# Digital Modulation MCQ || Introduction To Digital Modulation Questions and Answers

Ques.1. The bit rate of the digital communication system is M kbps. The modulation used is 16 QAM. The minimum bandwidth required for ideal transmission is _________.

1. M/2 kHz
2. M/16 kHz
3. M kHz
4. M/8 kHz

Explanation:-

Given

Bit rate = M kbps

Number of levels = N = 16

${\left( {BW} \right)_{min}} = \frac{{2{R_b}}}{{{{\log }_2}N}}Hz$

Where,

Rb = bit rate in bps

N = number of levels in M-Array scheme

$\begin{gathered} \therefore {\left( {BW} \right)_{{\text{min}}}} = \frac{{2{R_b}}}{{{{\log }_2}N}}Hz \hfill \\ \hfill \\ = \frac{{2M}}{{{{\log }_2}16}}kHz \hfill \\ \hfill \\ = \frac{{2M}}{{{{\log }_2}{2^4}}}kHz \hfill \\ \hfill \\ = \frac{{2M}}{{4\: \times \:{{\log }_2}2}}kHz \hfill \\ \end{gathered}$

(BW)min = M/2 KHz

So. The minimum bandwidth will be M/2 kHz, for ideal transmission.

Ques.2. Which of the following signaling Scheme the least noise immunity?

1. QAM
3. FSK
4. PSK

Explanation:-

ASK, PSK and FSK are signaling schemes used to transmit binary sequences through free space. In these schemes, bit-by-bit transmission through free space occurs.

Among all ASK has the least noise immunity due to the following:

1. FSK is less susceptible to errors than ASK – the receiver looks for specific frequency changes over a number of intervals, so voltage (noise) spikes can be ignored.

2. PSK allows information to be transmitted in the radio communication in a way more efficiently as compared to that of FSK and it is also less prone to error when we compare to ASK modulation.

3. ASK technique is not suitable for high-bit rate data transmission and has poor bandwidth efficiency.

Hence it is highly susceptible to noise and other external factors.

Ques.3. The following type of multiplexing cannot be used for Digital signaling?

1. FDM
2. TDM
3. WDM
4. Both 1 and 3

Explanation:-

Multiplexing is the process of combining multiple signals into one signal, over a shared medium.

If the analog signals are multiplexed, then it is called analog multiplexing. Similarly, if the digital signals are multiplexed, then it is called digital multiplexing.

Types of Multiplexing:

(1) Analog Multiplexing

• Frequency Division Multiplexing
• Wavelength Division Multiplexing

Digital Multiplexing:
The term digital represents discrete bits of information. Hence, the available data is in the form of frames or packets, which are discrete.

Time Division Multiplexing

• In Time Division Multiplexing (TDM), the time frame is divided into slots. This technique is used to transmit a signal over a single communication channel, by allotting one slot for each message.
• Time Division Multiplexing (TDM) can be classified into Synchronous TDM and Asynchronous TDM.

Synchronous TDM

• In Synchronous TDM, the input is connected to a frame. If there are ‘n’ number of connections, then the frame is divided into ‘n’ time slots. One slot is allocated for each input line.
• In this technique, the sampling rate is common for all signals and hence the same clock input is given. The MUX allocates the same slot to each device at all times.

Asynchronous TDM

• In Asynchronous TDM, the sampling rate is different for each of the signals and a common clock is not required.
• If the allotted device for a time slot transmits nothing and sits idle, then that slot can be allotted to another device, unlike synchronous
• This type of TDM is used in Asynchronous transfer mode networks.

Ques.3. Comparing Delta Modulation (DM) with PCM systems, DM requires:

1. a lower sampling rate

2. a higher sampling rate

3. least bandwidth

4. simpler hardware

1. 1, 2 and 4 only
2. 1, 2 and 3 only
3. 2, 3 and 4 only
4. 1, 3 and 4 only

Answer.3. 2, 3 and 4 only

Explanation:-

• In PCM an analog signal is sampled and encoded into different levels before transmission
• The bandwidth of PCM depends on the number of levels If each sample is encoded into n bits, then the bandwidth of PCM is nfs
• However, in the case of Delta modulation, each sample is sent using only 1 bit which is +Δ or -Δ Hence there is a bandwidth saving in Delta modulation
• DM has a simple hardware requirement in comparison to PCM.

Ques.4. Which modulation technique does not use past information for modulation?

1. Delta modulation
2. Pulse Code Modulation
3. Adaptive  Differential Pulse Code Modulation

Explanation:-

“In PCM, the modulating signal is sampled and the amplitude of each sample is rounded off to the nearest one of the finite set of allowable values, then a binary digit is generated, which is proportional to the rounded-off value, and the binary, digits are transmitted.

• It is used in voice, audio, and CD recording
• PCM system doesn’t use the previous samples in its processing of information.

Ques.5. Four voice signals, each limited to 4 kHz and sampled at Nyquist rate, are converted into binary PCM signal using 256 quantization levels. The bit transmission rate for the time-division multiplexed signal will be

1. 8 kbps
2. 64 kbps
3. 256 kbps
4. 512 kbps

Explanation:-

The bitrate of a PCM system for an encoded signal sampled at a frequency of fs is given by:

Rb = m.n.fs

fs = Sampling frequency

n = number of bits used for encoding.

m = number of message signals

n is related to the number of quantization levels (L) as:

L = 2n

or n = log2L

Calculation:

Since the sampling frequency is not mentioned, we’ll assume it to be sampled at the Nyquist rate, i.e.

fs = 2fm

fm = Maximum frequency present at the modulating signal.

∴ For the given band-limited signal with a frequency 4 kHz, the sampling frequency will be:

fs = 2 × 4 = 8 kHz

With L = 256, the number of bits will be:

n = log256 = log2 28

n = 8 bits

Now for 4 Voice signals

Rb = 4 × n × fs = 4 × 8 × 8000

Rb = 256 kbps

Ques.6. If the number of bits per sample in a PCM system is increased from 8 to 16, then the bandwidth will be increased by

1. 2 times
2. 4 times
3. 8 times
4. 16 times

Explanation:-

The number of levels for an n-bit PCM system is given by:

L = 2n

Also, the bandwidth of PCM is given by:

BW = nfs

n = number of bits to encode

fs = sampling frequency

For n = 8, the bandwidth will be:

B.W. = 8 fs

Similarly, For n = 16, the bandwidth will be:

B.W. = 16 fs

We observe that the Bandwidth is increased by 2 times.

Ques.7. The bandwidth of an N-bit binary-coded PCM signal for modulating a signal sampled at ‘f’ Hz is

1. f/N Hz
2. f/N2 Hz
3. Nf Hz
4. N2f Hz

Explanation:-

The bandwidth of PCM is given by:

BW = nfs

N = number of bits to encode

f = sampling frequency

Note:

The number of levels for an N-bit PCM system is given by:

L = 2N

We can also state that the number of bits for a given quantization level will be:

N = log2 L

Ques.8. ________ is mostly preferred for telegraphy.

1. Single-tone modulation
2. On-off keying
3. Frequency shift keying
4. Pulse code modulation

Explanation:-

Frequency shift keying is mostly preferred for telegraphy.

• FSK is a system of frequency modulation in which the nominal unmodulated carrier frequency corresponds to the mark condition, and space is represented by a downward frequency shift
• In the FSK generator, the frequency shift may be obtained by applying the varying dc output of the telegraph machine to a varactor diode in a crystal oscillator
• At the receiving end, the signal is demultiplexed and applied to a standard phase discriminator
• This is how a telegraph works with FSK modulation

Ques.9. Which circuit is called a regenerative repeater?

1. Analog circuits
2. Digital circuits
3. Amplifiers
4.  A/D converters

Explanation:-

A regenerative repeater amplifies and reconstructs such a badly distorted digital signal and develops a nearly perfect replica of the original at its output. Regenerative repeaters are an essential key to digital transmission in that we could say that the “noise stops at the repeater.” Digital circuits are called regenerative repeaters.

Ques.10. In a 10-bit PCM system, a message signal having a maximum frequency of 4 KHz is to be transmitted. If the bit rate of this PCM system is 60 Kbits / sec, the appropriate sampling frequency is

1. 6 kHz
2. 7 kHz
3. 8 kHz
4. 9 kHz

Explanation:-

The bandwidth of a PCM system for an encoded signal sampled at a frequency of fs is given by:

Bit rate = n fS

fS = Sampling frequency

n = number of bits used for encoding.

n is related to the number of quantization levels (L) as:

L = 2n

n = log2 L

Calculation:

The maximum frequency for the signal will be 4 kHz.

n = 10 bits, Rb = 60 Kbits/sec

Bit rate = n fS

fs = 6 kHz

And fm is given as 4 kHz i.e

fs ≥ 2fm

fs ≥ 8 kHz

So, fs = 6kHz leads to undersampling.

Hence, options 1 and 2 can’t be correct.

Option 4: fs = 9 kHz leads to oversampling

Hence, option 3 is correct fs = 8 kHz

∴ The appropriate value of the sampling frequency (fs) will be 8 kHz

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