GATE 2018 EC Q.54: solution and answer

GATE 2018 · ECQ.54 · 2 marks · NAT

A band limited low-pass signal $x (t)$ of bandwidth 5 kHz is sampled at a sampling rate $f_{s}$ . The signal $x (t)$ is reconstructed using the reconstruction filter $H (f)$ whose magnitude response is shown below:

[Figure]

The minimum sampling rate $f_{s}$ (in kHz) for perfect reconstruction of $x (t)$ is ______.

GATE 2018 EC question 54: Plot of |H(f)| (vertical axis) against f (kHz) (horizontal axis). — Figure from the GATE 2018 EC paper. It is part of the question.

Answer and solution

ANSWER

The minimum sampling rate for perfect reconstruction is $f_{s} = 13 kHz$ .

Official answer: 13Source: Official answer key, GATE 2018 EC (IIT Guwahati)

Spectrum of the sampled signal (ideal impulse sampling): $X_{s} (f) = f_{s} n = - \infty \sum \infty X (f - n f_{s})$ The baseband copy ( $n = 0$ ) occupies $∣ f ∣ \leq 5 kHz$ , and the $n$ -th copy occupies $[n f_{s} - 5, n f_{s} + 5] kHz$ .
Condition 1 (the wanted copy is not distorted): the whole baseband $∣ f ∣ \leq 5 kHz$ must lie where the gain is constant. The flat region of $H (f)$ is $∣ f ∣ \leq 6 kHz$ , so this holds whatever $f_{s}$ is.
Condition 2 (no image leaks through): every shifted copy must lie where $H (f) = 0$ , that is $∣ f ∣ \geq 8 kHz$ . In the sloped band $6 < ∣ f ∣ < 8 kHz$ the gain is between $0$ and $K$ , so any image content falling there would appear, attenuated, in the output.
The first image ( $n = 1$ , and by symmetry $n = - 1$ ) is the closest one, so it sets the requirement. Its lower edge must satisfy $f_{s} - 5 \geq 8 kHz \Rightarrow f_{s} \geq 13 kHz$ For any $f_{s} < 13 kHz$ the band from $f_{s} - 5$ up to $8 kHz$ falls on the slope, where $∣ H (f) ∣ > 0$ , so part of the image passes and reconstruction is not exact.
Check at $f_{s} = 13 kHz$ : the first image spans $[8, 18] kHz$ (and $[- 18, - 8] kHz$ for $n = - 1$ ), where $H (f) = 0$ , so it is completely rejected. Higher-order images lie even further out. The filter output is $Y (f) = H (f) X_{s} (f) = K f_{s} X (f)$ , a scaled copy of $x (t)$ , so reconstruction is exact.
Minimum sampling rate: $f_{s} = 13 kHz$ . This is above the Nyquist rate of $2 \times 5 = 10 kHz$ because the reconstruction filter has a $2 kHz$ transition band instead of a brick-wall cut-off.

KEY IDEA

With a non-ideal reconstruction filter you need two conditions. First, the signal band $W$ must sit inside the flat passband ( $W \leq f_{pass}$ ; here $5 \leq 6$ ). Second, the first spectral image must start at or beyond the stopband edge: $f_{s} - W \geq f_{stop}$ , so $f_{s} \geq W + f_{stop} = 5 + 8 = 13 kHz$ .

WATCH OUT FOR

Answering the Nyquist rate $2 W = 10 kHz$ ignores the filter's transition band. Other wrong routes: using the passband edge ( $f_{s} - 5 \geq 6$ , which gives $11 kHz$ ), which lets the image into the sloped band where $∣ H (f) ∣ > 0$ ; or taking twice the filter's stopband edge ( $2 \times 8 = 16 kHz$ ), which treats the filter as if it were the signal's bandwidth.

From the GATE 2018 EC paper. Every question of this paper · Download the full paper · All 8 Sampling questions, with notes

GATE 2018 EC Q.54: solution and answer

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