1. 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
a) 5 W
b) 10 W
c) 20 W
d) 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.
2. In PCM system, if the quantization levels are increased form 2 to 8, the relative bandwidth requirement will
a) Remain same
b) Be doubled
c) Be tripled
d) 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.
3. 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 is 8 kHz and each sample is quantized into 256 levels, then the output bit the rate will be
a) 3 kb/s
b) 8 kb/s
c) 64 kb/s
d) 256 kb/s
Explanation: fs = 8 kHz, 2n = 256 or n = 8. Bit Rate = 8 x 8k = 64 kb/s.
4. 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?
a) 120 kbps
b) 200 kbps
c) 240 kbps
d) 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.
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
a) 8 kbps
b) 64 kbps
c) 256 kbps
d) 512 kbps
Explanation: Nyquist Rate 2 x 4k = 8 kHz
Total sample 4 x 8 = 32 k sample/sec
256 = 28, so that 8 bits are required
Bit Rate 32k x 8 = 256 kbps.
6. 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
a) 1180 K bits/sec
b) 1280 K bits/sec
c) 1180 M bits/sec
d) 1280 M bits/sec
Explanation: Total sample 8000 x 20 = 160 k sample/sec
Bit for each sample 7 + 1 = 8
Bit Rate = 160k x 8 = 1280 kilobits/sec.
7. 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
a) 5 kHz
b) 20 kHz
c) 40 kHz
d) 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.
8. 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
a) 8
b) 10
c) 12
d) 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.
9. 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
a) 1.12 KBytes
b) 140 KBytes
c) 168 KBytes
d) None of the mentioned
Explanation: (SNR)q = 1.76 + 6.02(n) = 40 dB, n = 6.35
We take the n = 7.
Capacity = 20 x 8k x 7 = 1.12 Mbits = 140 Kbytes.
10. 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
a) 2.04 V
b) 1.08 V
c) 4.08 V
d) 2.16 V
Explanation: Amax = (0.1 x 68k)/(2000p) or 1.08V.