## Why Digital Communication is preferred over Analog Communication?

Introduction:-

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

Examples of different types of communications are

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

Why Digital?

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

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

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

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

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

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

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

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

3.Digital Circuits Vs Analog Circuits:-

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

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

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

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

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

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

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

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

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

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

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

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

i.e, The magneto motive force ($\because&space;\oint_{l}\overrightarrow{H}.\overrightarrow{dl}\rightarrow$ 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&space;}X\overrightarrow{E}=-\frac{\partial&space;\overrightarrow{B}}{\partial&space;t}$$\Rightarrow&space;\oint_{l}\overrightarrow{E}.\overrightarrow{dl}=-\int_{s}\frac{\partial&space;\overrightarrow{B}}{\partial&space;t}.\overrightarrow{ds}$.

The electro motive force ($\because&space;\oint_{l}\overrightarrow{E}.\overrightarrow{dl}\rightarrow$ 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.$\overrightarrow{\bigtriangledown&space;}.\overrightarrow{D}=\rho&space;_{v}$  $\Rightarrow&space;\oint_{s}\overrightarrow{D}.\overrightarrow{ds}=\int_{v}\rho&space;_{v}dv$.

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

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

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 $\frac{\partial&space;\overrightarrow{B}}{\partial&space;t}$ is known as magnetic current, e.m.f as electric voltage and m.m.f as magnetic voltage.

the first two Maxwell’s Equations can be stated as

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

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

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

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

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

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

$\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{H}=\overrightarrow{J}$

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

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

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

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

then by taking the Divergence of the above equation

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

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

$\overrightarrow{\bigtriangledown&space;}.\overrightarrow{G}=-\overrightarrow{\bigtriangledown&space;}.\overrightarrow{J}$

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

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

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

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

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

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

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 $\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}$ 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 $\overrightarrow{J_{D}}$.

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

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

$\overrightarrow{J_{D}}=\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}$

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

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

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

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

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

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

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

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

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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 $v_{o}(T_{b})$

$v_{o}(t)=v(t)*h(t)$

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

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

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

$h(T_{b}-\tau&space;)=x_{2}(T_{b}-T_{b}+\tau&space;)-x_{1}(T_{b}-T_{b}+\tau&space;)$

$h(T_{b}-\tau&space;)=x_{2}(\tau&space;)-x_{1}(\tau&space;)------Equation(2)$

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

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

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

Now by replacing $\tau$ with $t$ the above equation becomes

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

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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## Relation between E and V

The potential difference between two points A and B is $V_{AB}=&space;V_{A}-V_{B}$ and $V_{BA}=&space;V_{B}-V_{A}$.

i.e, $V_{AB}=-V_{BA}$

$V_{AB}+V_{BA}=0$.

This equation implies that the total work done in moving a charge from A to B $(V_{AB})$and then from B to A $(V_{BA})$ is zero.

i.e, $\oint_{l}\overrightarrow{E}.\overrightarrow{dl}=0$

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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 $r_{A}and&space;r_{B}$ units from the origin O,from Coulomb’s law the force acting on a test charge $Q_{t}$ is $\overrightarrow{F}=&space;Q_{t}\overrightarrow{E}$

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

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

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

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

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, $\overrightarrow{E}&space;=&space;\frac{Q}{4\pi&space;\epsilon&space;_{o}r^{2}}\overrightarrow{a_{r}}$

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

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

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

$V_{AB}=&space;-\frac{Q}{4\pi&space;\epsilon&space;_{o}}\left&space;[&space;\frac{-1}{r}&space;\right&space;]_{r_{A}}^{r_{B}}$

$V_{AB}=&space;-\frac{Q}{4\pi&space;\epsilon&space;_{o}}[\frac{1}{r_{B}}-\frac{1}{r_{A}}]$

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

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 $V_{A}&space;=0$ 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

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

$\therefore&space;V&space;=&space;\frac{Q}{4\pi&space;\epsilon&space;_{o}r}$ 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, $V&space;=&space;-\int_{\infty&space;}^{r}\overrightarrow{E}.\overrightarrow{dl}$

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

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

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

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

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

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

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

$ACF\leftrightarrow&space;ESD$

$R_{xx}(t)\leftrightarrow&space;S(f)$

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

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

$\frac{1}{\sqrt{2\pi&space;}}e^{\frac{-t^{2}}{2}}\leftrightarrow&space;\frac{1}{\sqrt{2\pi&space;}}.\sqrt{2\pi&space;}e^{-2&space;\pi&space;^{2}f^{2}}$

$\frac{1}{\sqrt{2\pi&space;}}e^{\frac{-t^{2}}{2}}\leftrightarrow&space;e^{-2&space;\pi&space;^{2}f^{2}}$

$x(t)\leftrightarrow&space;X(f)$

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

$S(f)&space;=&space;e^{-4\pi&space;^{2}f^{2}}$

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

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

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

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

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

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