# Pulse Code Modulation (PCM) System MCQ || Pulse code Modulation Questions And Answers

1. Quantizing noise occurs in

1. PCM
2. TDM
3. FDM
4. PPM

Explanation:-

• Quantization is the process through which a range of continuous analog values are quantized or rounded off to a single value, thereby forming samples of a discrete digital signal.
• Quantization Error occurs when there is a difference between an input value and its quantized value.
• Quantization occurs when an analog signal is converted into its digital form, thus it occurs in Pulse Code Modulation (PCM).

2. A telephone signal with a cut-off frequency of 4 kHz is digitalized into 8-bit samples at a Nyquist sampling rate of s = 2 W. Calculate the baseband transmission bandwidth.

1. 8 kHz
2. 64 kHz
3. 16 kHz
4. 2 kHz

Explanation:-

Number of samples produced each sec : 2 × 4000 = 8000 (Nyquist criteria)

Each sample is encoded into 8 bits

Bit rate = 8 × 8000 = 64 kilo bits/sec

For PCM bit rate = Transmission bandwidth

Hence transmission bandwidth is 64 kHz

3. The main advantage of PCM is:

1. Less bandwidth
2. Better performance in presence of a noise
3. Possibility of multiplexing
4. Less power

Answer.2. Better performance in presence of a noise

Explanation:-

• PCM (Pulse Code Modulation) is a digital scheme for transmitting analog data.
• The amplitude of an analog signal can take any value over a continuous range, i.e. it can take on infinite values.
• But, digital signal amplitude can take on finite values.
• Analog signals can be converted into digital by sampling and quantizing.

• Encoding is possible in PCM.
• Very high noise immunity, i.e. better performance in the presence of noise.
• Convenient for long-distance communication.
• Good signal-to-noise ratio.

4. The bit rate of a digital communication system is R Kbits/sec. The modulation used is 32-QAM. The minimum bandwidth for inter-symbol interference-free transmission is –

1. R/10 Hz
2. R/10 KHz
3. R/5 Hz
4. R/5 KHz

Explanation:-

For M-ary signaling B.W ≥ $\frac{{{R_b}}}{{{{\log }_2}M}}$

Bit Rate Rb = R k bits/sec

32-ary signaling is used:

$\frac{R}{{{{\log }_2}32}} \cdot kHz$

Minimum Bandwidth = R/5 KHz

5. The number of bits used in a 4096 level PCM system is:

1. 12
2. 16
3. 20
4. 10

Explanation:-

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

Calculation:

The number of levels given = 4096, i.e. L = 4096

The number of bits used will be:

n = log2 (4096)

= log2 (212)

= 12 log2 (2)

n = 12

6. The disadvantage of FM over AM is that ________

1. The noise is very high for high-frequency signal
2. High modulating power is required
3. Larger bandwidth is required
4. High output power is required

Explanation:-

Advantages of FM over AM are:

• Improved signal to noise ratio.
• Smaller geographical interference between neighbouring stations.
• Well-defined service areas for given transmitter power.

• Much more Bandwidth
• More complicated receiver and transmitter.

7. To avoid slope overload error in delta modulation, the maximum amplitude of the input signal is

1. A ≤ 2πfm
2. A ≤ cos 2πfm
3. $A \le \frac{{2\pi {f_m}}}{{{\rm{\Delta }}{f_s}}}$
4. $A \le \frac{{{\rm{\Delta }}{f_s}}}{{2\pi {f_m}}}$

Answer.4. $A \le \frac{{{\rm{\Delta }}{f_s}}}{{2\pi {f_m}}}$

Explanation:-

Two types of distortions (Errors) occurs in Delta modulation system i.e.

2) Granular Error

To avoid the slope overload error, the optimum or desired condition is:

$\frac{{\rm{\Delta }}}{{{T_s}}} = \frac{{d\;m\left( t \right)}}{{dt}}$

i.e. the step rise of the quantized output must follow the input

$\frac{{\rm{\Delta }}}{{{T_s}}} < \frac{{d\;m\left( t \right)}}{{dt}}$

To prevent/avoid slope overload error, the condition that shall be satisfied is:

$\frac{{d\;m\left( t \right)}}{{dt}} \le \frac{{\rm{\Delta }}}{{{T_s}}}$

m(t) is a sinusoidal waveform given as:

m(t) = Am sin ωm t

The condition to avoid slope overload, therefore, becomes:

$\frac{d}{{dt}}({A_m}\sin {\omega _m}t) \le \frac{{\rm{\Delta }}}{{{T_s}}}$

${A_m}\cos {\omega _m}t\;\left( {{\omega _m}} \right) \le \frac{{\rm{\Delta }}}{{{T_s}}}$

Am cos (2π (mt)).2πfm ≤ Δ⋅fs

${A_m} \le \frac{{{\rm{\Delta }}{f_s}}}{{2\pi {f_m} \cdot \cos \left( {2\pi {f_m}f} \right)}}$

For maximum Amplitude, the above condition becomes:

${A_m} \le \frac{{{\rm{\Delta }} \cdot {f_s}}}{{2\pi {f_m}}}$

8. Consider the AM signal A(1 + μ cos (ωmt)) cos (ωct). The efficiency η of the AM signal for 75% modulation is approximate:

1. 33%
2. 15%
3. 22%
4. 50%

Explanation:-

The transmission efficiency of an AM signal is given by the expression:

$\eta = \frac{{{\mu ^2}}}{{2 + {\mu ^2}}} \times 100$

With μ = 0.75, the transmission efficiency will be

$\eta = \frac{{{{\left( {0.75} \right)}^2} \times 100}}{{2 + {{\left( {0.75} \right)}^2}}}$

μ = 21.9% ≅ 22%

9. In TV, an electrical disturbance (noise) affects

1. neither the video nor the audio signals
2. only the audio signals
3. both the video and audio signals
4. only the video signals

Explanation:-

In TV transmission:

• A video is Vestigial sideband modulated, which is a type of amplitude-modulated waveform
• The Video signals are thus encoded in amplitude variations of the carriers
• The Audio signal is encoded in FM waveform
• Thus, the audio signals are encoded as frequency variations of the carrier

Noise is the signal that affects amplitude majorly. Thus, the video signal is distorted from amplitude variations.

10. A signal contains components at 400 Hz and 2400 Hz This signal modulates a carrier of frequency 100 MHz. However, after demodulation, it is found that the 400 Hz signal component is present. The channel bandwidth is 15 kHz. What is the reason for the higher frequency signal not being detected properly?

1. Modulation used in FM and BW is insufficient
2. Modulation used in AM and BW is insufficient
3. Modulation used in FM but pre-emphasis is not used
4. Modulation used in AM but detector is for FM

Answer.3. Modulation used in FM but pre-emphasis is not used

Explanation:-

In FM broadcasting, pre-emphasis improvement is the improvement in the signal-to-noise ratio of the high-frequency portion of the baseband. The higher frequency signal is not to be detected properly because modulation is used in FM but pre-emphasis is not used.

In the process of detecting a frequency modulated signal, the receiver produces a noise spectrum that rises in frequency (a so-called triangular spectrum). This is the reason why a preemphasis is required.

Without pre-emphasis, the received audio would sound unacceptably noisy at high frequencies, especially under conditions of low carrier-to-noise ratio, i.e., during fringe reception conditions.

Preemphasis increases the magnitude of the higher signal frequencies, thereby improving the signal-to-noise ratio.

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