Answer:
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[tex]_s[/tex] must be twice the bandwidth of the signal.
so
F[tex]_s[/tex] = 2B
F[tex]_s[/tex] = 2( 3 kHz )
F[tex]_s[/tex] = 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[tex]_s[/tex]/L ≤ 50
L ≥ F[tex]_s[/tex]/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[tex]_s[/tex]
T = 120 / ( 6 × 1000 Hz )
T = 120 / 6000
T = 0.02 s
Therefore, the minimum length of the analog signal is 0.02 s