Region of Convergence (ROC)

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

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

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

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

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

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

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

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

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

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

Properties of ROC/ constraints (or) Limitations:-

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

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


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

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

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

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

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

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

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

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

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

But is does not include any pole.

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

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Analogy between Vectors and Signals

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

we always search for analogies while studying a new problem.

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

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

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


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

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

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

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

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

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

There exists two other possibilities.

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

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

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

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

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

\therefore (1) = (2) .

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

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

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

i.e, C_{12} =0.


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

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

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

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

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

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

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

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

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

But the process of averaging gives a false indication.

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

  f_{1}(t) =C_{12}f_{2}(t).

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

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

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

f_{e}(t) = \sin t .

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

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

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

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

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


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

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

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

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

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


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Relation between Laplace and Fourier Transform

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

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

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

we know that s=\sigma + j\omega 

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

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

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

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

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

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

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


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System in Continuous Time-domain

A system is defined as an entity that acts on an input signal x(t) and Transforms it into an output signal y(t)

In a system, there exists Cause and effect relationship between two (or) more signals

ex:- Electrical systems, mechanical systems etc.

A system is represented by a block, which accepts signals as inputs and produces signals as outputs.

input signal:- is called as excitation , source 9or) driving function -x(t)

output signal:- is called as response-y(t)

systems may be single input single output systems (or) Multi input Multi output systems.

Here we discuss only about SISO systems only.


T- Transformation of x(t) (Or) operation on x(t).

In general systems are either Continuous -Time (or) Discrete Time systems.

continuous-Time systems (CT system) Discrete-Time systems (DT system)
A system that operates/produces on Continuous Time signals is a CT system. A system that operates/produces on Discrete Time signals  and produces a DT signals is a  DT system.

Classification of systems:-

Systems are classified into

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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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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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.


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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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 


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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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 .

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 .

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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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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Fourier Series and it’s applications

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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few problems on Auto correlatioon Function(ACF) and Energy Spectral Density(ESD)

  1. Find the Auto correlation function of x(t) = \frac{1}{\sqrt{2\pi }}\exp ^{\frac{-t^{2}}{2}}.

Ans. We know that  Auto correlation function forms fourier transform pair with Energy Spectral Density function

ACF\leftrightarrow ESD

R_{xx}(t)\leftrightarrow S(f)

the Fourier Transform of  e^{-ct^{2}}\leftrightarrow \frac{\sqrt{\pi }}{c}e^{-\pi ^{2}f^{2}}

\frac{1}{\sqrt{2\pi }}e^{\frac{-t^{2}}{2}}\leftrightarrow \frac{1}{\sqrt{2\pi }}.\sqrt{\frac{\pi }{(1/2)}}e^{\frac{-\pi ^{2}f^{2}}{(1/2)}} here c = \frac{1}{2}

\frac{1}{\sqrt{2\pi }}e^{\frac{-t^{2}}{2}}\leftrightarrow \frac{1}{\sqrt{2\pi }}.\sqrt{2\pi }e^{-2 \pi ^{2}f^{2}}

\frac{1}{\sqrt{2\pi }}e^{\frac{-t^{2}}{2}}\leftrightarrow e^{-2 \pi ^{2}f^{2}}

x(t)\leftrightarrow X(f)

\therefore the Fourier Transform of x(t) is X(f) and is X(f) = e^{-2\pi ^{2}f^{2}} and the Energy Spectral Density S(f) = \left | X(f) \right |^{2}

S(f) = e^{-4\pi ^{2}f^{2}}

By finding the inverse Fourier Transform of S(f) gives the Auto Correlation Function

S(f) = e^{\frac{-\pi ^{2}f^{2}}{(1/4)}}

e^{\frac{-\pi ^{2}f^{2}}{(1/4)}}\leftrightarrow \frac{e^{\frac{-t^{2}}{4}}}{\sqrt{4\pi }}

e^{\frac{-\pi ^{2}f^{2}}{(1/4)}}\leftrightarrow \frac{e^{\frac{-t^{2}}{4}}}{2\sqrt{\pi }}

\therefore the ACF of the given signal is inverse Fourier Transform of S(f) which is R_{xx}(t) = \frac{e^{\frac{-t^{2}}{4}}}{2\sqrt{\pi }}.


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Compensators are corrective sub systems to compensate the deficiency in the performance of the plant or system, so given a plant and a set of specifications suitable compensators are to be designed so that the overall system will meet given specifications. Proper selection of performance specifications is the most important step in the design of compensators.

i.e, All the control systems are designed to achieve specific objectives that is the requirements are defined for the control system. A good control system has less error, good accuracy, good speed of response, good relative stability, good damping which will not cause unusual Overshoots etc.

For stationary performance of the system, gain is adjusted first but gain adjustment alone can not provide satisfactory results. When gain increases, Steady-state behavior of system improves but results into poor Transient response (or) even instability.

The desired behavior of a system is specified in terms of 

  • Transient response.
  • Steady-state error(e_{ss}).

e_{ss} →is usually specified in terms of constants k_{a},k_{v} and k_{p} for u(t),r(t) and p(t) as inputs.

Transient response → it measures relative stability and speed of response which are specified in time or frequency domain.

In time domain the measure of relative stability is in terms of \xi or M_{p}, while the speed response is measured in terms of rise time t_{r} , settling time t_{s} or natural frequency \omega _{n}. Where as in frequency-domain the measure of relative stability is given by Resonant peak M_{r} or Phase margin \varphi _{pm} and the speed response is measured by Resonant frequency \omega _{r} or band width \omega _{b}.

Once a set of performance specifications have been selected, the next step is to chose the appropriate compensator. There exists Electrical,hydraulic,pneumatic and mechanical compensators and in this context we prefer Electrical compenators.

An external device which is used to alter the behavior of the system so as to achieve given specifications is called as compensator .

Compensators can be added to the system in series or in parallel or in combination of both.

Series Compensation:-

The flow of signal in series scheme is from lower energy level towards higher energy level. This requires additional amplifiers to increase the gain and also provide necessary isolation. The number of components required in series scheme is more than in parallel scheme.


Parallel Compensation:-

In this compensation technique energy flow is from higher energy level to lower energy level.  As there is no need of any amplifiers additional components required are less.

Series-Parallel Compensation:-

This is a compensation technique which utilizes the advantages of both series and parallel compensation techniques.


Electronics Circuits Lab Viva questions



  1. What are the applications of clippers?
  2. Diode cut in voltage value for si and Ge diodes?
  3. Why the output of the clipper is taken in d.c mode?
  4. Other names of clippers are—— –.
  5. Voice frequency range is——– –.
  6. Positive peak clipper means
  7. Negative peak clipper means
  8. Positive base and negative base clippers means
  9. What is the difference between clipping and clamping operations?
  10. And the working procedures of each clipping circuit
  11. How clippers are applicable in noise reduction in audio applications?
  12. Draw the output waveforms of different clippers with different input and reference voltages—–
  13. Why clippers are called as diode clippers?
  14. What is the difference between slicing and clipping of a voltage waveform?
  15. Clippers comes under linear or non linear wave shaping?
  16. Explain linear and nonlinear wave shaping meanings?
  17. Names some linear and non linear wave shaping circuits?
  18. Define diode threshold and cutin voltages?
  19. An ideal diode acts as———– under forward
  20. An ideal diode acts as———— under reverse
  21. Practical diode equivalent circuit in forward bias——–
  22. Practical diode equivalent circuit in reverse bias——–
  23. Draw diode d.c characteristics.


  1.  Define clamping operation? And how it is different from clippers?
  2. What are the other names of clampers?
  3. Difference between clipping and clamping circuits
  4. Why capacitors are used in clampers?
  5. What are different types of clampers?
  6. Working Operation of each clamper circuit
  7. Practical applications of clampers are
  8. What happens in the output waveform if the polarity of capacitor is changed in the clampers
  9. Capacitor acts as———- for D.C input
  10. Capacitor acts as————- for A.C
  11. Fully charged capacitor acts as——– –.
  12. What do u mean by transient and steady state response of a clamper?
  13. DC restorers are used in CRO’s yes or no
  14. DC inserter means————
  15. Voltage or amplitude slicers means———–
  16. Voltage or amplitude limiters means———– –.
  17. Explain the operation of positive peak clamper?
  18. What is the working principle of a capacitor/
  19. Output waveforms of clampers with different input and different reference voltages—
  20. The voltage across a capacitor remains———— once the capacitor is fully


  1. What is the phase difference between input and output wave forms of a CE amplifier?
  2. What type of biasing is used in the given circuit?
  3. What is the effect of emitter bypass capacitor on frequency response of a CE amplifier?
  4. What is the effect or importance of coupling capacitor?
  5. Why source resistance Rs is used in the input side?
  6. What are the different regions of operation of a BJT?
  7. The phase difference of input and output in CB amplifier?
  8. CE amplifier is voltage or current amplifier?
  9. Draw the equivalent h-parameter model of a CE amplifier?
  10. Why NPN transistor is preferred over PNP transistor?
  11. What is the effect of bypass capacitor over stability of the CE amplifier?
  12. The CE amplifier is in voltage divider bias or in fixed bias configuration?
  13. The values of hfe,hie,hoe and hre in CE configuration are———– –.
  14. hfe,hie,hoe and hre are called as————— –.
  15. What do you mean by loading effect?
  16. What are the practical applications of single stage CE amplifier?
  17. How do you know the amplifier is single stage CE amplifier?
  18. Why this circuit is called RC coupled amplifier?
  19. What is the operating point in CE amplifier from the design specifications?
  20. Quiescent conditions means——–
  21. C conditions are for———
  22. What is the relationship between collector current and base current?
  23. Does β and hfe are one and the same?
  24. Why coupling capacitor Cb is connected in reverse (-,+) at the input side and    Cc is connected in forward(+,-) at the output side?
  25. How do u know a transistor is working in cut off region?
  26. Write VBE(active),VBE(sat),VBE(cutoff) values of a Si and Ge transistors
  27. Transistor meaning is———
  28. Define frequency response of a CE amplifier?
  29. Define voltage gain of a CE amplifier?
  30. Units of gain as a ratio are———-
  31. Units of gain in logarithim is———
  32. Magnitude response and frequency response are one and the same say yes or no?
  33. Define bandwidth ?
  34. What are 3 dB frequencies?
  35. Define cutoff frequencies?
  36. BW is approximately equal to fH justify?
  37. Gain is constant in————-
  38. Semi logarithmic graph is linear or non linear graph?
  39. Stability factor is a function of—–
  40. In the circuit R1 and R2 are used for——–
  41. The feedback type that Re and Ce introduces in the circuit is———–(Ans: negativefeedback).
  42. Negative feedback increases stability yes or no(Ans:yes).
  43. CB amplifier provides more band width than CE amplifier justify?


  1. What is the need for cascading in amplifiers?
  2. Why do we need two stage CE amplifier?
  3. Why this circuit is called two stage CE amplifier?
  4. What is the relationship between gain and bandwidth?
  5. Cascading of number of stages in amplifiers increases——— –.
  6. Cascading of number of stages linear amplifiers decreases—– –.
  7. Why do we use a coupling capacitor between two stages?
  8. Why RC coupling is preferred in Audio frequency range?
  9. Explain various types of capacitors?
  10. What do you mean by loading effect? How to avoid it?
  11. Identify the stages of the amplifier?
  12. CE-CE combination is called as———-
  13. Cascade configuration means——
  14. What is the overall gain of two satge amplifier—–
  15. Is two stage amplifier is a multistage amplifier or not?
  16. Why this circuit is called RC coupled amplifier?
  17. Does CB-CB combination is possible?
  18. CE-CC combination is known as——-
  19. CB-CC combination is called as——–
  20. There exist an inverse proportionality between gain and BW yes or no?


  1. What do u mean by feedback?
  2. What are the two types of feedback?
  3. What are the advantages of negative feedback over positive feedback?
  4. Voltage series feedback means?
  5. Input of a voltage series combination is———–
  6. Output of a voltage series combination is———–
  7. For voltage series amplifier Ri , Ro,gain and BW changes as——-
  8. How does u know the feedback is voltage series in the given circuit?
  9. What are different types of feedback topologies?
  10. Applications of voltage series amplifier?
  11. What is the difference between two stage and voltage series amplifier?
  12. Which is the best topology among four?
  13. How feedback amplifiers are different from ordinary amplifiers?
  14. Define feedback gain?


  1.  Mention the conditions for oscillations in RC phase shift oscillator?
  2. Give the formula for frequency of oscillations in RC phase shift oscillator?
  3. The phase produced by a single RC network is RC phase shift oscillator?
  4. RC phase shift oscillator uses positive feedback or negative feedback?
  5. The phase produced by basic amplifier circuit in RC phase shift oscillator is?
  6. What is the difference between damped oscillations un damped oscillations?
  7. What are the applications of RC phase shift oscillator?
  8. How many resistors and capacitors are used in RC phase shift feedback network?
  9. How the Barkhausen’s criterion is satisfied in RC phase shift oscillator
  10. Mention the basic reason for any oscillations?
  11. What is meant by Barkhausen’s criterion?
  12. Audio frequency range is————
  13. RC phase shift oscillator is ——–
  14. Oscillator is a circuit operates on internal input power supply yes or no?
  15. Define an oscillator?
  16. Oscillator did not take any external input yes or no?
  17. Type of feedback used in oscillators is——-
  18. Radio frequency range is———–
  19. Show that single RC section provides a phase shift of 60
  20. Positive feedback causes instability yes or no?
  21. In oscillators loop gain must be————–
  22. In oscillators And overall phase shift produced by the circuit is————
  23. Feedback gain 𝛽 must be less than———–


  1. What are the advantages of hartley’s and colpitts oscillators over RC phase shift oscillator?
  2. Applications of C and H oscillators are—–.
  3. C oscillator uses—- as feedback network.
  4. H oscillator uses—– as feedback
  5. Define mutual inductance—–
  6. These are used at———–


  1. Differentiate symmetrical and asymmetrical networks?
  2. Define image impedance?
  3. Define characteristic or nominal impedance?
  4. Theoretical Calculations of open circuit and short circuit impedances for all the
  5. What is the significance of image impedance?
  6. Is there any difference between image and inverse impedance?
  7. Why char impedance is called as nominal impedance?
  8. Does char impedance and image impedance are same yes or no?


  1. Define a filter?
  2. What is the difference between ordinary LPF and constant –k-LPF?
  3. Why this filter is called as constant-k-low pass filter?
  4. What does u mean by a LPF?
  5. Draw the gain characteristics of ideal and practical Low pass filters?
  6. Define attenuation? units of attenuation are——
  7. Gain and attenuation are——-
  8. Define cut off frequency of a constant k low pass filter?
  9. Draw the attenuation characteristic and explain that graph?
  10. Draw the output voltage vs frequency curve and explain it?’
  11. What are the applications of constant-k low pass filter?
  12. Advantages and disadvantages of constant k low pass filter?


  1. Define high pass filter?
  2. Define the constant m?
  3. How m derived filter is different from ordinary HPF?
  4. Why we will go for m derived filters?
  5. Draw the model graphs of m derived HPF and explain?
  6. Applications of m derived HPF?
  7. Advantages and disadvantages of m derived HPF?
  8. Define infinity frequency and cut off frequency?
  9. Are these filters are passive filters or active filters?

NOTE: read applications and advantages and disadvantages for all the circuits.