5-1 Determine the minimum Nyquist rate for a maximum analog input frequency of t

Question

5-1
Determine the minimum Nyquist rate for a maximum analog input frequency of the following:
a. 4 kHz
b. 10 kHz
c. 7 kHz

5-2
For the following samples rates, determine the maximum analog input frequency:
a. 20 kHz
b. 8 kHz
c. 6 kHz


5-5
Determine the minimum number of PCM bits (include the sign bit) for a dynamic range of 80 dB.

Explanation / Answer

Nyquist Criteria : Sample rate > 2*highest signal frequency using this 5-1 minimum sample rate = 2*highest frequency a. = 2*4 kHz = 8 kHz b. = 2*10 kHz = 20 kHz c. = 2*7 kHz = 14 kHz 5-2 maximum signal frquency = sample rate / 2 a. = 20/2 kHz = 10 kHz b. = 8/2 kHz = 4 kHz c. = 6/2 kHz = 3 kHz 5-3 Q = no. of bits in PCM Dynamic Range < 20 log(2^Q) [log has base = 10] 80 < 20 log(2^Q) 4

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