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.


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.

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  • $\begingroup$ 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

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