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A signal m(t) = 3 cos (6000 πt) + 4 cos (7000 πt) + 5 cos (10000 πt) to be sampled. What is the minimum sampling rate for bandpass sampling for truth full representation:

A signal m(t) = 3 cos (6000 πt) + 4 cos (7000 πt) + 5 cos (10000 πt) to be sampled. What is the minimum sampling rate for bandpass sampling for truth full representation: Correct Answer 10000 Hz

Concept:

Sampling Theorem:

1) A continuous signal band limited to fm hertz can be converted to its sample equivalent without information loss provided fs ≥ 2 × fm samples/sec.

2) Aliasing is a form of overlapping between signal components that occurs when fs < 2 × fm.

3) The minimum sampling rate and maximum sampling rate interval allowed to avoid aliasing is called the Nyquist rate.

Calculation:

m(t) = 3 cos (6000 πt) + 4 cos (7000 πt) + 5 cos (10000 πt)

fm1 = 3000 Hz

fm2 = 3500 Hz

fm3 = 5000 Hz

Minimum Sampling Rate (S):

(S) = 2 × max.(fm1,fm2,fm3)

(S) = 2 × 5000

(S) = 10000 Hz.

Important Points

Consider a modulating signal:

m(t) = (cos(2πf1t)  ×  cos(2πf2t)) + (cos(2πf3t)  ×  cos(2πf4t))

The maximum sampling rate will be:

(S) = 2 × max((f1 + f2), (f3 + f4))     

Related Questions

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