This is not a homework problem, I am solving practice problems for my exam.
Consider the analog-to-discrete-to-analog system shown in figure 1. The CT signal $x_a(t)$ is sampled at a frequency of $F_s = 2000$ Hz ($T_s = 0.5$ msec). The resulting impulse train is then converted to a discrete time sequence $x_d[n]$. The Lowpass DT filter $H_d(e^{j\Omega}$ is subsequently used to filter $x_d[n]$ giving $y_d[n]$. Finally, a CT version of the output $y_a(t)$ is created, using an ideal DT-to-CT converter (at the same sampling frequency $F_s = 2000$ Hz).
Note : $H_d(e^{j\Omega}$, which is obviously periodic, is shown for only one period.
a) For the CTFT of $x_a(t)$ given by $X_a(\omega)$ in the figure with $B = 2000\pi$ rad/sec, sketch the $X_d(e^{j\Omega})$, the DTFT of the DT sequence $x_d[n]$.
b) Sketch $Y_a(\omega)$, the CTFT of the CT signal $y_a(t)$. Again, assume that we are using the same frequency as that of sampling $F_s = 2000$ Hz.
I followed the following steps:
Step 1: Multiplied $x_a(t)$ by an infinite pulse train and then switched to frequency domain so I can have an expression for $X_d(\omega)$ in CTFT.
Step 2: After conversion, the "envelope" (which in this case is the triangular shapes) are not drawn but instead we have pulses, something like this:
(Excuse my very bad drawing skills) However, only the red lines exist in this case since we are now in discrete time.
Step 3: I believe in step 2 I have found how $X_d(e^{j\Omega})$ should look like. But then here I get stuck about how should I filter the signal. I get that the filter is periodic but how do I use this in here? Isn't the width of the filter much smaller than the width of the signal?
