UNIT-IV

Noise in analog modulation: Introduction, Receiver Model, Noise in DSB-SC receivers,

Noise in AM receivers, Threshold effect, Noise in FM receivers, Capture effect, FM
threshold effect, FM threshold reduction, Pre-emphasis and De-emphasise in FM.

QUESTIONS (PDF: 10_03_2020_1292903273) PAGES FROM 104-122. UNIT-IV NOTES

AVAILABLE)

1.A)Define noise and write the types of noises.

B)Explain noise in receiver model.

2.Explain the noise models in DSB-SC that figure of merit for DSBSC-AM is

unity.
B) Identify the significance and importance of Pre-emphasis in
analog communication.(HINT: Pre-emphasis works by boosting the high-frequency portion of
the signal. This compensates for the high-frequency loss in the cable. De-emphasis works by
cutting the low-frequency portion of the signal. This may be coupled with an increased transmit
voltage)

4 A)Illustrate the importance of FM threshold effect.(HINT: Below the FM threshold point, the
noise signal (whose amplitude and phase are randomly varying) may instantaneously have
amplitude greater than that of the wanted signal. When this happens, the noise will produce a
sudden change in the phase of the FM demodulator output)

FM Threshold Effect

Introduction:

This section explains the threshold effects of the FM system in detail.

Theory:

 The defining the output signal-to-noise ratio of an FM receiver, is valid only if the carrier-to-
noise ratio, measured at the discriminator input, is high compared with unity.
 It is found experimentally that as the input noise power is increased so that the carrier-to-
noise ratio is decreased, the FM receiver breaks.
 At first, individual clicks are heard in the receiver output, and as the carrier-to-noise ratio
decreases still further, the clicks rapidly merge into a crackling or sputtering sound.
 This phenomenon is known as the threshold effect.
 The threshold is defined as the minimum carrier-to-noise ratio yielding an FM improvement
that is not significantly deteriorated from the value predicted by the usual signal-to-noise
formula assuming a small noise power.
 For a qualitative discussion of the FM threshold effect, consider first the case when there is a
no signal present, so that the carrier wave is unmodulated.
 Then the composite signal at the frequency discriminator input is

 where nr (t) and nQ(t) are the in-phase and quadrature components of the narrowband noise
n(t) with respect to the carrier wave.
 The phasor diagram of Figure 2.43 displays the phase relations between the various
components of x(t)
 As the amplitudes and phases of nr(t) and nQ(t) change with time in a random manner, the
point Pi [the tip of the phasor representing x(t)} wanders around the point P2 (the tip of the
phasor representing the carrier).
 When the carrier-to-noise ratio is large, Kj(t) and nQ(t) are usually much smaller than the
carrier amplitude Ac , and so the wandering point P, in Figure 2.43 spends most of its time
near point P2.
 Thus the angle θ(t) is approximately nQ(t)/Ac to within a multiple of 2π.
 When the carrier-to-noise ratio is low, on the other hand, the wandering point Px occasionally
sweeps around the origin and θ(t) increases or decreases by 2π radians.
B) Identify the significance and importance of Pre-emphasis & De-emphasis in
analog communication. .(HINT: Pre-emphasis works by boosting the high-frequency portion of
the signal. This compensates for the high-frequency loss in the cable. De-emphasis works by
cutting the low-frequency portion of the signal. This may be coupled with an increased transmit
voltage)

5.a) Illustrate the importance of capture effect.

b) Short note on how to reduce the threshold in FM.

UNIT-V
Transmission of Binary Data in Communication Systems: Digital Codes, Principles of Digital
Transmission, Transmission Efficiency, Modem Concepts and Methods – FSK, BPSK, Error
Detection and Correction

1.Write a short notes on a) Digital Codes. b) Principles of Digital Transmission

2. Explain the any two Modem Concepts and Methods

3.Generate and Regenerate the FSK- with neat sketch of wave forms.

4 Generate and Regenerate the BPSK- with neat sketch of wave forms.

5.Explain the error detection and correction concepts in

 a)Cyclic redundancy check
b)Hamming code method.

Encoding the message with hamming code

Selecting the number of redundant bits
The hamming code uses the number of redundant bits depending on the number of information
bits in the message.

Let n be the number of information or data bits, then the number of redundant bits P is
determined from the following formula,

For example, if 4-bit information is to be transmitted, then n=4. The number of redundant bits is
determined by the trial and error method.

Let P=2, we get,

The above equation implies 4 not greater than or equal to 7. So let’s choose another value of
P=3.

Now, the equation satisfies the condition. So number of redundant bits, P=3.
In this way, the number of redundant bits is selected for the number of information bits to be
transmitted.

Choosing the location of redundant bits

For the above example, the number of data bits n=4, and the number of redundant bits P=3. So
the message consists of 7 bits in total that are to be coded. Let the rightmost bit be designated as
bit 1, the next successive bit as bit 2 and so on.

The seven bits are bit 7, bit 6, bit 5, bit 4, bit 3, bit 2, bit 1.
In this, the redundant bits are placed at the positions that are numbered corresponding to the
power of 2, i.e., 1, 2, 4, 8,… Thus the locations of data bit and redundant bit are D4, D3, D2, P3,
D1, P2, P1.

Assigning the values to redundant bits

Now it is time to assign bit value to the redundant bits in the formed hamming code group. The
assigned bits are called a parity bit.

Each parity bit will check certain other bits in the total code group. It is one with the bit location
table, as shown below.

Bit Location 7 6 5 4 3 2 1

Bit designation D4 D3 D2 P3 D1 P2 P1

11 10 01
Binary representation 111 101 011 001
0 0 0

Information / Data
D4 D3 D2 D1
bits

Parity bits P3 P2 P1 Parity bit P1 covers all

data bits in positions
whose binary representation has 1 in the least significant position(001, 011, 101, 111, etc.). Thus
P1 checks the bit in locations 1, 3, 5, 7, 9, 11, etc..

Parity bit P2 covers all data bits in positions whose binary representation has 1 in the second
least significant position(010, 011, 110, 111, etc.). Thus P2 checks the bit in locations 2, 3, 6, 7,
etc.

Parity bit P3 covers all data bits in positions whose binary Each parity bit checks the
corresponding bit locations and assign the bit value as 1 or 0, so as to make the number of 1s as
even for even parity and odd for odd parity.

representation has 1 in the third least significant position(100, 101, 110, 111, etc.). Thus P3
checks the bit in locations 4, 5, 6, 7, etc.

Example problem 1
Encode a binary word 11001 into the even parity hamming code.

Given, number of data bits, n =5.

To find the number of redundant bits,

Let us try P=4.

The equation is satisfied and so 4 redundant bits are selected.

So, total code bit = n+P = 9

The redundant bits are placed at bit positions 1, 2, 4 and 8.

Construct the bit location table.

Bit Location 9 8 7 6 5 4 3 2 1

Bit designation D5 P4 D4 D3 D2 P3 D1 P2 P1

100 001
Binary representation 1000 0111 0110 0101 0100 0010 0001
1 1

Information bits 1 1 0 0 1

Parity bits 1 1 0 1
To determine the parity bits

For P1: Bit locations 3, 5, 7 and 9 have three 1s. To have even parity, P1 must be 1.

For P2: Bit locations 3, 6, 7 have two 1s. To have even parity, P2 must be 0.

For P3: Bit locations 5, 6, 7 have one 1s. To have even parity, P3 must be 1.

For P4: Bit locations 8, 9 have one 1s. To have even parity, P2 must be 1.

Thus the encoded 9-bit hamming code is 111001101.

How to detect and correct the error in the hamming code?

After receiving the encoded message, each parity bit along with its corresponding group of bits
are checked for proper parity. While checking, the correct result of individual parity is marked as
0 and the wrong result is marked as 1.

After checking all the parity bits, a binary word is formed taking the result bits for P1 as LSB. So
formed binary word gives the bit location, where there is an error.

If the formed binary word has 0 bits, then there is no error in the message.

Example problem 2
Let us assume the even parity hamming code from the above example (111001101) is
transmitted and the received code is (110001101). Now from the received code, let us detect
and correct the error.

To detect the error, let us construct the bit location table.

Bit Location 9 8 7 6 5 4 3

Bit designation D5 P4 D4 D3 D2 P3 D1

Binary representation 1001 1000 0111 0110 0101 0100 0011

Received code 1 1 0 0 0 1 1
Checking the parity bits

For P1 : Check the locations 1, 3, 5, 7, 9. There is three 1s in this group, which is wrong for even
parity. Hence the bit value for P1 is 1.

For P2 : Check the locations 2, 3, 6, 7. There is one 1 in this group, which is wrong for even
parity. Hence the bit value for P2 is 1.

For P3 : Check the locations 3, 5, 6, 7. There is one 1 in this group, which is wrong for even
parity. Hence the bit value for P3 is 1.

For P4 : Check the locations 8, 9. There are two 1s in this group, which is correct for even parity.
Hence the bit value for P4 is 0.

The resultant binary word is 0111. It corresponds to the bit location 7 in the above table. The
error is detected in the data bit D4. The error is 0 and it should be changed to 1. Thus the
corrected code is 111001101.