Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

The step response can be determined by recalling that the response of an LTI to any input signal is found by computing the convolution of that signal with the impulse response of the system. Therefore we can write

$s(t) = u(t)∗h(t) =\int_{−∞}^{∞} u(τ)h(t−τ) dτ$

The convolution is commutative, meaning that $u(t)∗h(t) = h(t)∗u(t)$

It is more convenient to write the step response in the following way:

$s(t) = h(t)∗u(t) =\int_{−∞}^{∞}h(τ)u(t−τ) dτ=\int_{−∞}^{t}h(τ) dτ$

How do you explain the last step? How did we get the following;

$\int_{−∞}^{∞}h(τ)u(t−τ) dτ=\int_{−∞}^{t}h(τ) dτ$

share|cite|improve this question

migrated from Feb 24 '13 at 2:22

This question came from our site for electronics and electrical engineering professionals, students, and enthusiasts.

What is the actual question? – AndrejaKo Feb 23 '13 at 10:03
what is the author saying? – Rajesh K Singh Feb 23 '13 at 10:05
how did we get \$\int_{−∞}^{∞}h(τ)u(t−τ) dτ=\int_{−∞}^{t}h(τ) dτ\$ – Rajesh K Singh Feb 23 '13 at 10:06
up vote 4 down vote accepted

Not exactly sure what the question is, but the equation given is basically the convolution between the impulse response $t \mapsto h(t)$ and the input signal, which in this case is a step $t \mapsto u(t)$.

The equation is simplified simply because of the nature of the step input (it steps from $0$ to $1$ at $t=0$), so the infinite integral can be simplified in the sense that the limits are changed so that they only cover the portion where $u(t-\tau) = 1$, therefore with the new integration range ($\tau \in (-\infty, t)$) you can replace it with $1$ (so it gets "removed" from the equation).

share|cite|improve this answer
@ apalopohapa:Do you mean both, t and \$\tau\$, are variable here. – Rajesh K Singh Feb 23 '13 at 10:23
the step function under the integral is \$u(-(\tau - t))\$. – Rajesh K Singh Feb 23 '13 at 10:31
@Rajesh Properly answering requires more space, but simply put: tao is your integration variable, the thing you change as you perform the infinite summation, and t is a "constant", because the whole integral is to calculate the response at a specific point in time (t). Thankfully, it is "parametrized", so you can evaluate the integral for any t of your choosing (meaning you can calculate the value of the response at any instant). – apalopohapa Feb 23 '13 at 10:35
@ apalopohapa : thank you, for the explanation. – Rajesh K Singh Feb 23 '13 at 10:39

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.