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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;

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

Robotbugs
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HOLYBIBLETHE
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1 Answers1

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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).

Alex M.
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apalopohapa
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  • @ apalopohapa:Do you mean both, t and $\tau$, are variable here. – HOLYBIBLETHE Feb 23 '13 at 10:23
  • the step function under the integral is $u(-(\tau - t))$. – HOLYBIBLETHE 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. – HOLYBIBLETHE Feb 23 '13 at 10:39