This is not a homework problem, I am solving practice problems for my exam.

enter image description here

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: enter image description here

(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?

  • 1
    $\begingroup$ Perhaps pedantic: but why aren't you using z-domain representation? There are a lot of tools available. BTW: I do agree with Schwarz's solution since it addresses your particular question. $\endgroup$
    – rrogers
    Jul 30, 2019 at 20:28

1 Answer 1


Your first step is correct. However, a signal which is discrete in time domain does not have a discrete frequency domain representation per se, as in theory the Discrete-Time Fourier Transform may be calculated. The signal spectrum would be discrete in combination with the Discrete Fourier Transform (DFT) only if a finite representation was required.

The main consequence of time-domain discretisation is periodicity in the frequency domain. Your solution for exercise 2 should be, as well. For this, you may need to check whether the spectrum will overlap at $\frac{F_s}{2}$.

Your assumption about the discrete filter transfer function being periodic in the frequency domain is correct. Since the filter is applied in frequency domain, you should multiply it with the signal spectrum to get the output spectrum. Should the filter be smaller than the signal, the signal components outside the passband are attenuated (or set to zero, as this filter is ideal). The computation looks like this:

$$y_d[n]=x_d[n] * h_d[n] = \mathcal{F}_\text{DTFT}^{-1}\{X(e^{j\Omega}) \cdot H(e^{j\Omega})\}$$


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