0
$\begingroup$

The periodic square wave, $$x(t) = 1, |t| < T_1 \\ x(t) = 0, T_1 < |t| < T/2$$

What I wonder is ..

(1) If $T_1$ is constant and $T \to \infty$, is x(t) a constant function?

(2) If $T_1$ is constant and $T \to 0$, is this non-sense?

(3) If $T$ is constant and $T_1 \to 2/T$, is x(t) a constant function?

(4) If $T$ is constant and $T_1 \to 0$, is x(t) a impulse function?

Actually, I am a undergraduate and it is my first class about signal. I don't know impulse function correctly. Could you explain impulse function in terms of (4) ?

The above questions are my intutitions. I don't know answers.

$\endgroup$
3
$\begingroup$

If $T_1$ remains constant and $T\rightarrow\infty$ then you get a single square pulse of length $2T_1$. There won't be another square pulse because the period is infinite.

$T_1$ constant and $T\rightarrow 0$ is indeed nonsensical because the definition of the signal assumes $T_1 < T/2$.

If $T$ is constant, you can get a constant function if $T_1=T/2$ (and not $2/T$).

If $T$ is constant and $T_1\rightarrow 0$ your function becomes 0. You will only get an impulse if you keep the area under the rectangle constant, i.e. if its height increases as its width decreases.

$\endgroup$
  • $\begingroup$ Thank you, but I don't understand the final case. What do you mean by "keep the area" is a additional particular condition? Is this right? $\endgroup$ – jakeoung Apr 22 '13 at 13:41
  • $\begingroup$ Yes, if the area is not kept constant you don't get an impulse but the signal simply disappears as the pulse width approaches zero. $\endgroup$ – Matt L. Apr 22 '13 at 15:08

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.