March 2023 - Page 2 of 3

Adaptive Delta Modulation(ADM)

Adaptive Delta Modulation, a modification of Linear Delta Modulation (LDM) is a scheme that circumvents the deficiency of DM. In ADM step size Δ of the Quantizer is not a constant but varies with time , we shall express Δ as $\Delta (n)= 2 \delta (n)$ .

where $\delta (n)$ increases during a steep segment of input and decreases for a slowly varying segment of input.

The adaptive step size control which forms the basis of an ADM scheme can be classified in various ways such as

Discrete or Continuous.
instantaneous (or) syllabic (fairly gradual change).
forward (or) backward.

we shall describe an adaptation scheme that is backward, instantaneous and discrete in practical implementation , the step size $\delta (n)$ is constrained in between some pre-determined minimum and maximum values.

$\delta _{min} \leq \delta (n)\leq \delta _{max}$

The upper limit $\delta _{max}$ controls the amount of Slope over load distortion and the lower limit $\delta _{min}$ controls the granular noise (or) Idle noise.

The adaptive rule for $\delta (n)$ can be expressed in the general form $\delta (n) = g(n) \delta (n-1)$

where the time varying gain g(n) depends on the present binary output b(n) and M previous values b(n-1),b(n-2) ……….b(n-M). The algorithm is usually initiated with $\delta _{min}$ .

when M=1, b(n) and b(n-1) are compared to detect probable slope over load {b(n) = b(n-1)} (or) probable granularity {b(n) ≠ b(n-1)} then g(n) is

$g(n) = P$ if $b(n) =b(n-1)$ .
$g(n)=\frac{1}{P}$ if $b(n) \neq b(n-1)$ .

when $b(n) =b(n-1)$ Slope overload distortion is detected and when $b(n) \neq b(n-1)$ Idle noise is detected.

where $P\geq 1$ , note that $P=1$ represents LDM. $P_{optimum}=1.5$ minimizes the Quantization noise for speech signal, where $1< P< 2$ is for broad class of signals.

Adaptive Delta Modulation System:-

ADM Transmitter:-

The figure shows ADM Transmitter and ADM Receiver which includes the logic for step size control both in Transmitter and receiver.

The logic for step size control is added in the diagram the step size increases (or) decreases according to specified rule depending up on one bit Quantizer output.

for example if one bit Quantizer output is high then the step size may be doubled for next sample.

if one bit Quantizer output is low, the step size may be reduced by one step.

ADM Receiver:-

In the Receiver of Adaptive Delta Modulator there are two parts logic for step size and accumulator.

The first block produces the step size from each incoming bit, which uses the same principle used in the Transmitter.

The previous input and present input decides the step size. It is then applied to an accumulator, which builds up stair case wave form. The LPF then smoothens out the stair case wave form to reconstruct the original signal.

Advantages of ADM:-

ADM has certain advantages over DM

The signal to Noise Ratio becomes better than ordinary Delta modulation because of the reduction in slope over load distortion and Idle noise.
Because of the variable step size the dynamic range of ADM is wider than simple DM.
Utilization of Band Width is better than DM.

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Pulse Code Modulation(PCM)

To describe Pulse Code Modulation (PCM) the idea about Sampling and Quantization are required in hand.

PCM is the most basic form of digital pulse modulation

PCM :- In PCM a message signal is represented by a sequence of coded pulses , which is accomplished by representing the signal in discrete form in both time and Amplitude.

PCM Block Diagram:-

The Block Diagram of a PCM system is given in the following figure. This Block Diagram consists of the basic elements of a PCM system.

Transmitter.
Transmission path.
Receiver.

PCM Transmitter:-

the basic operations performed in the transmitter are Sampling, Quantization and Coding -Analog to Digital Conversion(ADC).

LPF:- The Low Pass Filter to sampler is inclined to prevent aliasing of the message signal.

Sampler:- Converts the analog band limited signal to discrete signal.

ADC:- It’s a device which performs two operations known as Quantization and Encoding.

Transmission path:-

Regenerative repeaters:- PCM encoded output is transmitted through the transmission path to the receiver, during transmission the PCM output get distorted in the channel by noise which can be regenerated by the device called regenerative repeater.

PCM Receiver:-

The essential operations in the receiver are decoding, regeneration of impaired PCM output .

the device which performs decoding and demodulation at the receiver is known as (DAC) Digital-to-Analog Converter.

Few points about PCM:-

we know that PAM, PWM and PPM are analog pulse modulation techniques, where as PCM is a pure Digital Modulation scheme.
It should be understood that the PCM is not a modulation technique in the conventional sense.
The output of PCM is in the coded digital form that is in the form of code words. A PCM system consists of PCM encoder as well as decoder.

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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

Memory consumed by Routing Tables.
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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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

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).
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 Seq.no. 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 Seq.no 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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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

A layer should be created where a different abstraction is required.
Each layer should perform a well defined function.
A layer boundaries should be chosen to minimize the information flow across the interface.
The function of each layer chosen by keeping an eye toward defining international standardized protocols.
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 switching and Routing.
Switching:- refers to make a temporary connection between physical links for N/w transmission.
Routing:- means choosing a best path for sending a packet from one node to another when number of paths are available. This can be done by Routing algorithms.
congestion control:- If too many packets are present in the subnet at the same time. They will from bottlenecks forming a congestion which can be controlled by using a congestion control algorithms.
when a packet has to travel from one n/w to another to get its destination many problems are like

i. The addressing used by second N/w may differ from first one.

ii. The second one may not accept the packet at all because it is too long.

iii. The protocols used may differ.

N/w layer may takes care of all the above issues in order to provide inter connection between heterogeneous structures.

In Broadcast N/ws the routing problem is simple often thin (or) non-existent.

Transport Layer:-

Transport layer (TL) accepts data from session layer and makes into smaller units called TPDU (Transport Protocol Data Units).
TL passes this TPDU into N/w layer and makes sure that these pieces are all arriving at the other end correctly.
TL determines what type of service to provide to the session layer.
The most popular type of transport connection is an error free point to point channel that derives messages (or) bytes as the order in which they were sent.
The other possible service is transporting isolated messages and broadcasting messages to multiple destinations in which layer 1 to 3 are chained and layers 4 to 7 are end-to-end.

Session layer:-

session layer allows users on different machines to establish sessions between them and offers services as follows
dialog control: – keeping track of whose turn it is to transmit.
token management: -If two parties attempting for the same operation at the some time.

ex:- Bank–>cash withdrawl–> giving some tokens.

Synchronization:- keeping check points when long transmission needs to transmit bulk data .

ex:-while sending 2000 pages of a file keep check points after every 100 pages makes transmission efficient.

Presentation layer:-

Mostly deals with syntax and semantics of data to be transmitted.
for communicating computers with different data structures can be possible through presentation layer.
for encryption and decryption.

Application layer:-

It contains different protocols that are commonly needed by users

accessing webpages using HTTP(or) TCP.
email.
directory services.
for N/w news.
FTAM- File Transfer Access and Management.
to access files in remote computer.
to retrieve files from remote computer.
to manage and control remote computer.

Differential Pulse Code Modulation (DPCM)

Differential Pulse Code Modulation (DPCM):-

When a voice signal (or) video signal is sampled at a rate higher than Nyquist rate , which is usually done in PCM.

The resulting sampled signal is found to exhibit a high degree of correlation between adjacent samples.

The meaning of this high correlation is that the signal does not change rapidly from one sample to the next.

As a result ,

The variance of adjacent samples < variance of the signal x(t) (or) m(t).

When these correlated samples are encoded, as in the standard PCM system, the resulting encoded signal contains redundant information.

As a result of encoding process, redundant samples that are not essential in the transmission of information are generated.

By removing the redundancy before encoding, we obtain a more efficient coded signal which is the basic idea behind DPCM.

If we know the past behavior of a signal x(t) up to certain point of time, we may use prediction to make an estimate of a future value of the signal.

The figure shows the DPCM transmitter

The fact that motivates the scheme DPCM is the possibility of prediction of future value of the signal x(t).

i.e, the DPCM works on the principle of prediction the value of the present sample is predicted from the past samples the prediction may not be exact but it is very close to that actual sample value.

Equations in DPCM:-

$e[n]=x[n]-\widehat{x[n]}$ .

Where $x[n]$ – is the input sample.

$\widehat{x[n]}$ -Prediction of input sample.

$e[n]$ – is the difference between un Quantized input sample and prediction of it.

$\widehat{x[n]}$ – The predicted value is produced by using a linear prediction filter the input to the prediction filter is a quantized version of input sample $x_{q}[n]$ .

$\therefore$ The difference signal $e[n]$ is the prediction error.

Prediction error:- it is the amount by which the prediction filter fails to predict the input exactly.

The Quantizer output may be expressed as $e_{q}[n] = e[n]+q[n]$ .

$e_{q}[n]$ – Quantized version of $e[n]$

$e[n]$ – is the prediction error

$q[n]$ – Quantization error.

$x_{q}[n]=e_{q}[n]+\widehat{x[n]}$ .

$e_{q}[n]$ – Output of Quantizer.

$\widehat{x[n]}$ – Prediction of $x[n]$ .

form equation(2)

$x_{q}[n]=\widehat{x[n]}+e[n]+q[n]$

$\therefore \widehat{x[n]} +e[n] =x[n]$ from equation (1)

$\therefore x_{q}[n]=x[n]+q[n]$ .

By encoding the Quantizer output, we obtain a variant of PCM known as differential pulse code Modulation (DPCM).

DPCM Receiver :-

The figure shows the block diagram of DPCM receiver

The decoder first reconstructs the Quantized error signal from incoming binary signal.

The prediction filter output and the quantized error signals are summed up to give the quantized version of the original signal.

Same prediction filter is used at the receiver as that of transmitter.

Thus the signal at the receiver differ from actual signal $x[n]$ by Quantization error $q[n]$ , which is introduced permanently in the reconstructed signal.

Comparison of DPCM with DM and PCM:-

DPCM includes M as a special case.

DPCM and DM are basically similar except for two important differences.

DM is the ‘1’ bit version of DPCM.

Note that unlike a standard PCM system, the transmitters of both the DPCM and DM involve the use of feedback.

Like DM , in DPCM also slope overload distortion exists whenever input signal changes too rapidly for the prediction filter to track it.

Like PCM,DPCM suffers Quantization error.

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Phase Shift Keying (PSK)

It is the most efficient of the 3 digital modulation techniques ASK,PSK and FSK. It is used for high bit rates.

Definition:-

In BPSK, the Phase of the carrier is shifted/ Changed according to the incoming Binary data sequence.

BPSK was developed during the early days of the deep space programs. PSK is now widely used in both Military and commercial communication systems.

we represent data sequence $\left \{ b_{k} \right \}$ in Bipolar-NRZ scheme.

Expressions of PSK:-

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

$S(t)=A \cos 2\pi f_{c}t$ .

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

$A=\sqrt{2P_{s}}$ .

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

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

Now according to PSK Binary ‘1’ is represented with carrier phase $0^{o}$ and Binary ‘0’ is represented with a phase $180^{o}$ .

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

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

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

i.e,

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

The wave forms of PSK modulation scheme are shown in the figure , where b(t) represents the polar NRZ representation of binary sequence $\left \{ b_{k} \right \}$

i.e, $b(t) = \left\{\begin{matrix} \ +v \ (or) \ 1\ volt \ when \left \{ b_{k} \right \}=1\\ \ -v \ (or) \ -1\ volt \ when \left \{ b_{k} \right \}=0 \end{matrix}\right.$ .

PSK Transmitter:-

The figure shows the PSK generator (or) PSK Transmitter

The incoming binary sequence $\left \{ b_{k} \right \}$ (in the form of a signal) into $\pm 1 \ volt$ by using NRZ level encoder.

$b(t)= \left\{\begin{matrix} +1V \ for \ symbol \1\\ -1V \ for \ symbol \0 \end{matrix}\right.$

Now b(t) and the carrier S(t) are applied to product modulator, to get the PSK modulated signal at the output.

i.e, $S_{PSK}(t)=b(t).S(t)$

finally,

Coherent PSK Detector:-

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

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

The product $S_{PSK}(t).C(t)$ is applied to the Integrator. The integrator eliminates the noise.

The Integrator integrates the input over one bit interval $T_{b}$ and the output is given to a threshold device. If the threshold voltage is set to 0 V.

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

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

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

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

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Magnetic Boundary conditions

Magnetic Boundary conditions:-

In order to find out the $\overrightarrow{B}$ , $\overrightarrow{H}$ and $\overrightarrow{M}$ at the interface between two different magnetic materials boundary conditions are required.

Consider two magnetic materials having permeabilities $\mu _{1}$ and $\mu _{2}$ as shown in the figure

We will find out these boundary conditions by using

$\oint_{s} \overrightarrow{B} .\overrightarrow{ds} =0$ – Gauss’s law in Magneto statics.
$\oint_{l} \overrightarrow{H} .\overrightarrow{dl} = I$ – Ampere’s law.

In order to apply $\oint_{s} \overrightarrow{B} .\overrightarrow{ds} =0$ , a surface is required which is nothing but Gaussian surface (or) a Pillbox.

$\oint_{s} \overrightarrow{B} .\overrightarrow{ds} = \oint_{side} \overrightarrow{B} .\overrightarrow{ds} + \oint_{top} \overrightarrow{B} .\overrightarrow{ds} + \oint_{bottom} \overrightarrow{B} .\overrightarrow{ds}$ .

( $\oint_{s} \overrightarrow{B} .\overrightarrow{ds} =0$ -Because on the boundary $\Delta h=0$ , so there exists no side of surface).

$B_{1n}\Delta S -B_{2n}\Delta S =0$ .

$B_{1n}=B_{2n}-----------(1)$

since $\overrightarrow{B} = \mu \overrightarrow{H}$ , $\mu _{1} H_{1n}= \mu _{2} H_{2n}$ .

$H_{1n}= \frac{\mu _{2}}{\mu _{1}} H_{2n} ---------(2)$ .

From equations (1) and (2) the normal component of $\overrightarrow{B}$ is continuous at the boundary and the normal components of $\overrightarrow{H}$ is discontinuous at the boundary.

we know that $\overrightarrow{B} = \mu _{o}(1+ \chi _{m})\overrightarrow{H}$ .

$\overrightarrow{B} = \mu _{o}\overrightarrow{H}+ \mu _{o}\overrightarrow{M}$ .

$\overrightarrow{M} = \chi _{m}\overrightarrow{H}$ .

$M_{2n} = \chi _{m2} H_{2n}$ .

$M_{1n} = \chi _{m1} H_{1n}$ .

$M_{2n} = \chi _{m2} H_{2n}$ since $H_{2n} = \frac{\mu _{1}}{\mu _{2}} H_{1n}$ .

$M_{2n}=\chi _{m2} \frac{\mu _{1}}{\mu _{2}} H_{1n}$ .

$M_{2n}=\frac{\chi _{m2}}{\chi _{m1}} \frac{\mu _{1}}{\mu _{2}} M_{1n}$ .

The magnetisation normal components are also discontinuous.

To apply $\oint_{l} \overrightarrow{H} .\overrightarrow{dl} = I$ a path is required, which is a closed one in a plane normal to the boundary surface.

Here a closed path is abcda , which encloses a surface current density k on the surface of the boundary

Then $\oint_{l} \overrightarrow{H} .\overrightarrow{dl} = I$ .

$H_{1t}\Delta w -H_{1n}\frac{\Delta h}{2}-H_{2n}\frac{\Delta h}{2}-H_{2t}\Delta w+H_{2n}\frac{\Delta h}{2}+H_{1n}\frac{\Delta h}{2}=I$ .

(since $\Delta h=0$ On the boundary)

$H_{1t}\Delta w-H_{2t}\Delta w=k \Delta w$ .

$H_{1t}-H_{2t}=k$ .

$\frac{B_{1t}}{\mu _{1}}-\frac{B_{2t}}{\mu _{2}}=k$ , So the tangential components of $\overrightarrow{B}$ and $\overrightarrow{H}$ are discontinuous.

$(\overrightarrow{H_{1t}} -\overrightarrow{H_{2t}})X \overrightarrow{a_{n12}} = \overrightarrow{k}$ .

(or) $(\overrightarrow{H_{1t}} -\overrightarrow{H_{2t}}) = \overrightarrow{k}X \overrightarrow{a_{n12}}$ .

When $\overrightarrow{k}=0$ , $\overrightarrow{H_{1t}}=\overrightarrow{H_{2t}}$ and $\frac{\overrightarrow{B_{1t}}}{\mu _{1}} =\frac{\overrightarrow{B_{2t}}}{\mu _{2}}$ .

Similar to normal components of magnetisation the tangential components are

$M_{2t} = \chi _{m2} H_{2t}$ .

$M_{1t} = \chi _{m1} H_{1t}$ .

$M_{2t} = \chi _{m2} H_{2t}$ since $H_{2t} = H_{1t}-k$ .

$M_{2t}=\chi _{m2} (H_{1t}-k)$ .

$M_{2n}=\frac{\chi _{m2}}{\chi _{m1}} M_{1t}-\chi _{m2} k$ .

Obtain the results for magnetisation by using the same procedure as that of normal components.

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Boundary conditions static electric fields

Boundary conditions static Electric fields:-

So far, we have considered the existence of the electric field in a Homogeneous medium. If the field exists in a region consisting of two different media, the conditions that the field must satisfy at the interface separating the media are called “Boundary conditions”.

These conditions are helpful in determining the field on one side of the boundary if the field on the other side is known.

The conditions will be dictated by the types of material the media are made of.

We shall consider the Boundary conditions at an interface separating

Di-electric $(\epsilon _{1})$ and Di-electric $(\epsilon _{2})$ .
Conductor $(\sigma )$ and Di-electric $(\epsilon )$ .
Conductor and free space.

To determine the boundary conditions, we need to use two Maxwell’s equations.

$\oint_{s} \overrightarrow{D} .\overrightarrow{ds} =Q_{enclosed}$ – Gauss’s law in Electrostatics.
$\oint_{l} \overrightarrow{E} .\overrightarrow{dl} = 0$ .

The Electric field intensity could be considered as result of two components tangential and normal components.

$\overrightarrow{E}=\overrightarrow{E_{t}}+\overrightarrow{E_{n}}$ .

Di-electric $(\epsilon _{1})$ and Di-electric $(\epsilon _{2})$ :-

In order to find out the $\overrightarrow{E}$ , $\overrightarrow{D}$ at the interface between two different magnetic materials boundary conditions are required.

Consider two Di-electric materials having permeabilities $\epsilon _{1}$ and $\epsilon _{2}$ as shown in the figure

To apply $\oint_{l} \overrightarrow{E} .\overrightarrow{dl} = 0$ a path is required, which is a closed one in a plane normal to the boundary surface.

Here a closed path is abcda on the surface of the boundary

Then $\oint_{l} \overrightarrow{E} .\overrightarrow{dl} = 0$ .

$E_{1t}\Delta w -E_{1n}\frac{\Delta h}{2}-E_{2n}\frac{\Delta h}{2}-E_{2t}\Delta w+E_{2n}\frac{\Delta h}{2}+E_{1n}\frac{\Delta h}{2}=0$ .

(since $\Delta h=0$ On the boundary)

$E_{1t}\Delta w-E_{2t}\Delta w=0$ .

$E_{1t}=E_{2t}$ .

$\frac{D_{1t}}{\epsilon _{1}}=\frac{D_{2t}}{\epsilon _{2}}$ , So the tangential components of $\overrightarrow{D}$ are discontinuous where as $\overrightarrow{E}$ are continuous.

In order to apply $\oint_{s} \overrightarrow{D} .\overrightarrow{ds} =Q$ , a surface is required which is nothing but Gaussian surface (or) a Pillbox enclosing some charge.

$\oint_{s} \overrightarrow{D} .\overrightarrow{ds} = \oint_{side} \overrightarrow{D} .\overrightarrow{ds} + \oint_{top} \overrightarrow{D} .\overrightarrow{ds} + \oint_{bottom} \overrightarrow{D} .\overrightarrow{ds}$ .

( $\oint_{s} \overrightarrow{D} .\overrightarrow{ds} =Q$ -Because on the boundary $\Delta h=0$ , so there exists no side of surface).

$D_{1n}\Delta S -D_{2n}\Delta S =\Delta Q$ .

$D_{1n}-D_{2n}=\frac{\Delta Q}{\Delta S}$ .

$D_{1n}-D_{2n}=\rho _{s}-----(1)$ .

since $\overrightarrow{D} = \epsilon \overrightarrow{E}$ ,

$\epsilon _{1}E_{1n}-\epsilon _{2}E_{2n}=\rho _{s}-----(2)$

From equations (1) and (2) the normal components of $\overrightarrow{D}$ and $\overrightarrow{E}$ are discontinuous at the boundary .

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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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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

\[ {\color{DarkGreen} V_{CE}} \]

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.

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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Comparison of Transistor Characteristics in three Configurations

Comparison of Transistor Characteristics in three Configurations :-

Transistor in three Configurations Common Emitter, Common Base and Common Collector can be compared in terms of Input and output impedance’s ,voltage and current gains and some other parameters. Those are tabulated as follows

characteristic	Common Base (CB)	Common Emitter (CE)	Common Collector (CC)
Input Impedance/Resistance ( $Z_{i}$ )	low ( $\approx 100\Omega$ )	Medium ( $750\Omega$ )	Very High ( $750K\Omega$ )
Output Impedance ( $Z_{o}$ )	Very High ( $450K\Omega$ )	Moderate ( $45K\Omega$ )	low ( $25\Omega$ )
Current gain ( $A_{I}$ )	Unity	High	High
Voltage gain ( $A_{v}$ )	about $150$ High ( $\approx 131$ )	about $500$ High	less than $1$
Phase shift b/w i/p and o/p	$0^{o}/360^{o}$	$180^{o}$	$0^{o}/360^{o}$
Applications	High Frequency Circuits	For Audio Circuits	For impedance matching

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Current equation of the diode

The diode under Forward bias is as follows

The current equation related to the voltage V and current I is given by

$I=I_{o}(e^{\frac{V}{\eta v_{T}}}-1) Amperes$

I- Diode current

I_o– Reverse saturation current of the diode at room temperature.

V-applied External voltage

$\eta$ – constant = 1 For Ge

= 2 For Si.

$v_{T}=\frac{kT}{q}$ – volt equivalent temperature 26 mV at room temp.

where k-Boltzmann constant = 1.38 X 10^-23 J/K.

T-Temperature of Diode in kelvin ^oK = ^o C + 273.

q- charge of electron = 1.6X10^-19C.

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Reverse Saturation Current (I_o) of PN-Diode:-

Comparision table HWR, FWR and Bridge Rectifier

Parameter	Half-Wave Rectifier	Full-wave Rectifier	Bridge Rectifier
No of diodes	1	2	4
Maximum Efficiency ( η )	40.6%	81.2%	81.2%
$V_{dc}$	$\frac{V_{m}}{\pi }$	$\frac{2V_{m}}{\pi }$	$\frac{2V_{m}}{\pi }$
$I_{dc}$	$\frac{I_{m}}{\pi }$	$\frac{2I_{m}}{\pi }$	$\frac{2I_{m}}{\pi }$
Output RMS voltage	$\frac{V_{m}}{2}$	$\frac{V_{m}}{\sqrt{2}}$	$\frac{V_{m}}{\sqrt{2}}$
Average current I_dc	$I_{dc}$	$\frac{I_{dc}}{{2}}$	$\frac{I_{dc}}{{2}}$
Ripple Factor ( $\Gamma$ )	1.21	0.48	0.48
Peak Inverse Voltage (PIV)	$V_{m}$	$2V_{m}$	$V_{m}$
Output Frequency	f	2f	2f
TUF(Transformer Utilization Factor)	0.287	0.693	0.812
Form Factor	1.57	1.11	1.11
Peak factor	2	$\sqrt{2}$	$\sqrt{2}$

Drift and Diffusion currents

The flow of charge (or) current through a semi conductor material is of two types. Similarly the net current that flows through a PN diode is also of two types (i) Drift current and (ii) Diffusion current.

Drift current:-

When an Electric field is applied across the semi conductor, the charge carriers attains certain velocity known as drift velocity $v_{d}= \mu E$ with this velocity electrons move towards positive terminal and holes move towards negative terminal of the battery. This movement of charge carriers constitutes a current known as ‘Drift current’.

Drift current is defined as the flow of electric current due to the motion of the charge carriers under the influence of an external field.

Drift current density due to free electrons $J_{n} = qn\mu _{n}E$ atoms/Cm² and the Drift current density due to free holes $J_{p} = qp\mu _{p}E$ atoms/Cm².

The current densities are perpendicular to the direction of current flow.

Diffusion Current:-

It is possible for an electric current to flow in a semi conductor even in the absence of the applied Electric field (or) voltage provided there exists a concentration gradient.

concentration gradient exists if the number of electrons (or) holes is greater in one region than other region in a semi conductors.

Now the charge carriers move from higher concentration to that lower concentration of same type charged regions.

The resulting current is known as diffusion current.

Diffusion current density ( $J_{P}$ ) due to holes is $J_{p} = -q D_{p}\frac{dp}{dx}$ A/Cm² .

Diffusion current density ( $J_{P}$ ) due to holes is $J_{n} = q D_{n}\frac{dn}{dx}$ A/Cm² .

$\therefore$ Total current in a semi-conductor is the sum of drift and diffusion currents

In P-type total current density is $J= qp\mu _{p} E-qD_{p}\frac{dp}{dx}$ .

In N-type total current density is $J= qn\mu _{n} E+qD_{n}\frac{dn}{dx}$ .

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Crystal Oscillator

Crystal Oscillator:-

In LC oscillators the frequency of oscillation f_o depends on the tank circuit parameters L & C, whereas L & C values change with respect to time, temperature, aging etc. Therefore f_o does not remain constant so at high frequencies LC oscillators are unsuitable because of instability. Crystal oscillators are more suitable at high frequencies and uses crystal as oscillatory element.

Piezo-electric effect:-

It is the ability of certain materials to generate an electric charge when mechanical stress is applied and vice-versa ( vice-versa is called Reverse Piezo electric effect).

i.e, If mechanical pressure is applied across x-axis, the electric charges appear perpendicular to x-axis that is along y-direction. similarly if electric field is applied along x-direction mechanical strain is produced along y-direction.

working of quartz crystal:-

In this circuit crystal is placed between two metal plates then it acts as a capacitor with dielectric material as crystal between two metal plates.

i.e, when a.c voltage is applied across these plates the crystal vibrates at a frequency of the applied a.c voltage . when f_i = f_o resonance takes place and crystal vibrates with it’s natural frequency almost of constant value.

Equivalent circuit of crystal:-

when crystal is not vibrating it is equivalent to a capacitance C_m.

when it is vibrating it is equivalent to series R-L-C circuit as shown below

and the series resonant frequency is given by $\left | X_{L} \right | = \left |X_{C} \right |$

$\omega L =\frac{1}{\omega C}$

$f_{s}=\frac{1}{2\pi \sqrt{LC}}$

Parallel resonance $\left | X_{L}+X_{C} \right |=\left | X_{Cm} \right |$

$\omega L-\frac{1}{\omega _{C}}=\frac{1}{\omega _{Cm}}$

$\omega ^{2}=\frac{1}{L}\sqrt{\frac{1}{c}+\frac{1}{C_{m}}}$

$f_{p}=\frac{1}{2\pi }\sqrt{\frac{1}{L}(\frac{1}{c}+\frac{1}{C_{m}})}$ Crystal Oscillator has two modes of operation Series $f_{s}$ resonance mode and parallel resonance mode $f_{p}$ .

Crystal Oscillator with BJT:-

using crystal oscillator can be built up as follows

In this circuit crystal acts as a parallel tuned circuit at parallel resonance, the crystal impedance is maximum that is maximum voltage drop is there across C₁ this allows that maximum energy transfer through feedback network through f_p. BJT offers a phase shift of 180^o further 180^o is produced by the capacitor voltage. Oscillations are possible only through f_p, which provides stable oscillations.

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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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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 $C_{1}=C_{2}$ .

working:-

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

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

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

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

therefore un damped oscillations results in the circuit.

Derivation for frequency of oscillations:-

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

Analysis(Qualitative):-

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

from the general condition for an Oscillator

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

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

find the real and imaginary parts,

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

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

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

after simplification

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

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

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

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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 $z=d$ .

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

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

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

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

$\phi = \frac{\mu _{o}I}{2\pi } ln d$

$L=\frac{\phi }{I} =\frac{\mu _{o}d}{2\pi } ln$ 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 $L=\frac{\mu_{o} N^{2}h}{2\pi }ln$ Henries.

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Current components of a PNP Transistor

The various Current components which flow across a PNP Transistor are as shown in the figure.

For Normal operation

Emitter Junction $J_{E}$ is Forward Biased.
collector Junction $J_{C}$ is Reverse Biased.

The current flows into Emitter is Emitter current $I_{E}$ , $I_{E} = I_{hE}+I_{eE}$ .

This current consists of two components

$I_{hE}$ or $I_{pE}$ – Current due to majority carriers(holes).
$I_{eE}$ or $I_{nE}$ – Current due to minority carriers(electrons).

since $I_{eE}$ is very small $I_{E} \simeq I_{hE}-----------Equation(1)$

All the holes crossing the Emitter junction $J_{E}$ do not reach the Collector junction because some of them combine with the electrons in the N-type Base.

$I_{hC}$ – is the hole current in the Collector.

∴ Base current = Total hole current in Emitter – hole current in Collector.

i.e, $I_{B} = I_{hE}-I_{hC}----------------Equation(2)$ .

If emitter were open circuited $I_{E} = 0$ Amperes which implies $I_{E} = I_{hE}$ from Equation(1) $I_{hE}\approx 0$ Amperes.

Under these conditions, Base-Collector junction acts as Reverse-Biased Diode and gives rise to a small reverse-Saturation current known as $I_{CO}$ .

when $I_{E} \neq 0$ , Total Collector current $I_{C}$ is the sum of current due to holes in the Collector and Reverse Saturation current $I_{CO}$ .

i.e, $I_{C} = I_{hC}+I_{CO}$ .

i.e, In a PNP Transistor $I_{CO}$ consists of holes moving across $J_{C}$ (from Base to Collector) that is $I_{hCO}$ and electrons crossing the junction $J_{C}$ (from Collector to Base) constitutes $I_{eCO}$ .

$I_{CO} = I_{hCO}+I_{eCO}$

i.e, $I_{E} = 0$ $\Rightarrow I_{C} = I_{CO}$ only

when $I_{E} \neq 0$ $\Rightarrow I_{C} = I_{hC}+I_{CO}$ .

$\therefore$ Total current in the transistor is given by $I_{E} = I_{B}+I_{C}$ .

$\therefore$ The general expression for Collector current is $I_{C} = \alpha I_{E}+I_{CO}$

$I_{C} =\frac{\alpha }{(1-\alpha )} I_{B}+\frac{1}{(1-\alpha )}I_{CO}$ .

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Working /Operation of NPN and PNP Transistor

NPN Transistor Working:-

For Normal operation of NPN Transistor Emitter junction J_E is Forward Biased and Collector junction J_C is Reverse Biased.

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

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

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

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

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

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

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

PNP Transistor Working:-

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

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

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

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

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

Majority carriers are holes in PNP Transistor.

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

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

Forward Bias:-

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

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

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

0.3 Volts in Ge diode and

0.7 Volts in Si diode.

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

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

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

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

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

PN-juction Diode in Reverse-Bias:-

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

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

that is when the PN-junction is Reverse Biased

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

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Biot-savart’s law

It states that the magnetic field intensity dH produced, at the point P by the differential current element I dl

is proportional to the product I dl and the $\sin \alpha$ the angle between the element and the line joining the point P to the element.
and is inversely proportional to the square of the distance R between P and the current element.

then the direction of $\overrightarrow{dH}$ can be determined by right hand rule with the right hand thumb pointing in the direction of the current and the fingers encircling the wire in the direction of $\overrightarrow{dH}$ .

i.e, $dH \propto \frac{I \ dl \ \sin \alpha }{R^{2}}$ .

$dH = \frac{ k\ I \ dl \ \sin \alpha }{R^{2}}$ .

where k is the constant of proportionality , $k=\frac{1}{4\pi }$ .

$\overrightarrow{dH} = \frac{ \ I \ dl \ \sin \alpha }{4\pi \ R^{2}}\ \overrightarrow{a_{R}}$ A/m.

$\overrightarrow{dH} = \frac{ \ I \ \overrightarrow{dl} X \ \overrightarrow{a_{R}} }{4\pi \ R^{2}}$ A/m.

$\overrightarrow{dH}$ is perpendicular to the plane that contains $\overrightarrow{dl}$ and $\overrightarrow{a_{R}}$ .

$\overrightarrow{dH} = \frac{ \ I \ \overrightarrow{dl} X \ \overrightarrow{R} }{4\pi \ R^{3}}$ A/m.

then the total magnetic field strength measured at a point P is given by

$\overrightarrow{H} = \oint \frac{ \ I \ \overrightarrow{dl} X \ \overrightarrow{a_{R}} }{4\pi \ R^{2}}$ A/m.

closed path is taken since the current can flow only in closed path and this is called as integral form of Biot-Savart’s law.

as similar to different charge distributions in electro-statics , there exists different current elements like line, surface and volume in the study of static magnetic fields.

$\overrightarrow{H} = \int_{l} \frac{ \ I \ \overrightarrow{dl} X \ \overrightarrow{a_{R}} }{4\pi \ R^{2}}$ A/m. —-for a line current element.

$\overrightarrow{H} = \int_{s} \frac{ \overrightarrow{k} \ ds X \ \overrightarrow{a_{R}} }{4\pi \ R^{2}}$ A/m. —-for a surface current element.

$\overrightarrow{H} = \int_{v} \frac{ \overrightarrow{J} \ dv \ X \ \overrightarrow{a_{R}} }{4\pi \ R^{2}}$ A/m. —-for a volume current element.

the dot and cross products between dl and I represents either H is out of (or) into the page(plane) .

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Transmission line equations

The transmission line will be analyzed in terms of voltage and current and also the relation between V and I of a line can be obtained by the simplest type of Transmission line.

i.e, a pair of parallel wires of uniform size which are spaced a small distance (S) apart in air.

Consider a short section of Transmission line $AA^{'}BB^{'}$ of length $\Delta x$ (i.e an infinitesimal section with a very small length).

at $AA^{'}$ – voltage is V and current is I and at $BB^{'}$ voltage is $V+\Delta V$ and current is $I+\Delta I$ .

the voltage drop between $AA^{'}$ and $BB^{'}$ is

$V-(V+\Delta V) = I(R+j\omega L) \Delta x---------EQN(I)$

similarly, the current difference between $AA^{'}$ and $BB^{'}$ is

$I-(I+\Delta I) = V(G+j\omega C) \Delta X-------------EQN(II)$

from EQNS (I) and(II)

$\therefore -\frac{\Delta V}{\Delta x}=I (R+j\omega L)$ and $-\frac{\Delta I}{\Delta x}=V (G+j\omega C)$ .

to neglect the transit time effect the condition to be applied is $\Delta x\rightarrow 0$ , then the two equations will become

$\lim_{\Delta x\rightarrow 0} -\frac{\Delta V}{\Delta x}=\lim_{\Delta x\rightarrow 0}I (R+j\omega L)$ $\Rightarrow -\frac{d V}{d x}=I (R+j\omega L)$ .

$\lim_{\Delta x\rightarrow 0} -\frac{\Delta I}{\Delta x}=\lim_{\Delta x\rightarrow 0}V (G+j\omega C)$ $\Rightarrow -\frac{d I}{d x}=V (G+j\omega C)$ .

choose $-\frac{d V}{d x}=I (R+j\omega L)$

differentiating the above equation w.r.to x

$\Rightarrow -\frac{d^{2} V}{d x^{2}}=\frac{dI}{dx} (R+j\omega L)$

replacing $\frac{dI}{dx}$ in the above equation will result

$\frac{d^{2} V}{d x^{2}}= (R+j\omega L)(G+j\omega C)V$ .

$\frac{d^{2} V}{d x^{2}}=\gamma ^{2} V----------EQN(1)$ , Let $\gamma ^{2} =(R+j\omega L)(G+j\omega C)$ .

where $\gamma$ is the propagation constant which is a complex number.

Similarly from $-\frac{d I}{d x}=V (G+j\omega C)$

differentiating the above equation w.r.to x

$\Rightarrow -\frac{d^{2} I}{d x^{2}}=\frac{dV}{dx} (G+j\omega C)$

replacing $\frac{dV}{dx}$ in the above equation will result

$\frac{d^{2} I}{d x^{2}}= (R+j\omega L)(G+j\omega C)I$ .

$\frac{d^{2} I}{d x^{2}}=\gamma ^{2} I------EQN(2)$

the solutions of Equations(1) and (2) are

$V=A e^{-\gamma x}+Be^{\gamma x}$ .

$I=Ce^{-\gamma X}+D e^{\gamma X}$ .

where A, B, C and D are arbitrary constants in which A and B have dimensions of voltage and C and D have current dimensions.

since $\gamma$ is complex that is by replacing $\gamma = \alpha +j\beta$ in the V and I equations

$V=A e^{-\alpha x} e^{-j\beta x}+Be^{\alpha x} e^{j\beta x}$ .

$I=C e^{-\alpha x} e^{-j\beta x}+D e^{\alpha x} e^{j\beta x}$ .

the term $e^{-\alpha x} e^{-j\beta x}$ represents waves travelling from source end to load end and are called as incident waves .

Similarly the term $e^{\alpha x} e^{j\beta x}$ represents reflected waves when a transmission line is terminated with any load impedance $Z_{R}$ at the output end.

$V=A e^{-\gamma x}+Be^{\gamma x}$ .

differentiating V w.r.to x

$\frac{dV}{dx}=-A\gamma e^{-\gamma x}+B\gamma e^{\gamma x}$ .

but $\frac{dV}{dx}=-I(R+j\omega L)$

$\therefore -I(R+j\omega L)=-A\gamma e^{-\gamma x}+B\gamma e^{\gamma x}$ .

$I=\frac{\gamma }{(R+j\omega L)}(Ae^{-\gamma x}-B e^{\gamma x})$ .

$I=\frac{\sqrt{(R+j\omega L)(G+j\omega C)}}{(R+j\omega L)}(Ae^{-\gamma x}-B e^{\gamma x})$

since $\gamma ={\sqrt{(R+j\omega L)(G+j\omega C)}}$ .

$I=\sqrt{\frac{(G+j\omega C)}{(R+j\omega L)}}(Ae^{-\gamma x}-B e^{\gamma x})$ .

$\therefore I=\frac{1}{Z_{o}}(Ae^{-\gamma x}-B e^{\gamma x})$ , where $Z_{o} =\sqrt{\frac{(R+j\omega L)}{(G+j\omega C)}}$ .

AS $\cos h\gamma x = \frac{e^{\gamma x}+e^{-\gamma x}}{2}$ and $\sin h\gamma x = \frac{e^{\gamma x}-e^{-\gamma x}}{2}$ .

this implies $\cos h\gamma x-\sin h\gamma x = e^{-\gamma x}$ , $\cos h\gamma x+\sin h\gamma x = e^{\gamma x}$ .

by substituting $e^{-\gamma x}$ and $e^{\gamma x}$ in the Voltage and current equations , V and I results to be

$V= (A+B)\cos h\gamma x-(A-B)\sin h\gamma x---EQN(3)$ .

$I= \frac{1}{Z_{o}}((A-B)\cos h\gamma x-(A+B)\sin h\gamma x)---EQN(4)$ .

at the input terminals $x=0$ , $V = V_{s}$ and $I = I_{s}$ , after substituting this condition in EQNs (1) and (2)

$V_{s} =(A+B)$ and $I_{s} =\frac{1}{Z_{o}}(A-B)$ .

then the EQNs (3) and (4)

$V= V_{s}\cos h\gamma x-I_{s}Z_{o}\sin h\gamma x$ .

$I= I_{s}\cos h\gamma x-\frac{V_{s}}{Z_{o}}\sin h\gamma x$ .

The above equations are known as general equations of a transmission line for voltage and current at any point which is located at x from the sending end.

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condition for minimum attenuation

The attenuation constant $\alpha$ of a transmission line is

$\alpha =\sqrt{\frac{(RG-\omega ^{2}LC)+\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}{2}}-----EQN(1)$

$\alpha$ depends up on the 4 primary constants along with frequency $\omega$ . In order to achieve the minimum attenuation , these primary constants should be varied in turn.

case 1:-

consider the case where the attenuation ( $\alpha$ ) variation depends only on Inductance(L) .

i.e by considering R, G , C and $\omega$ as constants

from EQN(1), $2\alpha ^{2} =(RG-\omega ^{2}LC)+\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}$

differentiating the above equation w.r. to L

$\therefore 4\alpha \frac{d\alpha }{dL} = -\omega ^{2}C + \frac{2L(\omega ^{2}G^{2}+\omega ^{4}C^{2})}{2 \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}$ .

$4\alpha \frac{d\alpha }{dL} =\frac{-\omega ^{2}C \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}+\omega ^{2}L(G^{2}+\omega ^{2}C^{2})}{\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}$ .

$\frac{d\alpha }{dL} =\frac{-\omega ^{2}C \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}+\omega ^{2}L(G^{2}+\omega ^{2}C^{2})}{4\alpha\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}$ .

now $\frac{d\alpha }{dL}=0$ .

$-\omega ^{2}C \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}+\omega ^{2}L(G^{2}+\omega ^{2}C^{2})=0$ .

$\omega ^{2}C \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}=\omega ^{2}L(G^{2}+\omega ^{2}C^{2})$ .

$C \sqrt{(R^{2}+\omega ^{2}L^{2})}=L\sqrt{(G^{2}+\omega ^{2}C^{2})}$ .

$C (R^{2}+\omega ^{2}L^{2})=L(G^{2}+\omega ^{2}C^{2})$ .

$\Rightarrow LG=RC$ .

$\therefore L=\frac{RC}{G}$ H/m.

The condition for L to get minimum attenuation is $\frac{RC}{G}$ H/m but in general L should be maintained at less than this value.

case 2:-

consider the case where the attenuation ( $\alpha$ ) variation depends only on Capacitance(C) .

i.e by considering R, L, G and $\omega$ as constants

from EQN(1), $2\alpha ^{2} =(RG-\omega ^{2}LC)+\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}$

differentiating the above equation w.r. to L

$\therefore 4\alpha \frac{d\alpha }{dL} = -\omega ^{2}L + \frac{2C(\omega ^{2}R^{2}+\omega ^{4}L^{2})}{2 \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}$ .

$4\alpha \frac{d\alpha }{dL} =\frac{-\omega ^{2}L \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}+\omega ^{2}C(R^{2}+\omega ^{2}L^{2})}{\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}$ .

$\frac{d\alpha }{dL} =\frac{-\omega ^{2}L \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}+\omega ^{2}C(R^{2}+\omega ^{2}L^{2})}{4\alpha\sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}}$ .

now $\frac{d\alpha }{dL}=0$ .

$-\omega ^{2}L \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}+\omega ^{2}C(R^{2}+\omega ^{2}L^{2})=0$ .

$\omega ^{2}L \sqrt{(R^{2}+\omega ^{2}L^{2})(G^{2}+\omega ^{2}C^{2})}=\omega ^{2}C(R^{2}+\omega ^{2}L^{2})$ .

$C \sqrt{(R^{2}+\omega ^{2}L^{2})}=L\sqrt{(G^{2}+\omega ^{2}C^{2})}$ .

$C (R^{2}+\omega ^{2}L^{2})=L(G^{2}+\omega ^{2}C^{2})$ .

$\Rightarrow LG=RC$ .

$\therefore C=\frac{LG}{R}$ F/m.

The condition for C to get minimum attenuation is $\frac{LG}{R}$ F/m but in general C should be maintained at more than this value.

open circuit line and Short Circuit line

open circuit line:-

In order to observe the properties of an open circuited Transmission Line, chose a Transmission line of length l and is open circuited at the load end.

from the figure

$V_{S}$ – Source voltage (at sending end (or) source end).

$I_{S}$ – Source current (at sending end (or) source end).

$V_{R}$ – Load voltage (at receiving end (or) Load end).

$I_{R}$ – Load current (at receiving end (or) Load end).

at Load end $I_{R} =0$ , $V_{R} -$ maximum voltage and $Z_{R}=\infty$ .

We knew that the input impedance of a Transmission line is given by $Z_{S} = Z_{O} \frac{(Z_{R}+Z_{O} \tan h\gamma l)}{(Z_{O}+Z_{R} \tan h\gamma l)}$ .

after substituting the conditions of a open circuit conditions that is $Z_{R}=\infty$

$Z_{S} =\frac{Z_{O}}{\tan h\gamma l}$ .

$Z_{S} =Z_{O}\cot h\gamma l$ .

if the Transmission line is a loss-less line then $\alpha =0$ and $\gamma = j\beta$ .

$Z_{S} =Z_{O}\cot (h j\beta l)$ .

$Z_{S} =-jZ_{O}\cot \beta l$ . since $\cot (hj\beta l)=-j\cot \beta l$ .

alternative method:-

The alternative way to derive the input impedance is by using voltage and current equations of a basic Transmission line

$V= V_{s}\cos h\gamma x-I_{s}Z_{o}\sin h\gamma x$ .

$I= I_{s}\cos h\gamma x-\frac{V_{s}}{Z_{o}}\sin h\gamma x$ .

at $x=l$ , $V=V_{R}$ and $I=I_{R}$ then the equations changes to

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

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

by substituting $I_{R} =0$ in equation(2) and also choosing the line as loss-less line

$0= I_{s}\cos (j\beta l)-\frac{V_{s}}{Z_{o}}\sin (j\beta l)$ .

$I_{s}\cos \beta l=\frac{V_{s}}{Z_{o}} j\sin \beta l$ . since $\cos (hj\beta l) = \cos \beta l$ and $\sin (hj\beta l) = j\sin \beta l$ .

$Z_{S} =\frac{V_{s}}{I_{s}}=Z_{o} \frac{\cos \beta l}{j\sin \beta l}$ .

$Z_{S}=Z_{OC} =-jZ_{o} \cot \beta l$ .

short circuit line:-

In order to observe the properties of an short circuited Transmission Line, chose a Transmission line of length l and is short circuited at the load end.

from the figure

$V_{S}$ – Source voltage (at sending end (or) source end).

$I_{S}$ – Source current (at sending end (or) source end).

$V_{R}$ – Load voltage (at receiving end (or) Load end).

$I_{R}$ – Load current (at receiving end (or) Load end).

at Load end $V_{R} =0$ , $I_{R} -$ maximum current and $Z_{R}=0$ .

We knew that the input impedance of a Transmission line is given by $Z_{S} = Z_{O} \frac{(Z_{R}+Z_{O} \tan h\gamma l)}{(Z_{O}+Z_{R} \tan h\gamma l)}$ .

after substituting the conditions of a short circuit conditions that is $Z_{R}=0$

$Z_{S} =Z_{O}\tan h\gamma l$ .

if the Transmission line is a loss-less line then $\alpha =0$ and $\gamma = j\beta$ .

$Z_{S} =Z_{O}\tan (h j\beta l)$ .

$Z_{S} =jZ_{O}\tan \beta l$ . since $\tan (hj\beta l)=j\tan \beta l$ .

alternative method:-

The alternative way to derive the input impedance is by using voltage and current equations of a basic Transmission line

$V= V_{s}\cos h\gamma x-I_{s}Z_{o}\sin h\gamma x$ .

$I= I_{s}\cos h\gamma x-\frac{V_{s}}{Z_{o}}\sin h\gamma x$ .

at $x=l$ , $V=V_{R}$ and $I=I_{R}$ then the equations changes to

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

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

by substituting $V_{R} =0$ in equation(1) and also choosing the line as loss-less line

$0= V_{s}\cos (j\beta l)-I_{s}Z_{o}\sin (j\beta l)$ .

$V_{s}\cos \beta l=I_{s}Z_{o} j\sin \beta l$ . since $\cos (hj\beta l) = \cos \beta l$ and $\sin (hj\beta l) = j\sin \beta l$ .

$Z_{S} =\frac{V_{s}}{I_{s}}=Z_{o} \frac{j\sin \beta l}{\cos \beta l}$ .

$Z_{S}=Z_{SC} =jZ_{o} \tan \beta l$ .

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Distortion in a Transmission line

Signals transmitted over Transmission lines are mostly complex and consists of high frequency components, while passing through the line distortion occurs in the signal , this is commonly known as line distortion.

In ideal Transmission line , it is intended that the waveform at the receiving end of the line must be identical to the wave form at the sending end then the line is said to be distortion less line (or) distortion free line.

It has been seen that distortion in the waveform exists if all frequencies in the complex wave form do not have the same attenuation and same delay during propagation.

when the received signal is not the exact replica of the transmitted signal then the signal is said to be distorted.

causes of distortion in a Transmission Line:-

distortion is caused due to the following three reasons mainly

variation of characteristic impedance with respect to frequency:-The characteristic impedance of the line varies with change in frequency and line should be terminated in an impedance that does not vary with frequency to avoid distortion.
variation of attenuation with respect to frequency:- the attenuation of the line varies with frequency. Hence waves of different frequencies are attenuated by different amounts known as frequency distortion.
variation of phase constant with respect to frequency:- phase distortion due to the variation of phase constant with the frequency which in turn varies the velocity of propagation with frequency . Therefore waves of different frequencies arrive at different times at the end of the line.

Thus for a line to be distortion less the characteristic impedance $Z_{o}$ , attenuation $(\alpha )$ and phase constant $(\beta )$ should be independent of frequency.

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

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

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

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

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

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

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

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

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

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

then by taking the Divergence of the above equation

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Propagation of plane EM wave in conducting medium (or) lossy dielectrics

A lossy dielectric medium is one which an EM wave as it propagates losses power owing to imperfect dielectric,that is a lossy dielectric is an imperfect conductor that is a partially conducting medium $(\sigma \neq 0)$ .

where as a lossless dielectric is a $(\sigma =0)$ perfect dielectric,then wave equations for conductors are also holds good here

i.e, $\bigtriangledown ^{2}\overrightarrow{E}=\mu \sigma \frac{\partial \overrightarrow{E}}{\partial t} +\mu \epsilon \frac{\partial ^{2}\overrightarrow{E}}{\partial t^{2}}$

$\frac{\partial }{\partial t}= jw$

then $\bigtriangledown ^{2}\overrightarrow{E}= j\omega \mu \sigma \overrightarrow{E}+\mu \epsilon(j\omega ) ^{2}\overrightarrow{E}$

$\bigtriangledown ^{2}\overrightarrow{E}= (\sigma + j \omega \epsilon )j\omega \mu \overrightarrow{E}$

$\bigtriangledown ^{2}\overrightarrow{E}= \gamma ^{2} \overrightarrow{E}$

$\bigtriangledown ^{2}\overrightarrow{E}- \gamma ^{2} \overrightarrow{E}=0---------Equation (1)$

Equation (1) is called helm holtz equation and $\gamma$ is called propagation constant.

$\gamma ^{2} =j\omega \mu (\sigma +j\omega \epsilon )$

$\gamma ^{2}= j\omega \mu \sigma -\omega ^{2}\mu \epsilon$

Since $\gamma$ is a complex quantity it can be expressed as $\gamma = \alpha +j\beta$

$\alpha$ – is attenuation constant measured in Nepers/meter.

$\beta$ -is phase constant measured in radians/meter.

$(\alpha +j\beta ) ^{2}= j\omega \mu \sigma -\omega ^{2}\mu \epsilon$

$\alpha^{2} +2j\alpha \beta-\beta ^{2} = j\omega \mu \sigma -\omega ^{2}\mu \epsilon$

by equating real and imaginary parts separately $\alpha ^{2}-\beta ^{2}= -\omega ^{2}\mu \epsilon------Equation(2)$

and $2\alpha \beta =\omega \mu \sigma$

$\alpha =\frac{\omega \mu \sigma}{2\beta }$

by substituting $\alpha$ value in the equation (2) $\frac{\omega ^{2}\mu ^{2}\sigma ^{2}}{4\beta ^{2}}-\beta ^{2}=-\omega ^{2}\mu \epsilon$

${\omega ^{2}\mu ^{2}\sigma ^{2}}-4\beta ^{4}=-4\omega ^{2}\beta ^{2}\mu \epsilon$

$4\beta ^{4}-4\omega ^{2}\beta ^{2}\mu \epsilon -{\omega ^{2}\mu ^{2}\sigma ^{2}}=0$

let $\beta ^{2}=t$

$4t^{2}-4\omega ^{2}t\mu \epsilon -{\omega ^{2}\mu ^{2}\sigma ^{2}}=0$

$t^{2}-\omega ^{2}t\mu \epsilon -\frac{\omega ^{2}\mu ^{2}\sigma ^{2}}{4}=0$

the roots of the above quadratic expression are

$t=\frac{\omega ^{2}\mu \epsilon \pm \sqrt{\omega ^{4}\mu ^{2}\epsilon ^{2}-4(-\frac{\omega ^{2}\mu ^{2}\sigma ^{2}}{4})}}{2}$

$t=\frac{\omega ^{2}\mu \epsilon \pm \sqrt{\omega ^{4}\mu ^{2}\epsilon ^{2}(1+\frac{\sigma }{\omega \epsilon })^{2}}}{2}$

$\beta ^{2}=\frac{\omega ^{2}\mu \epsilon \pm \sqrt{\omega ^{4}\mu ^{2}\epsilon ^{2}(1+\frac{\sigma }{\omega \epsilon })^{2}}}{2}$

$\beta =\sqrt{\frac{\omega ^{2}\mu \epsilon \pm \sqrt{\omega ^{4}\mu ^{2}\epsilon ^{2}(1+\frac{\sigma }{\omega \epsilon })^{2}}}{2}}$

$\beta =\sqrt{\frac{\omega ^{2}\mu \epsilon (1+ \sqrt{(1+\frac{\sigma }{\omega \epsilon })^{2}})}{2}}$

similarly,

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Poynting theorem

Introduction:-

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.

Proof:-

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

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

then

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

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.

Maxwell’s First Equation in Electrostatics

From the Divergence theorem, we have

$\overrightarrow{\bigtriangledown }.\overrightarrow{D}= div \overrightarrow{D}=( \frac{\partial D _{x}}{\partial x}+ \frac{\partial D _{y}}{\partial y}+ \frac{\partial D _{z}}{\partial z})$

$\lim_{dv->0}\frac{\oint_{s}\overrightarrow{D}.\overrightarrow{ds}}{dv}=\lim_{dv->0}( \frac{\partial D _{x}}{\partial x}+ \frac{\partial D _{y}}{\partial y}+ \frac{\partial D _{z}}{\partial z})$

${\oint_{s}\overrightarrow{D}.\overrightarrow{ds}}=\( \frac{\partial D _{x}}{\partial x}+ \frac{\partial D _{y}}{\partial y}+ \frac{\partial D _{z}}{\partial z})dv$

from Gauss’s law $\oint_{s}\overrightarrow{D}.\overrightarrow{ds} =Q_{enclosed}$

${\oint_{s}\overrightarrow{D}.\overrightarrow{ds}}=\( \frac{\partial D _{x}}{\partial x}+ \frac{\partial D _{y}}{\partial y}+ \frac{\partial D _{z}}{\partial z})dv = Q_{enclosed}$

dividing it by $dv(or) \Delta v$ differential volume on both sides

$\frac{{\oint_{s}\overrightarrow{D}.\overrightarrow{ds}}}{dv} = \frac{Q_{enclosed}}{dv}$

by applying limit on both sides

$\lim_{dv->0}\frac{{\oint_{s}\overrightarrow{D}.\overrightarrow{ds}}}{dv} = \lim_{dv->0} \frac{Q_{enclosed}}{dv}$ .

$div\overrightarrow{D} = \rho _{v}$

$\overrightarrow{\bigtriangledown }.\overrightarrow{D} = \rho _{v}$

This equation is known as Maxwell’s first equation and is also known as point form of Gauss’s law /Differential form of Gauss’s law.

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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 $P_{n}$ $=V_{n1}Q_{n}+V_{n2}Q_{n}+V_{n3}Q_{n}+.......$

$\therefore$ Total 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.

simillarly,

$W_{E} = \frac{1}{2}\int_{l}\rho _{l}dl. V$ 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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Electric Potential (V)

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

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

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

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

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

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

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

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

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

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

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

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

$V_{AB}= -\int_{r_{A}}^{r_{B}}\frac{Q}{4\pi \epsilon _{o}r^{2}}\overrightarrow{a_{r}}.dr \overrightarrow{a_{r}}$ since $\overrightarrow{dl}=dr.\overrightarrow{a_{r}}$

$V_{AB}= -\int_{r_{A}}^{r_{B}}\frac{Q}{4\pi \epsilon _{o}r^{2}}dr$

$V_{AB}= -\frac{Q}{4\pi \epsilon _{o}}\left _{r_{A}}^{r_{B}}$

$V_{AB}= -\frac{Q}{4\pi \epsilon _{o}}$

$V_{AB}=V_{B}-V_{A}$

similarly, $V_{BA}=V_{A}-V_{B}$

where $V_{A}$ and $V_{B}$ are the scalar potentials at the points A and B respectively. If A is located at $\infty$ with respect to origin ,with zero potential $V_{A} =0$ and B is located at a distance r with respect to origin. then the work done in moving a charge from A (infinity) to B is given by

$V_{AB} = \frac{Q}{4\pi \epsilon _{o}r_{B}}$ here $r_{B} = r$

$\therefore V = \frac{Q}{4\pi \epsilon _{o}r}$ volts.

hence the potential at any point is the potential difference between that point and a chosen point at which the potential is zero. In other words assuming Zero potential at infinity .

The potential at a distance r from a point charge is the work done per unit charge by an external agent in transferring a test charge from infinity to that point.

i.e, $V = -\int_{\infty }^{r}\overrightarrow{E}.\overrightarrow{dl}$

So a point charge $Q_{1}$ located at a point P with position vector $\overline{r_{1}}$ then the potential at another point Q with a position vector $\overline{r}$ is

$V_{at \overline{r}} = \frac{Q_{1}}{4\pi \epsilon _{o}\left | \overline{r} -\overline{r_{1}}\right |}$

As like $\overrightarrow{E}$ superposition principle is applicable to V also that is for n point charges $Q_{1},Q_{2},Q_{3},Q_{4},........Q_{n}$ located at points with position vectors $\overline{r_{1}},\overline{r_{2}},\overline{r_{3}},.......\overline{r_{n}}$

then the potential at $\overline{r}$ is

$V_{at \overline{r}} = \frac{Q_{1}}{4\pi \epsilon _{o}\left | \overline{r} -\overline{r_{1}}\right |}+\frac{Q_{2}}{4\pi \epsilon _{o}\left | \overline{r} -\overline{r_{2}}\right |}+\frac{Q_{3}}{4\pi \epsilon _{o}\left | \overline{r} -\overline{r_{3}}\right |}+........+\frac{Q_{n}}{4\pi \epsilon _{o}\left | \overline{r} -\overline{r_{n}}\right |}$

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Electric field Intensity-E

Electric field Intensity( $\overrightarrow{E}$ ):-

Let us suppose a point charge Q is placed somewhere in space and if any other charge q is brought near to it, q experiences a force on Q and vice-versa. Thus there exists a region around a charge in which it experience a force on any other charge located in that region.

∴ The region around a charge distribution is called as Electric field produced by that charge distribution.

The electric field intensity (or) Electric field strength is defined as the force per unit charge (test charge)

i.e, $\overrightarrow{E} = \frac{\overrightarrow{F}}{q} Newtons/Coulomb$ —-> Equation (1)

∴ The expression for $\overrightarrow{E_{at P}}=\frac{Q}{4\pi \epsilon _{o}R^{2}}\widehat{a_{R}}$

q is small test charge +Q is a positive charge placed in free space. $\overrightarrow{E}$ is the electric field produced around +Q charge, then after placing a small test charge q in a field $\overrightarrow{E}$ , $\overrightarrow{E}$ exerts some force on this test charge q , given by

$F = E q$

and simillarly +Q and q experiences force on each other given by $F = \frac{Qq}{4\pi \epsilon r^{2}}$ —>Equation(2) (magnitude)

Force (as a vector) $\overrightarrow{F} = \frac{Qq}{4\pi \epsilon r^{2}}\widehat{a_{R}}$

from equations (1) and (2) F = F (magnitudes)

$Eq = \frac{Qq}{4\pi \epsilon r^{2}}$

$E= \frac{Q}{4\pi \epsilon r^{2}}$ —> Scalar magnitude of $\overrightarrow{E}$ produced by the charge Q.

where as $\overrightarrow{E}= \frac{Q}{4\pi \epsilon r^{2}}\widehat{a_{R}}$ which acts along the direction of Coulomb’s force $\overrightarrow{F}$ .

Types of charge distributions:-

In order to find out the electric field strength due to different types of charge distributions, first of all one must know how many types of charge distributions are there? the positive and negative charges can be distributed into 3 types of distributions.

Properties:-

charge is conserved , i.e, charge is neither be created nor destroyed.
charges are surrounded by electric and magnetic fields.

Point charge distribution:-

The name itself indicates that the charge confined to a point is known as point charge distributions. In practical point charges my not exists and point charge does not occupy any space.

Example:- an electron with a charge of 1.6X10^-19 C is a point charge.

Line charge distribution(ρ_L):-

If the charge is distributed along the length of the line, then it is known as line charge distribution. It may be a uniform (or) non-uniform distribution as shown in the figures.

If the charges are distributed uniformly along the line then it is a uniform charge distribution ρ_L is constant through out the line, otherwise it is Non-uniform type.

The line charge density $\rho _{L}=\lim_{\Delta l->0}\frac{\Delta Q}{\Delta l}$

$\rho _{L}=\frac{dQ}{dl}=\frac{Q}{l} Coulomb/m$ —> $Q=\int_{l}\rho _{l}dl Coulomb$

$\rho _{L}$ is defined as the charge per unit-length.

Surface charge distribution(ρ_s):-

If the charge is distributed uniformly over a two dimensional surface. Then it is called uniform surface charge distribution otherwise non-uniform.

then the surface charge density $\rho _{s}= \lim_{\Delta s->0}\frac{\Delta Q}{\Delta s}$

$\rho _{s}=\frac{dQ}{ds}=\frac{Q}{S} c/m^{2}$

$Q=\int_{s}\rho _{s}ds$ Coulomb

ρ_s is defined as charge per unit surface area and is measured in terms of C/m².

Volume charge distribution (ρ_v):-

If the charge is distributed uniformly in a volume then it is called as uniform volume distribution. Sphere represents a volume here

$\rho _{v}=\lim_{\Delta v->0}\frac{\Delta Q}{\Delta v}$

$\rho _{v}=\frac{dQ}{dv}=\frac{Q}{V} c/m^{3}$

$Q=\int_{v}\rho _{v}dv$ Coulombs

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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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Electrostatics-coulomb’s law

Electrostatics:-

Electrostatics deals with static electric fields ( charges which are at rest).

i.e, Electric fields which are independent of time, these are the fields produced by the charges at rest. A charge can be either concentrated at a point or distributed in some fashion like line, surface, volume and the charge distribution is assumed to be constant with respect to time.

Coulomb’s law:-

Let us suppose there exists two charged bodies placed apart at a distance ‘R’ as shown in the figure

then there exists a force between the two charges, if the two charges are like charges , the force is repulsive in nature. where as for unlike charges the force is attractive in nature, that is the force is either the force of attraction (or) the force of repulsion which is given by

$F \propto Q_{1}Q_{2}$

$\propto \frac{1}{R^{2}}$

Statement:-

The force of attraction or repulsion between the two charged bodies i. is directly proportional to the product of the two charges. ii. is inversely proportional to the square of the distance between them. iii. and acts along the line joining the two point charges.

i.e, $F =\frac{k Q_{1}Q_{2}}{R^{2}}$ where k is constant of proportionality

$F = \frac{Q_{1}Q_{2}}{4\pi \epsilon R^{2} }$ , $k=\frac{1}{4\pi \epsilon }$

where $\epsilon$ is the absolute permittivity of the medium given by $\epsilon = \epsilon _{o}\epsilon _{r}$

here $\epsilon _{o}$ is free space permittivity

$\epsilon _{r}$ is the relative permittivity of the medium.

$\epsilon _{o} =8.854 X 10^{-12}F/m$ or $\epsilon _{o} = \frac{10^{-9}}{36\pi } F/m$

$k= \frac{1}{4\pi \epsilon _{o}}= 9X10^{9}m/F$

permittivity (or) capacitivity (∈) :-

This is defined as the ability (or) Capacity to store electrical energy ∈_r : 8 to 9 for alumina. ∈_r : 2 to 3 for Teflon fibre glass.

the force is scalar one in the previous equation, Now the vector form is

$\overrightarrow{F} = \frac{Q_{1}Q_{2}}{4\pi \epsilon _{o}R^{2}}\hat{a_{R}}$ where $\hat{a_{R}}$ is the unit vector in the direction of R

If there exists two charges, Q₁ and Q₂ then force acting on 1 due to 2 is given by ${\overrightarrow{F}_{12}} = \frac{Q_{1}Q_{2}}{4\pi \epsilon _{o}R_{21}^{2}}\hat{a_{21}}$

simillarly Force acting on 2 by 1 is given by

$\overrightarrow{F}_{21}= \frac{Q_{1}Q_{2}}{4\pi \epsilon _{o}R_{12}^{2}}\hat{a_{12}}$

$\therefore \overrightarrow{F_{12}}=-\overrightarrow{F_{21}}$

both are equal in magnitude, they differ in their directions.

Q1. Two point charges $0.7mC$ and $4.9uC$ are situated in free space at (2,3,6) and(0,0,0) . Calculate the force acting on the 0.7mC charge.

Ans: Let Q₁ = 0.7mC and Q₂ = 4.9uC

Force acting on 1 by 2 is

$\overrightarrow{F}_{12} = \frac{Q_{1}Q_{2}}{4\pi\epsilon _{o}R_{21}^{2}}\hat{a_{21}}$ $= \frac{0.7X10^{-3}4.9X10^{-6}(2\overrightarrow{a_{x}}+3\overrightarrow{a_{y}}+6\overrightarrow{a_{z}})}{4\pi \epsilon _{o}X7^{2}(\sqrt{4+9+36})}$

$\overrightarrow{F}_{12}= 0.18\overrightarrow{a_{x}}+0.27\overrightarrow{a_{y}}+0.54\overrightarrow{a_{z}} Newtons$

Force due to number of charges:-

Imagine a situation when there exists more than two charges, then each will experience a force on the other, then the resultant force on any charge can be obtained by the principle of superposition (i.e linear addition).

the total force on Q_o in such case is vector sum of all forces acting on Q_o by each of the charges Q₁ , Q₂ & Q₃.

force acting on Q_o due to Q₁ is

$\overrightarrow{F}_{o1}= \frac{Q_{o}Q_{1}}{4\pi \epsilon _{o}R_{1o}^{2}}\hat{a_{1o}}$

force acting on Q_o due to Q₂ is

$\overrightarrow{F}_{o2}= \frac{Q_{o}Q_{2}}{4\pi \epsilon _{o}R_{2o}^{2}}\hat{a_{2o}}$

force acting on Q_o due to Q₃ is

$\overrightarrow{F}_{o3}= \frac{Q_{o}Q_{3}}{4\pi \epsilon _{o}R_{3o}^{2}}\hat{a_{3o}}$

then the Resultant force is

$\overrightarrow{F}=\overrightarrow{F_{o1}}+\overrightarrow{F_{o2}}+\overrightarrow{F_{o3}}$ , This is for taking four charges into account, if there exists ‘n’ number of charges, then the force action on Q_o by the remaining (n-1) charges on it is given by

then $\overrightarrow{F}=\frac{Q_{o}}{4\pi\epsilon _{o}}\sum_{i=1}^{n}\frac{Q_{i}\hat{a_{io}}}{R_{io}^{2}}$ .

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Need for modulation

Need for modulation:-

Modulation is the fundamental need for communications, the following are the basic needs for modulation

Distance:-

As low frequency signals can not travel longer distances, low frequencies can be translated into higher frequencies by using modulation schemes.

Improved Signal to Noise Ratio:-

Signal to Noise Ratio has been improved because of modulation at the Receiver.

Practicability of antennas:-

If the communication medium is free space , then messages are transmitted and received with the help of antennas.

∴ The height of the antenna is of the order of the wavelength ( λ ) of the signal being transmitted, when the signals from transmitter are transmitted without modulation then the height required for the antenna is very high, for example to transmit a message signal of frequency f = 5 KHz

height of the antenna required would be

$h \simeq \frac{\lambda }{4} \simeq \frac{c}{4f}$

$h \simeq \frac{3 X 10^{8}}{4 X 5X10^{3}}$

$h\simeq 15000 meters$

Designing an antenna with 15 Km length is almost impractical. To reduce the height of the antenna , instead of sending low frequency (5 KHz) signal as it is modulation is preferred , modulation technique reduces the height of the antenna and makes the antenna to be more practical both at Tx^r and Rx^r.

Narrow Banding:-

There is a problem caused by direct transmission of base band signal which can be explained as follows , suppose a base band signal spectrum ranges from (50 Hz-10 KHz) then the height of the antenna must be 1.5 X 10⁶ meters to receive 50 Hz signal at the receiving end , where as for 10 KHz , the height would be 7500 meters that means height of the antenna is not same for all frequencies .

∴ A wide band antenna which can operate for band edge ratio of 200 is required which is impractical, so modulation is required to use same antenna at the receiver to receive certain range of frequencies.

∴ A wide band message signal from 50 Hz to 10 KHz gets converted to a narrow band signal by a carrier frequency of 1 MHz.

This Narrow banding of Base band signal is possible with modulation which in turn eliminates the complexity of antenna height at the receiver .

Multiplexing:-

Simultaneous transmission of multiple messages over a channel is known as multiplexing. suppose number of messages from different transmitters are transmitted without modulation then there is a possibility of interference (one with other) since the base band spectrum is identical for all the messages. Hence the transmitted messages will not be received properly at the receiver.

one technique to eliminate interference is by the use of modulation and the other technique is multiplexing.

Frequency division multiplexing (FDM) which uses analog modulation techniques.
Time division multiplexing (TDM) uses pulse modulation techniques.

Multiplexing reduces number of channels needed and reduce the cost of installation and maintenance.

Radio Frequency spectrum from ELF to EHF:-

frequency Range	Description of frequency band
upto 300 Hz	Extreme Low frequency(ELF)
300 Hz-3 KHz	Voice frequency (VF)
3 KHz-30 KHz	Very Low frequency (VLF)
30 KHz-300 KHz	Low Frequency (LF)
300 KHz-3 MHz	Medium frequency (MF)
3 MHz-30 MHz	High frequency (HF)
30 MHz-300 MHz	Very High frequency (VHF)
300 MHz-3 GHz	Ultra High frequency (UHF)
3 GHz-30 GHz	Super High frequency (SHF)
30 GHz-300 GHz	Extreme High frequency (EHF)

*Audio Frequency range: 20 Hz-20 KHz.

*UHF,SHF and EHF are Micro wave frequencies.

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Properties of Z-Transforms (Bi-lateral)

1.Linearity Property:-

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

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

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

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

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

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

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

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

2.Time-shifting Property:-

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

$x[n-k]\leftrightarrow \ ?$ .

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

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

Here n is a variable and k is a constant.

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

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

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

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

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

3. Scaling in-Z-domain property:-

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

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

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

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

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

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

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

4. Time-reversal property:-

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

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

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

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

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

5. Differentiation in Z-domain:-

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

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

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

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

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

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

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

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

6. Convolution in Time-domain:-

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

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

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

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

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

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

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

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

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

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

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Derivation of Z and inverse Z-Transforms

Derivation of Z-Transform:-

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

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

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

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

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

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

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

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

Inverse Z-Transform:-

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

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

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

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

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

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Initial and Final Value Theorems of Laplace Transforms

Initial-value Theorem:-

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

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

Proof:-

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

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

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

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

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

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

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

Hence proved.

final-value Theorem:-

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

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

Proof:-

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

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

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

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

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

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

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

Hence proved.

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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

In the analyzation of signals and LTI systems.
Designing of Signals & Systems.
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

The theory of Integration.
Point-set topology.
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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Gauss’s law and its applications

Gauss’s law:-

Statement:- Gauss’s law states that the total flux coming out of a closed surface is equal to the net charge enclosed by that surface.

i.e, $\oint_{s}\overrightarrow{D}.\overrightarrow{ds}= Q_{enclosed}$ —–> i.e, $\psi _{e} = Q_{enclosed}$

Proof:-

Consider a closed surface of any shape with charge distribution as shown in the given figure. Now this surface is enclosed by a charge of Q coulombs , then a flux of Q coulombs will pass through the enclosing surface.

Now divide the entire area S into small pieces of differential areas ds or Δs.

Now at point P , the flux density is D_s and is direction is as shown in the figure. The surface S is chosen is an irregular surface and $\overrightarrow{D}$ is also not uniform. i.e, $\overrightarrow{D}$ direction as well as it’s magnitude is going to change from point to point.

now for the surface area ds the normal vector to the surface is $\overrightarrow{ds_{n}}= ds.\overrightarrow{a_{n}}$ the direction of $\overrightarrow{D}$ at a point P on the surface ds is making an angle θ w.r.to $\overrightarrow{ds_{n}}$ , then flux density at point ‘P’ is $D = \frac{d\psi }{ds}$

$d\psi = D.ds$ —>equation 1

to get maximum flux out of the surface $\overrightarrow{D}$ and $\overrightarrow{ds_{n}}$ should be in the same direction, there is a need to find out the component of $\overrightarrow{D}$ along $\overrightarrow{ds_{n}}$ is $D_{s}cos\theta = D_{s normal}$

then equation 1 becomes $d\psi = D_{snormal} ds$

$d\psi =D _{s}cos\theta ds$ —> $d\psi = \overrightarrow{D}.\overrightarrow{ds}$

Total flux is $\psi = \oint_{s}d\psi =\oint_{s}\overrightarrow{D}.\overrightarrow{ds}$

$\therefore \psi =\oint_{s}\overrightarrow{D}.\overrightarrow{ds}$

∴ Total flux $\psi$ = net charge enclosed Q

$\therefore Q _{enclosed}=\oint_{s}\overrightarrow{D}.\overrightarrow{ds}$

If there are n number of charges Q₁, Q₂,Q₃ …..Q_n then $Q= \sum Q_{n}$

i. for a line charge distribution $Q=\int_{l}\rho _{l}dl$ ii. for a surface charge distribution $Q=\int_{s}\rho _{s}ds$ iii. for a volume charge distribution $Q=\int_{v}\rho _{v}dv$ .

Closed Gaussian surface:-

The Gauss’s law is used to find out $\overrightarrow{E} or \overrightarrow{D}$ for symmetrical charge distributions and is used to find out $\psi or Q$ of any closed surface.

$\overrightarrow{D}$ is every where either normal (or) tangential to the closed surface, that is θ = π/2 or θ = 0^o / π so that $\overrightarrow{D}.\overrightarrow{ds}$ is maximum or zero.
$\overrightarrow{D}$ is constant over the portion of the closed surface for which $\overrightarrow{D}.\overrightarrow{ds}$ is not zero.

To apply Gauss’s law to a charge distribution, surface is required to be a Gaussian surface that encloses the charge distribution or charged body.

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Application of Ampere’s circuit law to infinite line current element

Consider as infinitely long straight conductor placed along z-axis carrying a current I .

In order to determine $\overrightarrow{H}$ at some point P. we allow a closed path which passes through the point P and encloses the current carrying conductor symmetrically such path is known as Amperian path.

To apply Ampere’s law the conditions t be satisfied are

The field $\overrightarrow{H}$ is either tangential (or) Normal to the path at each point of the closed path.
The magnitude of $\overrightarrow{H}$ must be same at all points of the path where $\overrightarrow{H}$ is tangential.

Now, $\overrightarrow{H}$ is given by $\overrightarrow{H} =H _{\rho } \overrightarrow{a}_{\rho }+H _{\phi } \overrightarrow{a}_{\phi }+H _{z} \overrightarrow{a}_{z}$ .

The path we are assuming is in the direction of $\phi$ so $\overrightarrow{dl} = dl \overrightarrow{a}_{\phi }$ .

$\overrightarrow{dl} = \rho \ d\phi \overrightarrow{a}_{\phi }$ .

Ampere’s law is used to find out $\overrightarrow{H}$ at P

i.e, from Ampere’s circuit law $\oint \overrightarrow{H} . \overrightarrow{dl} = I_{enc}$ .

$\oint \overrightarrow{H} . \overrightarrow{dl} = I$ .

$\oint \overrightarrow{H} . \overrightarrow{dl} =\oint(H _{\rho } \overrightarrow{a}_{\rho }+H _{\phi } \overrightarrow{a}_{\phi }+H _{z} \overrightarrow{a}_{z}) . \rho \ d\phi \overrightarrow{a}_{\phi }$ .

$=\oint (H _{\phi } \overrightarrow{a}_{\phi }) . \rho \ d\phi \overrightarrow{a}_{\phi }$ .

$=\int_{\phi = 0}^{2\pi } H _{\phi } \rho \ d\phi$ .

$= H _{\phi } \ \rho \ 2\pi$ .

from Ampere’s law $H _{\phi } \ \rho \ 2\pi = I$ .

$H _{\phi } =\frac{ I}{\ 2\pi\ \rho }$ .

$\therefore \overrightarrow{H _{\phi } } = \frac{ I}{\ 2\pi\ \rho } \overrightarrow{a_{\phi }}$ .

Ampere’s law is applied to find the value of $\overrightarrow{H}$ at any point P in it’s field.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$I= 2H_{o} \ b$ .

$k_{y}\ b = 2H_{o} \ b$ .

$H_{o} =\frac{1}{2} k_{y}$ .

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

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

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

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

Maxwell’s Equations for time-varying fields in point and Integral form are:

$\overrightarrow{\bigtriangledown }X\overrightarrow{H}=\overrightarrow{J}+\frac{\partial \overrightarrow{D}}{\partial t}$ $\Rightarrow \oint_{l}\overrightarrow{H}.\overrightarrow{dl}=\oint_{s}\overline{J}.\overrightarrow{ds}+\int_{s}\frac{\partial \overrightarrow{D}}{\partial t}.\overrightarrow{ds}$ .
$\overrightarrow{\bigtriangledown }X\overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}$ $\Rightarrow \oint_{l}\overrightarrow{E}.\overrightarrow{dl}=-\int_{s}\frac{\partial \overrightarrow{B}}{\partial t}.\overrightarrow{ds}$ .
$\overline{\bigtriangledown }.\overrightarrow{D}=\rho _{v}$ $\Rightarrow \oint_{s}\overrightarrow{D}.\overrightarrow{ds}=\int_{v}\rho _{v}dv$ .
$\overrightarrow{\bigtriangledown }.\overrightarrow{B}=0$ $\Rightarrow \oint_{s}\overrightarrow{B}.\overrightarrow{ds}=0$ .

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

word statements of the field Equations:-

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

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

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

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

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

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

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

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

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

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

the first two Maxwell’s Equations can be stated as

The magnetic voltage around a closed path is equal to the electric current through the path.
The electric voltage around a closed path is equal to the magnetic current through the path.

Maxwell’s Equations for static fields in point and Integral form are:

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

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

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input impedance of Transmission Line

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

In order to find out the impedance at the input terminals of a Transmission line choose the Basic Transmission Line equations

$V= V_{s}\cos h\gamma x-I_{s}Z_{o}\sin h\gamma x$ .

$I= I_{s}\cos h\gamma x-\frac{V_{s}}{Z_{o}}\sin h\gamma x$ .

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

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

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

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

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

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

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

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

$Z_{S} \ (or) \ Z_{in}=Z_{o} \frac{(Z_{R}\cos h\gamma l+Z_{O}\sin h\gamma l)}{(Z_{o}\cos h\gamma l+Z_{R}\sin h\gamma l)}$ .

$Z_{S}=Z_{o} \frac{(Z_{R}+Z_{O}\tan h\gamma l)}{(Z_{o}+Z_{R}\tan h\gamma l)}$ represents the source (or) input impedance of a basic Transmission Line.

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H due to finite long straight conuctor

Consider a conductor of finite length placed along z-axis as shown in the figure

The conductor has a finite length AB , where A and B are located at distances Z₁ and Z₂ above the origin with it’s upper and lower ends respectively subtending angles $\alpha _{2}$ and $\alpha _{1}$ at P.

P is the point at which $\overrightarrow{H}$ is to be determined.

Consider a differential element $\overrightarrow{dl}$ along the Z-axis at a distance Z from the origin.

where $\overrightarrow{dl} =dl \ \overrightarrow{a_{z}}$ .

$\overrightarrow{R} = -Z \overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}$ .

$\widehat{a_{R}} = \frac{\overrightarrow{R} = -Z \overrightarrow{a_{z}}+\rho \overrightarrow{a_{\rho }}}{\sqrt{(Z^{2}+\rho ^{2})}}$ .

$\overrightarrow{dl} X \ \widehat{a_{R}} = \frac{\rho dz \ \overrightarrow{a_{\phi }}}{\sqrt{(Z^{2}+\rho ^{2})}}$ .

$\overrightarrow{H} = \oint \frac{ I \ \rho \ dz \ \overrightarrow{a_{\phi }}}{4\pi (Z^{2}+\rho ^{2})^{\frac{3}{2}}}$ .

$\overrightarrow{H} = \oint_{Z_{1}}^{Z_{2}} \frac{ I \ \rho \ dz \ \overrightarrow{a_{\phi }}}{4\pi (Z^{2}+\rho ^{2})^{\frac{3}{2}}}$ .

as $z=\rho \ cot\alpha$ , $Z_{2} = \rho \ \cot \alpha _{2}$ and $dz = - \rho \ cosec^{2} \alpha \ d\alpha$ .

$\overrightarrow{H} =\frac{-I}{4\pi } \int_{\alpha _{1}}^{\alpha _{2}} \frac{\rho ^{2} \ cosec^{2}\alpha \ d\alpha \ \overrightarrow{a_{\phi }}}{ (\rho ^{2}+\rho ^{2} \cot ^{2}\alpha )^{\frac{3}{2}}}$ .

$\overrightarrow{H} =\frac{-I}{4\pi\ \rho } \int_{\alpha _{1}}^{\alpha _{2}} \rho ^{2} \ \sin \alpha \ d\alpha \ \overrightarrow{a_{\phi }}$ .

$\overrightarrow{H} =\frac{I}{4\pi\ \rho } \left \overrightarrow{a_{\phi }}$ .

Case 1 :-

when the conductor is semi-finite that is A is located at origin and B at $\infty$ .

i.e, $Z_{1} =0 \ \Rightarrow \ \alpha _{1} = 90^{o}$ and $Z_{2} =\infty \ \Rightarrow \ \alpha _{2} = 0^{o}$ .

then $\overrightarrow{H} = \frac{I}{4\pi \ \rho } \overrightarrow{a_{\phi }}$ .

Case 2:-

when conductor is infinite in length A is at $-\infty$ and B at $\infty$ implies $Z_{1} = -\infty \Rightarrow \alpha _{1} = 180 ^{o}$ and $Z_{2} = \infty \Rightarrow \alpha _{2} = 0 ^{o}$ .

$\overrightarrow{H} = \frac{I}{2\pi \ \rho } \overrightarrow{a_{\phi }}$ .

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Half, Quarter and Eight-wave lines

Eight-wave line (or) $\frac{\lambda }{8}$ -line:-

Consider a eight-wave line with length of the wave length as $\frac{\lambda }{8}$ .

from the input impedance of a Transmission line

$Z_{s}$ (or) $Z_{input} = \frac{Z_{O}(Z_{R}+Z_{O}\tan h\gamma l)}{(Z_{O}+Z_{R}\tan h\gamma l)}$ .

for a loss-less line $\alpha =0$ , $\gamma = j\beta$ .

then the input impedance changes to $Z_{input} = \frac{Z_{O}(Z_{R}+Z_{O}\tan (hj\beta l))}{(Z_{O}+Z_{R}\tan (hj\beta l))}$ .

$Z_{input} = \frac{Z_{O}(Z_{R}+jZ_{O}\tan \beta l)}{(Z_{O}+jZ_{R}\tan \beta l)}$ . since $\tan (j\beta l) =j \tan \beta l$ .

now by substituting $\beta l=\frac{2\pi }{\lambda }.\frac{\lambda }{8}\Rightarrow \beta l=\frac{\pi }{4}$ .

$Z_{input} = \frac{Z_{O}(Z_{R}+jZ_{O}\tan \frac{\pi }{4} )}{(Z_{O}+jZ_{R}\tan \frac{\pi }{4} )}$ .

$\therefore Z_{input} = Z_{O}$ .

i.e, The eight wave line , the input impedance is the Characteristic impedance ( $Z_{O}$ ) .

Half-wave line (or) $\frac{\lambda }{2}$ -line:-

Consider a Half-wave line with length of the wave length as $\frac{\lambda }{2}$ .

from the input impedance of a Transmission line

$Z_{s}$ (or) $Z_{input} = \frac{Z_{O}(Z_{R}+Z_{O}\tan h\gamma l)}{(Z_{O}+Z_{R}\tan h\gamma l)}$ .

for a loss-less line $\alpha =0$ , $\gamma = j\beta$ .

then the input impedance changes to $Z_{input} = \frac{Z_{O}(Z_{R}+Z_{O}\tan (hj\beta l))}{(Z_{O}+Z_{R}\tan (hj\beta l))}$ .

$Z_{input} = \frac{Z_{O}(Z_{R}+jZ_{O}\tan \beta l)}{(Z_{O}+jZ_{R}\tan \beta l)}$ . since $\tan (j\beta l) =j \tan \beta l$ .

now by substituting

i.e, The Half wave line , the input impedance is the load impedance () , this line repeats it’s terminating impedance, hence it is called as one-to-one transformer.

the main application of line is to connect a load to source where both of them can’t made adjacent in such case , we may connect a parallel Half-wave line at the load end.

Quarter-wave line (or) -line:-

Consider a Quarter-wave line with length of the wave length as .

from the input impedance of a Transmission line

$Z_{s}$ (or) $Z_{input} = \frac{Z_{O}(Z_{R}+Z_{O}\tan h\gamma l)}{(Z_{O}+Z_{R}\tan h\gamma l)}$ .

for a loss-less line $\alpha =0$ , $\gamma = j\beta$ .

then the input impedance changes to $Z_{input} = \frac{Z_{O}(Z_{R}+Z_{O}\tan (hj\beta l))}{(Z_{O}+Z_{R}\tan (hj\beta l))}$ .

$Z_{input} = \frac{Z_{O}(Z_{R}+jZ_{O}\tan \beta l)}{(Z_{O}+jZ_{R}\tan \beta l)}$ . since $\tan (j\beta l) =j \tan \beta l$ .

now by substituting $\beta l=\frac{2\pi }{\lambda }.\frac{\lambda }{4}\Rightarrow \Rightarrow \beta l=\frac{\pi }{2}$ .

$Z_{input} = \frac{Z_{O}(\frac{Z_{R}}{\tan \beta l}+jZ_{O})}{(\frac{Z_{O}}{\tan \beta l}+jZ_{R})}$ .

$Z_{input} = \frac{Z_{O}(\frac{Z_{R}}{\tan \frac{\pi }{2}}+jZ_{O})}{(\frac{Z_{O}}{\tan \frac{\pi }{2}}+jZ_{R})}$ .

$Z_{input} = \frac{Z_{O}(jZ_{O})}{(jZ_{R})}$ .

$Z_{input} = \frac{Z_{O}^{2}}{Z_{R}}$ .

$\therefore Z_{O} = \sqrt{Z_{input}Z_{R}}$ .

if $Z_{O}$ is constant then $Z_{s}\propto \frac{1}{Z_{R}}$ .

i.e, The Quarter wave transformer transforms a load impedance $Z_{R}< Z_{O}$ into a value $Z_{S}$ which is larger than $Z_{O}$ and vice-versa.

sometimes it is called as impedance inverter if $Z_{R}$ is pure resistive $Z_{S}$ also becomes resistive.

This Quarter wave transformer is useful when it is desired to transform a resistance into a different resistance value either larger (or) smaller.

The desired value of $Z_{O}$ can be obtained by choosing proper value of the ratio of spacing of the line conductors to their diameters because the physical dimensions can not practically have an infinite range of values where as the transformer ratio of the $\frac{\lambda }{4}$ transformer is subject to the practical limitation.

it is very useful device because of its simplicity and the ease with which it’s behavior is calculated .

if the length is an odd multiple of $\frac{\lambda }{4}$ will have same transformation properties but when the length is made larger the sensitivity to a small change of frequency becomes larger.

since input impedance is inversely proportional to $Z_{R}$ . If $Z_{R}$ is high , $Z_{S}$ will be low and vice-versa.

if $Z_{R}$ is capacitive , $Z_{S}$ will be inductive and vice-versa. if $Z_{R}$ is resistive , $Z_{S}$ will be resistive and vice-versa.

Thus depending up on $Z_{R}$ , a $\frac{\lambda }{4}$ transformer acts as step-up (or) step-down impedance transformer and that is why it is being called as impedance inverter.

a $\frac{\lambda }{4}$ transformer is disadvantageous , as it is sensitive to change in frequency because a new frequency section will no longer be $\frac{\lambda }{4}$ in length.

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Correlation Receiver (Special case of Optimum Receiver)

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

$v(t)=x(t)+n(t)$

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

$v_{o}(t)=v(t)*h(t)$

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

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

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

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

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

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

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

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

Now by replacing $\tau$ with $t$ the above equation becomes

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

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

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

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Comparison of Transistor Characteristics in three Configurations :-

Crystal Oscillator:-

Electric field Intensity():-

Types of charge distributions:-

Electrostatics:-

Need for modulation:-

Radio Frequency spectrum from ELF to EHF:-

Gauss’s law:-

Maxwell’s Equations for time-varying fields in point and Integral form are:

word statements of the field Equations:-

Maxwell’s Equations for static fields in point and Integral form are:

Electric field Intensity( $\overrightarrow{E}$ ):-