# Does the unit step response not affect the transfer function?

I am still confused, this question says that this figure shows three unit step responses and was asked to find the transfer function. Doesn't the input step responses only affect the output? So to look for the transfer function does not matter how many step unit response?

• This appears to be a three part question, and you are missing the other 2 graphs. This graph is labelled "a", implying more than one. Can you solve for this output, assuming one unit step input? Mar 1, 2020 at 7:58
• In case @Mattman944's hint gets unnoticed: there are three graphs to help you better with determining the time constants involved. None of them affect the transfer function, but all three of them will help the eye make better readings. Jun 22, 2021 at 8:33

In general the output of a linear system is the convolution of the input signal with the impulse response of the system.

The "Step Response" is just the system output when the input is the unit step function.

Also remember that convolution in the time domain corresponds to multiplication in the laplace domain. Also note that the unit step function has a laplace transfomr of 1/s, and that multiplying by 1/s in the laplace domain corresponds to integration in the time domain.

It appears that they are giving you (in the graphs) the integral of the impulse response, and they want you to find the impulse response itself given that integral.

http://www-control.eng.cam.ac.uk/gv/p6/Handout2.pdf

this question says that this figure shows three unit step responses and was asked to find the transfer function.

The picture you include is a single step input response, maybe there are different plots nearby with the other two step responses.

Doesn't the input step responses only affect the output?

Not sure what you mean by this, but for LTI system, you usually find the step response by having a system that is at zero $$\y(0^-)\$$ and with $$\u(0^-)\$$ a zero input before $$\t=0\$$, which then "steps" to be $$\u(t)=1, \; t\geq 0\$$. So yes, the output only depends on the input.

So to look for the transfer function does not matter how many step unit response?

Again, you could be a bit more clear on your question, but if you know the input $$\u(t)\$$ and the output $$\y(t)\$$ you can find the transfer function $$\H(s) = \frac{Y(s)}{U(s)}\$$.

So, suppose that by "having three step inputs" you mean that you had one step at t=0, another at t=1 and a last one at t=2, that would definitely not look like a single unit step input, nor like a single 3-units step. But, due to linearity and time invariance, $$\y(t)\$$ would look like having the unit step response $$\g(t)\$$ and some shifted versions of it, that is, $$\y(t)=g(t)+g(t-1)+g(t-2)\$$.