GATE 2025 IN Q.41: solution and answer

GATE 2025 · INQ.41 · 2 marks · MCQ

A chopper amplifier shown in the figure is designed to process a biomedical signal $v_{in} (t)$ to generate conditioned output $v_{o u t} (t)$ . The signals $v_{in} (t)$ and $v_{os} (t)$ are band limited to 50 Hz and 10 Hz, respectively. For the system to operate as a linear amplifier, choose the correct statement from the following options.

GATE 2025 IN question 41: Block diagram titled 'Chopper Amplifier' inside a dash-dot box. — Figure from the GATE 2025 IN paper. It is part of the question.

(A)
The minimum frequency of $s (t)$ required is 100 Hz and $v_{os} (t)$ gets attenuated by the system
(B)
The minimum frequency of $s (t)$ required is 100 Hz and $v_{os} (t)$ also gets amplified by the system by a factor $\frac{200}{π}$
(C)
The minimum frequency of $s (t)$ required is 80 Hz and $v_{os} (t)$ gets attenuated by the system
(D)
The minimum frequency of $s (t)$ required is 80 Hz and $v_{os} (t)$ also gets amplified by the system by a factor $\frac{200}{π}$

Answer and solution

ANSWER

Option (A): the minimum frequency of s(t) is 100 Hz, and v_os(t) is attenuated (removed) by the system.

Official answer: ASource: Official answer key, GATE 2025 IN (IIT Roorkee)

Signal before the filter. The first multiplier gives $v_{in} (t) s (t)$ . The summer adds $v_{os} (t)$ , the amplifier multiplies by $100$ , and the second multiplier multiplies by $s (t)$ again: $x (t) = 100 [v_{in} (t) s (t) + v_{os} (t)] s (t) = 100 v_{in} (t) s^{2} (t) + 100 v_{os} (t) s (t)$
Use $s^{2} (t) = 1$ . $s (t)$ is $+ 1$ or $- 1$ at every instant, apart from the switching instants, which take up zero time and so do not affect the output. That gives $s^{2} (t) = 1$ and $x (t) = 100 v_{in} (t) + 100 v_{os} (t) s (t)$ The input comes back at baseband with gain $100$ , whatever the chopping frequency.
Fourier series of $s (t)$ . $s (t)$ is a $\pm 1$ square wave with a 50% duty cycle and period $T$ . Its average (DC term) is zero and it contains only odd harmonics: $s (t) = \frac{4}{π} n = 1, 3, 5, \dots \sum \frac{1}{n} sin (n ω_{0} t), ω_{0} = \frac{2 π}{T}, f_{s} = \frac{1}{T}$
Where the offset term lands (modulation property). Multiplying by $s (t)$ copies the $0$ to $10 Hz$ spectrum of $v_{os} (t)$ into bands centred on $f_{s}, 3 f_{s}, 5 f_{s}, \dots$ . Because $s (t)$ has no DC term, no copy stays at baseband. The lowest offset frequency in $x (t)$ is therefore $f_{s} - 10 Hz$ .
The filter cutoff in hertz. $H_{2}$ is an ideal low-pass filter with cutoff $ω_{c} = π / T$ , so $f_{c} = \frac{π / T}{2 π} = \frac{1}{2 T} = \frac{f _{s}}{2}$
Condition 1: pass $v_{in}$ undistorted. The filter has to pass the whole $50 Hz$ band of $v_{in}$ : $f_{c} \geq 50 Hz \Rightarrow \frac{f _{s}}{2} \geq 50 \Rightarrow f_{s} \geq 100 Hz$ At $f_{s} = 80 Hz$ the cutoff would be $40 Hz$ , and the $40$ to $50 Hz$ part of $v_{in}$ would be lost. So $80 Hz$ is not enough.
Condition 2: reject the offset. The lowest offset component has to lie above the cutoff: $f_{s} - 10 > \frac{f _{s}}{2} \Rightarrow f_{s} > 20 Hz$ At $f_{s} = 100 Hz$ the offset occupies $90$ to $110 Hz$ and higher odd-harmonic bands. All of that is above $f_{c} = 50 Hz$ .
Conclusion. Both conditions hold when $f_{s} \geq 100 Hz$ , so the minimum chopping frequency is $100 Hz$ . At that frequency $v_{o u t} (t) = 100 v_{in} (t)$ . The offset $v_{os} (t)$ is removed (attenuated) by the system, not amplified. This is option (A).

KEY IDEA

The input is multiplied by the ±1 square wave twice. Since $s^{2} (t) = 1$ , it comes back at baseband with gain $100$ . The offset enters between the two multipliers, so it is multiplied only once. Because $s (t)$ has no DC term, the offset ends up near the odd harmonics $f_{s}, 3 f_{s}, \dots$ . The output filter's cutoff $π / T$ equals $f_{s} /2$ in hertz. It must pass the $50 Hz$ signal band, which gives $f_{s} \geq 100 Hz$ . At that frequency the offset band ( $90$ to $110 Hz$ and up) lies outside the passband.

WATCH OUT FOR

One mistake is to take the fundamental amplitude $4/ π$ of $s (t)$ and conclude that the offset reaches the output with gain $100 \cdot \frac{4}{π} \cdot \frac{1}{2} = \frac{200}{π}$ (options B and D). That component sits at $f_{s} \pm 10 Hz$ , above the filter cutoff $f_{s} /2$ , so it never reaches $v_{o u t}$ . Another mistake is to read the cutoff $π / T$ as a frequency in hertz, or never to check it against the signal band. At $f_{s} = 80 Hz$ the cutoff is only $40 Hz$ , so part of the $50 Hz$ signal band is cut off (options C and D).

From the GATE 2025 IN paper. Every question of this paper · Download the full paper · All 11 Fourier series questions, with notes

GATE 2025 IN Q.41: solution and answer

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