# How to determine the step response using convolution of the signal's impulse response?

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;

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

-

## migrated from electronics.stackexchange.comFeb 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

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).
@ 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