## BPUT sample paper of Digital Communication Technique

SAMPLE PAPER OF DIGITAL COMMUNICATION TECHNIQUE

**DIGITAL COMMUNICATION TECHNIQUE**

**Full marks:-70**

*Q.1 contains 20 marks and*

*Q.2- Q.8 contains 10 marks each (out of which 5 has to be done)*

**1.**

**(a) What is the need of signalling in a PCM system?**

Ans:- Following are needs of using signalling:

(i) Increases the transmission bandwidth.

(ii) Increase the power efficiency.

(iii) Increases the error detection capability.

(iv) Better Transparency.

**(b) State the properties of MSK modulation.**

Ans:- Properties of MSK modulation

(i) Band width=1.5 fb

(ii) Distance between signal points=2√Eb

(iii) Phase is continuous

(iv) MSK does not have amplitude variations

**(c) What is the need of an anti aliasing filter?**

Ans:- When a continuous signal is sampled at a rate lower than *Nyquist rate,fs < 2fm *(Where, fs = sampling frequency and fm= the maximum frequency present in the band limited or message signal.) then the successive samples of the signal are overlapped with each other. Aliasing is a phenomenon in which a high frequency component of the signal takes identify of a lower frequency component in the spectrum of the sampled signal and for this reason we can’t recover the message signal as the output of the LPF. In order to avoid all these disadvantages we are using anti aliasing filter or pre-aliasing filter.

**(d) If the data transmission rate is R of a communication system, how much band width is required if you use BPSK modulation?**

Ans:- The transmission rate of a communication system =R.

B.W. of BPSK= 2 f*b* (where f*b* =maximum frequency of the bit sequence.)

The transmission rate of a system is the frequency at which the data is transmitted, among all the frequency is preferred because the noise encounter is very less. Therefore f*b*=R and B.W.of BPSK =2 R.

**(e) What is white noise? Where is it encountered?**

Ans:- A noise whose power spectral density is uniform over the entire frequency range of interest and occurs due to the superposition of all visible light spectral components is called “white noise”.

Gn(f) =ᵑ/ 2 ,where ᵑ =constant

It is encountered i the communication channel. It is uniform for both +ve and –ve frequency components. This noise is always additive in nature. That why the channel is calld “AWGN channel”.

**(f) Why pulse shaping is done? Is it a base band or pass band operation?**

Ans:- When the rectangular pulses are passed through a band limited channel, the pulse will spread in times and the pulse for each symbol will smear into the time intervals of succeeding symbols. This causes inter symbol interference and which leads to increase Pe at the receiver.

To minimize the ISI it is obvious to increase the BW of the channel and the technique which is used to out of band radiation, reduce the bandwidth, reducing the ISI is highly desirable is called “Pulse Shaping Techniques”.

**(g) What is the advantages of channel coding?**

Ans:- Advantages of channel coding are:-

(i) Improves the efficiency of operation of a communication channel.

(ii) It increases the rate at which information may be transmitted over channel while maintaining a fixed error rate.

(iii) It allows us to design a communication system in which both information bit rate and error rate are independently and arbitrarily specified but subject to a constraint on BW.

**(h) Bring out 2 differences between thermal noise and quantization noise.**

Ans:- *Thermal noise *

(i) It is introduced due to the resistive component of the system.

(ii) N*th*=M^2/1+4Pe. 2^2n

*Quantization noise*

(i) This is introduced in the transmitter and is carried along to the receiver output

(ii) Nq=S^2 /12

**(i) The information is a speech wave form with maximum frequency content of 3.5 kHz. The quantization error is not exceeded +-1% of the peak to peak analogue signal. What is the sampling rate,minimum number of bits per sample and the minimum transmission rate?**

Ans:- A speech waveform with maximum frequency,fm = 3.5 kHz.

Maximum quantization error, Emax = +-1% of the peak to peak analogue signal = +-1% of Xmax

Emax = 1% Xmax = (0.01) Xmax = (0.01)Xmax

As we know that maximum quantization Noise = Δ/2, where Δ=step size

or,Emax = Δ/2

or,(0.01) Xmax = Δ/2

or,Δ = 0.02×max,

but step size, Δ = (2×max) / 2^q, where 2^q = number of levels, q= number of bits

or, 0.02×max = (2×max) / 2^q

or, 2^q = 1 / 0.01

or,2^q = 100

or,q = log 2100 = 6.625

or,q = 7 bit

(i) The sampling rate > or =2fm = 2×3.5 kHz =7kHz

(ii) The minimum number of bits/sample=7 bits

(iii) The minimum transmission rate r = q.fs = q.fm

=7. [2×3.5] kHz = 49000 bits/sec

**(j) Can signal-to-noise ratio and channel bandwidth of a bandlimited, Gaussian channel be traded keeping channel capacity at optimum level?support your answer with suitable example.**

Ans:- Yes, SNR and channel BW of a band limited, Gaussian channel can be traded keeping channel capacity at optimum level.

C=B log2 (1 + s /n B)

=(s/ᵑ). (ᵑB / s) log2 [1+(s / ᵑB) ]

=(s/ᵑ) log2 [1+(s/ᵑ B)] ^ᵑ B/s

or,C = s/ᵑ log 2e………(1)

As C→∞ i.e. B→∞

C= lim (s/ᵑ) log 2e = 1.44 (s/ᵑ)…………..(2)

Eg:- If s/N =7 and B =4kHz we find C =12× 10^3 bits/sec

If S/N =15 and B decreased by 3 kHz and the channel capacity remains the same.With the 3 kHz band width the noise power will have to be increased by factor

(3/4) × (15×7) = 1.6

So,the 25% reduction in BW requires a 60 % increase in signal power.

**2. The speech signal is band limited to 3.5 kHz. What should be the desirable sampling rate? The maximum acceptable error in the sample amplitude is 0.45 % of the peak amplitude mp . The quantized samples are binary coded. Find the minimum bandwidth of a channel required to transmit the encoded binary signals. 24 such signals are time-division multiplexed. Determine the minimum transmission bandwidth required to transmit the multiplexed signal.**

Ans:- The speech signal is band limited to 3.5 kHz.

fm= 3.5 kHz

Therefore, the desirable sampling rate,fs = 2 fm

= 2× 3.5 kHz

=7 kHz

Max. Quantization error, Emax =0.45 % Mp= 0.45/100 of Mp

As we know that maximum quantization Noise= Δ/2

Δ= step size

Or, Emax = Δ/2

Or, 0.0045 Mp = Δ/2

Or, Δ =0.009 Mp

And, Step size=2Mp / 2^q

Or, 0.009 Mp = 2 Mp / 2^q

Or, 2^q = 1 / 0.0045

Or, 2^q = (1 / 45) × 10^4 =222.2

Or, q = log*2 *(222.2) = 7.79 = 8 bit (approx)

(i) The minimum number of bits/sample = 8 bit

(ii) The minimum band width = q × fm = 8 × 3.5 kHz =28 kHz

(iii) A TDM contains 24 channels.

Therefore, minimum transmission bandwidth = N × fm = 24 × 3.5 kHz = 84 kHz

**3. (a) For what type of noise is Shanon’s channel capacity theorem valid? Hence compute the capacity of a typical telephone channel.**

Ans:- White Gaussian Noise is preferred because it’s spectral density is uniform.

C=B log2(S/N)

**(b) For what kind of modulation, can you employ envelope detection? What is the advantages of this kind of digital modulation?**

Ans:-Envelope detector can be employed as:-

(i) Envelope detectors are used in Non-coherent Binary Amplitude shift Keying(ASK)

(ii) In the binary ASK case, the transmitted signal is defined as.

S(t)=√2Ps cos(2π fc t)……….(1)

(iii) Binary ASK signal can also be demodulated non-coherently using envelope detector

Advantages

(i) Simplifies the design requirement.

(ii) Don’t require phase-coherent local oscillator.

**4. Explain Shanon’s theorem and capacity of a Gaussian channel theorem.**

Ans:- **Shanon’s theorem**

It’s the most fundamental theorem of communication, is concerned wit the rate of transmission of information with an arbitrarily small probability of error provided that the information rate R is less thsn or equal to a rate C called the “channel capacity” ; this technical approach is called “coding”.

Definition :- A source M equally likely messages, with M >> 1, which is generating information at a rate R. Given by a channel capacity C, If R <= c , then there exist a coding techniques such that the output of the source may be transmitted over the channel with a probability of error in the received message which may be made arbitrarily small.

*Important features*

(a) When R<=C then transmission accomplished by without any error in the presence of noise.

(b) Since probably the noise is Gaussian noise, having a probability density tends to infinite.

Negative statement of Shannon’s theorem

Given a source of M equally messages with M>>1, which is generating information at a rate R; then R > C.

The probability of error is close to unity fo every possible set of M transmitter signals”.

*Problems*

(a) Error of probability will increase towards unity as M increases.

(b) Increasing the complexity of the coding results in the probability of error.

*Capacity of a Gaussian channel theorem*

The channel capacity of a white, band limited Gaussian channel is

C= B log*2* (1 + S/N) bits / sec…………(1)

B= band width of the channel

Where,

S= signal power

N= Total noise within the channel bandwidth = ᵑ B

ᵑ/2 = double sided power spectral density.

*Features *

- First, we find that channels are encountered in physical systems generally, are at least approximatelt, Gaussian.
- Second, it turns out that the results obtained for a Gaussian often provide a lower bound on the performance of a system operating a non Gaussian channel.
- If a particular encoder-decoder is used for with a Gaussian channel and an probability of error results, then with a non Gaussian channel another encoder-decoder can be designed so thst the Pe is smaller.
- Eq (1) can also be designed for number of non- Gaussian channel.

**5. 24 speech signals are to be time multiplexed. Make a reasonable choice of the highest frequency component of the speech signal. The signals are sampled quantized, PCM encoded and then transmitted. Each sample is coded into 8 bits. Find the bit rate in the channel and the channel bandwidth. If 4 such channels coming from widely separated locations are to be time multiplexed ,what problems may be experienced and how to overcome it?**

Ans:- Number of speech signal= 24

Each sample is coded into 8 bits.

Therefore, total no of bits = 24×8 = 192 bits

In PCM, for synchronization we need 1 extra bit.

Therefore, Bit rate = 193 bits/channel duration

Channel duration = 1/8000 = 125 μs

= 193/ [125 × 10 ^(-6) ] bits / sec = 1.544 Mbps………….(1)

Bandwidth = 1/2(Bit rate)

= ½ (1.544 Mpps) = 772 kHz………………(2)

*PROBLEMS*

(i) The distortion introduced by channel.

(ii) The signal is interfered with each other.

*SOLUTIONS*

(i) Introduce repeater between specific distances.

(ii) Introduce synchronization bit after every 192 bits.

**6. (a) Binary data is transmitted over a band pass channel at a rate of 33 bps using non-coherent FSK signalling scheme with tone frequency 1070 and 1270 Hz. Calculate the probability of error Pe assuming A^2 / No = 8000?**

Ans:- Here, we have Pe =1/2 exp (-Eb /2 No)

But , therefore, we write Pe = ½ exp (-A^2 T / 4 No)

And, T = 1/ r*b *= 1/300

Hence, Pe = ½ exp[ (-8000 × 1) / (4 × 300) ]

= ½ exp (-20/3)

Therefore, Pe = ½ exp(-6.67)

= 6.34 × 10 ^-4

**(b)For an FSK system, the following data are observed. Transmitted binary data rate = 2.5 × 10^6 bits/sec power spectral density (psd) of noise = 10 ^-20 watts/Hz. Amplitude of received signal = 1 μv. Dtermine the average probability of symbol error assuming coherent detection.**

Ans:- Here, we have Pe = ½ erfc √Eb / 2 No

T= T*b* =1/r*b*

Eb = A^2 T/2

=[ (1 × 10^-6)^2 / 2] × 1/ (2.5 × 10^6)

= (1/5) × 10^-18

Also, psd = No/2 = 10^-20 W/Hz

Eb/ 2No =[ (1/5) × 10 ^-18] = (1/20) × 10^2 =5

Pe = (1/2) erfc (√5) = (1/2) erfc (2.23)

Or, Pe = (1/2) × 1.84 × 10^-3 (from erfc table)

**7. (a) Shanon’s capacity formula is valid for any kind of noise. Justify**

Ans:- The shanon’s capacity formula .

C= B log2 [ 1+ (S/N)] bits/s……………..(1)

This theorem is restricted to white Gaussian channel because

* We find that channels encountered in physical systems generally are, at least approximately, Gaussian.

* It turns out that the results obtained for a Gaussian channel often provide a lower bound on the performance of a system operating over a non- Gaussian channel.

* But by Gaussian channel the Pe increase and by using non Gaussian channel the Pe becomes smaller.

* The eqn (1) is applicable for number of non- Gaussian channels.

* Gaussian channel will consider under following considerations,

(i) the messages are represented by fixed voltages.

(ii) Messages are generated in equal interval of time i.e., one after another.

**(b) One internet service provider gives dial-up connections at 56 kbps. Assume that the telephone connection provides a usable bandwidth of 3.5 kHz. What is the minimum SNR required to support this?**

Ans:- Given:- data rate, R= 56 kbps , BW = 3.5 kHz, Channel capacity, C = Blog*2* [ 1 + (S/N) ]

By shanon theorem R<= C, R = C = 56 kbps

Or, 56 × 10^3 =3.5 × 10^3 log*2* [1+ (S/N)]

Or, (56/3.5) = log*2* [1+(S/N)]

Or, 16 = log*2* [1+ (S/N)]

Or, 2^(16) = 1+ (S/N)

Or, (S/N) = 2^(16) – 1 = 65536 – 1 =65535

(SNR)db =10 log*10 *(65535)

=48.16 dB

**8.Write short notes on:-**

**(i) Minimum Shift Keying**

Ans:- The bandwidth requirement of QPSK is high. Filters of other methods can overcome these problems, but they have other side effects. **MSK **overcomes these problems. In MSK the output waveform is continuous in phase hence these no abrupt changes in amplitude. The side lobes of MSK are very small hence BPF isn’t required to avoid interchannel interference.

The transmitted MSK signal is represented as under

S(t) = √2Ps [b*e* (t) sin (2πt/4T*b*) ] cos (2π fc t) + √2 Ps [ b*o* (t) cos(2πt / 4Tb) ] sin (2π fc t)

*Block diagram of MSK*

*Signal comparison*

**(ii) Adaptive Delta Modulation**

Ans:- * To overcome the disadvantages of DM, the step size is made adaptive to variations in the input signal x(t).

* In the steep segment of the signal x(t), the step size is increased. Also life the input is varying slowly, the step size is reduced. This method is called ADM.

* The ADM can take continuous changes in step size or discrete change in step size. The step size increase or decreases according to a specified rule depending upon one bit quantizer output.

* If one bit quantizer output is high, then the step size may be doubled or next sampled. If one bit quantizer output is low, the step size may be reduced by one step.

*Block diagram of ADP*

Advantages:-

(i) The SNR is better than ordinary delta modulation.

(ii) Because of the variable step size, the dynamic range of ADM is wider than simple DM.

(iii) Better utilization of BW.

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