# Signals - question due to sampling

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

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

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:

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

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

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

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

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