Ques.11. A speech signal is sampled at 8 kHz and encoded in PCM format using 8-bit/sample PCM data is transmitted through a baseband channel via 4-level PAM. The minimum Bandwidth required for transmission is
16 kHz
8 kHz
24 kHz
10 kHz
Answer.1. 16 kHz
Explanation:-
Given
fs = 8 kHz
n = 8 bits/sample
M = 4
Rb = nfs
Minimum Bandwidth will be required if transmission of data takes place using sinc pulses and the bandwidth is given by:
Ques.12. The bandwidth of a video signal is 4.5 MHz. The signal is to be transmitted using PCM with 1024 quantization levels. The minimum bit rate required for transmission is
90 Mbps
45 Mbps
4.5 Mbps
10 Mbps
Answer.1. 90 Mbps
Explanation:-
For a minimum bit rate, the modulating signal must be sampled at the minimum sampling rate, i.e. at the Nyquist Rate.
fs = 2fm
fm = Maximum frequency present at the modulating signal.
∴ For the given bandlimited signal with a maximum frequency of 4.5 MHz, the sampling frequency will be:
fs = 2 × 4.5 M = 9 MHz
With L = 1024, the number of bits will be:
n = log2 1024 = log2 210
n = 10 bits
Now, the required minimum bit rate will be:
Rb = n fs = 10 × 9 Mbps
Rb = 90 Mbps
Ques.13. What are the advantages of digital circuits?
Less noise
Less interference
More flexible
All of the above
Answer.4. All of the above
Explanation:-
Digital circuits are more reliable.
Digital circuits are easy to design and cheaper than analog circuits.
The hardware implementation in digital circuits is more flexible than analog.
The occurrence of cross-talk is very rare in digital communication.
Digital circuits are less subject to noise, distortion, and interference as it works on digital pulses, and also the pulses can be regenerated.
Ques.14. In digital transmission, the modulation technique that requires the minimum bandwidth is:
PCM
PAM
DPCM
Delta modulation
Answer.4. Delta modulation
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
The bandwidth of DPCM is almost the same as that of PCM signal, the only difference between PCM and DPCM is that the dynamic range is reduced in the DPCM signal
However, in the case of Delta modulation, each sample is sent using only 1 bit which is +Δ or -Δ
Hence there is bandwidth saving in Delta modulation
A comparison of different modulation schemes is as shown in the table below:
Parameter
PCM
DM
DPCM
Number of bits
It can use 4, 8
or 16 bits per sample
It uses only
one bit for one sample
Bits can be more than one but are less than PCM
Level/Step size
Step size
is fixed
Step size is fixed
and
cannot be varied
A fixed number of levels are used.
Quantization error or Distortion
Quantization error depends
on the number
of levels used
Slope
overload distortion and granular
noise is
present
Slope overload distortion
and quantization noise is
present
Bandwidth
of the transmission channel
The highest bandwidth is required since the number of bits is high
The lowest bandwidth is required
The bandwidth required is lower than PCM
Signal to Noise ratio
Good
Poor
Fair
Area of Application
Audio and
Video
Telephony
Speech and images
Speech and video
Ques.15. Multiplexing scheme which uses carrier phase shifting and synchronous detection to permit two DSB signals to occupy the same frequency band is called
NBFM
CDMA
QAM
FDMA
Answer.3. QAM
Explanation:-
QAM (quadrature amplitude modulation) is a method of combining two amplitude modulation (AM) signals into a single channel. QAM is used to achieve high levels of spectrum usage efficiency. This is accomplished by utilizing both the amplitude and phase components to provide a form of modulation.
1) QAM scheme uses carrier phase shifting and synchronous detection to permit two DSB signals to occupy the same frequency band.
2) QAM uses two carrier signals with the same frequency but with which are in quadrature. Quadrature here means out phase by 90°.
3) Since the two signals have the same frequency, they are detected using synchronous detection.
4) Hence, amplitude and the phase of the carrier frequency both vary with the message signal.
A constellation diagram of a QAM signal with 2 different amplitude levels and 8 different phases is shown:
Ques.16. In QAM, both ________ of a carrier frequency vary.
frequency and amplitude
phase and amplitude
phase and frequency
Phase and frequency
Answer.2. phase and amplitude
Explanation:-
QAM (quadrature amplitude modulation) is a method of combining two amplitude modulation (AM) signals into a single channel. QAM is used to achieve high levels of spectrum usage efficiency. This is accomplished by utilizing both the amplitude and phase components to provide a form of modulation.
QAM is a mixture of ASK and PSK. Hence, amplitude and the phase of the carrier frequency both vary with the message signal.
Ques.17. The main advantage of PCM system is lower
Bandwidth
Power
Noise
None of these
Answer.3. Noise
Explanation:-
PCM (Pulse Code Modulation):
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.
Advantages of PCM:
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.
Disadvantage of PCM:
The circuitry is complex.
It requires a large bandwidth.
Synchronization is required between transmitter and receiver.
Ques.18. An analog voltage is in the range of 0 to 8 V is divided in eight equal intervals for conversion to 3-bit digital output. The maximum quantization error is
0 V
0.5 V
1 V
2 V
Answer.2. 0.5 V
Explanation:-
The maximum quantization is given as:
Qe(max) = Δ/2
Δ = (Vmax − Vmin)/L
L = Number of levels
Calculation:
With n = 3, the number of levels will be:
L = 23 = 8
With analog input in the range 0 to 8 V and L = 8, the step size will be:
Δ = (8 − 0)/8 = 1
Now, the maximum quantization error will be:
Qe(max) = 1/2 = 0.5
Ques.19. To transmit N signals each band limited to fm Hz by time-division multiplexing will require a minimum bandwidth of
fm
2fm
N fm
fm/N
Answer.3. N fm
Explanation:-
Given:
N signal with fm Bandwidth.
TDM minimum B.W = Nnfm
Assume n = 1 Bit
BWmin = Nfm
Ques.20. Which block or device does the data compression?
Channel encoder
Source encoder
Modulator
None of the mentioned
Answer.2. Source encoder
Explanation:-
Source encoding aims to convert information waveforms (text, audio, image, video, etc.) into bits, the universal currency of information in the digital world. Source encoder converts the digital or analog signal to a sequence of binary digits. This process is called source encoding or compression.