Let xa(t)be an analog signal with bandwidth B=3kHz. We wishto use an =2m–pointDFT to compute the spectrum ofthe signal with a resolution less than or equal to 50 Hz. Determine
(a) the minimum sampling rate,
(b) the minimum number of required samples, and
(c) the minimumlength of the analog signal record(in seconds).

a) the minimum sampling rate is 6 kHz

b) the minimum numbers of required samples are 120

c) the minimum length of the analog signal is 0.02 s

Explanation:

Given the data in the question;

(a) the minimum sampling rate;

band width of analog signal xₐ(t) is;

bandwidth B = 3kHz

Now, according to sampling theorem, minimum sampling rate F must be twice the bandwidth of the signal.

so

F = 2B

F = 2( 3 kHz )

F = 6 kHz

Therefore, the minimum sampling rate is 6 kHz

(b) the minimum number of required samples;

Let L represent the minimum number of samples required,

given that; required resolution of the spectrum of the signal is less than or equal to 50 Hz

F/L ≤ 50

L ≥ F/50

L ≥ ( 6 × 1000 Hz ) / 50

L ≥ 6000 / 50

L ≥ 120

Therefore, the minimum numbers of required samples are 120

(c) the minimum length of the analog signal record(in seconds).

minimum number of samples required is 120

T = L /  F

T = 120 / ( 6 × 1000 Hz )

T = 120 / 6000

T = 0.02 s

Therefore, the minimum length of the analog signal is 0.02 s

answer:

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