What is the relationship between an amplifier's step input and the frequency sweep response?

Question

What is the relationship between step input response and the frequency response of a system such as an amplifier?

I see similarities between the step input response of an amplifier and the frequency response of an amplifier(how gain changes with the frequency). Of course this could apply to a filter as well.

Does that mean I extract the same information about an amplifier's gain versus freq by 1-) applying a step input and recording its response 2-) sweeping the amplifier input between zero to max freq. and plot the frequency versus gain?

Im not into the subjects much but I would like to have an overview between these two concepts and similarities. Even though I sense some similarities there is still some fog in my mind related to some missing math or concepts. Would be great to link or relate them as a big picture.

Is that because the sharp step input includes many sines already embedded in it?

Answer 1

The frequency response is the Fourier transform of the impulse response in the time domain. This is the response when the input is a Dirac-distribution, which is defined by the following properties:

$$\delta(t) = \begin{cases} +\infty &\mbox{if }t = 0 \\ 0 &\mbox{else} \end{cases}$$ and $$\int_{-\infty}^{+\infty}\delta(t)dt = 1$$

Integrating this Dirac distribution (as per the second property), will give you the step function, which is equivalent in the Fourier-domain to multiplying with \$\frac{1}{j\omega}\$. Also notable is that the Fourier transform of this Dirac distribution is exactly 1 over all frequencies.

So you can use these properties (\$u(t)\$ is the step function) for any linear system \$f\$:

$$\begin{align} \mathcal{F}\{f\circ u\} &= \mathcal{F}\{f\}\cdot\mathcal{F}\{u\} \\ &= \mathcal{F}\{f\}\cdot\frac{1}{j\omega}\mathcal{F}\{\delta\} \\ &= \frac{1}{j\omega}\mathcal{F}\{f\} \end{align}$$

So if you ever measured a step response and want to derive the frequency response of the system from that, you can simply multiply the Fourier transform of the step response by \$j\omega\$.