# Recovering a signal after nonuniform sampling

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 $$p_1(t) = \sum_{k = -\infty}^{+\infty}\delta\left(t - \frac{2\pi k}{M}\right), \quad \text{with}\quad \Delta= \frac{\pi}{2M}$$

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 $$z(t)$$.

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.

• 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. – robert bristow-johnson Dec 26 '20 at 20:53
• @robertbristow-johnson Why is recovering DC component impossible? – S.H.W Dec 26 '20 at 20:58
• because the DC component of $x_p(t)$ is always zero, independent of the DC in $x(t)$. – robert bristow-johnson Dec 26 '20 at 21:06
• @robertbristow-johnson I see. Do you have any idea for recovering $x(t)$ from $z(t)$? – S.H.W Dec 26 '20 at 21:26
• 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. – robert bristow-johnson Dec 26 '20 at 21:53

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.

• 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 $X(j\omega)$. – S.H.W Dec 27 '20 at 19:18
• @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... – Fat32 Dec 27 '20 at 20:21
• @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 $\cos(Mt+\pi/4)$. – Fat32 Dec 27 '20 at 21:01
• @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. – Matt L. Dec 27 '20 at 22:53
• @S.H.W: I've updated and corrected my original answer. – Matt L. Dec 28 '20 at 11:24