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Assertion (A): Nyquist rate of sampling is the theoretical minimum sampling rate at which the signal can be sampled and still be reconstructed from its samples. Reason (R): When the Nyquist rate sampling is used, only an ideal low pass filter can be used to extract signal x(t) from sampled signal xs(t). Code:

Assertion (A): Nyquist rate of sampling is the theoretical minimum sampling rate at which the signal can be sampled and still be reconstructed from its samples. Reason (R): When the Nyquist rate sampling is used, only an ideal low pass filter can be used to extract signal x(t) from sampled signal xs(t). Code: Correct Answer Both (A) and (R) are true and (R) is the correct explanation of (A).

Nyquist Sampling Theorem:

A continuous-time signal can be represented in its samples and can be recovered back when sampling frequency fs is greater than or equal to twice the highest frequency component of the message signal, i.e.

fs ≥ 2fm

Therefore when we want to convert continuous signals to discrete signals, the sampling must be done at the Nyquist rate.

In the sampling process, if the sampling frequency is greater than the Nyquist frequency for the reconstruction of the original signal, LPF is used

Hence, Both (A) and (R) are true and (R) is the correct explanation of (A).

Analysis:

We have assumed spectrum x(t) is band limited to ωm.

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The sampling frequency can take three possible values with respect to spectrum width (2 ωm)

(i) ωs > 2ωm

(ii) ωs = 2 ωm

(iii) ωs < 2ωm

Case 1:

ωs > 2ωm

Let ωs = 3ωm

The spectrum of the sampled signal will be:

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We see for ωs > 2ωm, there is no overlap between the shifted spectrum of X(ω). Thus, as long as the sampling frequency ωs is greater than the twice the signal bandwidth (2 ω­m), x(t) can be recovered by passing the sampled signal xs(t) through an ideal or practical low pass filter having bandwidth between ωm and (ωs - ωm) rad/sec.

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