# Ideal low pass filter output at given sampling frequency

Consider the signal $$\cos(30t)$$ sampled at $$w_s=40 rad/s$$ using a unit impulse train. The sampled signal is filtered with an ideal low pass filter with unity gain and cutoff frequency $$w_c = 40rad/s$$. Find the resulting output.

The solution is given as $$\frac{20}{\pi}(cos(30t)+cos(10t))$$

Attempt:

Taking the Fourier transform of $$cos(30t)$$ gives $$\frac{\pi}{2} (\delta(\omega-30)+\delta(\omega+30))$$.

Using the unit impulse train: $$X(\omega)_{T_s} = \frac{1}{T_s}\sum X(\omega-k\omega_s)$$ but this gives $$\frac{1}{T_s}\sum \frac{\pi}{2} (\delta(\omega-70)+\delta(\omega+10))$$.

If this is an acceptable approach, it's unclear where the cosine arguments and factor of $$\frac{20}{\pi}$$ is coming from. Help is appreciated.

That $$\frac{1}{T_s}$$ factor in your sampled signal expression is $$\frac{20}{\pi}$$.

$$\omega_s = 40 \ rad/sec$$ $$2\pi f_s = 40$$ $$f_s = \frac{40}{2\pi} = \frac{20}{\pi}$$

The sampled signal is basically given by : $$X_{sampled}(f) = f_s \sum^{\infty}_{k=-\infty} X(f - kf_s)$$ Meaning : Sampling a signal with impulse train at sampling frequency $$f_s$$ gives you, images of $$X(f)$$ scaled by $$f_s$$ and centered at $$kf_s$$.

You can check this answer to understand why : Sampling with Impulse Train explained pictorially

The original signal is at frequency $$\frac{15}{\pi}$$ and being sampled at sampling frequency $$\frac{20}{\pi}$$. And hence the frequency components in the sampled signal will be : $$\pm k\frac{20}{\pi} \pm \frac{15}{\pi}$$.

Notice that for $$k=0$$, you get the original signal and for $$k=\pm 1$$, you have an alias term $$\cos{(2\pi (\frac{20}{\pi} - \frac{15}{\pi})t)} = cos(10t)$$ which will get inside the lowpass filter cutoff frequency.

And, that will give you the final result as : $$f_s (\cos{(2\pi \frac{15}{\pi}t)} + \cos{(2\pi \frac{5}{\pi}t)})$$ $$= \frac{20}{\pi} (\cos(30t) + \cos(10t))$$

The given solution is correct. You should convolve the main signal (cos(30t)) with the impulse train in frequency domain. So, there would be some freqs like: 10rad/s, 30, 70 and so on. The out-freq 30 comes from the effect of impulse at freq=0 and the out-freq 10 comes from the result of convolution by impulse at freq=40. After filtering the rest of out-freq are discarded. So, the output involve freqs 10 and 30. I think you would have missed the effect of impulse at freq=0.