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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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A signal m(t) with bandwidth 500 Hz is first multiplied by a signal g(...
Signal m(t) with bandwidth 500 Hz
Let's start by understanding the characteristics of the signal m(t). Given that its bandwidth is 500 Hz, it means that the signal contains frequency components ranging from -250 Hz to +250 Hz. This signal can be represented as:
m(t) = A(t)cos(2πf(t)t + φ(t))
Where:
- A(t) represents the amplitude of the signal, which can vary with time.
- f(t) represents the instantaneous frequency of the signal, which can also vary with time.
- φ(t) represents the instantaneous phase of the signal, which can vary with time.
Multiplication with signal g(t)
Next, the signal m(t) is multiplied by another signal g(t). The multiplication of two signals in the time domain is equivalent to their convolution in the frequency domain. Therefore, the frequency content of the resulting signal will be the convolution of the frequency content of m(t) and g(t).
Ideal low pass filter with bandwidth 1 kHz
The resulting signal is then passed through an ideal low pass filter with a bandwidth of 1 kHz. An ideal low pass filter completely eliminates frequency components above its cutoff frequency and allows all frequency components below the cutoff frequency to pass through without any distortion. In this case, the cutoff frequency of the low pass filter is 1 kHz, which means that it allows frequency components ranging from -500 Hz to +500 Hz to pass through.
Output of the low pass filter
Since the bandwidth of the low pass filter is lower than the bandwidth of the multiplied signal, the output of the low pass filter will be the frequency components within the bandwidth of the filter, which is -500 Hz to +500 Hz. Therefore, the output of the low pass filter will be:
Output = m(t) * g(t) * LPF(1 kHz)
where LPF(1 kHz) represents the output of the low pass filter with a bandwidth of 1 kHz.
Answer: Option C) 0
Based on the characteristics of the low pass filter, the output will be zero outside its bandwidth. Therefore, the output of the low pass filter will be 0 for frequency components outside the range of -500 Hz to +500 Hz.
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A signal m(t) with bandwidth 500 Hz is first multiplied by a signal g(...
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.
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