## Matched Filter, impulse response h(t)

Matched Filter can be considered as a special case of Optimum Filter. Optimum Filter can be treated as Matched Filter when the noise at the input of the receiver is White Gaussian Noise.

Transfer Function of Matched Filter:-

Transfer Function of Optimum filter is $H(f)=\frac{k&space;X^{*}(f)e^{-j2\pi&space;fT}}{S_{ni}(f)}$

if input noise is white noise , its Power spectral density (Psd) is $S_{ni}(f)=\frac{N_{o}}{2}$.

then H(f) becomes $H(f)=\frac{k&space;X^{*}(f)e^{-j2\pi&space;fT}}{\frac{N_{o}}{2}}$

$H(f)=\frac{2k}{N_{o}}X^{*}(f)e^{-j2\pi&space;fT}-----Equation(I)$

From the properties of Fourier Transforms , by Conjugate Symmetry property  $X^{*}(f)&space;=&space;X(-f)$

Equation (I) becomes

$H(f)=\frac{2k}{N_{o}}X(-f)e^{-j2\pi&space;fT}------Equation(II)$

From Time-shifting property of Fourier Transforms

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

From Time-Reversal Property  $x(-t)\leftrightarrow&space;X(-f)$

By Shifting the signal $x(-t)$ by T Seconds in positive direction(time) ,the Fourier Transform is given by  $x(T-t)\leftrightarrow&space;X(-f)e^{-j2\pi&space;ft}$

Now the inverse Fourier Transform of the signal from the Equation(II) is

$F^{-1}[H(f)]=F^{-1}[\frac{2k}{N_{o}}X(-f)e^{-j2\pi&space;fT}]$

$h(t)=\frac{2k}{N_{o}}x(T-t)$

Let the constant $\frac{2k}{N_{o}}$ is set to 1, then the impulse response of Matched Filter will become $h(t)&space;=&space;x(T-t)$.

(2 votes, average: 5.00 out of 5)

## Mutual Information I(X ; Y) Properties

Property 1:- Mutual Information is Non-Negative

Mutual Information is given by equation $I(X&space;;&space;Y)&space;=\sum_{i=1}^{m}\sum_{j=1}^{n}P(x_{i},&space;y_{j})\log&space;_{2}&space;\frac{P(\frac{x_{i}}{y_{j}})}{P(x_{i})}---------Equation(I)$

we know that $P(\frac{x_{i}}{y_{j}})=\frac{P(x_{i},&space;y_{j})}{P(y_{j})}-------Equation(II)$

Substitute Equation (II) in Equation (I)

$I(X&space;;&space;Y)&space;=\sum_{i=1}^{m}\sum_{j=1}^{n}P(x_{i},&space;y_{j})\log&space;_{2}\frac{P(x_{i},&space;y_{j})}{P(x_{i})P(y_{j})}$

The above Equation can be written as

$I(X&space;;&space;Y)&space;=-\sum_{i=1}^{m}\sum_{j=1}^{n}P(x_{i},&space;y_{j})\log&space;_{2}\frac{P(x_{i})P(y_{j})}{P(x_{i},&space;y_{j})}$

$-I(X&space;;&space;Y)&space;=\sum_{i=1}^{m}\sum_{j=1}^{n}P(x_{i},&space;y_{j})\log&space;_{2}\frac{P(x_{i})P(y_{j})}{P(x_{i},&space;y_{j})}------Equation(III)$

we knew that $\sum_{k=1}^{m}&space;p_{k}\log&space;_{2}(\frac{q_{k}}{p_{k}})\leq&space;0---Equation(IV)$

This result can be applied to Mutual Information $I(X&space;;&space;Y)$ , If $p_{k}&space;=&space;P(x_{i},&space;y_{j})$ and $q_{k}$ be $P(x_{i})&space;P(&space;y_{j})$, Both $p_{k}$ and $q_{k}$ are two probability distributions on same alphabet , then Equation (III) becomes

$-I(X&space;;&space;Y)&space;\leq&space;0$

i.e, $I(X&space;;&space;Y)&space;\geq&space;0$  , Which implies that Mutual Information is always Non-negative (Positive).

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## Delta modulation and Demodulation

DM is done on an over sampled message signal in its basic form, DM provides a stair case approximated signal to the over sampled version of message signal.

i.e, Delta Modulation (DM) is a Modulation scheme in which an incoming  message signal is over sampled (i.e, at a rate much higher than the Nyquist rate $f_{s}>&space;2f_{m}$) to purposely increase the correlation between adjacent samples of the signal. Over sampling is done to permit the use of a sample Quantizing strategy for constructing the encoded signal.

Signaling rate and Transmission Band Width are quite large in PCM. DM is used to overcome these problems in PCM .

DM transmits one bit per sample.

The process of approximation in Delta Modulation is as follows:-

The difference between the input ($x[nT_{s}]$) and the approximation ($x[(n-1)T_{s}]$) is quantized into only two levels $\pm&space;\Delta$ corresponding to Positive and negative differences.

i.e, If the approximation  ($x[(n-1)T_{s}]$) falls below the signal ($x[nT_{s}]$)at any sampling epoch(the beginning of a period)output signal level is increased by $\Delta$.

On the other hand the approximation   ($x[(n-1)T_{s}]$) lies above the signal ($x[nT_{s}]$) , output signal level is diminished by $\Delta$ provided that the input signal does not change too rapidly from sample to sample.

it is observed that the  change in stair case approximation lies with in  $\pm&space;\Delta$ .

This process can be illustrated in the following figure

Delta Modulated System:- The DM system consists of Delta Modulator and Delta Demodulator.

Delta Modulator:-

Mathematical equations involved in DM Transmitter are

error signal: $e[nT_{s}]=x[nT_{s}]-x_{q}[(n-1)T_{s}]$

Present sample of the (input) sampled signal: $x[nT_{s}]$

last sample approximation of stair case signal: $x_{q}[(n-1)T_{s}]$

Quantized  error signal( output of one-bit Quantizer): $e_{q}[nT_{s}]$

if         $x[nT_{s}]\geq&space;x_{q}[(n-1)T_{s}]&space;\Rightarrow&space;e_{q}[nT_{s}]&space;=&space;\Delta$.

and  $x[nT_{s}]<&space;x_{q}[(n-1)T_{s}]&space;\Rightarrow&space;e_{q}[nT_{s}]&space;=&space;-\Delta$.

encoding has to be done on the after Quantization that is when the output level is increased by $\Delta$ from its previous quantized level, bit ‘1’ is transmitted .

similarly when output is diminished by $\Delta$ from the previous level  a ‘0’ is transmitted.

from the accumulator $x_{q}[nT_{s}]=x_{q}[(n-1)T_{s}]+&space;e_{q}[nT_{s}]$

$e_{q}[nT_{s}]=e[nT_{s}]+&space;q[nT_{s}]$

where $q[nT_{s}]$ is the Quantization error.

(1 votes, average: 5.00 out of 5)

## Block Diagram of Digital Communication System/Elements of DCS

A General Communication System can be viewed as a Transmitting unit and a Receiving Unit connected by a medium(Channel). Obviously Transmitter and Receiver consists of various sub systems (or) blocks.

Our basic aim is to understand the various modules and sub systems in the system. If we are trying to understand the design and various features of DCS , it is plus imperative that we have to understand how we should design a transmitter and we must understand how to design a very good quality Receiver. Therefore one must know the features of the channel to design a good Transmitter as well as receiver that is the channel and it’s contribution will come repeatedly in digital Communications.

Source:- the primary block (or) the starting point of a DCS is an information source, it may be an analog/digital source , for example the signal considered is analog in nature, then the signal generated by the source is some kind of electrical signal which is random in nature. if the signal is a speech signal (not an electrical signal) that has to be converted into electrical signal by means of a Transducer, which can be considered as a part of source itself.

Sampling & Quantization:- the secondary block involves the conversion of analog to discrete signal

this involves the following steps

Sampling:- it is the process that involves in the conversion of Continuous Amplitude Continuous Time (CACT) signal into Continuous Amplitude Discrete Time (CADT) signal.

Quantization:- it is the process that involves in the conversion of Continuous Amplitude Discrete Time (CADT) signal into Discrete Amplitude Discrete Time (DADT) signal.

Source Encoder:-  An important problem in  Digital Communications is the efficient representation of data generated by a Discrete Source, this is accomplished by source encoder.

” The process of representation of incoming data  from a Discrete source into a more suitable form required for Transmission is known as source encoding”

Note:-The blocks Sampler, Quantizer followed by an Encoder constructs ADC (Analog to Digital Converter).

∴ the output of Source encoder is a Digital Signal, the advantages of Source coding are

• It reduces the Redundancy.
• Minimizes the average bit rate.

Channel encoder:-Channel coding is also known as error control coding. Channel coding is a technique which reduces the probability of error $P_{e}$ by reducing Signal to Noise Ratio at the expense of Transmission Band Width.The device that performs the channel coding is known as Channel encoder.

Channel encoding increases the redundancy of incoming data , this also involves error detection and error correction  along with the channel decoder at the receiver.

Spreading Techniques:- Spread Spectrum techniques are the methods by which a signal generated with a particular Band Width is deliberately spread in the frequency domain, resulting in a signal with a wider Band width.

There are two types of spreading techniques available

1. Direct Sequence Spread Spectrum Techniques.
2. Frequency Hopping Spread Spectrum Techniques.

The output of a spreaded signal is very much larger than incoming sequence. Spreading increases the BW required for transmission, which is a disadvantage even though spreading is done for high security of data.

SS techniques are used in Military applications.

Modulator:- Spreaded sequence is modulated by using digital modulation schemes like ASK, PSK, FSK etc depending up on the requirement, now the transmitting antenna transmits the modulated data into the channel.

Receiver:- Once you understood the process involved in transmitter Block. One should perform reverse operations in the receiver block.

i.e the input of the demodulator is demodulated after that de-spreaded and then the channel decoder removes the redundancy added by the channel encoder ,the output of channel decoder is then source decoded and is given to Destination.

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## Why Digital Communication is preferred over Analog Communication?

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.

(1 votes, average: 3.00 out of 5)

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.

(2 votes, average: 4.00 out of 5)

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 $\Delta&space;(n)=&space;2&space;\delta&space;(n)$ .

where $\delta&space;(n)$ 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 $\delta&space;(n)$ is constrained in between some pre-determined minimum and maximum values.

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

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

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

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 $\delta&space;_{min}$.

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

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

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

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

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## 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.

## 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.

(1 votes, average: 5.00 out of 5)

## 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.

(1 votes, average: 4.00 out of 5)

## 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

The Received signal is delayed by one bit time, then compared with the next signaling element in the Balanced Modulator, here Balanced Modulator acts as Phase Comparator .
If the Phases are same a logic ‘1’ is identified and for different phases a logic ‘0’. If the reference phase is incorrectly assumed only the first demodulated bit is in error .

Differential Encoding can be implemented with higher than Binary digital Modulation schemes. Although the Differential algorithms are much more complicated than for DPSK.

The Primary Advantage of DPSK is the Simplicity with which it can be implemented. No carrier recovery circuit is needed. A Disadvantage of DPSK is that it requires 1 dB and 3 dB more Signal to Noise Ratio to the same bit error rate as that of absolute PSK.

DPSK Generator and Detector are Collectively called as DPSK system.

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## Probability of error of QPSK

### Probability of error of QPSK :-

In order to understand the probability of error of a QPSK system, Let us consider a coherent QPSK system, the received signal x(t) is defined by   X(t)= Si(t) + w(t)  0≤ t ≤ T where i=1,2,3,4,  where w(t) is the sample function of a white Gaussian Noise process of Zero mean and power spectral density No/2.

Correspondingly, the observation vector x has 2 elements x1 and x2 defined by $x_{1}&space;=&space;\int_{0}^{T}x(t)\phi_{1}&space;(t)dt$  and $x_{2}=&space;\int_{0}^{T}x(t)\phi&space;_{2}(t)dt$

$x_{1}=&space;\int_{0}^{T}x(t)\phi&space;_{1}(t)dt$

$=&space;\sqrt{E}cos(2i-1)\frac{\pi&space;}{4}&space;+&space;W_{1}$

$=&space;\pm&space;\sqrt{\frac{E}{2}}&space;+&space;W_{1}$

simillarly $x_{2}=&space;\int_{0}^{T}x(t)\phi&space;_{2}(t)dt$

$=&space;-&space;\sqrt{E}sin(2i-1)\frac{\pi&space;}{4}&space;+&space;W_{2}$

$=&space;\mp&space;\sqrt{\frac{E}{2}}&space;+W_{2}$

Thus the observable elements x1 and x2 are sample values of independent Gaussian random variables with mean values equal to $\pm&space;\sqrt{\frac{E}{2}}$ and $\mp&space;\sqrt{\frac{E}{2}}$, respectively and with a common variance.

The decision rule is to simply to decide that

x1 falls inside the region z1 —–> S1 was transmitted.

x2 falls inside the region z2 —–> S2 was transmitted.

x3 falls inside the region z3 —–> S3 was transmitted.

x4 falls inside the region z4 —–> S4 was transmitted.

The regions and the signals were easily identified in the signal space diagram.

An erroneous decision will be made if for example signal S4(t) is transmitted . but the noise W(t) is such that the received signal point falls outside the region z4.

To calculate the  average probability of symbol error from the figure.

the coherent QPSK Receiver is in fact equivalent to two coherent binary PSK systems working in parallel and using two carriers that are in-phase and Quadrature-phase. x1 and x2 may be viewed as the individual outputs of the ‘2’ coherent BPSK systems may be characterized as follows.

The signal energy per bit = Eb/2 and the Noise power spectral density = No/2.

from the signal space representation , reference carriers are Φ1(t) and Φ2(t) all the signal vectors S1, S2, S3 and S4 are at 45o (or multiple of 45o) to these reference carriers. Suppose if S1 is transmitted. If the phase-shift of reference carrier is more than 45o (either Φ1 or Φ2)  it will be detected as some other symbol apart from S1.

The probability of error of BPSK system due to imperfect phase is given by

$P^{1}_{e&space;BPSK}=&space;\frac{1}{2}erfc\sqrt{\frac{E_{b}cos^{2}\phi&space;}{N_{o}}}$

$=&space;\frac{1}{2}erfc\sqrt{\frac{E_{b}}{2N_{o}}}$ when Φ = 45o.

or  Pe of a coherent BPSK system is Pe

$P_{e&space;BPSK}&space;=&space;\frac{1}{2}erfc(\sqrt{\frac{E_{b}}{N_{o}}})$ here Eb = Eb /2

there are two BPSK systems $P^{'}_{e1}&space;=&space;P^{'}_{e2}&space;=&space;\frac{1}{2}erfc\sqrt{\frac{E_{b}}{2N_{o}}}$ the bit errors in the in-phase and Quadrature-phase channels of QPSK system are stastically independent.

The in-phase channel makes a decision on one of the two bits constituting a symbol.(di-bit) of the QPSK signal and the Q-channel takes care of the other bit.

The Average probability of a correct decision resulting from the combined action of the two channels working together is

$P_{c}&space;=&space;(1-P^1_{e1})(1-P^1_{e2})$

$=(1-P^{1}_{e1})&space;^{2}&space;as&space;P^{1}_{e1}&space;=&space;P^{1}_{e2}$.

$=&space;1+&space;P^{2}_{e1}&space;-&space;2P_{e1}$

Normally $P_{e1}$ is very very small,$P_{e1}$ <<1 —–> $P^{2}_{e1}$ is negligible

$P_{c}=&space;1-2P_{e&space;BPSK}$

$P_{e}$ of coherent QPSK is $P_{eQPSK}&space;=&space;1-P_c{}$

$=&space;1-(1-2P_{e1})$

$=&space;2P_{e1}$

$P_{eQPSK}=&space;2X\frac{1}{2}erfc\sqrt{\frac{E_{b}}{2N_{o}}}$

$P_{eQPSK}=&space;erfc\sqrt{\frac{E_{b}}{2N_{o}}}$

here $E_{b}$ is the energy per bit, in terms of symbol energy $E_{s}=2E_{b}$ in QPSK

$P_{eQPSK}=&space;erfc\sqrt{\frac{E_{s}}{4N_{o}}}$ where $E_{s}$ is the energy per symbol in QPSK.

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## Digital Communications Slip Test Questions

Slip test-1

1. In a PCM system, if the Quantization levels are increased from 2 to 8,                 i. Find the change in Signal to Quantization Noise ratio.                                ii.Find the change in Transmission Bandwidth.
2.  i. Convert the following signal x(t) = 10 cos (200πt) to a discrete signal x[n] if sampling rate is 1000Hz.                                                                                       ii. Write the advantages of Digital Communication system over Analog Communication system.

PART-B

1. Draw the Basic Block Diagram of Digital Communication System and explain each block in detail.
2. Deduce the expression for Signal to Quantization noise ratio in PCM system.                                                                  Slip test-2
1. A PCM source transmits four samples (messages) with a rate 2B samples /second. The probabilities of occurrence of these 4 samples (messages) are equally likely Find out the information rate of the source.
2. A source produces 26 symbols with equal probability what is the average information produced by this source?

PART-B

1. State Mutual information. Prove any three properties of Mutual information.
2. A source alphabet has 10 symbols with the given probabilities 0.02, 0.04, 0.17, 0.02, 0.16, 0.06, 0.03, 0.27, 0.20, 0.03 construct Shannon-Fano coding and calculate the efficiency.

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## Digital Communications Quiz with solutions

UNIT-1 –QUIZ

1. The band width  needed to transmit Television video plus audio signals of bandwidth 4.2 MHz using Binary PCM  quantization level of 512 is (  c     )         a. 2 MHz            b. 25.6MHz                        c.37.8MHz                    d.75.6MHz
2. A signal m(t)has a bandwidth of 1.0 KHz and exhibits a maximum rate of change of 2.0 volts/sec. The signal is sampled at a sampling frequency of 20 KHz and quantized using delta modulator. The minimum step size to avoid slope overload is (  b    )                                                                                                                      a. 1.0mV        b. 0.1mV                c.10.0mV                      d.0.01mV
3. For a 10 bit PCM system, the signal to Quantization noise ratio is 62 dB. If the number of bits are increased by 2, then the signal to Quantization noise ratio will be (  c      )                                                                                                                    a. Increased by 6 dB                                   b.  Decrease by 6 dB                                       c. Increased by 12 dB                                  d. Decrease by 12 dB
4. Write the condition to eliminate slope overload error in a Delta modulation system—————————
5. Write the condition to eliminate slope overload error in a Delta modulation system if the input is a single-tone signal—————————
6. For which value of A , A-law has linear transfer characteristics——————–.
7. For which value of µ , µ-law has linear transfer characteristics——————–.
8. In a PCM system if the step size is 5V , then the Quantization noise in dB is   (  d    )                                                                                                                                             a. -5 dB.          b. -3.18dB       c. -10dB                      d. 3.18dB.
9. Draw the characteristics of Mid-rise and Mid-tread type Quantizers ————–
10. The maximum slope of the signal ———————-
11. The sampling rate of the signal ————————–
12. The step size in a 8- bit pcm system if the input signal to PCM is oscillating between [+4V,-4V] is—————————–

UNIT-2-QUIZ 2

1. Entropy is a measure of ——————————– 3M.
2. The capacity of a band – limited AWGN channel in terms of kbps if the average received signal power to noise power spectral density is 1000 and the bandwidth is approximately infinite is (  a    )   [Hint: Shannon’s bound Cinfinity =  S/No]                                                                                                                        a.  1.44        b. 1.08                                c. 0.72                           d. 0.36
3. If Y= g(X) where g denotes a deterministic function, then the conditional entropy H(Y/X) is (  b ) 3M.                                                                                                           a. ≠ 1         b.  = 0                           c. = 1                           d. ≠ 0
4. A Source generates three symbols with probability of 0.25, 0.25, and 0.50 at a rate of 3000 symbols per second. Assuming the symbols are generated independently from the source, the most efficient source encoder would have average bit rate of ( b ) 4M.                                                                                             a. 6000 bits/sec                                                         b. 4500 bits/sec                                 c. 3000 bits/sec                                                         d. 1500 bit/sec.
5. A source generates four equi-probable symbols. If the source coding is adopted, the average length of the code for 100% efficiency is ( c )                       a. 6 bits / symbol                                                          b.   3 bits / symbol                           c.  2 bits / symbol                                                        d.  4 bits / symbol
6. Draw the channel diagram of Binary Erasure channel and write its channel matrix ………………………..
7. For a Binary Symmetric Channel the entropies of X & Y are if the conditional probability p=0.5, where X & Y are input & output random variables of BSC  (   d     )3M                                                                                                                          a. 1,0           b. 0,0               c. 0,1                           d. 1,1
8. Given a Binary Symmetric Channel, the expression for Entropy is (pis conditional probability of error   ——————- 3M.
9. Match the following   (  d    )                                                                                                       a. Lossless channel                                           1.  only one non-zero element in each row                                                                                                                                        b. Deterministic channel                                2. Only one non-zero element in each row and each column.                                                                                                   c. Noiseless channel                                           3. Only one Non-zero element in each column.                                                                                                                                d. for a noiseless channel                                      4. H(Y/X)=0 and H(X/Y)= 0.

a. none of these                                                  b.   b-3,c-4,a-1,d-2                                    c. c-4,d-3,a-1,b-2                                             d.  b-1,c-2,a-3,d-4.

Digital Communications Unit-3 Quiz with Solutions

UNIT3 –QUIZ3[CO3]

1. The number of Parity check bits in an (n, k) Linear Block codes are (    b      )      a.n                b. (n-k)                       c.  (n+k)                       d. k
2. The Hamming Weight of the following code words 10011101 & 00111100 is      (   c   ) 2M.                                                                                                                                       a. None of these      b. 4, 5              c.5, 4                           d.3,4
3. A cyclic code can be generated using———————— and A block can be generated using——————-.(   c   )                                  a.Generator matrix & Generator polynomial.                                                          b.Generator matrix & Generator matrix.                                                            c.Generator polynomial & Generator matrix.                                                            d.None of the above.
4. The rate of a Block code is the ratio of(    c     )                                                                a.Message length to Block length.                 b.Block length to message length.  c.Message weight to Block length.                 d.None of the mentioned.
5. The syndrome in LBC is calculated using , where Y represents received code word   (     a       ) 2M                                                                                                                      a. S= Y HT          b. S = YH                    c. S= YT H                  d. S= YT HT
6.  A non-Zero value of Syndrome in a Block code represents ( b    ) 2M.                  a.No error during transmission.     b.An error occurred during transmission.   c.Both a and  b                                         d.None of the above.
7. The transmitted code word(X) in an LBC can be obtained from received code word( Y) by using the equation (  a   ), where E represents error vector.              a. X= E + Y           b.  X = X.Y                  c. X= E.Y                    d. X= X/Y
8. The parity check matrix of a (6,3) block code if Generator matrix is G  3M.             ——————-.
9. In the above question find the Code words corresponding to message vectors [110] and [111]——————-. Ans: [110110], [111000].
10. For the Q.8. Find the syndrome value when the received code word is 001111—-4MAns: syndrome value [0 0 1]
11. For a (7,4) cyclic code , the generator polynomial is given as find the codeword  for the data 1100
• Non systematic codeword is—————- .Ans:  1011100
• Systematic codeword is——————–.Ans: 1011100.

UNIT-4

QUIZ-4

1. The modulation technique that provides minimum probability of error is (  b   ). a. ASK                  b. PSK                   c. DPSK                        d.FSK
2. At a given probability of error, binary coherent FSK is inferior to binary coherent PSK by (  c  )                                                                                                         a. 6 dB                b. 2 dB           c.  3 dB           d.  0 dB             e. None of these.
3. If Eb, the energy per bit of a binary digital signal, is 10-5watt-sec and the one-sided power spectral density of the white noise, No= 10-6 W/Hz, then the output SNR of the matched filter is  (   d    )                                               a. 26 dB         b.  20 dB          c. 10 dB          d. 13 dB          e. None of these
4. In which system, bit stream is portioned into even and odd stream (   c   )  a. BPSK                  b.  MSK            c. QPSK                     d.  FSK
5. Optimum filter can be called as———when the input noise is white noise (  a   ) 2M.                                                                                                                        a. Matched filter   b.  High pass filter        c. Low pass filter      d. None of these
6. The probability of error of ASK is ————————————.
7. The probability of error of FSK is ————————————.
8. Write the expression for QPSK modulation —————————————-2M.
9. Draw the block diagram of DPSK system————————————.
10. A pulse g(t) = A cos(πt/2T) for 0 ≤ t ≤ T is transmitted over an AWGN channel with two sided noise power spectral density No/2 Watts/Hz. The impulse response of the matched filter is (  a     )                                                           a. A sin (πt/2T)                                                            b.   A rec (πt/2T)                           c. A rec (π(T-t)/2T)                                                    d. A sin (π(T-t)/2T)
11. If the probability of error function of a modulation scheme is Pe = (½) erfc(x) thenthe same Pe interms of Q-function is ( b   )                                             a. Q ( 31/2 * x)                    b. Q (21/2 * x)                c. Q(x)             d. none of these.

## Digital Modulation schemes Viva Questions

2. Draw the waveforms of ASK scheme.
3. Draw the Phasor diagrams of ASK.
4. Write the equation of ASK wave.
5. What are the applications of ASK?
7. Write the expression for Probability of error of ASK.
8. What is the BW of ASK?
9. Draw the signal constellation diagram of ASK.
10. Draw the block diagram of ASK generator. Explain its working.
11. Draw the block diagram of coherent ASK demodulator. Explain its working.
12. Draw the block diagram of non-coherent ASK demodulator. Explain its working.
13. How ASK is different from AM?
14. If bit rate is 10µsec then calculate the band width of ASK.
15. Which is the efficient modulation scheme among ASK and PSK give the reason.
16. Write the relation between complementary error function and Q-Function.
17. Write the Pe of ASK in terms of Q function.
18. What is meant by Signal space diagram?
19. A source generates three symbols with probability 0.25,0.25 and 0.5 at a rate of 3000 symbols/sec Assuming the symbols are generated independently from the source the most efficient source encoder would have average bit rate of——————————.

FSK Generation & Detection:

1. Define FSK. Write the practical applications of FSK.
2. Draw the waveforms of FSK scheme.
3. Draw the Phasor diagrams of FSK.
4. Write the equation of FSK wave.
6. Write the expression for Probability of error of FSK.
7. What is the BW of FSK?
8. List some applications of FSK.
9. Draw the block diagram of FSK generator and explain its working.
10. Draw the block diagram of coherent FSK demodulator and explain its working.
11. Draw the block diagram of non-coherent FSK demodulator and explain its working.
12. Show that FSK is inferior to PSK by 3dB for a given probability of error.
13. Define digital modulation schemes.
14. Identify the differences between Pass band and Base band Communication.
15. What is a matched filter?
16. What are the properties of matched filter?
17. What is MSK? What is the need for Minimum Shift Keying?
18. For the given 8 bit data 10111010 draw the FSK output waveform.
19. Draw the constellation diagram of FSK.
20. What will happen if the same frequency is used for both the carriers?
21. Draw FM signal and FSK signal differentiate those two signals.
22. What are the applications of FM?
23. How FSK is different from FM?
24. Write the probability of error expression of FSK in-terms of Q function.
25. Write the relation between complementary error function and Q-Function.
26. Among FSK and PSK which is the most efficient scheme and why?

PSK Generation & Detection:

1. Define PSK. Write the practical applications of PSK.
2. Draw the waveforms of PSK scheme.
3. Draw the signal space diagram of PSK.
4. Write the equation of PSK wave.
5. What are the applications of PSK?
7. Write the expression for Probability of error of PSK in terms of erfc()
8. Write the expression for Probability of error of PSK in terms Q().
9. Write the difference between erf(x) and erfc(x)..
10. What is the BW of PSK?
11. What are antipodal signals?
12. Give the equation for average probability of symbol error for coherent binary PSK.
13. Explain how QPSK differs from PSK in terms of transmission bandwidth and bit information it carries.
14. What are the advantages of M-ary modulation scheme?
15. Draw the constellation diagram for QPSK.
16. Compare FSK and PSK.
17. What do we infer from constellation diagrams of various modulation schemes?
18. Draw the block diagram of PSK generator. Explain its working.
19. Draw the block diagram of coherent PSK detector. Explain its working.
20. How non coherent PSK is advantageous over coherent PSK?
21. Other name of Non-coherent PSK is———–.
22. Write the expression for a Gaussian function———————.

DPSK Generation & Detection:

1. Define DPSK.
2. Draw the waveforms of DPSK scheme.
3. Write the equation of DPSK.
4. What are the applications of DPSK?
6. Write the expression for Probability of error of DPSK.
7. What is the BW of DPSK?
8. What are the applications of DPSK?
9. Draw the DPSK system and explain its working.
10. If the input sequence to a DPSK modulator is 10011100 write the following sequences, Deferentially encoded sequence if previous bit is 0 And the phase of the encoded sequence.
11. The capacity of a band limited AWGN channel in terms of KBPS if the average received signal to noise power spectral density is 1000 and the BW is approximately infinite is…………..
12. For error free channel conditional probability is ————-.
13. The modulation technique that provides minimum probability of error is———-.

QPSK Generation & Detection:

1. Define QPSK.
2. Draw the waveforms of QPSK scheme.
3. Draw the Phasor diagrams of QPSK.
4. Write the equation of QPSK wave.
5. What are the applications of QPSK?
7. Write the expression for Probability of error of QPSK.
8. Draw the block diagram of QPSK modulator. Explain its working.
9. Draw the block diagram of coherent QPSK de modulator. Explain its working.
10. Write the table for four di-bits with phases used in a QPSK modulation scheme.
11. Write the equation of QPSK wave.
12. What is the BW of QPSK?
13. What are the applications of QPSK?
14. Draw the constellation diagram for QPSK.
15. In Minimum Shift Keying the relation between the signal frequencies and bit rate are————.
16. A pulse for is passed through a matched filter then the impulse response of matched filter is—————————.
17. A lossless channel means—————–.
18. A distortion less channel means————————-.
19. Write the channel matrix for a Binary symmetric channel——————–.

Losses in fibre optics:

1. Define Transmission loss of Optical fibre?
2. What are various losses in fibre optics?
3. What are step index graded index optical fibers?

Data Formats:

1. Compare NRZ-I and NRZ- L.
2. What is the use of data formatting?
3. Compare NRZ and biphase encoding.
4. What is the relationship between quantization levels and number of bits in a code word?
5. Give the advantages of Manchester encoding.
6. Assume a data stream is made of ten 0s. Encode this stream using the following encoding schemes.
7. How much change can you find for each scheme? a)NRZ b)RZ c)Bi phase
1. List out the merits & demerits of each data formats.
2. Represent the given data 11010100 in Manchester encoding and NRZ –M scheme.
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## Delta Modulation and Demodulation Viva Questions

Delta Modulation & Demodulation:

1. What is Delta Modulation?
2. A message signal m(t) has a BW of 1KHz and exhibits a maximum rate of change of change of 2 Volts/sec the signal is sampled at a sampling frequency of 20 KHz and Quantized using a delta modulator the minimum step size to avoid slope overload is———-
3. Which type of Quantization is used in DM?
4. What are two types of quantization errors?
5. What is granular noise?
6. What is slope overload distortion?
7. Draw the block diagram of Delta Modulation system.
9. What happens to the output signal if the variation of the message signal is
• greater than the step size
• less than the step size
1. What is the advantage of delta modulation over PCM?
2. Compare DPCM, PCM& Delta modulation.
3. How to reduce the quantization noise that occurs in DM?
4. A band pass signal has a spectral range that extends from 20 to 82 KHz. Find the acceptable sampling frequency.
1. Find the Fourier series expansion of an Impulse train.
2. Mention the applications of DM.
3. Write the condition to eliminate Slope over load distortion in Delta Modulation for a single tone message signal.
4. Write the condition to eliminate Slope over load distortion in Delta Modulation for some arbitrary signal.
5. We chose oversampling in DM system. True or False?
6. How DPCM is different from DM?
7. Draw the waveforms of DM demodulation and Demodulation.
8. Differentiate Coherent and Non coherent modulation schemes.
9. Explain the working DM system.
10. Draw the Wave forms and explain two distortions in DM system.

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## Pulse Code Modulation Viva Questions

Pulse Code Modulation:

1. What is PCM? Does PCM come under Digital Modulation technique?
2. What are the applications of PCM?
3. What is the need for Regenerative repeaters in PCM?
4. Draw the block diagram of PCM system.
5. Define µ-law and A-law. And draw the characteristics of two laws.
6. What is meant by Quantization?
7. Differentiate Uniform and Non uniform Quantization techniques.
8. What is meant by companding in PCM?
10. What is the difference between ADC & PCM?
13. What is slope overload distortion?
14. Draw the waveforms of PCM modulation and demodulation.
15. State sampling theorem.
16. What is aliasing?
17. Give the expression for aliasing error and the bound for aliasing error.
18. What is quantization?
19. What are the various steps involved in A/D conversion.
20. Define step size.
21. What is the importance of regenerative repeater?
22. List out the three basic functions of regenerative repeater.
23. What is companding?
24. Write the mean square quantization error if the step size is S.
25. What is a mid tread Quantizer?
26. What is a mid rise Quantizer?
27. What do you mean by quantizing process?
28. What will happen when sampling rate is greater than Nyquist rate?
29. What will happen when sampling rate is less than Nyquist rate?
30. Find the A/D Converter output for input DC voltage of 3.6V.
31. In digital telephony, (a) what kind of modulation is used? (b)Give the typical sampling rate, output data rate and speech signal Bandwidth.
32. Mention some applications of PCM.
33. What is the function of Sample and Hold circuit?
34. Write the expression for SNR in decibels of a PCM system.
35. For a DAC the required resolution is 25mv and the total input is 5V. the number of bits required would be——————————————.
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