Let $x(t)$ be a bandlimited signal such that $X(j\omega) =0 $ when $|\omega|>M$. Also $p(t) = p_1(t) - p_1(t-\Delta)$ is a nonuniformly spaced periodic pulse train where $$p_1(t) = \sum_{k = -\infty}^{+\infty}\delta\left(t - \frac{2\pi k}{M}\right), \quad \text{with}\quad \Delta= \frac{\pi}{2M}$$ enter image description here

Let $x_p(t) = x(t)p(t)$ and apply an ideal low-pass filter with cutoff frequency $\omega_c = M$ to $X_p(j\omega)$. The result is $z(t)$. Design a system which recovers $x(t)$ from $z(t)$.

enter image description here

My try:

It's easy to see that $$P(j\omega) = \left(1 - e^{-j\Delta \omega}\right)\left(M\sum_{k = -\infty}^{+\infty}\delta(\omega - kM)\right) = M\sum_{k = -\infty}^{+\infty}\left(1 - e^{-j\Delta kM}\right)\delta(\omega - kM) $$

Also we have

$$ X_p(j\omega) = \frac{1}{2\pi}X(j\omega)\star P(j\omega) = \frac{M}{2\pi}\sum_{k = -\infty}^{+\infty}\left(1 - e^{-j\Delta kM}\right)X\big(j(\omega - kM)\big) $$

Since $\Delta= \frac{\pi}{2M}$ we have $$ X_p(j\omega) = \frac{M}{2\pi}\sum_{k = -\infty}^{+\infty}\left(1 - e^{\frac{-jk\pi}{2}}\right)X\big(j(\omega - kM)\big) $$

After low-pass filter we get

$$ Z(j\omega) = \begin{cases} \frac{M}{2\pi}\bigg[(1+j)X\big(j(\omega - M)\big) + (1-j)X\big(j(\omega + M)\big)\bigg], & \lvert \omega\rvert < M\\ 0, & \text{O.W} \end{cases} $$

I've got stuck here. How can we recover $x(t)$? I tried to use phase shifter and $z(t)\cos(Mt)$ but it didn't work.

  • $\begingroup$ all's i can say, by just glancing at this, is that you will not be able to recover the DC component of the signal being sampled. $\endgroup$ Commented Dec 26, 2020 at 20:53
  • $\begingroup$ @robertbristow-johnson Why is recovering DC component impossible? $\endgroup$
    – S.H.W
    Commented Dec 26, 2020 at 20:58
  • $\begingroup$ because the DC component of $x_p(t)$ is always zero, independent of the DC in $x(t)$. $\endgroup$ Commented Dec 26, 2020 at 21:06
  • $\begingroup$ @robertbristow-johnson I see. Do you have any idea for recovering $x(t)$ from $z(t)$? $\endgroup$
    – S.H.W
    Commented Dec 26, 2020 at 21:26
  • $\begingroup$ it requires that i look at this more. as best as i can tell, you can recovered the spectrum of $x(t)$, (which is $X(j\omega)$) everywhere except where the spectrum was multiplied by zero. you just can't divide by zero. $\endgroup$ Commented Dec 26, 2020 at 21:53

1 Answer 1


I agree with your result for $Z(j\omega)$. Apart from scaling, what you have is the original spectrum with positive and negative frequencies swapped and multiplied with factors $1-j$ and $1+j$, respectively. In order to restore the original signal, we need to add right-shifted and left-shifted versions of the spectrum, while getting rid of the complex factors. Final lowpass filtering removes the redundant components in the frequency range $[M,2M]$.

The necessary shifting and complex scaling is achieved by multiplication with


Modulation with $(1)$, lowpass filtering with cut-off frequency $M$, and appropriate scaling restores the original signal.

  • $\begingroup$ Thanks. I think we need to construct $z(t)\sin(M t)$ as well, because $Z(j\omega)$ is limited to $|\omega|<M$. I mean $X(j(\omega + M))$ and $X(j(\omega - M))$ are one-sided spectrum in $Z(j\omega)$, so $z(t)\cos(M t)$ contains two one-sided spectrums of $X(j\omega)$, one of them is multiplied by $1+j$ and the other one is multiplied by $1-j$. Therefore their sum doesn't give $X(j\omega)$. $\endgroup$
    – S.H.W
    Commented Dec 27, 2020 at 19:18
  • $\begingroup$ @S.H.W I've also observed the sin / cos multiplied terms and an original term too. But the sin / cos terms can be combined into a single cosine with phase... $\endgroup$
    – Fat32
    Commented Dec 27, 2020 at 20:21
  • 1
    $\begingroup$ @S.H.W oops my $-z(t)$ is reduntant, so yours seems to be correct, hence apart from a linear scaling factor, the multiplication on $z(t)$ seems like $\cos(Mt)-\sin(Mt)$ or similarly $\cos(Mt+\pi/4)$. $\endgroup$
    – Fat32
    Commented Dec 27, 2020 at 21:01
  • 1
    $\begingroup$ @S.H.W: You're right about the sine component, I went a bit too quick. I'll come back later to update and correct my answer. $\endgroup$
    – Matt L.
    Commented Dec 27, 2020 at 22:53
  • 1
    $\begingroup$ @S.H.W: I've updated and corrected my original answer. $\endgroup$
    – Matt L.
    Commented Dec 28, 2020 at 11:24

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