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The impulse response of the LTI system is

$$h(t)=e^{-4t} u(t)$$

The expression for the step response is

$$\frac14 \left(1-e^{-4t}\right)u(t)$$

My question is how $u(t)$ appears in the answer.

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  • $\begingroup$ just like the integral of the impulse is the step, the integral of the impulse response is the step response. $\endgroup$ – robert bristow-johnson Dec 9 '15 at 5:33
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Since your question is not about how to obtain the answer, but specifically about the reason why the unit step $u(t)$ appears in it, I assume that you know that the response to a step input is computed by convolving the impulse response with a step:

$$y(t)=\int_{-\infty}^{\infty}h(\tau)u(t-\tau)d\tau=\int_{-\infty}^{t}h(\tau)d\tau\tag{1}$$

Since $h(t)$ is causal, i.e., $h(t)=0$ for $t<0$, the lower integration limit can be changed to $0$. This implies that if in $(1)$ the upper integration limit $t$ is less than $0$, the result must be zero. Consequently, the response to a step input can be written as

$$y(t)=u(t)\cdot\int_0^{t}h(\tau)d\tau\tag{2}$$

Evaluating $(2)$ gives you the final answer, which includes the step function $u(t)$.

EDIT: Triggered by Dilip Sarwate's comment I add some extra explanation. Note that the step function in $(2)$ is actually redundant because for $t<0$ the integral in $(2)$ is zero because for $t<0$ we have

$$\int_0^{t}h(\tau)d\tau=-\int_{-|t|}^0h(\tau)d\tau=0,\quad t<0$$

because $h(t)=0$ for $t<0$. However, if $h(t)=f(t)u(t)$ with some function $f(t)$ that is not zero for $t<0$ (in the given example we have $f(t)=e^{-4t}$), we cannot just replace the lower integration limit in $(1)$ by $0$ and leave out the step function:

$$\int_{-\infty}^th(\tau)d\tau=\int_{-\infty}^tf(\tau)u(\tau)d\tau=\int_{0}^tf(\tau)u(\tau)d\tau\neq \int_{0}^tf(\tau)d\tau$$

because now we generally have for $t<0$

$$\int_{0}^tf(\tau)d\tau=-\int_{-|t|}^0f(\tau)d\tau\neq 0$$

The correct way to evaluate the integral for $h(t)=f(t)u(t)$ is

$$\int_{-\infty}^th(\tau)d\tau=u(t)\cdot\int_0^tf(\tau)d\tau$$

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  • $\begingroup$ I have edited your answer to include (what I think is) the correct argument as to why $\int_0^t h(\tau)d\tau = 0$ when $t < 0$. Ordinarily, $\int_0^{-1}f(x)dx= - \int_{-1}^0 f(x)dx$ and so setting the lower limit to $0$ early on requires using this, and then saying that $f(x) = 0$ for $x \in (0,1)$ etc. Feel free to roll back if you like your version better. $\endgroup$ – Dilip Sarwate Dec 9 '15 at 14:32
  • $\begingroup$ @DilipSarwate: I think I agree with everything you said in your comment, but I also think it's all there in my answer. Note that as long as you leave $h(t)$ as the integrand (and not the given expression without $u(t)$), then the output can be written as $\int_0^th(\tau)d\tau$ because for $t<0$ the integral is zero for the given $h(t)$. Just if you replace $h(t)$ by the given expression leaving out $u(t)$ you'll get into trouble. So even though in (2) $u(t)$ is strictly speaking redundant, it's easier to put it in because you'll need it later when you do the integration. $\endgroup$ – Matt L. Dec 9 '15 at 15:09

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