same no.of times and use the average.

while measuring delay one question that arises is to consider the load (or) not? If load is considered, the round trip timer must be started when ECHO packet is queued.

when load is ignored the timer shouted be started when the ECHO packet reaches the front queue.

when a Router has a choice between 2 lines with the same Band width one of which is loaded all the time and the other one  is not loaded at all.

Then Router will chose the time with less load as the shortest path, this will result in better performance.

Consider a Sub net which is divided into 2 parts X and Y an is connected by 2 lines CF and EI.

Suppose the line CF is heavily loaded with long delays(including Queuing delay) after the new Routing tables have been installed most of the traffic will now go on EI.

Consequently in the nest update CF will appear as best path. Routing Tables may oscillate wildly causing some potential problems.

One solution to this is to divide the load equally among the lines but that may disturb the concept of best path.

S3:- Building link state Packets

Once the transformation needed for the exchange has been collected the next step is for each Router is to build a link state packet.

link state packet consists of information regarding to sender , sequence no, Age, neighbors delays.

for example consider the Sub net 

These link state packets have to build periodically and also when a Router going down etc.

S4:- Distributing the Link State Packets

The net thing is to distribute the  Ls packets reliably. in order to distribute the packets we may use flooding , to make flooding more efficient we use sequence numbers to packets.

The main problem is with sequence no’s repetition of Seq.nos one solution is to use a 32 bit which may take 137 years to repeat the same no.

If a Router crashes the sequence no becomes a zero then there is a possibility a Router may discards it.

To avoid all the above problems we use a parameter called Age whenever Age=0 the Router discards a packet .

after distribution process we use refinements to this distribution process(flooding).

whenever a packet arrives it first placed in a holding area later on another packet arrives the 2 Seq.nos are compared , if they are equal the duplicate is discarded.

The figure shows the buffer space at Router B

Suppose a packet is coming from source A with 21 and Age as 60 

we may expect an Acknowledgement from C and F but not from A.

Computing the new Routes for a Router :-

after constructing the LS packets to all the Routers 

we may use Dijkstra’s algorithm to construct the shortest path to all the destinations and this can be updated from time to time. 


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Distance Vector Routing(dynamic)

Modern Computer Networks uses dynamic algorithms rather than static algorithms.

dynamic algorithms may consider the current  traffic (or) load on the Network.

Two types of  dynamic routing algorithms are there

  1. Distance Vector Routing (DVR).
  2. Link state Routing.

Distance Vector Routing operates by the following way

each Router maintains a table (gives the information about distance to other routers) and updates these routing tables by exchanging information with it’s neighbors.

It is also known as Bellman-Ford (or) Ford Fulkerson algorithm.

DVR is used in ARPANET  and also as RIP.

In DVR each Router will maintain a Routing Table regarding to each Router in the subnet and the estimate of the time (or) distance to the destination.

one can use different design metrics like no.of hops, time delay in (milli Seconds), no.of packets Queued etc.

Here time delay is used as a metric.

Therefore a Router knows a delay to each of it’s neighbors and once every T milli Seconds these delays get updated by exchanging information with it’s neighboring Routers.

Consider a subnet with Routers A,B,…..L . Now choose a Router J with immediate neighbors (directly connected to J) are A, I, H and K.

Now the estimated delay of J to A, I, H & K are 8, 10, 12 & 6 milli Seconds respectively.

Suppose J wants to calculate a new route from J to G this is possible  by finding the delay from J to G using the neighbors to J.

i.e, J to G delay (through A) = J to A delay +A to G delay = 8+18=26 mSec.

J to G delay (through I) = J to I delay +I to G delay = 10+31=41 mSec.

J to G delay (through H) = J to H delay +H to G delay = 12+6=18 mSec.

J to G delay (through K) = J to K delay +K to G delay = 6+31=37 mSec.

The best among the 4 possibilities is through H with less delay 18 mSec and makes as entry in it’s Routing table.

In this way Router J computes all possible delays to each router and updates it in it’s Routing table.

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Shortest Path Routing (static)

The idea of this shortest path routing is simple, which is used to build a graph with each node as router and arc represents a communication line (or) link.

This algorithm just finds the shortest path between them on the graph.

There exists many design metrics to choose to get the shortest path are no.of hops, queue length,transmission delay etc.

for example if we choose no.of hops as metric, the paths ABC, ABE have equal no of hops means that those are equally long but ABC is much larger than ABE.

The labels on the above graph (2,2,7) are computed as a function of the distance, Band width, average traffic, cost etc.

one of the algorithm used for computing the shortest path between 2 nodes is Dijkstra’s algorithm.

it is as follows, Initially all nodes are labeled with infinite distance.

Let us consider the figure as shown below

to find the shortest path from A to D.

step 1:- choose the source node as A and mark it as permanent node.

step 2:- find the adjacent nodes to A  those are B and G then choose the node with the smallest label as the permanent node.

Now this node B becomes the new working node.

step 3:- Now start at B and repeat the same procedure

by following above procedure two paths are available ABEGHD with a distance of 11 from A and ABEFHD with a distance of 10 from A.

so the second path is chosen as a shortest path.

therefore the final shortest path is ABEFHD with nodes A,B,E,F,H and D as permanent nodes.

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Broadcast Routing(dynamic)

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

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

 Different methods of Broadcasting:-

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

so this is least desirable one.

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

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

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

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

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

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

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

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

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

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

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

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

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

advantages of reverse path forwarding:-

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

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

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


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

This is reverse path forwarding.

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Asynchronous Transfer Model(ATM)


Asynchronous Transfer Model is another important connection oriented Network.

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

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

Design goals:-

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

The problems associated with existing networks:-

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

Frame Networks:-

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

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

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

Mixed N/w Traffic:-

If there exists variable frame sizes

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

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

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

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

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

Cell Networks:-

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

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

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

Asynchronous TDM:-

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

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

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

ATM architecture:-

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

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

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

Virtual Connection:-

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

ATM Virtual Circuits:-

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

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



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Parallel Polarization

Parallel Polarizaton:-

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

then the incident and reflected fields is given by

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

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

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

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

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

now the reflected wave is 

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

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

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

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

and the transmitted fields in the second  medium is  

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

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

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

Transmission coefficient:-

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


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

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

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

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

(1)+(2) implies 

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

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

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

Reflection coefficient:-

from EQN (1)

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

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

after simplification

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

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

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

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Surface impedance

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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




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oblique incidence

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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

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

the wave propagates 

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

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

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

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

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

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

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




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Poynting theorem


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

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

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

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

i.e, it gives the direction of propagation .

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


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

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

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

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

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

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

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

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

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

by using the vector identity 

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


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

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

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

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

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

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

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

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

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

by integrating the above equation by over  a volume 

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

by converting the volume integral to surface integral

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

the above equation gives the statement of Poynting theorem.

Poynting theorem:-  

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






Lag compensator

Lag compensator:-

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

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

Pole-Zero Plot of Lag compensator:-

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

Realization of Lag compensator as Electrical Network:-

The lag compensator can be realized by an electrical Network.

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

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

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

after simplification

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

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

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

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

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

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

using the EQN(II)   

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

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

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

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

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

Frequency-response of Lag compensator:-

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Choice of \beta :-

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

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

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

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

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

atypical value of \beta =10.

Procedure for bode-plot of a lead compensator:-

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

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

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

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

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

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

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

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

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

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Lead Compensator

Lead compensator:-

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

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

Pole-zero plot of Lead compensator:-

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

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

Realization of Lead compensator as an Electrical Network:-

The lead compensator can be realized by an electrical Network.

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

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

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

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

after simplification

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

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

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

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

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

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

using the EQN(II)   

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

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

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

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

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

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

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

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

Frequency-response of a lead compensator:-

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

the frequency response of lead compensator is 

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Choice of \alpha :-

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

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

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

Procedure for bode-plot of a lead compensator:-

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

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

Step 2:- using the relation 

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

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

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

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

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

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

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

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

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

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Introduction to Root Locus

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

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

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

A stable system with low damping is also unwanted.

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

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

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

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

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

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

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

Root Locus:-

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


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

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



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

They have to satisfy ‘2’ criteria known as

  1. Magnitude Criterion.
  2. Angle Criterion.

Magnitude criterion:-

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

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

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

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

Angle criterion:-

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

where q=0,1,2……….

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

Root Locus definition:-

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

It is a graphical method 

-\infty to 0       \rightarrow     Inverse Root Locus

0  to \infty    \rightarrow  Direct Root Locus

generally Root Locus means Direct Root Locus.

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

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

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

m= no .of zeros

n= no.of poles

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

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

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

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

from the Angle criterion,

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

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

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

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

where q=0,1,2………


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Inductance of a Co-axial cable, Solenoid & Toroid

Inductance of a Co-axial cable:-

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

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

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

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

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

where \phi-Total flux coming out of the surface

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

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

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

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

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

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

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

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


Inductance of a Solenoid:-

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

\phi – Total flux coming out of solenoid.

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

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

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

from EQN(I) ,

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

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

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

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

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

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

Inductance of a Toroid:-

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

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

then flux linkage  \lambda = N\phi

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

\lambda = NBA

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

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

L=\frac{\lambda }{I}

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

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

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Classification (or) topologies of feedback Amplifiers

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

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

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

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

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

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

For any Amplifier circuit we require 

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

Classification of feedback Amplifiers is also known as feedback Topologies.

Voltage-Series feedback Connection:-

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

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

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

Current-Series feedback Connection:-

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

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

Voltage-Shunt feedback Connection:-

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

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

Current-Shunt feedback Connection:-

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

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

Effect of negative feedback on different topologies:-

Type of f/b

Voltage gain


Band Width with f/b i/p impedance



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

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

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



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continuity equation

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

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

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

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

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

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

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

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

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

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

for a constant surface the derivative becomes the partial derivative 

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

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

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

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

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

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

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


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Effect of negative feedback on Band width of an Amplifier

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

Proof:-  Consider an amplifier with gain ‘A’

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

the frequency response of an amplifier consists of three regions

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

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

A_{v} -open loop gain,

f– frequency,

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

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

In low-frequency region:-

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

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

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

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

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

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

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

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

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

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

In High-frequency region:-

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

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

Substituting A_{vh} in the above equation 

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

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

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

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

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

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

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

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

In Mid-frequency region:-

Gain with out feed back is A_{v}

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

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

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

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


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Full Wave Rectifier

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

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

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

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

Working of Full Wave Rectifier:-

During positive half cycle of applied i/p signal

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

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

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

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

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

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

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

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

Ripple Factor (\Gamma):-

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

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

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

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

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

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

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

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

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

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

                                                                        \Gamma = 0.482


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Time-domain representation of SSB-SC signal

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

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

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

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

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

By finding the Fourier Transform of in-phase component

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

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

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

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

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

By finding the Fourier Transform of Q-phase component

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

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

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

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


from the above figure,

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

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

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

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


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

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

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

since Signum function is 

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

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

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

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

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

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

similarly, SSB signal when only LSB has been transmitted 

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


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Analog ElectronicCircuits Lab

Expt-1-i PN-Junction Diode Characteristics

Expt-1-ii-Zener Diode Characteristics

Expt-2-Full Wave Rectifier with and without filters

Expt-3-Common Emitter Characteristics

Expt-4- Clippers


Expt-6-single stage CE Amplifier-frequency response

Expt-7-Inverting amplifier

Expt-8-Non-inverting Amplifier

Expt-9-Differentiator and Integrator using op-amp

Expt-10-square and triangular wave generators 


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Input and Output characteristics of transistor in Common Base Configuration

Input Characteristics:-

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

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

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

common base input and output characteristics

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

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

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

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

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

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

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

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

Output Characteristics:-

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

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

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

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

active region:-

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

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

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

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

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

saturation region:-

here both junctions are Forward Biased.

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

cut-off region:-

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

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

punch through/reach through region:-

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

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

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Early effect in Common Base Configuration

Early effect (or) Base-width modulation:-

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

The early effect has three consequences:-

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

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

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Hall effect

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

The figure shows the  experimental arrangement to observe Hall effect  Now 

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

B\rightarrow Applied Magnetic field (z-direction)

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

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

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

under equilibrium F_{E} = F_{B}

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

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

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

and d- is the distance between the two plates.

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

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

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

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

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

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

V_{H} = Bd v_{d}

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

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

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

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

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

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

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


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E due to infinite line charge distribution

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Energy Density in Electrostatic Fields

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

EQN (I)+EQN(II) gives

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

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

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

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

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

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


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

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

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

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Switches-Circuit Switches

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

Circuit Switches (or) Structure of Circuit Switch:-

The Switches used in Circuit Switching are called Circuit-Switches

Space-Division Switch:-

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

Cross-bar Switch:-

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

The number of cross points required = n X m.

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

Multi stage Switch:-

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

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

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

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

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

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

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

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

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Solved Example problems in Electro Magnetic Theory

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

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

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

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

                                \fn_cm =75^{o}

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

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

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

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

                      \fn_cm r= \sqrt{35}

                    \fn_cm r=5.91

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

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

\fn_cm \theta =32.31^{o}

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

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

\fn_cm \phi = 75^{o}

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


Phase Locked Loop (PLL)

Demodulation of an FM signal using PLL:-

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

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

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

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

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

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

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

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

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

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

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

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

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

the over all output of a loop filter is 


Frequency domain representation of a Wide Band FM

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Continuous Fourier Series  expansion of  

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

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

Now substituting this in the Equation (I)

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

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

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

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

The  Frequency spectrum  can be obtained by taking Fourier Transform 

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

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

From the above Equation it is clear that 

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






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Matched Filter, impulse response h(t)

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

Transfer Function of Matched Filter:-

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

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

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

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

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

Equation (I) becomes

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

From Time-shifting property of Fourier Transforms

x(t)\leftrightarrow X(f)

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

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

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

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


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

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Mutual Information I(X ; Y) Properties

Property 1:- Mutual Information is Non-Negative

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

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

Substitute Equation (II) in Equation (I)

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

The above Equation can be written as

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

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

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

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

-I(X ; Y) \leq 0

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

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Example Problems in Electro Magnetic Theory Wave propagation

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

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

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

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

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

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

So given medium is a Conductor (Copper)

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

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

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

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

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

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

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

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

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

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

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

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

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

\bg_white \gamma =?,\eta =?

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

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

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

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

    \bg_white = 18.86.

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

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

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

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

\bg_white \eta = 40 Ω.


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Delta modulation and Demodulation

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

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

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

DM transmits one bit per sample. 

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

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

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

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

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

This process can be illustrated in the following figure

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

Delta Modulator:- 

Mathematical equations involved in DM Transmitter are 

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

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

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

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

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

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

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

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

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

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

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


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Fourier Series and it’s applications

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Block Diagram of Digital Communication System/Elements of DCS

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

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

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

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

this involves the following steps

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

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

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

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

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

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

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

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

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

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

There are two types of spreading techniques available

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

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

SS techniques are used in Military applications.

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

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

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

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Why Digital Communication is preferred over Analog Communication?


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

Examples of different types of communications are

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

Why Digital?

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

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

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

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

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

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

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

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

3.Digital Circuits Vs Analog Circuits:-

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

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

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

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

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

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

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

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

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


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