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

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?

• That's a broad question, related to "signals and systems" field and involving much of math (Laplace and Fourier transforms). But in short - step signal is consisting of infinite frequencies and theoretically covering the frequency response fully. – Eugene Sh. Sep 13 '18 at 19:18
• A hint, complementary to what @EugeneSh. said: the derivative of the step response is the impulse response. – a concerned citizen Sep 13 '18 at 20:33
• Try this interactive RLC filter to get a feel about how the step response and frequency response interact. There is also the math below the tool and it gives you the math for the step response if you press the appropriate button. – Andy aka Sep 14 '18 at 7:09
• A hint, complementary to what @a concerned citizen. said: The derivative in time domain is a multiplication by $s$ in Laplace domain, or $j\omega$ in Fourier domain. – Harry Svensson Sep 14 '18 at 7:41

$$\delta(t) = \begin{cases} +\infty &\mbox{if }t = 0 \\ 0 &\mbox{else} \end{cases}$$ 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 $j\omega$.