A chopper amplifier shown in the figure is designed to process a biomedical signal \( v_{in}(t) \) to generate conditioned output \( v_{out}(t) \). 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.
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The minimum sampling frequency in a chopper amplifier is determined by the highest frequency component of the input signal. The modulation and demodulation processes, along with the low-pass filter, are crucial for separating and amplifying the desired signal while attenuating unwanted components like the offset voltage.
The minimum frequency of $s(t)$ required is 100 Hz and $v_{os}(t)$ gets attenuated by the system
The minimum frequency of $s(t)$ required is 100 Hz and $v_{os}(t)$ also gets amplified by the system by a factor $\frac{200}{\pi}$
The minimum frequency of $s(t)$ required is 80 Hz and $v_{os}(t)$ gets attenuated by the system
The minimum frequency of $s(t)$ required is 80 Hz and $v_{os}(t)$ also gets amplified by the system by a factor $\frac{200}{\pi}$
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The Correct Option isA
Solution and Explanation
For the chopper amplifier to act as a linear amplifier without aliasing the input signal \( v_{in}(t) \) (band limited to 50 Hz), the sampling frequency, which is the frequency of the switching signal \( s(t) \) (\( f_s = 1/T \)), must satisfy the Nyquist criterion: \( f_s \ge 2 \times f_{max}(v_{in}) = 2 \times 50 \, \text{Hz} = 100 \, \text{Hz} \). Thus, the minimum frequency of \( s(t) \) required is 100 Hz. This eliminates options (C) and (D).
Now consider the effect on the offset voltage \( v_{os}(t) \) (band limited to 10 Hz). The signal after the first multiplier is \( v_{in}(t) s(t) \). After adding the offset, we have \( v_{in}(t) s(t) + v_{os}(t) \). This is amplified by \( A = 100 \), resulting in \( 100 v_{in}(t) s(t) + 100 v_{os}(t) \). The final multiplication by \( s(t) \) gives \( 100 v_{in}(t) s^2(t) + 100 v_{os}(t) s(t) = 100 v_{in}(t) + 100 v_{os}(t) s(t) \) (since \( s(t) = \pm 1 \), \( s^2(t) = 1 \)).
The Fourier series of \( s(t) \) is \( s(t) = \frac{4}{\pi} \sum_{n=1, 3, 5, \dots}^{\infty} \frac{1}{n} \sin(n \omega_s t) \). The multiplication of \( v_{os}(t) \) (0-10 Hz) by \( s(t) \) shifts its spectrum to frequencies around multiples of \( f_s = 100 \) Hz. The low-pass filter with cutoff \( \frac{\pi}{T} = \pi f_s / \pi = f_s / 2 = 50 \) Hz will pass the baseband component of \( 100 v_{in}(t) \) and attenuate the high-frequency components of \( 100 v_{os}(t) s(t) \). Therefore, \( v_{os}(t) \) gets attenuated by the system.
