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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

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    $\begingroup$ What is the $\operatorname{Sa}(\cdot)$ function? How is it defined? $\endgroup$ – robert bristow-johnson Jan 23 '19 at 0:36
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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:

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

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

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

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Hint:

(1) $sinc(\frac{t}{T})$ is the inverse fourier transform of a $rect$ function in frequency domain with $f=[-\frac{1}{2T}, \ \frac{1}{2T}]$. Time-shift will only result in a phase term in Fourier transform which does not impact the bandwidth.

(2) $z(t)$ is product of two $sinc(\frac{t}{T})$ in time-domain. Hence, its frequency domain representation will be a convolution of two $rect$ functions given in (1). Calculate convolution of two $rect$ function with $f=[-\frac{1}{2T}, \ \frac{1}{2T}]$ and focus on its maximum frequency.

(3) $m(t)$ is sum of two time-domain functions, so, its frequency domain representation is also a sum of frequency-domain representation of the two signals.

(4) Sampling frequency should be at least twice of the maximum frequency in Fourier Transform of $m(t)$.

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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 $\operatorname{\Pi}(\cdot)$

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Peter K. Jan 30 '19 at 22:28

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