Module 5

1.Principles of Communication Systems – 15EC45 June/July 2017

10(a) With relevant diagrams, explain the generation and reconstruction of PCM signal 6
Solution: Refer section 5.12 pages 4 & 5 of Module 5 study material

2.Principles of Communication Systems – 15EC45 DEC 2017/JAN2018

10(a) Explain the generation and reconstruction of pulse code modulation (PCM) signal 8
Solution: Refer section 5.12 pages 4 & 5 of Module 5 study material
10(b) What is quantization? Derive the output signal to noise ratio of a uniform quantizer 8
Solution : Amplitude quantization is the process of transforming sampled amplitude values
x(nTs) of the message signal x(t)at times t = nTs into discrete levels xq(nTs) taken from finite
set of possible amplitudes and the device which does this function is called Quantizer.
Refer section quantization noise pages 3 and 4 of module 5 study material

3.Principles of Communication Systems – 15EC45 June/July 2018

9© For a sinusoidal modulating signal, show that the signal to quantization noise ratio is 4
1.8 + 6 N dB where N is the number of bits per sample
Solution : Refer section quantization noise (for sinusoidal signals) page 4 of module 5 study
material
10(b) What is quantization process? Explain the different types of quantization with their 5
input output characteristics.
Solution : Refer pages 1 and 2 of module 5 study material
10(c) Represent the binary data : 10011101 in polar NRZ and bipolar RZ formatting 4
Solution : Refer page 21 of module 5 study material

4.Principles of Communication Systems – 15EC45 Dec 2018/Jan 2019

10(b) With a neat block diagram, explain the concept of PCM 10
Solution: Refer section 5.12 pages 4 & 5 of Module 5 study material

5.Principles of Communication Systems – 15EC45 June/July 2019

10(b) Explain the channel Vocoder with a neat block diagram 8

Solution: Refer section 5.14(a) – pages12 and 13 in Module 5 of study material. Write up to
basic speech model (included)

6.Principles of Communication Systems – 17EC44 June/July 2019

9(b) A TV signal with a bandwidth of 4.2 MHz is transmitted using binary PCM. The 6
 number of quantization levels is 512. Calculate (i) Code word length (ii) Transmission
BW (iii) Final bit rate.
Solution : Refer problem in page 20 of Module 5 study material
9(c) With a neat diagram explain the generation and detection of pulse code modulation 8
(PCM) signals
Solution : Refer section 5.12 pages 4 and 5 of Module 5 study material
10(a) A PCM system uses a uniform quantizer followed by a V bit encoder. Show that rms 6
signal to quantization noise is approximately given by (1.8 + 6V) dB
Solution : Refer section Quantization noise in pages 3 and 4 of Module 5 study material

7.Principles of Communication – 17EC44 Dec 2019 / Jan 2020

9(a) A Continuous time signal x(t) has a BW f3 = 10 kHz and it is sampled at fS = 22 kHz 8
using 8 bit sampler. The signal is properly scaled so that |X(n)|<128 for all n. Determine
your best estimate of the variance of the quantization error e2 (ii) We want to increase
the sampling rate by 16 times. How many bits per samples you would use in order to
maintain the same level of quantization.
Solution: Refer problem in page 21 of Module 5 study material.

8. Principles of Communication – 15EC45 Dec2019 / Jan2020

10(b) Discuss briefly quantization noise and show the output signal to noise ratio of a 8
𝟑𝐏
uniform quantizer is [𝐒𝐍𝐑]𝐨 = 𝐱 𝟐 𝟐𝟐𝐑
𝐦𝐚𝐱
Solution : Refer section quantization noise pages 3 and 4 of module 5 study material

9. Principles of Communication – 15EC45 Aug / Sept - 2020

10(a) Explain PCM system 8
Solution: Refer section 5.12 pages 4 & 5 of Module 5 study material
10(b) Write short notes on (i) Regeneration (ii) VOCODER 8
Solution: (i) Refer section 5.12(d) page 7 and (ii) 5.14(a) – pages12 and 13 in Module 5 of
study material. Write up to basic speech model (included)

10. Principles of Communication – 17EC44 Aug / Sept- 2020

10(b) Write on application to vocoders, considering speech model used in voice coders and 8
block diagram of vocoder.
Solution: Refer section 5.14(a) – pages12 and 13 in Module 5 of study material. Write up
to basic speech model (included)
10(c) An analog waveform with BW 15 Hz is to be quantized with 200 levels and transmitted 4
via binary PCM signal. Find the rate of transmission and bandwidth required if 10
such signals are to be multiplexed and find the bandwidth requirement.
Solution : W = 15 Hz; L = 200  N = 20 Log2(L) = 20 Log2(200) = 7.64  8 bits
Rate of transmission rb = N(fs) = N(2W) = 8(2)(15) = 240 bps
 If 10 such signals are to be multiplexed
BW = n (NW) = 10(8)(15) = 1200 Hz

11. Principles of Communication Systems – 17EC44 Jan / Feb 2021

9(b) Design a PCM multiplexing system using 256 levels signal quantizer for transmission 10
of 3 signals m1, m2 and m3 band limited to 5 kHz, 10 kHz and 5 kHz respectively.
Assuming that each signal is sampled at Nyquist rate and 8 bits are transmitted
simultaneously. Compute (i) Maximum bit duration (ii) Channel bandwidth required
to pass PCM signal (iii) The commutator speed in rpm
Solution: Given L = 256 = 2N or N = Log2(L) = Log2(256)= 8
Bandwidth of three input signals,
W1 = 5 kHz, W2 = 10 kHz, W3 = 5 kHz
Since, the bandwidth of signals are harmonically related, the overall bandwidth of
multiplexed signals as
W = W1+W2+W3 = 5 + 10 + 5 = 20 kHz
So, the Nyquist rate for multiplexed signal is given by
fS = 2W = 2(20) = 40 kHz
Transmission rate rb = Nfs = 8(40) = 320 kbps
The maximum bit duration for the transmitted signal Tb = (1/rb) = (1/320k) = 3.125µsec
Channel bandwidth BT = (rb/2) = (320/2) = 160 kHz
There are n = 4 signals of band with W = 5 kHz and
speed of the commutator = 2W = 2(5) = 10 kHz
Commutator speed in rps = n(speed of the commutator) = 4(10) = 40 kbps
10(a) Describe the basic elements of a PCM system 10
Solution: Refer section 5.12 pages 4 & 5 of Module 5 study material
10(c) Represent the binary data given below interms of (i) unipolar NRZ signaling (ii) split 4
phase. Binary data 0 1 1 0 1 0 0 1
Solution : Refer section 5.12(b) in module 5 study material

12. Principles of Communication Systems – 18EC53 Jan / Feb 2021

9(a) With a neat block diagram, explain the basic elements of a PCM 08
Solution: Refer section 5.12 pages 4 & 5 of Module 5 study material
9(b) Discuss the concept and operation of delta modulation in detail 8
9(c) A PCM system uses a uniform quantizer followed by a 7 bit encoder. The bit rate of 4
the system is 50Mbps. What is the minimum message BW?
Solution: given rb = 50 Mbps and N = 7  fs = (rb/N) = (50/7) = 7.14 MHz
BW = N(fs/2) = 7(7.14/2) = 25 MHz
10(a) Write note on MPEG + Video 10
Solution : Refer section digitization of video and MPEG – pages 21 and 22 in module 5
study material
10(b) Draw the resulting waveform for 01101001 using NRZ, Polar NRZ, Uniform RZ and 6
bipolar RZ
Solution : Refer section 5.12(b) in module 5 study material
10(c) A TV signal with a bandwidth of 4.2 MHz is transmitted using binary PCM. The 4
 number of quantization levels is 512. Calculate (i) Code word length (ii) Transmission
BW (iii) Final bit rate.
Solution: Refer problem in page 20 of module 5 tudy material
13. Principles of Communication Systems – 15EC45 July / Aug 2021
10(a) Explain the quantization process quantization noise and show that the output SNR of 10
an uniform quantizer increases exponentially with increasing number of bits per
sample.
Solution : Amplitude quantization is the process of transforming sampled amplitude values
x(nTs) of the message signal x(t)at times t = nTs into discrete levels xq(nTs) taken from finite
set of possible amplitudes and the device which does this function is called Quantizer.

The noise appears at the output of the quantizer due to random errors produced in the
quantization process is called Quantization noise.
Refer section quantization noise pages 3 & 4 of module 5 study material

14. Principles of Communication Systems – 18EC53 July / Aug 2021

9 a. Explain the construction and regeneration of PCM signal 10
Solution: Refer section 5.12 pages 4 & 5 of Module 5 study material
b. Explain the construction of Delta modulated signal 6
Solution :
Delta modulation provides a staircase
approximation to the oversampled version
continuous time message signal. The
difference between the input and the
approximation is quantized into only two
levels + corresponding to positive and
negative differences, respectively If the approximation falls below the signal at any
sampling epoch, it is increased by +. On the other hand, if the approximation lies above
the signal, it is decreased by -.
Mathematical Analysis
Figure shows the delta modulator used for generating DM signal. The error between the
sampled value of x(t) and last approximated
sample is given as: 𝑒𝑝 𝑛𝑇𝑠 = 𝑥 𝑛𝑇𝑠 −
𝑥 𝑛𝑇𝑠
Where ep(nTs) = error at present sample;
x(nTs) = sampled signal of x(t);
𝑥 𝑛𝑇𝑠 is the last sample approximation of staircase waveform.
If we assume u(nTs) as the present sample approximation of staircase output, then
𝑢( 𝑛−1)𝑇𝑠 = 𝑥 𝑛𝑇𝑠 ------- (last sample approximation of staircase waveform)
Let us define a quantity v(nTs) in such a way that, v(nTs) =sgn[ep(nTs)]
This means that depending on the sign of error ep(nTs) , the sign of step size Δ is decided.
In other words

if v(nTs) = +Δ then a binary ‘1’ is transmitted ; if v (nTs) = –Δ then a binary ‘0’ is

transmitted
Where Ts = sampling interval
The output of the second summer is given by
𝑢(𝑛𝑇𝑠 ) = 𝑥 (𝑛𝑇𝑠 ) + 𝑣(𝑛𝑇𝑠 ) = 𝑥 (𝑛𝑇𝑠 ) + 𝑒𝑞 (𝑛𝑇𝑠 )
The input to the Quantizer is given by
𝑒𝑝 (𝑛𝑇𝑠 ) = 𝑥 (𝑛𝑇𝑠 ) − 𝑥 (𝑛𝑇𝑠 ) = 𝑥 (𝑛𝑇𝑠 ) − 𝑢[(𝑛 − 1)𝑇𝑠 ]
𝑒𝑝 (𝑛𝑇𝑠 ) = 𝑥 (𝑛𝑇𝑠 ) − [𝑥 [(𝑛 − 1)𝑇𝑠 ] + 𝑣[ 𝑛 − 1 𝑇𝑠 ]]
The output of the Quantizer is given by
𝑣(𝑛𝑇𝑠 ) = 𝑒𝑞 (𝑛𝑇𝑠 ) = ∆ 𝑠𝑔𝑛[𝑒𝑝 (𝑛𝑇𝑠 )]
Where 𝑥 (𝑛𝑇𝑠 ) − Previous value of the delay circuit
𝑒𝑞 (𝑛𝑇𝑠 ) = 𝑣(𝑛𝑇𝑠 ) − Quantizer Output
𝑢(𝑛𝑇𝑠 ) = 𝑣 (𝑛𝑇𝑠 ) + 𝑢[(𝑛 − 1)𝑇𝑠 ]
Assuming zero initial condition of Accumulation;
𝑛

𝑢(𝑛𝑇𝑠 ) = 𝑣 (𝑛𝑇𝑠 ) = ∆ 𝑠𝑔𝑛[𝑒𝑝 𝑖𝑇𝑆 ]

𝑖=0
𝑛

Accumulated version 𝑜𝑓 𝐷𝑀 𝑜𝑢𝑡𝑝𝑢𝑡 = 𝑣 𝑖𝑇𝑆 − − − − − (𝐴)

𝑖=0
𝑛−1

𝑥 (𝑛𝑇𝑠 ) = 𝑢 𝑛 − 1 𝑇𝑠 = 𝑣 𝑖𝑇𝑆 − − − − − (𝐵)

𝑖=0

Equation (A) and (B) describes the structure of Accumulator.

A Stair-case approximated waveform will be the output of the delta modulator with the
 step-size as delta (Δ). The output quality of the waveform is moderate.
c Write short notes on vocoder 4
Solution: Refer section 5.14(a) – pages12 and 13 in Module 5 of study material. Write up
to basic speech model (included)
OR
10 a. What is quantization? Derive the output SNR of the uniform quantizer. 7
Solution : Amplitude quantization is the process of transforming sampled amplitude values
x(nTs) of the message signal x(t)at times t = nTs into discrete levels xq(nTs) taken from finite
set of possible amplitudes and the device which does this function is called Quantizer.
Refer section quantization noise pages 3 and 4 of module 5 study material
b. To transmit a bit sequence 10011011. Draw the relevant waveform using (i) Unipolar 6
NRZ (ii) Polar NRZ (iii) Bipolar RZ (iv) Unipolar RZ (v) Manchester (Split phase)
Solution : Refer section 5.12(b) in module 5 study material
c Explain how digitization of video and MPEG is achieved with relevant diagrams 7
Solution : Refer section digitization of video and MPEG – pages 21 and 22 in module 5
study material