Communications Systems-

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UNIT-V

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 Digital Communication Systems
Transmitter

A/D Source Channel

Source Modulator
converter encoder encoder

Absent if
source is Noise Channel
digital

D/A Source Channel

User Detector
converter decoder decoder

Receiver

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 Topics to be Covered
Analog signal

Sampling
Discrete-time continuous-valued signal

Quantization
Discrete-time discrete-valued signal

Encoding

Bit mapping

Pulse Code Digital transmission

Modulation

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 Sampling Process

∞ ∞

xδ (t ) = ∑ x (nTs )δ (t − nTs ) = x (t )∑ δ (t − nT
s )
n =−∞ n =−∞

The sampling process can be regarded a modulation process with

carrier given by periodic impulses. It’s also called pulse modulation

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 Sampling Process

∞ ∞
1 n 1
Xδ ( f ) = X ( f ) ∗
Ts
∑ δ f −
Ts
=
Ts
∑ X f −n
Ts
n =−∞ n =−∞

X(f )
1
. If fs = < 2W or Ts > 1
T 2W
s -W W f
aliasing error, reconstruction is not
possible, Xδ ( f )

-1/Ts 1/Ts f

. The minimum sampling rate is known

Xδ ( f )
as Nyquist sampling rate

-1/Ts 1/Ts f

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 Reconstruction
. LPF with frequency response

Ts f <W
H( f ) = 1
0 f ≥ −W
Ts

. Ideal LPF

where

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 Reconstruction
. With this choice, we have

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 Topics to be Covered
Analog signal

Sampling
Discrete-time continuous-valued signal

Quantization
Discrete-time discrete-valued signal

Encoding

Bit mapping

Pulse Code Digital transmission

Modulation

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 Quantization

sampling

quantization
(rounding)

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 Quantization
Quantization is a non linear transformation which
maps elements from a continuous set to a f inite
set. It is also the second step required by A/D
conversion.
 Classification of Quantization
Process
Quantization

Uniform Non-Uniform
Quantization Quantization

Midtread Type Midrise type

Uniform Quantizer:
A Uniform quantizer is a type of quantizer in
wh ich th e s tep s ize remain s con s tan t
through out the input range.
Non-Uniform Quantizer:
A Uniform quantizer is a type of quantizer in
which the step size varies according to the
input signal values
Types of Uniform Quantizer:
1.Symmetric quantizer of Midtread type
2.Symmetric quantizer of Midrise type
In the input-output characteristics of midtread type the
origin lies in the middle of a tread of staircase like
graph.
In the input-output characteristics of midrise type the
origin lies in the middle of rising part of staircase like
graph.
 Pulse Code Modulation
 Pulse-code modulation or PCM is known as a digital pulse modulation
technique .

 In PCM an analog signal or information is converted into a binary

sequence, i.e.,’ 1’s and ‘0’s. The output of a PCM resembles a binary
sequence.

 PCM produces a series of numbers or digits instead of a pulse train. In

Pulse Code Modulat ion, t he m essage signal is represent ed by a
sequence of coded pulses.

PCM produces a series of numbers or digits instead of a pulse train.

Elements of a PCM System

It consists of three main parts

i.e. ,
1. Transmitter
2. Transmission path
3. Receiver
PCM Transmitter
PCM Transmission Path
 Regenerative Repeater
 The re pe at e r s pe r for m t hre e basic ope rat ions such as :
quantization, timing and decision making.
 The amplitude quantizer shapes the distorted PCM wave so as
t o com pe nsa t e f or t h e e f fe ct s of a m pl i t ude a nd ph a se
distortions. The timing circuit produces a periodic pulse train
which is derived from the input PCM pulses.
 The decision device makes a decision about whether the
equalized PCM wave at its input has 0 value or 1 value at the
instant of sampling.
PCM Receiver
 Problems on PCM
1.A Television signal having a B.W of 4.2MHz
is transmitted using binary PCM system.
Given that the no.of quantization levels is
512.Determine
(i) Code word length
(ii) Transmission B.W
(iii) Final Bit rate
(iv) Output signal to quantization noise ratio
2. The information in an analog signal voltage
waveform is to be transmitted over a PCM system
with an accuracy of ±0.1%(Full scale).The analog
voltage waveform has a B.W of 100 Hz and
amplitude range of -10 to +10 volts.
(i)Find the minimum sampling rate required
(ii) Find the no.of bits in each PCM word
(iii) Find minimum bit rate required in the PCM
signal
(iv) Find the minimum absolute channel bandwidth
required for the transmission of the PCM signal.
3. The B.W of an input signal to the PCM is restricted
to 4 KHz. The input signal varies in amplitude from
-3.8 v to +3.8v and has the average power of
30mw.The required signal to noise ratio is given as
20 dB. The PCM modulator produces binary output.
Assuming uniform quantization,
(i)Find the no.of bits required per sample.
(ii)Output of 30 such PCM coders are time
multiplexed. What would be the the minimum
required transmission bandwidth for this multiplexed
signal
4. A PCM system uses an uniform quantizer
followed by a 7-bit binary encoder. The bit
rate of the system is equal to 50 x 106
bits/sec.
(i) What is the maximum message signal B.W
for which the system operates satisfactorily?
(ii) Calculate the output signal to quantization
noise ratio when a full load sinusoidal
modulating wave of frequency 1MHz is
applied to the input.
5. The information in an analog waveform with
maximum frequency fm =3KHz is to be transmitted
over an M-level PCM system where the no.of
quantization levels is M=16.The quantization
distortion is specified not to exceed 1% of peak to
peak analog signal.
(i)What would be the maximum no.of bits per
sample that should be used in this PCM system?
(ii) What is the minimum sampling rate & what is
the resulting bit transmission rate?
6. Consider an audio signal consisting of the
sinusoidal term given as x(t)=3 cos(500πt).
(i) Determine the signal to quantization noise ratio
when this is quantized using 10 bit PCM.
(ii) How many bits of quantization are needed to
achieve a signal to quantization noise ratio of
atleast 40 dB.
7.A signal having B.W equal to 3.5kHz is sampled ,
quantized and coded by a PCM system. The coded
signal is then transmitted over a transmission
channel of supporting a transmission rate of 50k
bit/s. Determine the maximum signal to noise ratio
that can be obtained by the system. The input
signal has peak to peak value of 4v and rms value
of 0.2v.
8.A telephone signal bandlimited to 4 kHz is to be
transmitted by PCM. The signal to quantization
noise is at least 40 dB for the sinusoidal input. Find
the no. of levels into which the signal is to be
encoded. Find the B.W required for the transmission.
9. A signal m1 is bandlimited to 5 kHz and 3 other
signals m2,m3,m4 each bandlimited to 4 kHz are
time multiplexed and fed to PCM encoder having
1024 quantization levels. Find the bit rate of PCM
and also B.W required for transmission.
Differential pulse code
Modulation (DPCM)
Differential pulse code Modulation (DPCM)
DPCM Transmitter
DPCM Receiver
Delta Modulation (DM)
 Delta Modulation (DM)
Delta Modulation transmits only one bit per
sample.
The Present sample value is compared with the
previous sample value.
Input signal is approximated to step signal. This
step size is kept fixed.
The difference between input signal x(t) and
stair case approximated signal is conf ined to
two levels. i.e +Δ and -Δ
 If step is increased ,’1’ is transmitted
If step is reduced ,’0’ is transmitted.
Delta Modulation (DM)
 Delta Modulation Transmitter
sampled
b

b (nTs) = +Δ if x(nTs) ≥ x^(nTs)

- Δ if x(nTs) < x^(nTs)
b (nTs) = +Δ then ‘1’ is transmitted
- Δ then ‘0’ is transmitted.
 Delta Modulation Transmitter
The summer in accumulator adds quantizer output
with previous sample approximation. This gives the
present sample approximation i.e,

u(nTs)=u[(n-1)Ts]+[±Δ]
(or)
u(nTs)=u[(n-1)Ts]+b(nTs)

Thus, depending on the sign of e(nTs),one bit

quantizer generates an output of +Δ or –Δ.
If step size is +Δ,then binary ‘1’ is transmitted.
If step size is -Δ,then binary ‘0’ is transmitted.
 Delta Modulation Receiver

Input
Advantages of Delta Modulation

DM transmits only one bit for one sample, the
signalling rate and transmission channel bandwidth
is quite small.
The transmitter and receiver implementation for DM
is very much simple.

Drawbacks of Delta Modulation
The DM has two major drawbacks
1. Slope overload distortion
2. Granular or Idle Noise.
Drawbacks of Delta
Modulation
 Slope Overload Distortion
 This distortion arises because of large dynamic range of
the input signal. The rate of rise of input signal x(t) is so
high that the staircase signal can not approximate it, the
step size ‘Δ’ becomes too small for staircase signal u(t) to
follow the step segment of x(t).

 Hence, there is a large error between the staircase

approximated signal and the original input signal x(t).

 This error or noise is known as slope overload distortion .

 To reduce this error a large step size is required, when

slope of signal x(t) is high.

 This can be achieved by using Adaptive Delta Modulation

Technique.
 Granular Noise
 Granular or Idle noise occurs when the step size is too
large compared to small variation in the input signal.

 This means that for very small variations in the input

signal, the staircase signal is changed by large amount (Δ)
because of large step size.

 Figure shows that when the input signal is almost flat , the
staircase signal u(t) keeps on oscillating by ±Δ around the
signal

 The error between the input and approximated signal is

called granular noise.

 The solution to this problem is to make the step size

 Solution
 In order to overcome the quantization errors due to slope
overload and granular noise, the step size (Δ) is made
adaptive to variations in the input signal x(t).

 Particularly in the steep segment of the signal x(t), the

step size is increased. And the step is decreased when
the input is varying slowly.

 This method is known as Adaptive Delta Modulation

(ADM).
Problems on Delta Modulation
1 .A Delta modulator system is designed to
operate at 5 times the Nyquist rate for a signal
having a bandwidth equal to 3 KHz bandwidth.
Calculate the maximum amplitude of a 2 KHz
input sinusoid for which the delta modulator
does not have slope overload. Given that the
quantizing step size is 250mV.
2. Determine the output signal to noise
ratio of a linear delta modulation system
for a 2 KH z s i n u s oi d al i n p u t s i g n al
s ampled at 64 KHz. Slope overload
dis tor tion is not pres ent & the pos t
reconstruction f ilter has a bandwidth of
4 KHz.
3.A sinusoidal voice signal x(t)= cos
(6000πt) is to be transmitted using either
PCM or DM. The sampling rate for PCM
system is 8 KHz & for the transmission
with DM, the step size is decided to be of
31.25mv.The slope overload distortion is
to be avoided. Assume that the no.of
quantization levels for a PCM system is
64.Detemine the signaling rate of both
the systems.
4. In a DM system, the voice signal is
sampled at a rate of 64,000 samples/sec.
The maximum signal amplitude Amax=1.
(i) Determine minimum value of step
size to avoid slope overload.
(ii) Determine quantization noise power
if voice signal bandwidth is 3.5 KHz.
(iii) Assuming voice signal to be a sine
wave, determine signal power s0 and
SNR.
5. A DM system is designed to operate at
3 times the Nyquist rate for a signal with
a 3 KHz bandwidth. The quantizing step
size is 250 mV.
(i) Determine the maximum amplitude
of a 1 KHz input sinusoid for which
the delta modulator does not show
slope overload.
(ii) Determine the post f il tered output
signal to quantizing noise ratio for the
signal of part (i).
6.In a single integration DM scheme, the
voice signal is sampled at a rate of
64KHz.The maximum signal amplitude
is 1 v olt, v oice s ign al ban dwidth is
3.5KHz.
(i) Determine minimum value of step
size to avoid slope overload.
(ii) Determine granular Noise N0.
(iii) Assuming signal to be sinusoidal,
calculate signal power and signal to
noise ratio.
(iv) A s s u m i n g t h a t n o i s e s i g n a l
amplitude is uniformly distributed in
Thank you