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A signal m(t) with bandwidth 500 Hz is first multiplied by a signal g(t). The resulting signal is passed through an ideal low pass filter with bandwidth 1 kHz. The output of the low pass filter would be :
- a)Impulse
- b)m(t)
- c)0
- d)m(t)del(t)
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A signal m(t) with bandwidth 500 Hz is first multiplied by a signal g(...
Explanation: m (t) g (t)->M (f)*G (f)
After low pass filtering with fc =1 kHz, hence the output is zero.
After low pass filtering with fc =1 kHz, hence the output is zero.
Most Upvoted Answer
A signal m(t) with bandwidth 500 Hz is first multiplied by a signal g(...
Signal Processing Problem
The output of the low pass filter.
Firstly, let's understand the concept of multiplication of two signals. The multiplication of two signals in the time domain is equivalent to the convolution of their Fourier transforms in the frequency domain. In this case, m(t) and g(t) are multiplied, and their Fourier transforms will be convolved.
Let's assume that the Fourier transform of m(t) is M(f) and the Fourier transform of g(t) is G(f). Therefore, the Fourier transform of the product of m(t) and g(t) is given by:
F(m(t)g(t)) = M(f) * G(f)
The ideal low pass filter is a filter that allows all frequencies below a certain cutoff frequency to pass through and attenuates all frequencies above that cutoff frequency. In this case, the cutoff frequency is 1 kHz, which means that all frequencies below 1 kHz will pass through, and all frequencies above 1 kHz will be attenuated.
Since the signal m(t) has a bandwidth of 500 Hz, its Fourier transform M(f) will have non-zero values only up to 500 Hz. Therefore, when M(f) is convolved with G(f), the resulting signal will have non-zero values only up to 500 Hz. When this signal is passed through the ideal low pass filter with a cutoff frequency of 1 kHz, all frequencies below 1 kHz will pass through, and all frequencies above 1 kHz will be attenuated. Hence, the output of the low pass filter will be 0.
Therefore, the correct answer is option (c) 0.
Given Information:
- Signal m(t) has a bandwidth of 500 Hz.
- m(t) is multiplied by a signal g(t).
- The resulting signal is passed through an ideal low pass filter with a bandwidth of 1 kHz.
To Find:
The output of the low pass filter.
Solution:
Firstly, let's understand the concept of multiplication of two signals. The multiplication of two signals in the time domain is equivalent to the convolution of their Fourier transforms in the frequency domain. In this case, m(t) and g(t) are multiplied, and their Fourier transforms will be convolved.
Multiplication of m(t) and g(t):
Let's assume that the Fourier transform of m(t) is M(f) and the Fourier transform of g(t) is G(f). Therefore, the Fourier transform of the product of m(t) and g(t) is given by:
Passing through an ideal low pass filter:
The ideal low pass filter is a filter that allows all frequencies below a certain cutoff frequency to pass through and attenuates all frequencies above that cutoff frequency. In this case, the cutoff frequency is 1 kHz, which means that all frequencies below 1 kHz will pass through, and all frequencies above 1 kHz will be attenuated.
Output of the low pass filter:
Since the signal m(t) has a bandwidth of 500 Hz, its Fourier transform M(f) will have non-zero values only up to 500 Hz. Therefore, when M(f) is convolved with G(f), the resulting signal will have non-zero values only up to 500 Hz. When this signal is passed through the ideal low pass filter with a cutoff frequency of 1 kHz, all frequencies below 1 kHz will pass through, and all frequencies above 1 kHz will be attenuated. Hence, the output of the low pass filter will be 0.
Answer:
Therefore, the correct answer is option (c) 0.
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