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

Answer.1. M/2 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.

Answer.2. ASK

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.

Answer.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.

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.

Answer.2. 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.

Answer.3. 256 kbps

Explanation:-

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

R_b = m.n.f_s

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.

f_s = 2f_m

f_m = Maximum frequency present at the modulating signal.

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

f_s = 2 × 4 = 8 kHz

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

n = log₂ 256 = log₂ 28

n = 8 bits

Now for 4 Voice signals

R_b = 4 × n × fs = 4 × 8 × 8000

R_b = 256 kbps

Answer.1. 2 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 = nf_s

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.

Answer.3. Nf Hz

Explanation:-

The bandwidth of PCM is given by:

BW = nf_s

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

Answer.3. Nf Hz

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

Answer.2. Digital circuits

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.

Answer.3. 8 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, R_b = 60 Kbits/sec

Bit rate = n fS

f_s = 6 kHz

And f_m is given as 4 kHz i.e

f_s ≥ 2f_m

f_s ≥ 8 kHz

So, f_s = 6kHz leads to undersampling.

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

Option 4: f_s = 9 kHz leads to oversampling

Hence, option 3 is correct f_s = 8 kHz

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