## Mass Action law

Mass-Action law:-

Under thermal equilibrium for any semi conductor the product of number of holes and the no. of electrons is constant( and is independent of the amount of donor and acceptor impurity doping)

$\large&space;np&space;=&space;n_{i}^{2}$

n- electron concentration (or) number of electrons

p- hole concentration (or) number of holes

ni is the intrinsic carrier concentration

If a pure semi conductor is doped with N-type impurities, the no. of electrons in the conduction band(CB) increases above a level and no. of holes decreases below a level in the valance band (VB).

Similarly the addition of P-type impurities to a pure semi conductor increases holes in the valance band(VB) above a level and decreases the no. of electrons in conduction band(CB) below a level .

Charge Densities in N-type Semi conductor:-

Under equilibrium in a pure-Semi conductor $np=n_{i}^2$

when an N-type impurity is added then in N-type Semi-conductor $n_{N}p_{N}=n_{i}^2$ , the suffix N represents type of Semi conductor.

Number of electrons in N-type semi conductor= Number of donor atoms added+ number of holes in N-type semi conductor.

i.e, $n_{N}=&space;N_{D}+p_{N}$

since number of holes are very less in N-type material $n_{N}=&space;N_{D}$ ($\because&space;p_{N}$ is negligible)

$p_{N}&space;=&space;\frac{n_{i}^{2}}{n_{N}}$

$p_{N}&space;=&space;\frac{n_{i}^{2}}{N_{D}}$

$n_{i}$ – intrinsic  carrier concentration.

$p_{N}$ – number of hole concentration in N-type material.

$n_{N}$– number of electrons in N-type Semi conductor/ number of donor atoms.

Charge Densities in P-type Semi conductor:-

Under equilibrium in a pure-Semi conductor $np=n_{i}^2$

when an P-type impurity is added then in P-type Semi-conductor $n_{P}p_{P}=n_{i}^2$ , the suffix P represents type of Semi conductor.

Number of holes in P-type semi conductor= Number of acceptor atoms added+ number of electrons in P-type semi conductor.

i.e, $p_{P}=&space;N_{A}+n_{P}$

since number of electrons are very less in P-type material $p_{P}=&space;N_{A}$ ($\because&space;n_{P}$ is negligible)

$n_{P}&space;=&space;\frac{n_{i}^{2}}{p_{P}}$

$n_{P}&space;=&space;\frac{n_{i}^{2}}{N_{A}}$

$n_{i}$ – intrinsic  carrier concentration.

$n_{P}$ – number of electron concentration in P-type material.

$p_{P}$– number of holes in P-type Semi conductor/ number of acceptor atoms.

(No Ratings Yet)

## Crystal Oscillator

### Crystal Oscillator:-

In LC oscillators the frequency of oscillation fo depends on the tank circuit parameters  L & C, whereas L & C values change with respect to time, temperature, aging etc. Therefore fo 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 fi = fo 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 Cm.

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

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

$\omega&space;L&space;=\frac{1}{\omega&space;C}$

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

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

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

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

$f_{p}=\frac{1}{2\pi&space;}\sqrt{\frac{1}{L}(\frac{1}{c}+\frac{1}{C_{m}})}$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 fp. BJT offers a phase shift of 180o further 180o is produced by the capacitor voltage. Oscillations are possible only through fp, which provides stable oscillations.

(No Ratings Yet)

## Noise

### Introduction:-

Noise is probably the only topic in electronics and tele communications with which everyone must be familiar, electrical disturbances that interfere with signals produces noise and this noise ever present and limits the performance of the most of the systems. Measuring noise is very controversial almost everybody has a different method of quantifying noise and its effects.

definition:- noise is unwanted energy that interfere with the required signal. In receivers :- noise is disturbance in electric nature.

• TV receivers —–> it appears a snow (or) colored snow pictures.
• In Pulse communication systems —->noise produces unwanted pulses.

In receivers noise effects sensitivity and band width and it decreases sensitivity as well as band width.

Basically noise can be classified as Internal and External noise .

 External Noise Internal Noise when noise sources are external to the receiver . i.e, noise source is located outside of the receiver. It is difficult to treat quantitatively external noise. Noise is created with in the receiver itself.i.e, noise source is internal to the receiver. internal noise can be treated quantitatively and reduction is also possible by appropriate receiver design.

### External Noise:-

Atmospheric noise:-

If we try to listen to short waves on a receiver which is not well equipped to receive them, an astonishing variety of strange sounds will be heard, all tending to interfere with the program. most of these sounds are the result of spurious sources of disturbance, which represents atmospheric noise generally called as “static”.

• Atmospheric noise is caused by lightning discharge in thunderstorms and other natural electric disturbances occurring in the atmosphere.
• It originates in the form of amplitude modulated impulses , and are spread over most of the RF spectrum normally used for broadcasting.
• i.e, It consists of spurious radio signal with components distributed over a wide range of frequencies. Atmospheric noise propagates over the earth in the same way as ordinary Radio waves of the same frequencies.
• Static is more severe in the case of Radio than that of Tele-vision and it becomes less severe at frequencies above 30 MHz. Since higher frequencies are limited to line of sight propagation.
• This noise is created in VHF range and above.

Extraterrestrial noise:-

This noise is generated in the earth’s outer space (atmosphere)

Extraterrestrial noise is divided into

1. Solar noise .
2. Cosmic noise.

Solar noise:-

• The sun radiates so many things our way noise is noticeable among them.
• Under “quiet” conditions , there is a constant noise radiation from the sun simply because its a large body at high temperature ≈ 6000o C.
• ∴ The radiation consists of the frequencies which we use for communications and interferes with them.
• However, the disturbances in the sun is variable and undergoes cycles at the peak of which electrical disturbances erupt. These additional disturbances are several orders of magnitude greater than the noise generated during periods of the quiet sun. The solar cycle repeats these period of great electrical disturbances approximately every 11 years, further these 11 year cycle peaks reach even a higher maximum peak every 100 years.
• Thus the noise generated by sun changes periodically with the solar disturbances.

Cosmic noise:-

• stars are also suns and have high temperatures, they radiate RF noise in the same manner as our sun, this refers to noise coming from distant stars other than sun.
• The noise received from such stars is also called “black-body noise” and is distributed fairly uniformly over the entire sky.
• Space noise is observable in the range from about 8 MHz to about 1.43 GHz this is the strongest component of noise in the range(20-120) MHz.

• This noise is strongest in Industrial areas and the frequency of Man made noise spans between 1 to 600 MHz.
• Man made noise is found in urban, sub-urban and industrial areas. The intensity of the noise made by human easily outstrips that created by any other source,  internal or external to the receiver.
• under this, sources such as Automobile, Aircraft ignition, electric motors and switching equipment leakage from high voltage lines and a multitude of other heavy electric machines are all included.
• Fluorescent lights are another powerful source of such noise and therefore should not be used where sensitive receiver is installed.

### Internal Noise:-

This noise is created by any of the active (or) passive devices found in receivers. It is created by various components used in processing the received signal and is completely internal to the system. The effect of this noise is significant at the front end of the receiver.This appears as  thermal and shot noise  caused by resistors, inductors and capacitors.

Thermal Noise:-

This noise is also known as agitation noise, Jhonson noise / white noise. thermal noise is random in nature, this mainly occurs due to random(or) rapid motion of molecules, atoms and electrons of which resistor is made up of.

from the theory of dynamics the noise generated by a resistor is proportional to it’s absolute temperature and BW over which the noise is to be measured.

$P_{n}\propto&space;T\Delta&space;f$ where B= BW= $\Delta&space;f$

$P_{n}=&space;kT\Delta&space;f$

• where k – boltzmann’s constant =1.38X10-23 J/K.
• T- Absolute temperature in Kelvin,   K = 273+ oC.
• $\Delta&space;f$ = BW of interest.
• $P_{n}$ is the maximum noise power output of a resistor.

In an ordinary resistor at the standard temp of 17oC is not connected to any voltage source and if we are measuring voltage using a DC volt meter to measure voltage across it shows a zero. Actually a resistor itself is a noise generator, if we use a very sensitive electronic volt meter it shows a very large voltage across R.

This noise voltage is caused by the random movement of electrons with in the resistor, which constitutes a current. The rate of arrival of electrons at either end of the resistor therefore varies randomly, and so does the potential difference exists between the two ends.

from the circuit diagram,

$P_{n}=\frac{V_{n}^{2}}{4R_{L}}$  —- equation(1).

The maximum power  is delivered to  load when R =RL.

$V_{n}=&space;i(R&space;+&space;R_{L})$

$V_{n}&space;=&space;i&space;2R$

$i&space;=&space;\frac{V_{n}}{2R}$

load voltage $V&space;=&space;i&space;R_{L}$

$V&space;=&space;\frac{V_{n}}{2&space;R}$

$V&space;=&space;\frac{V_{n}}{2}$ volts

Vn -source noise voltage.

V- ouput voltage measured across RL.

from equation (1) $V_{n}^{2}=P_{n}4R_{L}$

$V_{n}^{2}=4k&space;T\Delta&space;fR_{L}$

$V_{n}=\sqrt{4k&space;T\Delta&space;fR_{L}}$

Vn is known as RMS noise voltage asross a resistor.

Shot Noise:-

This occurs due to shot effect, it occurs in all active and amplifying devices (diodes/transistors).

It is caused by random variations in the arrival of electrons (or holes) at the output electrode of an amplifying device and appears as a randomly varying noise current super imposed on the output.It sounds like a shower of a lead shot were falling on a metal plate. Hence named it as shot noise.

• In electronic tubes shot noise is caused because of random emission of electrons from cathode.
• In semi-conductors shot noise occurs due to random diffusion of minority carriers .
• $i(t)&space;=&space;I_{o}&space;+&space;i_{n}(t)$

Total current = Mean DC constant current + shot noise current.

Shot noise is given by $i_{n}&space;=&space;\sqrt{2qi_{p}\Delta&space;f}$

• in -RMS shot-noise current.
• q-charge of an electron 1.6 X10-19 C.
• ip– direct diode current.
• $\Delta&space;f$– BW of the system.

This formula for shot noise is valid for vaccum tube diode under so called temp-limited conditions.

In all other cases we use the concept of equivalent noise resistance intead of shot noise formula.

Transit-time noise (or) High-frequency noise:-

It is generally observed in semi conductor devices when the transmit time of charge carriers crossing a junction is comparable with the time-period of the signal, some charge carriers diffuse back to the source (or) emitter.

this gives rise to input admittance Y

conductance G = 1/Y , this G increases with frequency which causes noise . This is also called as high frequency noise.

(No Ratings Yet)

## Gauss Divergence Theorem (or) Application of Gauss’s law to a differential volume element

### Gauss Divergence theorem:-

A vector field changes from point to point throughout space, this change can be identified by two things Divergence and curl,the second is curl, which will be examined when magnetic fields are discussed, so Divergence , which is a scalar function and is somewhat similar to derivative of a function will be discussed in Gauss-Divergence theorem.

Generally, when the divergence ≠ 0 ⇒ that there exists sources or sinks in a particular region.

If divergence is positive ⇒ a source                                                                                    If divergence is negative ⇒ a sink

In static electric fields, there is a correspondence between positive divergence, sources and positive electric charge ‘Q’, since electric flux ψ by definition originates on a positive charge, thus a region which consists of positive charges contains the source of flux ψ.

∴ The divergence of electric flux density $\overrightarrow{D}$ will be positive in that region.Divergence of a vector field $\overrightarrow{A}/\overrightarrow{D}$ at a point ‘P’ is defined by

$div\overrightarrow{A}=&space;\lim_{\Delta&space;v->0}\frac{\oint&space;\overrightarrow{A}.\overrightarrow{ds}}{\Delta&space;v}$            or      $div\overrightarrow{D}=&space;\lim_{\Delta&space;v->0}\frac{\oint&space;\overrightarrow{D}.\overrightarrow{ds}}{\Delta&space;v}$

here, the integration is over the surface of an infinitesimal volume $\Delta&space;v$ that shrinks at point P

Gauss-Divergence theorem:- (statement)

for a continuously differentiable vector field the net outward flux from a closed surface equals the volume integral of the divergence throughout the region bounded by that surface.

i.e, $\oint_{s}\overrightarrow{D}.\overrightarrow{ds}=\int_{v}(\overrightarrow{\bigtriangledown}.&space;\overrightarrow{D})dv$   where $\overrightarrow{D}$ be the vector field.

It relates volume integral with surface integral.

Proof of Divergence theorem:-

Now consider a slightly different type of situation to which Gauss’s law is to be applied, that is on the surface $\overrightarrow{D}$ is not constant (or) zero and this surface doesn’t have any symmetry at all. For such type of surface the change in $\overrightarrow{D}$ can be represented by Taylor series expansion by assuming $\overrightarrow{D}$ as constant.

i.e, $f(x+\frac{\Delta&space;x}{2})=f(x)+\frac{\frac{\Delta&space;x}{2}}{1!}{f}'(x)+\frac{\frac{\Delta&space;x}{2}^{2}}{2!}{f}''(x)+.....$ over the surface.

This change in $\overrightarrow{D}$ can be expressed by only first two terms

i.e, $f(x+\frac{\Delta&space;x}{2})=f(x)+\frac{\frac{\Delta&space;x}{2}}{1!}{f}'(x)$

for the differential volume element $\large&space;\Delta&space;x&space;=dx,\Delta&space;y=dy&space;and&space;\Delta&space;z=dz$

Let us consider a point ‘P’ which is located at the center of the volume element, which is a differential volume with an assumption that $\overrightarrow{D}$ is almost constant over this small volume dv and at P $\overrightarrow{D}$ is given as

$\overrightarrow{D}=D_{xo}\overrightarrow{a}_{x}&space;+D_{yo}\overrightarrow{a}_{y}+D_{zo}\overrightarrow{a}_{z}$

Now, from Gauss’s law $\oint_{s}\overrightarrow{D}.\overrightarrow{ds}&space;=Q_{enclosed}$ , the surface integral is divided into ‘6’ integrals as $\oint_{s}\overrightarrow{D}.\overrightarrow{ds}&space;=\int_{front}+\int_{back}+\int_{left}+\int_{right}+\int_{top}+\int_{bottom}$

$\int_{front}&space;=&space;\overrightarrow{D}_{front}.\overrightarrow{d&space;s}_{front}$

$\large&space;\overrightarrow{D}_{front}&space;=&space;\overrightarrow{D}_{xfront}\overrightarrow{a}_{x}&space;+\overrightarrow{D}_{yfront}\overrightarrow{a}_{y}&space;+\overrightarrow{D}_{zfront}\overrightarrow{a}_{z}$

$\large&space;\overrightarrow{d&space;s}_{front}&space;=&space;\Delta&space;y\Delta&space;z&space;\overrightarrow{a}_{x}$

$\large&space;\overrightarrow{D}_{front}.\overrightarrow{ds}_{front}=&space;(D_{xfront}\overrightarrow{a}_{x}+D_{yfront}\overrightarrow{a}_{y}+D_{zfront}\overrightarrow{a}_{z}).dydz\overrightarrow{a}_{x}$

$\large&space;=&space;D_{xfront}dydz$

now the component $\large&space;D_{xfront}=D_{xo}+\frac{dx}{2}\frac{\partial&space;D_{x}}{\partial&space;x}$

$\large&space;\therefore&space;\int_{front}=(D_{xo}+\frac{dx}{2}\frac{\partial&space;D_{x}}{\partial&space;x})dydz$

simillarly,  $\large&space;\therefore&space;\int_{back}=(D_{xo}-\frac{dx}{2}\frac{\partial&space;D_{x}}{\partial&space;x})(-dydz)$

$\large&space;\int_{front}+\int_{back}&space;=&space;dxdydz&space;\frac{\partial&space;D&space;_{x}}{\partial&space;x}$

then the remaining two integral values $\large&space;\int_{left}+\int_{right}&space;=&space;dxdydz&space;\frac{\partial&space;D&space;_{y}}{\partial&space;y}$

$\large&space;\int_{top}+\int_{bottom}&space;=&space;dxdydz&space;\frac{\partial&space;D&space;_{z}}{\partial&space;z}$

$\large&space;\oint_{s}\overrightarrow{D}.\overrightarrow{ds}=dxdydz&space;\frac{\partial&space;D&space;_{x}}{\partial&space;x}+dxdydz&space;\frac{\partial&space;D&space;_{y}}{\partial&space;y}+&space;dxdydz&space;\frac{\partial&space;D&space;_{z}}{\partial&space;z}$ —————>Equation(1)

According to the definition of div $\large&space;\overrightarrow{D}$

$\large&space;\lim_{\Delta&space;v->0}\frac{\oint&space;\overrightarrow{D}.\overrightarrow{ds}}{\Delta&space;v}=div\overrightarrow{D}=\overrightarrow{\bigtriangledown&space;}.\overrightarrow{D}$

from Equation(1) $\large&space;\oint_{s}\overrightarrow{D}.\overrightarrow{ds}=dv(&space;\frac{\partial&space;D&space;_{x}}{\partial&space;x}+&space;\frac{\partial&space;D&space;_{y}}{\partial&space;y}+&space;\frac{\partial&space;D&space;_{z}}{\partial&space;z})$

by dividing the above equation with dv=Δv

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

apply the limit on both sides

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

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

hence proved.

The divergence of the  flux density $\large&space;\overrightarrow{D}$ (vector ) is the flow of flux from a small closed surface per unit volume as the volume shrinks to zero $\large&space;(\lim_{dv->0})$.

(No Ratings Yet)

## Electricfield 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, $\overrightarrow{E}&space;=&space;\frac{\overrightarrow{F}}{q}&space;Newtons/Coulomb$ —-> Equation (1)

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

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

$F&space;=&space;E&space;q$

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

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

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

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

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

where as $\overrightarrow{E}=&space;\frac{Q}{4\pi&space;\epsilon&space;r^{2}}\widehat{a_{R}}$ 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:-

1. charge is conserved , i.e, charge is neither be created nor destroyed.
2. 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(ρL):-

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 ρL is constant through out the line, otherwise it is Non-uniform type.

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

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

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

Surface charge distribution(ρs):-

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 $\rho&space;_{s}=&space;\lim_{\Delta&space;s->0}\frac{\Delta&space;Q}{\Delta&space;s}$

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

$Q=\int_{s}\rho&space;_{s}ds$ Coulomb

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

Volume charge distribution (ρv):-

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

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

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

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

(No Ratings Yet)

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

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 Ds 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 $\overrightarrow{ds_{n}}=&space;ds.\overrightarrow{a_{n}}$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 $D&space;=&space;\frac{d\psi&space;}{ds}$

$d\psi&space;=&space;D.ds$—>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 $D_{s}cos\theta&space;=&space;D_{s&space;normal}$

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

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

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

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

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

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

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

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

Closed Gaussian surface:-

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

1. $\overrightarrow{D}$ is every where either normal (or) tangential to the closed surface, that is θ = π/2 or θ = 0o / π so that $\overrightarrow{D}.\overrightarrow{ds}$ is maximum or zero.
2. $\overrightarrow{D}$ 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.

(No Ratings Yet)

## Frequency Shift Keying(FSK)

### Frequency Shift keying:-

In a Binary FSK system , Symbols ‘1’ and ‘0’ are distinguished from each other by transmitting one of two sinusoidal waves that differ in frequency by a fixed amount. (or) the frequency of the carrier signal is shifted to two frequencies fc1 (or)fH and fc2/fL for symbols ‘1’ and ‘0’ transmission.

The equation for FSK signal  is

$S_{FSK}(t)=\sqrt{\frac{2E_{b}}{T_{b}}}cos(2\pi&space;f_{ci})(t),0\leq&space;t\leq&space;T_{b}$

$=&space;0&space;elsewhere$

and i= 1,2.

$S_{FSK}(t)=\sqrt{\frac{2E_{b}}{T_{b}}}cos(2\pi&space;f_{H})(t),&space;for&space;symbol'1'$

$S_{FSK}(t)=\sqrt{\frac{2E_{b}}{T_{b}}}cos(2\pi&space;f_{L})(t),&space;for&space;symbol'0'$

where $f_{c1}=f_{H}$ is generally a high-frequency, $f_{c2}=f_{L}$ is a low-frequency. Vice-versa is also true.

$S_{FSK}(t)=\sqrt{2P_{s}}cos(2\pi&space;\omega&space;_{H})(t)---->&space;binary'1'$

$S_{FSK}(t)=\sqrt{2P_{s}}cos(2\pi&space;\omega&space;_{L})(t)---->&space;binary'0'$

where                   $\omega&space;_{H}=&space;\omega&space;_{c}&space;+&space;\Omega$                                          and                $\omega&space;_{L}=&space;\omega&space;_{c}&space;-&space;\Omega$

$2\pi&space;f_{H}&space;=&space;2\pi&space;f_{c}&space;+&space;\Omega$                                                      $2\pi&space;f_{L}&space;=&space;2\pi&space;f_{c}&space;-&space;\Omega$

$f_{H}&space;=&space;f_{c}+\frac{\Omega&space;}{2\pi&space;}$                                                              $f_{L}&space;=&space;f_{c}-\frac{\Omega&space;}{2\pi&space;}$

### FSK/BFSK  Generator :-

The input to the FSK generator is a binary sequence 1  0  1  0 …..etc.

• FSK generator uses two product modulators upper modulator and lower modulator with carriers     $\phi&space;_{1}(t)=\sqrt{\frac{2}{T_{b}}}cos&space;2\pi&space;f_{H}t$  and  $\phi&space;_{2}(t)=\sqrt{\frac{2}{T_{b}}}cos&space;2\pi&space;f_{L}t$
• A level shifter is there in which the output of level-shifter is $\sqrt{P_{s}T_{b}}$ volts when the input is a binary ‘1’ and ‘0’ volts for the input ‘0’.
 Input to the level shifter output of the level shifter ‘1’ $\sqrt{P_{s}T_{b}}$ ‘0’ 0
• The working of FSK generator is as follows, when input binary sequence ‘1’ on the upper modulator 1 has been shifted to a voltage $\sqrt{P_{s}T_{b}}$ so that the output of product modulator 1 is

$S_{1}(t)&space;=&space;\sqrt{P_{s}T_{b}}&space;X&space;\phi&space;_{1}(t)$

$=\sqrt{P_{s}T_{b}}&space;X\sqrt{\frac{2}{T_{b}}}&space;cos&space;2\pi&space;f_{H}t$

$=&space;\sqrt{2P_{s}}&space;cos&space;2\pi&space;f_{H}t$

and on the lower modulator input is ‘1’ is passed through an inverter and the output of inverter is ‘0’ then the output of level shifter will not change it remains at ‘0’ volts itself, then the  output of the second product modulator is  $S_{2}(t)&space;=&space;0&space;X&space;\phi&space;_{2}(t)$

$=&space;0$

∴ the over all output      $S_{FSK}(t)&space;=&space;S_{1}(t)&space;+&space;S_{2}(t)$

$=S_{1}(t)$   since $S_{2}(t)=0$

$=&space;\sqrt{2P_{s}}&space;cos&space;2\pi&space;f_{H}t$.

similarly, when the input sequence is a binary ‘0’ , intermediate outputs at the product modulators $S_{1}(t)=0$ and $s_{2}(t)=&space;\sqrt{2P_{s}}&space;cos&space;2\pi&space;f_{L}t$

∴ the over all output $S_{FSK}(t)=&space;\sqrt{2P_{s}}&space;cos&space;2\pi&space;f_{L}t$ which is the required output for binary ‘0’ transmission.

## EMT-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

$F&space;\propto&space;Q_{1}Q_{2}$

$\propto&space;\frac{1}{R^{2}}$

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, $F&space;=\frac{k&space;Q_{1}Q_{2}}{R^{2}}$    where k  is constant of proportionality

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

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

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

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

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

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

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

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

simillarly Force acting on 2 by 1 is given by

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

$\therefore&space;\overrightarrow{F_{12}}=-\overrightarrow{F_{21}}$

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

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

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

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

force acting on Qo due to Q1 is

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

force acting on Qo due to Q2 is

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

force acting on Qo due to Q3 is

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

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 Qo by the remaining (n-1) charges on it is given by

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

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## Digital Communications Unit-4 FAQ

Digital Communications

Unit-4 (FAQ)

1. What is meant by Probability of Error Pe?
2. what is meant by Bit Error Rate BER? how BER and Eb/No ( approximately known as SNR) related to each other?
3. what is matched filter? give the impulse response of it.
4. Differentiate the following.                                                                                                     i. Base band transmission( or Communication) systems and Pass band Transmission systems.                        ii. coherent and Non-coherent systems.
5. what is the need for MSK modulation?
6. Draw the signal space diagrams of  i. ASK  ii. PSK  iii. FSK    iv. QPSK   v. DPSK
7. what is meant by gaussian error probability?
8. write a short note on M-ary QPSK?
9. Define error function, complementary error function and Q-function. Give the relation between Q- function and  complementary error function.
10. Compare digital modulation schemes in terms of Bandwidth and Pe.
11. Give the meaning of QPSK. how it is different from PSK?
12. write a short note about QPSK.
13. what is meant by singal space or signal constellation diagram . give its significance
14. draw the wave forms of ASK ,PSK and FSK when data input is 1100101.
15. problems on the impulse response of matched filter.
16. advantages of pass band over base band transmission?
17. applications of ASK, PSK and FSK ,QPSK and DPSK.

1. compare different digital modulation schemes.
2. Draw the block diagram of Correlation receiver. Explain the working of it in detail.
3. Draw the block diagram of integrate and dump receiver (base band Receiver). Explain the working of it in detail.
4. Explain the working of DPSK system with an example.
5. Derive the expression for Pe of the following     i. QPSK      ii. PSK      iii. matched filter    iv.  DPSK     v. Optimum receiver     vi. Integrate and dump receiver.
6. write a short note on synchronization methods.
7. derive the transfer function of optimum receiver.
8. derive the impulse response of matched filter.
9. draw block diagrams of modulator and de-modulator  for the following  coherent modulation schemes and explain its working in detail       i. FSK     ii. QPSK     iii. PSK        iv.ASK
10. draw block diagrams of modulator and de-modulator  for the following  non-coherent modulation schemes and explain its working in detail                i. FSK     ii.  non coherent PSK or DPSK       iii.ASK
12. Derivations of Pe of   i.Non- coherent ASK/Non- coherent BASK        ii  non-coherent PSK /Non- coherent BPSK         iii.coherent FSK/ Non- coherentBFSK.
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Q1. what is hamming distance? mention it’s significance.

Ans:Hamming weight:- number of non zero elements in a code word.

Hamming distance:- It is defined as the number of bits in which two code vectors differ

i.e, X = [1 0 1 0 1] and Y=[1 1 1 0 1] then the hamming distance between code vectors X and Y is ‘ 1 ‘.

i.e, X1 = [1 0 1 0 ] and X2 =[1 1 1 1] then the hamming distance between code vectors X1 and X2  is ‘ 2’.

Significance:- It is important because error detection and correction is possible if         $t\leq&space;\frac{1}{2}(d_{min}-1)$

where t is the hamming weight and dmin is the minimum hamming distance between two code words.

Q2. what is Bit Error rate?

Ans: In telecommunication transmission, the bit error rate (BER) is the percentage of bits that have errors relative to the total number of bits received in a transmission, usually expressed as ten to a negative power.

Bit Error Rate (BER) = (number of bits transmitted in error/ total number of bits transmitted).

For example, a transmission might have a BER of 10 to the minus 4, meaning that, out of 1,0,000 bits transmitted, one bit was in error. The BER is an indication of how often a packet or other data unit has to be re transmitted because of an error.

Q3. What is Probability of error in Digital Communications?

Ans:In digital Communications data to be transmitted is digital either a ‘0’ or ‘1’, when a digital signal is being transmitted during the transmission the signal being corrupted by some sort of random process( noise), these processes can be quantified by their Probability Density Function (pdf), such as Gaussian, Uniform, Rayleigh etc. Depending on the medium through which a signal travels, it is attacked by these random process, akin to adding a random signal to the one transmitted, Now the bit that was of say amplitude V(‘1’) may be changed to either V+(‘1’) or V-(‘0’). This may or may not cause an error at the receiver. This depends on the way the bit is mapped to symbols, a process called de-modulation.

Now probability of error is defined as the probability of receiving the bit incorrect at the receiver.

i.e, when a ‘0’ is transmitted at the transmitter if that bit ‘0’ is received at the receiver there is no error occurred during the transmission , instead of receiving a ‘0’ if a ‘1’ has been received then there is an error occurred in the transmission. Now, Probability of error gives the extent to which a ‘1’ has to be received in place of a ‘0’ and vice-versa. Pe gives the rate of making a bad decision.

By Baye’s rule of probability, Pe states the overall rate of making a bad decision

$P_{e}=&space;P(e/0)P(0)&space;+&space;P(e/1)P(1)$

$P_{e}&space;-->$ Overall Probability of error

$P(e/0)=P(1/0)-->$ Probability of receiving ‘1’ when a ‘0’ has been transmitted also known as Conditional probability.

$P(e/1)=P(0/1)-->$Probability of receiving ‘0’ when a ‘1’ has been transmitted also known as Conditional probability.

$P(0)-->$ Probability of ‘0’ and $P(1)-->$probability of ‘1’

$P_{e}&space;=&space;P(1/0)P(0)&space;+&space;P(0/1)P(1)$

If equal number of 0’s and 1’s are sent (0’s and 1’s are equi probable)

then $P(0)=P(1)=\frac{1}{2}$  , which gives $P_{e}&space;=&space;\frac{1}{2}P(1/0)+\frac{1}{2}P(0/1)$ , To evaluate this expression , an assumption is needed that is  $P(1/0)=P(0/1)$ provided that the threshold is exactly in the middle  then

$P_{e}=&space;P(1/0)=P(0/1)$

Q3. What is the need for MSK(Minimum Shift Keying) in Digital Communications?

Ans: Need for MSK:-

Linear modulation schemes without memory like QPSK, OQPSK, DPSK and
FSK exhibit phase discontinuity in the modulated waveform. These phase transitions
cause problems for band limited and power-efficient transmission especially in an
interference limited environment. The sharp phase changes in the modulated signal cause relatively prominent side-lobe levels of the signal spectrum compared to the main lobe. In a cellular communication system, where frequency reuse is done extensively, these side lobe levels should be as small as possible. Further in a power-limited environment, a nonlinear power amplifier along with a band pass filter in the transmitter front-end results in phase distortion for the modulated signal waveform with sharp phase transitions. The abrupt phase transitions generate frequency components that have significant amplitudes. Thus the resultant power in the side-lobes causes co-channel and inter-channel interference.

Consequently, in a practical situation, it may be necessary to use either a linear power amplifier or a non-linear amplifier using extensive distortion compensation or selective pre-distortion to suppress out-of-band frequency radiation. However, high power amplifiers may have to be operated in the non-linear region in order to improve the transmission power. Continuous phase modulation schemes are preferred to counter these problems.

Continuous Phase Frequency Shift Keying (CPFSK) refers to a family of
continuous phase modulation schemes that allow use of highly power-efficient non-linear power amplifiers. Minimum Shift Keying (MSK) modulation is a special subclass of CPFSK modulation and MSK modulation is free from many of the problems mentioned  above.

Minimum Shift Keying is a form of Continuous Phase -Frequency Shift Keying that is used in a number of  applications, A version of MSK modulation known as GMSK – Gaussian filtered Minimum Phase Shift Keying is used for a number of radio Communication applications including GSM cellular Tele Communication systems.

In MSK output wave form is continuous in phase here there are no phase discontinuities because the frequency changes occur at the carrier zero crossing points. hence there are no abrupt changes in Amplitude. The side lobes of MSK are very small hence Band pass filtering is not required to avoid inter channel interference.

MSK Ouput looks like no phase distortion existing.

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## Inter symbol Interference (ISI) and Eye pattern

### Inter Symbol Interference (ISI):-

In Tele-communications, Inter symbol Interference (ISI) is a form of distortion of a signal in which one symbol interferes with subsequent symbols. This is an unwanted phenomenon as the previous symbols have similar effect as noise. Thus making the communication less reliable.

i.e, the spreading of the pulse beyond its allotted time interval causes it to interfere with neighboring pulses.

ISI is usually caused by the two factors

•  Multi path propagation.
• The inherent Non-linear frequency response of the channel.

These two factors  causes successive symbols to “blur” together.i.e, the channel is always band-limited. A band limited channel disperses (or) spreads a pulse wave form passing through it.

when the channel Band width is much larger than the Pulse width the spreading of the pulse will be slight.

when the channel band width is close to the signal band width, the spreading will exceed a symbol duration and causes signal pulses to over lap. This over lapping is called  as ISI.

ISI causes system degradation, it is particularly insidious form of interference because raising the signal power to overcome the interference will not always improve the error performance. when considering the filtering aspects of  a typical digital communication system there are various filter throughout the system in the transmitter, Receiver and in the channel.

At the transmitter :- the information symbols x1,x2,x3 are characterized as impulses (or) voltage levels, are filtered to comply with some bandwidth constraint.

Channel:-

if the system is a Base band system—–> channel has distributed reactances(acts as  a filter) that distort the pulses .

if the system is Pass band system—–> wireless systems are characterized by fading channels, that behave like undesirable filters causes distortion in the pulses.

At the Receiver:- when the receiving filter is configured to compensate for the distortion caused by both the transmitter and channel it is often referred to as equalizing filter.

The equivalent model for the system, lumping all the filtering effects into one overall equivalent system transfer function

$H(f)&space;=&space;H_{t}(f)*H_{c}(f)*H_{r}(f)$

is given in the figure.

now, due to filtering effects, the received pulses can overlap one another as shown the tail of a pulse can “smear” into adjacent symbol intervals, there by interfering with the detection process and degrading the error performance, such interference is termed (ISI). Even in the absence of noise , the effects of filtering and channel- induced distortion leads to ISI.

Nyquist investigated the problem of specifying a received pulse shape so that no ISI occurs at the detector. The minimum band width required to detect Rs symbols/Sec without ISI is BWmin = Rs /2 Hz. This occurs when H(f) is rectangular.

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## QPSK equation, wave forms and Signal space diagram

### QPSK equation:-

The meaning of QPSK is  that the carrier signal takes on different phases Π/4, 3Π/4, 5Π/4 and 7Π/4 based on incoming di-bit combination  or symbol.

$\large&space;S_{QPSK}(t)=&space;\sqrt{\frac{2E_{s}}{T_{s}}}cos(2\pi&space;f_{c}t&space;+(2i-1)\frac{\pi}{4}),&space;0\leq&space;t\leq&space;T_{s}$

= 0, elsewhere, where  i =  1,2,3,4.

Eb and Tb are the bit energy and bit-interval , Es and Ts are the energy per symbol  and symbol duration. Ts = 2 Tb

The carrier frequency fc = nc /Ts. where nc is a fixed integer.

each possible value of phase corresponds to a unique di-bit. then the foregoing phase values to represent the gray encoded set of di-bits 11,01,10 and 00, where only a single bit is changed from one di-bit to the next.

QPSK equation can be represented in another format as follows

$\large&space;S_{QPSK}(t)&space;=&space;\sqrt{\frac{2E_{s}}{T_{s}}}cos&space;(2\pi&space;f_{c}t+(2i+1)\frac{\pi&space;}{4}&space;),&space;0\leq&space;t\leq&space;T_{s}$

= 0, elsewhere   ,where i=0,1,2,3.

The above two equations are same, there is a change in i values. alternately the equation can be represented as follows.

$S_{QPSK}(t)=&space;\sqrt{\frac{2E_{s}}{T_{s}}}cos&space;(2i-1)\frac{\pi&space;}{4}cos2\pi&space;f_{c}t&space;-&space;\sqrt{\frac{2E_{s}}{T_{s}}}sin&space;(2i-1)\frac{\pi&space;}{4}sin2\pi&space;f_{c}t$where i= 1,2,3,4.

The above equation can be expanded cos(A+B). There are two orthogonal functions Φ1(t) and Φ2(t) where

$\Phi&space;_{1}(t)=\sqrt{\frac{2}{T_{s}}}cos&space;2\pi&space;f_{c}t,&space;0\leq&space;t\leq&space;T_{s},&space;\Phi&space;_{2}(t)=\sqrt{\frac{2}{T_{s}}}sin&space;2\pi&space;f_{c}t,&space;0\leq&space;t\leq&space;T_{s}$

$S_{QPSK}(t)=\sqrt{E_{s}}cos(2i-1)\frac{\pi&space;}{4}&space;*&space;\Phi&space;_{1}(t)&space;-&space;\sqrt{E_{s}}sin(2i-1)\frac{\pi&space;}{4}&space;*&space;\Phi&space;_{2}(t)$

Let    $b_{o}(t)=&space;\sqrt{E_{s}}cos(2i-1)\frac{\pi&space;}{4}$     and   $b_{e}(t)=&space;-\sqrt{E_{s}}sin(2i-1)\frac{\pi&space;}{4}$

then the resultant equation is:     $S_{QPSK}(t)=&space;b_{o}(t)&space;*&space;\Phi&space;_{1}(t)&space;+&space;b_{e}(t)&space;*&space;\Phi&space;_{2}(t)$.

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

### Meaning of QPSK:-

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

### Generation of QPSK/ QPSK transmitter:-

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

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

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

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

x2>0—-> a symbol ‘1’ is decided.

x2<0—-> a symbol ‘0’ is decided.

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

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## Differences between Base band data Transmission and Pass Band Data Transmission

There are basically two types of transmission systems depending on the range of frequencies over which transmission takes place.
[table id=2 /]

### Requirements of Pass band Transmission:-

1. Maximum data transmission rate.
2. Minimum transmitted power.
3. Minimum probability of error Pe.
4. Minimum Circuit complexity.
5. Maximum Channel Bandwidth.

### Advantages of Pass band Transmission over Base band Transmission:-

1. Long distance transmission is possible.
2. Analog channels can be used for transmission.
3. Multiplexing techniques can be used for Band width Conservation.
4. Problems such as inter symbol interference and Cross talk are absent.
5. Pass band transmission can take place over wireless channels also. therefore Large variety of modulation techniques are available.

### Drawbacks of pass band Modulation:-

Modulation and demodulation equipment imparts transmitting & receiving antennas which increase the system complexity. i.e, pass band systems are complex systems.

[table id=3 /]

### Advantages of Digital Communication systems over analog Communication systems:-

 Analog Communication Systems(ACS) Digital Communication Systems(DCS) 1. Generation of analog signals involves a complex mechanism over Digital signals 1. Generation of digital signals is much easier when compared to Analog signals. 2. Regenerative repeaters are not used in ACS 2. DCS has a flexibility in terms of usage of regenerative repeaters at regular intervals along Transmission system. 3. Analog circuits are more subjected to distortion and noise 3. Digital circuits are less subjected to distortion and interference than are analog circuits, because of two state operation of digital circuits( i.e, fully on /fully off). 4. Error detection and error correction methods are not available in ACS. 4. Error detection and error correction methods makes DCS preferable over ACS, with this one can expect low error rates and high signal fidelity in DCS. 5. Reliability is less for analog circuits. cost is high for analog hardware. 5. Digital circuits are more reliable and can be produced at lower cost. 6. Multiplexing of analog signals using Frequency Division Multiplexing is complex. 6.Combining of digital signals using Time Division Multiplexing  is simpler. 7. Encryption and privacy are not possible for analog  modulation techniques 7. Encryption and privacy can be possible for digital techniques. 8. analog systems are not signal processing intensive. 8. Digital systems are very signal processing intensive. 9. AC is non-adaptive to DSP,DIP etc. 9. DC adaptive to other advanced branches of data processing such as DSP, DIP, data compression etc. 10. wide dynamic range is not possible with analog signals. 10. wide dynamic range is possible since the data is converted to the digital form. 11. low speed data transmission is possible with analog communication.      (No Ratings Yet) Loading... 11. High speed and reliable data transmission is possible with digital communication.

## Differential Phase Shift Keying (DPSK)

### Differential Phase Shift Keying:-

It is Non Coherent version of Phase Shift Keying. It eliminates the need for a Coherent reference signal at the Receiver, by combining the following two basic operations at the Receiver.
1. Differential encoding
2. PSK
Hence given the name DPSK.

### DPSK Generator:-

The Figure shows the Block Diagram of DPSK Transmitter. It consists of a logic Network (we can use EX-OR also) EX-NOR gate and a one bit-delay element so as to convert the Raw Binary sequence {bk} into a differentially encoded sequence {dk}.Now this sequence is amplitude level encoded and then used to modulate a carrier of frequency fc(similar to PSK Generator), there by producing the desired DPSK signal.

Now the generation of DPSK signal is explained with an example sequence

Input sequence : {bk} : 1 0 0 1 0 0 1 1
One bit delayed bit {dk-1} : 1
Differentially encoded sequence {dk}: 1 1 0 1 1 0 1 1 1
Transmitted Phase : 0 0 π 0 0 π 0 0 0
:0o 0o 180o 0o 0o 180o 0o 0o 0o

Now the corresponding phases has been transmitted from the Transmitter in the form of a PSK signal, Now this transmitted DPSK signal will be received at the Receiver

Now At the Receiving end
DPSK input to the Receiver : 0o 0o 180o 0o 0o 180o 0o 0o 0o

now the phase at the current instance is compared with its next phase , if both are same the received bit is a ‘1’ otherwise a ‘0’.
Therefore the Received sequence is : 1 0 0 1 0 0 1 1.

From the DPSK generator {dk} = {bk} Xnor {dk}.
The Transmitted phase has been chosen based on PSK scheme
for binary ‘1’ ——–> 0o
for binary ‘0’———> 180o