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Consider a sampled signal
y(t) = 5 × 10– 6 x(t) ∑∞n = – ∞ δ (t – nTs)
where x(t) = 10 cos (8π × 103)t and Ts = 100μ sec. When y(t) is passed through an ideal low pass filter with a cut-off frequency of 5 kHz, the output of the filter is
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- 5 × 10– 6 cos (8π × 103)t
- 5 × 10– 5 cos (8π × 103)t
- 5 × 10– 1 cos (8π × 103)t
- 10 cos (8π × 103)t
Correct Option: B
Given sampled signal,
y(t) = 5 × 10– 6 x(t) ∑∞n = – ∞ δ(t – nTs)
Ts = 100 μ sec
x(t) = 10 cos (8π × 103t)
| fs = | = 10kHz | |
| 100 ×10−6 |
and
fm= 4 kHz(given)
Since, fs > 2fm and cut-off frequency fc > fm. Hence the original signal can be recovered in frequency domain by passing its sampled version through a low pass filter (LPF).
The output of the filter = 5 × 10–6 × 10 cos (8π × 103t)
= 5 × 10–5 cos (8π × 103t)
