4
$\begingroup$

We have: $$ s(t) = 4u(t)−u(t−1)−2u(t−2)−2u(t−3.5)+2u(t−4) $$ where $u$ is the Heaviside function.

I had to draw the graph and find the minimum sampling period so we wont have loss of signal. I have drawn the graph.

$t=kT$ where $T$ is the sampling period.So $T=t/k$ so how do I find T?

enter image description here

Improve this question
$\endgroup$
3
  • $\begingroup$ What is k? what is t? $\endgroup$
    – willpower2727
    Feb 17 '16 at 13:05
  • $\begingroup$ @willpower2727, k is just a constant, the t is time. $\endgroup$
    – Mahendra Gunawardena
    Feb 17 '16 at 13:11
  • $\begingroup$ Pass from the time domain to frequency. You will see several impulses, and then you can see their frequencies and evaluate the needed from Nyquist statement. $\endgroup$
    – leCrazyEngineer
    Jul 17 '16 at 5:01
4
$\begingroup$

The smallest feature that you need to capture is the dip between 3.5 and 4. If you use a sampling period which is greater than 0.5, then you may end up not capturing this feature. The sampling period must therefore be $T\leq 0.5$ in order not to lose any features of the signal.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ In addition, you'd want to consider the issue of the nyquist or folding frequency. I'd suggest a sampling frequency >= 4x faster than the fastest feature of your signal. But this might be a consideration beyond the scope of the problem. $\endgroup$
    – willpower2727
    Feb 17 '16 at 14:54
2
$\begingroup$

To prevent loss of signal you need to sample at twice the highest frequency presented in the signal. This is also known as Nyquist rate.

Assuming that time t is in seconds, the smallest time period is about 0.5s or the highest time frequency is 2Hz. So if sampled between 2Hz and 4Hz the processed signal will be subjected to what is known as aliasing. If the signal is sampled above 4Hz, there will no data loss. Below are few references that gives better detailed explanation.

References:

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Since the signal is a step function rather than a sine wave, you should be able to get away with a lower sampling rate. $\endgroup$
    – Mark
    Feb 18 '16 at 2:33

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.