SNR  will be

1.  298

2.  1.75x103

3.  4.46x103

4.  201

View Answer  
4
Correct Answer :

4.46x103


Explanation :
No Explanation available for this question

The output signal to quantization noise ratio of a 10 bit PCM was found to be 30dB. The desired SNR is 42 dB. It can be increased by increasing the number of quantization level. In this wave fractional increase in the transmission bandwidth would be (assume log2 10=0.3]

1.  20%

2.  30%

3.  40%

4.  50%

View Answer  
4
Correct Answer :

20\%


Explanation :
No Explanation available for this question

Four signal each band limited to 5kHZ are sample at twice the  Nyquist rate. The resulting PAM samples are transmitted over a single channel after time division multiplexing. The theoretical maximu transmission bandwidth of the channel should be equal to

1.  5kHz

2.  20 kHz

3.   40 kHz

4.  80 kHz

View Answer  
4
Correct Answer :

80 kHz


Explanation :
No Explanation available for this question

Treating signal 10 sin 2π, 100t cos2 π, 300 t volt (with t in sec) to be a low signal, the Nyquist rate for this signal will be

1.  1400 s/sec

2.  1200 s/sec

3.  700 s/sec

4.  500 s/sec

View Answer  
4
Correct Answer :

1400 s/sec


Explanation :
No Explanation available for this question

Pulse rate in DM system is 56, 000 per second. Input signal is 5 cos (2π1000 t) + 2 cos (2π 2000 t) V, with t in second. The minimum value of step size which will avoid stope overload distortion should

1.  be less than 0.56 V

2.  be equal to 0.56 V

3.  not be less than 0.56 V

4.  none of these

View Answer  
4
Correct Answer :

not be less than 0.56 V


Explanation :
No Explanation available for this question

A low pass signal m(t) band limited to BHz is sampled by a periodic rectangular pulse train, Pr(t) of period T2 =1/3B second. Assuming natural sampling and that the pulse amplitude and pulse width are A volts arid 1/30B sec respectively, an expression for frequency spectrum of the sampled signal M2(t) is

1.  A sin C (2πf/B)

2.   A sin C (2πf/30B)

3.  A sin C (2πf/60B)

4.  A cos C (2πf/B)

View Answer  
4
Correct Answer :

A sin C (2πf/60B)


Explanation :
No Explanation available for this question

A received binary NRZ signal assume the voltage level of 500 milli volts and 500 volts respectively for 1 and ‘0’ transmission with a it rate or r bits/sec. The signal is corrupted by AWGN with a two-sided spectral density of 10-6 volt2/Hz. The received signal is processed by an integrate and Dump’s circuit in every bit interval and compared with a zero threshold to take a bit decision. Assuming '1' and 0 transmission to be equally likely, maximum value of r such that the bit error probability ≤ 10-5, will be

1.  12 kb/sec

2.  13.7kb/sec

3.  6.3kb/sec

4.  none of these

View Answer  
4
Correct Answer :

13.7kb/sec


Explanation :
No Explanation available for this question

An analog signal (low pass) of 4 kHz BW is sampled at Nyquist rate, subsequently quantized and encoded into a binary PCM wave with 128 quantization levels. The work synchronization is effected by adding a synchronizing pulse at the end of each code word. The bit rate to the resulting PCM signal will be

1.  8 kHz

2.  32 kHz

3.  16 kHz

4.  64 kHz

View Answer  
4
Correct Answer :

64 kHz


Explanation :
No Explanation available for this question

Band width of the input to PCM system is restricted to 4 kHz. The input varies from 3.8 V to a-3.8 V and has average power of 30 mW. The required signal M quantization noise power ratio is 20 dB. The modulator produces binary output. Assume uniform quantization. The number of bits required per sample will be

1.  8

2.  7

3.   6

4.  5

View Answer  
4
Correct Answer :

7


Explanation :
No Explanation available for this question

A speech signal occupying the bandwidth of 300Hz to 3kHz is converted into PCM format for use in digital communication. If sampling  frequency is 8kHz and each sample is quantized into 2561 levels, then output bit the rate will be

1.  3 kb/s

2.  8 kb/s

3.  64 kb/s

4.  256 kb/s

View Answer  
4
Correct Answer :

64 kb/s


Explanation :
No Explanation available for this question
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