Say we have a continuous LTI system of which we know the analytical expression of its step response. Let's call it $y_{\text{step}}(t)$. Having $y_{\text{step}}(t)$ (and therefore also $Y_{\text{step}}(\omega)$) how can we find the impulse response $y_{\text{impulse}}(t)$?

I read on this question that "With an LTI system, the impulse response is the derivative of the step response. Because the impulse function is the derivative of the step function. Derivative in, derivative out.", while it makes sense to me intuitively, how would one go about proving this?

Any help would be greatly appreciated!

My effort:

We know that: \begin{equation} \frac{df}{dt} = \lim_{h \rightarrow 0}\frac{f(t+h) - f(t)}{h} \quad \quad \text{and} \quad \delta(t)=\frac{du}{dt}, \end{equation} where $\delta(t)$ and $u(t)$ are the dirac distribution and step function respectively.

Due to our system being linear we know that the additivity and homogeneity properties hold. That is, if $S[\cdot]$ is the output of the system for the input $\cdot$, then: \begin{align*} S[x_1(t) + x_2(t)] &= S[x_1(t)] + S[x_2(t)] &\text{(Additivity)} \\ S[ax(t)] &= aS[x(t)] &\text{(Homogeneity)} \end{align*}

Therefore, using the fact that: \begin{equation} \delta(t)=\frac{du}{dt} = \lim_{h \rightarrow 0}\frac{u(t+h) - u(t)}{h} \end{equation} we write: \begin{align*} S[\delta(t)] &= S\left[\lim_{h \rightarrow 0}\frac{u(t+h) - u(t)}{h} \right] \\[10pt] &= \lim_{h \rightarrow 0} \frac{S[u(t+h)] - S[u(t)]}{h} \\[10pt] &= \frac{d\left( S[u(t)]\right)}{dt} \end{align*}

Is my effort correct? Is it true in general for L.T.I systems that \begin{equation} S\left[\frac{dx}{dt}\right] = \frac{d\left(S[x]\right)}{dt} \quad ? \end{equation}


1 Answer 1


Since the system is LTI, the input-output relationship is given by the convolution integral. Hence, for any signal $x(t)$, the output is given by

$$ y(t) = (x*h)(t) = \int_{-\infty}^{+\infty}x(t-\tau)h(\tau)d\tau $$

where $*$ denotes convolution and $h(t)$ is the impulse response of the system.

Let's take the derivative with respect to time from both sides when the input signal is the unit step signal $u(t)$

$$ \begin{eqnarray} \frac{\partial}{\partial t}y_{\text{step}}(t) & = & \frac{\partial}{\partial t}\left(\int_{-\infty}^{+\infty}u(t-\tau)h(\tau)d\tau\right) \\ & \stackrel{(a)}{=} & \int_{-\infty}^{+\infty}\frac{\partial u(t-\tau)}{\partial t}h(\tau)d\tau \\ & \stackrel{(b)}{=} & \int_{-\infty}^{+\infty}\delta(t-\tau)h(\tau)d\tau \\ & = & (\delta * h)(t) \\ & \stackrel{(c)}{=} & h(t) = y_{\text{impulse}}(t) \end{eqnarray} $$


  • equality (a) is based on the fact that order of linear operators (integrator and differentiator) can be swapped
  • equality (b) is based on the fact that derivative of the step function is a delta function.
  • equality (c) is based on the sifting property of $\delta$ function.
  • 1
    $\begingroup$ @Nyquist-er, yes, that's correct. In general, the convolution and differentiator operator commute, i.e., $D[f*g] = D[f]*g = f*D[g]$ if $f$ and $g$ are absolutely integrable and at least one of them has an absolutely integrable (L1) weak derivative. In signal processing, these conditions are typically hold. Perhap, you may want to say that in the second last equality, the time-invariance property is used. $\endgroup$
    – AHT
    Commented Apr 22 at 23:09

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