Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is a question from Bertsekas' Data Networks. It is question 2.2 on page 141.

It is asking for the convolution of the following 2 functions.

Function 1: $ s(t) = 1 $, when $0 \leq t \leq T$. It is $0$ elsewhere.

Function 2: $h(t) = \alpha e^{-a t}$, when $t \geq 0$. It is $0$ elsewhere.

My work:

The output $r(t) = \int_{-\infty}^{\infty} s(\tau) h(t-\tau) d\tau$. Then, we can rewrite $s(t) = 1 \times u(t) \times u(T-t)$, where $u(t)$ is the standard step function. Also, we can rewrite $h(t) = \alpha e^{-a t} u(t)$.

Then, $$ r(t) = \int_{-\infty}^{\infty} u(\tau) u(T-\tau) \alpha e^{-a (t-\tau)} u(t-\tau) d\tau$$

I am confused at this point because there are three separate step functions in the integral. The first step function $u(\tau)$ says the lower limit of $\tau$ in the integral must be 0.

The second step function $u(T-\tau)$ says $\tau \leq T$ or else the integral evalutes to zero. In the examples I am looking at this step function is not present, hence I have not seen this complication before.

The third step function $u(t-\tau)$ says the upper limit of $\tau$ in the integral must be $t$. I have seen this before in the examples.

Does the second step function have much influence on the result of the integral? I have two upper bounds, $\tau = T$ and $\tau = t$. I assume $\tau=t$ wins since it is more general.

Note that: $r(t)$ is defined in the following time range: $[0+0, T+\infty]$. That is, $r(t)$ exists for all non-negative times.

So, $r(t) = \int_0^t \alpha e^{-a (t-\tau)} d\tau$


share|cite|improve this question
Part of the confusion is: suppose I remove the $u(T-t)$ term from $s(t)$. According to my logic, the result $r(t)$ doesn't change. This seems strange to me. – jrand May 25 '11 at 21:48
$s(t)=u(t)-u(t-T)$ is an option as well. – Raskolnikov May 25 '11 at 22:01
@Raskolnikov: Yes, but after distributing, that still leaves the question of who wins in the upper bound: $\tau = T$ or $\tau = t$ or something else. – jrand May 25 '11 at 22:03
$t$, what else? You know that $t \leq T$, so once $\tau$ exceeds $t$ you get zero. – Raskolnikov May 25 '11 at 22:06
up vote 1 down vote accepted

Split into the cases $0 < t \leq T$ and $t > T$, to conclude that $$ (s*h)(t) = \int_0^{\min \{ t,T\} } {1 \cdot h(t - \tau )d\tau } = \int_0^{\min \{ t,T\} } {\alpha e^{ - \alpha (t - \tau )} d\tau } . $$

EDIT: It follows that $$ (s*h)(t) = 1 - e^{-\alpha t}, \;\; 0 < t \leq T, $$ $$ (s*h)(t) = e^{ - \alpha (t - T)} - e^{ - \alpha t} ,\;\; t > T. $$ (Since $(s*h)(t) = 0$ for $t \leq 0$, we see that $(s*h)(t)$ is continuous on $\mathbb{R}$.)

Relation to probability theory: With $s$ and $h$ as above, let $X$ be an exponential random variable with density function $h$, and $Y$ an independent uniform$[0,T]$ random variable, so that $Y$ has density function $\tilde s = s/T$. By the law of total probability, conditioning on $Y$, we have $$ {\rm P}(X + Y \le t) = \int_0^T {{\rm P}(X \le t - \tau )\frac{1}{T}d\tau }. $$ It follows that if $0 < t \leq T$, then $$ {\rm P}(X + Y \le t) = \int_0^t {{\rm P}(X \le t - \tau )\frac{1}{T}d\tau } = \frac{1}{T}\int_0^t {(1 - e^{ - \alpha (t - \tau )} )d\tau } = \frac{{t - (1 - e^{ - \alpha t} )/\alpha }}{T}, $$ while if $t > T$, then $$ {\rm P}(X + Y \le t) = \int_0^T {{\rm P}(X \le t - \tau )\frac{1}{T}d\tau } = \frac{1}{T}\int_0^T {(1 - e^{ - \alpha (t - \tau )} )d\tau } = \frac{{T - (e^{ - \alpha (t - T)} - e^{ - \alpha t} )/\alpha }}{T}. $$ Hence, the density function of $X+Y$ is given by $$ f_{X+Y} (t) = \frac{{1 - e^{ - \alpha t} }}{T} ,\;\; 0 < t \leq T, $$ $$ f_{X+Y} (t) = \frac{{e^{ - \alpha (t - T)} - e^{ - \alpha t} }}{T}, \;\; t > T. $$ On the other hand, since $X$ and $Y$ are independent with respective densities $h$ and $\tilde s \,(=s/T)$, $$ f_{X+Y} (t) = (h*\tilde s)(t) = (\tilde s * h)(t) = \frac{{(s*h)(t)}}{T}, $$ from which it follows that $$ (s*h)(t) = 1 - e^{ - \alpha t} ,\;\; 0 < t \leq T, $$ $$ (s*h)(t) = e^{ - \alpha (t - T)} - e^{ - \alpha t} ,\;\; t > T $$ (as we have already seen above).

share|cite|improve this answer
Hello Shai Covo, Am I correct in stating the following math statement: $\tau \leq T \wedge \tau \leq t \Rightarrow \tau \leq \min(T, t)$ – jrand May 26 '11 at 20:17
jrand, this is correct, and more generally $a \le \min \{ x_1 , \ldots ,x_n \} \Leftrightarrow a \le x_1 , \ldots ,a \le x_n $ (where "," stands for "and"). By the way, the symbol $\wedge$ is sometimes used instead of "min". – Shai Covo May 26 '11 at 20:35

It seems a shame that you can get into so much mathematical detail on this question without noticing that it is the perhaps the very simplest non-trivial example of signal processing in electrical circuit theory, namely: the charging of a capacitor (through a resistor) when you close a switch.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.