Ideal low pass filter output at given sampling frequency

Question

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.

Answer 1

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 : $$f_s (\cos{(2\pi \frac{15}{\pi}t)} + \cos{(2\pi \frac{5}{\pi}t)})$$ $$ = \frac{20}{\pi} (\cos(30t) + \cos(10t))$$

Answer 2

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.