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$\textbf{Question:}$ An analog-to-discrete is designed as,

$$x[n] = x_a(nT)$$

A_D Converter

In an attempt to recover the analog signal from its samples x[n], a D/A converter is designed as ,

D-A Converter

where $x_1(t)$ is defined as:

$$x_1(t) = x[n]e^{-a(t-nT)} \ \text{ if } \ nT \leq t <(n+1)T$$

Design the LTI filter which would get $x_r(t)=x_a(t)$ if the Nyquist rate is satisfied.

Hint: Hint

Here is what I know about the design:

  1. It must be an ideal LPF filter with cutoff frequencies at: $\omega = |\frac{\pi}{T}|$
  2. Gain does not really make a difference since the amplitude is insignificant at this stage

What other things must the design constitute? And what does the hint suggest?

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  • $\begingroup$ looks like homework. $\endgroup$
    – robert bristow-johnson
    Dec 26, 2022 at 17:39
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    $\begingroup$ Old Exam question actually. I have a decent grip on Shannon-Kotelnikov sampling but I don't know how to approach this and what would be sufficient in terms of the design. $\endgroup$
    – Ahson Yousef
    Dec 26, 2022 at 17:51
  • $\begingroup$ Here is another mathematical description (and more complete) for $x_1(t)$ $$x_1(t) = \sum\limits_{n-=\infty}^{\infty} x[n]e^{-a(t-nT)} \Pi\left( \frac{t-nT}{T} - \tfrac12\right)$$ where $$ \Pi(u) \triangleq \begin{cases} 0, & \text{if } |u| > \frac{1}{2} \\ \frac{1}{2}, & \text{if } |u| = \frac{1}{2} \\ 1, & \text{if } |u| < \frac{1}{2} \\ \end{cases} $$ $\endgroup$
    – robert bristow-johnson
    Dec 26, 2022 at 18:35
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    $\begingroup$ Thank you for your response but I do not really understand what you have done with this more complete definition of $x_1(t)$. Also I am aware that it is supposed to be a LPF, however, this question alone is worth 1/4th of the grade of that entire exam so just writing that we need a brickwall LPF would not suffice, I believe. $\endgroup$
    – Ahson Yousef
    Dec 27, 2022 at 6:46
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    $\begingroup$ From what I can gather from the hint: $x_1(t)$ is not a typical sample-and-hold signal but rather after every sample there is a decay until the next one. But I have no idea what $e^{-at}$ has to do with the filter... $\endgroup$
    – Ahson Yousef
    Dec 27, 2022 at 7:08

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Simply speaking, there is an effect of $e^{-at}$ on your sampled signal, that is your original signal multiplied by an impulse train. Then in the Fourier domain you need to get rid of the effect of the exponential. In essence, you need to find $\mathcal{F} [e^{-at}]$ and then multiply by its inverse in the Fourier domain (I guess you know that the inverse in the Fourier domain is the $\frac{1}{H(j\omega)}$ where $H(j\omega$) is equal to $\mathcal{F} [e^{-at}]$). Then, passing the resultant signal through an ideal LPF will be enough to get your original signal's Fourier transform back if the Shannon-Kotelnikov sampling conditions are satisfied. I assume you know these conditions and you know how to pass the signal through such a LPF as you mentioned that.

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  • $\begingroup$ Thank you! I understood. $\endgroup$
    – Ahson Yousef
    Dec 27, 2022 at 13:43
  • $\begingroup$ this answer is technically not correct. $\endgroup$
    – robert bristow-johnson
    Dec 27, 2022 at 19:03

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