2
$\begingroup$

I don't have an idea how to start it. Help pls :)

$$m = z(t) + \frac{\operatorname{Sa}(t-T)}{4}$$

We have some signal $z(t)=\left(\operatorname{Sa}(\frac{t}{T})\right)^2$. How often the m signal should be sampled to be sure of correct signal reconstruction

$\endgroup$
  • 1
    $\begingroup$ What is the $\operatorname{Sa}(\cdot)$ function? How is it defined? $\endgroup$ – robert bristow-johnson Jan 23 at 0:36
0
$\begingroup$

This is not an answer. I am just trying to restate the question so that the equation is dimensionally homogeneous:


I don't have an idea how to start it. Help pls :)

$$m(t) = z(t) + \tfrac{1}{4}\operatorname{sinc}\left(\tfrac{t-T}{T}\right)$$

We have some signal $z(t)=\left(\operatorname{sinc}(\tfrac{t}{T})\right)^2$. How often should the $m(t)$ signal be sampled to insure correct signal reconstruction?


this question can be meaningfully answered.

let's define:

$$ \operatorname{sinc}(u) \triangleq \begin{cases} \frac{\sin(\pi u)}{\pi u} \qquad & u \ne 0 \\ 1 \qquad & u = 0 \\ \end{cases} $$

$$ \Pi(u) \triangleq \begin{cases} 1 \qquad & |u| < \tfrac12 \\ \tfrac12 \qquad & |u| = \tfrac12 \\ 0 \qquad & |u| > \tfrac12 \\ \end{cases} $$

$$ \Lambda(u) \triangleq \begin{cases} 1 - |u| \qquad & |u| \le 1 \\ 0 \qquad & |u| > 1 \\ \end{cases} $$

$\endgroup$
-1
$\begingroup$

So in order to reconstruct a signal without loosing information you have to sample with $\frac{1}{T}=f_\mathrm{s}>2B$. This is known as the Nyquist rate.

Therefore in order to determine the minimal sampling rate we need to examine the function in the frequency domain.

$m(t) = \operatorname{sinc}\left(\tfrac{t}{T}\right)^2 +\dfrac{\operatorname{sinc}({t-T})}{4}$

So The Fourier transform would be

$M(f) = (T\operatorname{\Pi}(T\pi f) \circledast T\operatorname{\Pi}(T\pi f)) + \dfrac14\operatorname{\Pi}(\pi f)e^{-j2\pi fT}=T\Lambda (\pi f T) + \dfrac14\operatorname{\Pi}(\pi f)e^{-j2\pi fT}$

in order to faithfully reconstruct the signal : $ f_\mathrm{s}=\max(4T,2)$

The exponent doesn't matter it's bounded by the $\operatorname{\Pi}(\cdot)$

$\endgroup$
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Peter K. Jan 30 at 22:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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