I am currently learning convolution (which is not the easiest topic for sure). Just as by many fields in electronics also here are differential equations involved. With this (what I have seen until now) there are two types in convolution, namely:

1.) y'(t)+ay(t)=bx(t) *

2.) y'(t)+ay(t)=bx'(t) *

*With x(t) is the "unit impulse function" and y(t) is the "impulse response"

Then in the book out of nowhere they give the solutions to these differential equations, namely:

1.) enter image description here

2.) enter image description here

1.So the first one is kind of standard and I know what they are doing, but where did the x(t) goes. is it treated as a 1 with the integral or something?

2.The second is some random formula for me ;) , can somebody please give an step by step solution by how to get from the differential equation to the solution of both the 2 functions?

P.S. The delta t in the second solution is equal to x(t)

  • \$\begingroup\$ This is why differential equations is usually a prereq for circuits. If you don't understand diff-eqs, the solution looks like it comes out of nowhere. If you understand them, it's clear, and if you vaguely remember them, you at least know that there is a process that can get you from point A to point B, and can look it up if you need to. \$\endgroup\$ Mar 1, 2016 at 13:58

1 Answer 1


What you're describing is not really convolution per se, although the results can be used with convolution to get solutions for more complicated input signals. Maybe you're slightly misunderstanding what the book is trying to do?

The usual method for solving this for arbitrary \$x(t)\$ is to first solve it for \$x(t) = \delta(t)\$, which is what the calculation in your book very likely does. This solution is called the impulse response. The Fourier transform of \$\delta(t)\$ is indeed \$1\$ (or \$1/\sqrt{2 \pi}\$, depending on the conventions used), so in that sense the right hand side "is treated as one".

Then convolution is used to represent the general case as $$ x(t) = \int_{-\infty}^\infty \mathrm{d}T\, x(T) \delta(t - T). $$ Since the differential equation is linear and shift invariant (a sum of shifted solutions is a solution), and here \$t\$ appears only inside the delta function, you can use this to write the solution in the general case as a convolution between \$x(t)\$ and the impulse response.

In the second case, the input is the derivative of some given function \$x(t)\$, whose Fourier transform you can relate to the FT of the function itself by \$\mathcal{F}\{x'(t)\}(\omega) = i \omega \mathcal{F}\{x(t)\}(\omega)\$. I think what your book does here is that it solves the problem for the derivative of the delta function, to make a general formula for this case.

  • \$\begingroup\$ Ah, so if I understand correctly it is more an general differential equations understanding than strictly convolution. If so, could you give an explanation on how to get the y(t) from the differential equation? \$\endgroup\$
    – user90392
    Mar 1, 2016 at 13:25
  • \$\begingroup\$ Yes, it's all about linear ordinary differential equations theory, of which convolution is a part of. I'll try to find the time to type up an example, although I think the derivation on how to use this to solve equations with a general \$x(t)\$ should also be found in any textbook worth its salt. \$\endgroup\$
    – Timo
    Mar 1, 2016 at 14:18

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