5
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ 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. $\endgroup$ Dec 26, 2020 at 20:53
  • $\begingroup$ @robertbristow-johnson Why is recovering DC component impossible? $\endgroup$
    – S.H.W
    Dec 26, 2020 at 20:58
  • $\begingroup$ because the DC component of $x_p(t)$ is always zero, independent of the DC in $x(t)$. $\endgroup$ 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
    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. $\endgroup$ Dec 26, 2020 at 21:53

1 Answer 1

3
$\begingroup$

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

$$e^{jMt}(1+j)+e^{-jMt}(1-j)=2\cos(Mt)-2\sin(Mt)\tag{1}$$

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

$\endgroup$
10
  • $\begingroup$ 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
    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
    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
    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.
    Dec 27, 2020 at 22:53
  • 1
    $\begingroup$ @S.H.W: I've updated and corrected my original answer. $\endgroup$
    – Matt L.
    Dec 28, 2020 at 11:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.