I tried to solve this question from basic
Here is my work
But the correct answer is Option $(B)$.What is the mistake i am doing?
I tried to solve this question from basic
Here is my work
But the correct answer is Option $(B)$.What is the mistake i am doing?
First note that: $$ \cos(2\pi 50 t) \longleftrightarrow 0.5 \delta(f+50) + 0.5\delta(f-50) $$
$$\sin(2\pi 150 t) \longleftrightarrow 0.5 j \delta(f+150) -j 0.5\delta(f-150)$$
Hence the baseband spectrum is: $$ X(f) = 0.5 j \delta(f+150) + 0.5 \delta(f+50) + 0.5 \delta(f-50) - 0.5 j\delta(f-150) $$
Then a shift to right by 100 Hz yields (without sampling weight): $$ X(f-100) = 0.5\left[j \delta(f+50) +\delta(f-50) + \delta(f-150) - j\delta(f-250) \right]$$
Shift left by 100 Hz: $$ X(f+100) = 0.5\left[ j\delta(f+250) +\delta(f+150) + \delta(f+50) - j\delta(f-50) \right]$$
Shift right by 200 Hz: $$ X(f-200) = 0.5\left[ j \delta(f-50) +\delta(f-150) + \delta(f-250) - j\delta(f-350) \right]$$
Shift left by 200 Hz: $$ X(f+200) = 0.5\left[ j\delta(f+350) +\delta(f+250) + \delta(f+150) - j\delta(f+50) \right]$$
The sum of all the shifts for $n=0,\pm 1,\pm2$ : $$X_s(f) = 0.5\left[ j\delta(f+350) +\delta(f+250) + \delta(f+150) - j\delta(f+50) + j\delta(f+250) +\delta(f+150) + \delta(f+50) - j\delta(f-50) + j \delta(f+150) +\delta(f+50) + \delta(f-50) - j\delta(f-150) + j \delta(f+50) +\delta(f-50) + \delta(f-150) - j\delta(f-250) + j \delta(f-50) +\delta(f-150) + \delta(f-250) - j\delta(f-350) \right] $$
After the 100 Hz lowpass filter only those impulses inside $-100<f<100$ band remain: $$LPF\{ X_s(f) \} = 0.5\left[ - j\delta(f+50) + \delta(f+50) - j\delta(f-50) + \delta(f+50) + \delta(f-50) + j \delta(f+50) +\delta(f-50) + j \delta(f-50) \right] $$
Which can be further simplied into: $$LPF\{ X_s(f) \} = \delta(f+50) + \delta(f-50) $$
Hence we arrive (without sampling weight) $$ \boxed{ z(t) = 2\cos(2 \pi 50 t)} $$
Therefore we can conclude that the output of the LPF will be proportional to $\cos(2\pi 50 t)$.
I thought I would add a few interesting thoughts I had while looking at this. I believe the answer is (A) and whoever told you the answer is (B) failed to consider the aliasing components resulting from the negative frequencies.
In addition to the very nice answer by Fat32 that involves taking things to the frequency domain and back. There is one very simple way to see that it must be (A) using only a time domain approach. If $t$ were discrete-valued, then the answers (A), (B), and (C) could all be seen as correct thanks to the ambiguity found in signals at the highest frequency representable. However, the low pass filter will remove the discrete nature of the samples (assuming it is an analog low pass filter anyway), and this makes the $\sin(\cdot)$ term incorrect.
To see this, consider your original signal $x(t) = \cos(2 \pi 50 t) + \sin(2 \pi 150 t)$ sampled with the impulse train provided at a rate of $100$Hz. This can be written as a discrete-valued signal where $t=n/f_s$ and $f_s=100$Hz, $$ x\left(\frac{n}{100}\right) = \cos\left(\pi n\right) + \sin\left(\pi n\right).$$ Now, after you realize that $\sin(\pi n) = 0$, you will also realize that this could be written as $$ x\left(\frac{n}{100}\right) = \cos\left(\pi n\right). $$ For that matter, it could have been written as $$ x\left(\frac{n}{100}\right) = \cos\left(\pi n\right) - \sin\left(\pi n\right).$$ Now, passing these samples through an appropriate low pass filter with a cutoff of $100$Hz will yield (A) since it only contains information about the $\cos(\cdot)$ (since the $\sin(\cdot)$ terms are zero).
This happened because both the aliased components and the original $50$Hz cosine violate Nyquist. The Nyquist Theorem says that we will have no loss of information if the sampling rate is strictly greater than twice the highest frequency in our signal. Here, we have it at exactly twice the highest frequency that we are trying to represent (after aliasing). This leads to a phase/amplitude ambiguity in that frequency, and in this case it cancels out the effects of the $\sin(\cdot)$ terms completely.