Electronic Devices and Circuits

Digital Transmission

Question 1
Marks : +2 | -2
Pass Ratio : 100%
A speech signal has a total duration of 20 sec. It is sampled at the rate of 8 kHz and then PCM encoded. The signal-to-quantization noise ratio is required to be 40 dB. The minimum storage capacity needed to accommodate this signal is
1.12 KBytes
140 KBytes
168 KBytes
None of the mentioned
Explanation:
(SNR)q = 1.76 + 6.02(n) = 40 dB, n = 6.35
Question 2
Marks : +2 | -2
Pass Ratio : 100%
Assuming that the signal is quantized to satisfy the condition of previous question and assuming the approximate bandwidth of the signal is W. The minimum required bandwidth for transmission of a binary PCM signal based on this quantization scheme will be
5 W
10 W
20 W
None of the mentioned
Explanation:
The minimum bandwidth requirement for transmission of a binary PCM signal is BW= vW. Since v 10, we have BW = 10 W.
Question 3
Marks : +2 | -2
Pass Ratio : 100%
Analog data having highest harmonic at 30 kHz generated by a sensor has been digitized using 6 level PCM. What will be the rate of digital signal generated?
120 kbps
200 kbps
240 kbps
180 kbps
Explanation:
Nyquist Rate = 2 x 30k = 60 kHz 2n should be greater than or equal to 6. Thus n 3, Bit Rate = 60×3 = 18 kHz.
Question 4
Marks : +2 | -2
Pass Ratio : 100%
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
8 kbps
64 kbps
256 kbps
512 kbps
Explanation:
Nyquist Rate 2 x 4k = 8 kHz
Question 5
Marks : +2 | -2
Pass Ratio : 100%
A TDM link has 20 signal channels and each channel is sampled 8000 times/sec. Each sample is represented by seven binary bits and contains an additional bit for synchronization. The total bit rate for the TDM link is
1180 K bits/sec
1280 K bits/sec
1180 M bits/sec
1280 M bits/sec
Explanation:
Total sample 8000 x 20 = 160 k sample/sec
Question 6
Marks : +2 | -2
Pass Ratio : 100%
A linear delta modulator is designed to operate on speech signals limited to 3.4 kHz. The sampling rate is 10 time the Nyquist rate of the speech signal. The step size is 100 m V. The modulator is tested with a this test signal required to avoid slope overload is
2.04 V
1.08 V
4.08 V
2.16 V
Explanation:
Amax = (0.1 x 68k)/(2000p) or 1.08V.
Question 7
Marks : +2 | -2
Pass Ratio : 100%
Four signals each band limited to 5 kHz are sampled at twice the Nyquist rate. The resulting PAM samples are transmitted over a single channel after time division multiplexing. The theoretical minimum transmissions bandwidth of the channel should be equal to
5 kHz
20 kHz
40 kHz
80 kHz
Explanation:
fm = 5 kHz, Nyquist Rate 2 x 5 = 10 kHz Since signal are sampled at twice the Nyquist rate so sampling rate 2 x 10 = 20 kHz. Total transmission bandwidth 4 x 20 = 80 kHz.
Question 8
Marks : +2 | -2
Pass Ratio : 100%
A speech signal occupying the bandwidth of 300 Hz to 3 kHz is converted into PCM format for use in digital communication. If the sampling frequency is8 kHz and each sample is quantized into 256 levels, then the output bit the rate will be
3 kb/s
8 kb/s
64 kb/s
256 kb/s
Explanation:
fs = 8 kHz, 2n = 256 or n = 8. Bit Rate = 8 x 8k = 64 kb/s.
Question 9
Marks : +2 | -2
Pass Ratio : 100%
In PCM system, if the quantization levels are increased form 2 to 8, the relative bandwidth requirement will
Remain same
Be doubled
Be tripled
Become four times
Explanation:
If L = 2, then 2 = 2n or n = 1 ND. If L = 8, then 8 = 2n or n = 3. So relative bandwidth will be tripled.
Question 10
Marks : +2 | -2
Pass Ratio : 100%
A sinusoidal massage signal m(t) is transmitted by binary PCM without compression. If the signal to-quantization-noise ratio is required to be at least 48 dB, the minimum number of bits per sample will be
8
10
12
14
Explanation:
3(L2)/2 = 48 db or L = 205.09. Since L is power of 2, so we select L = 256 Hence 256 = 28, So 8 bits per sample is required.