1
$\begingroup$

Prove that the convolution of two even functions is an even function.

I have my own proof which I have included as an answer, but it assumes a linear time-invariant system. I want to know if there is a general way to prove this (or if it even holds for a general convolution) ?

$\endgroup$
  • 1
    $\begingroup$ you don't have to assume an LTI system to work with convolution operator? just use $z(t) = x(t) \star y(t) = y(t) \star x(t)$ as the convolution operator and use $z(t) = \int_{-\infty}^{\infty}x(\tau)y(t-\tau) d\tau = \int_{-\infty}^{\infty}y(\tau)x(t-\tau) d\tau$ Nothing about LTI has to be mentioned here. You have to show that $z(t) = z(-t)$ whenever both $x(t)$ and $y(t)$ are even, as you did in your solution. $\endgroup$ – Fat32 Oct 12 '17 at 20:45
4
$\begingroup$

Just do a change of variables:

\begin{align} z(t) &= x\star y \big|_{t}\\ &=\int_{\tau=-\infty}^{\tau = \infty} x(\tau)y(t-\tau) \,\mathrm d\tau &\scriptstyle{\text{Set}~\tau=-\lambda, \mathrm d\tau=-\mathrm d\lambda} \tag{1}\\ &=-\int_{-\lambda=-\infty}^{-\lambda = \infty} x(-\lambda)y(t+\lambda) \,\mathrm d\lambda &\scriptstyle{\text{Now remember that}~ x~\text{and}~y ~\text{are even functions}}\\ &=\int_{\lambda=-\infty}^{\lambda=\infty} x(\lambda)y((-t)-\lambda) \,\mathrm d\lambda &\scriptstyle{\text{Same as (1) except with }\lambda~\text{instead of }\tau ~\text{and with}~-t}\\ &= z(-t)\end{align} with nary a mention of LTI systems.

For a more DSPish proof, consider that the Fourier transform of a real-valued even function of time $t$ is a real-valued even function of frequency $f$, and thus if $X(f)$ and $Y(f)$ both are real-valued even functions of $f$, then so is $Z(f) =X(f)Y(f)$ also a real-valued even function of $f$, implying in turn that $z = x\star y$ is a real-valued even function of $t$ whenever $x(t)$ and $y(t)$ are real-valued even functions of $t$.

$\endgroup$
  • $\begingroup$ A downvote without any comment explaining why the down-voter believes the answer is not useful. Oh, well.... $\endgroup$ – Dilip Sarwate Oct 13 '17 at 11:56
3
$\begingroup$

This may be a lot easier in the frequency domain. Something like this:

  1. Even (and real) function in time transforms into a real function in frequency (and vice versa)
  2. Convolution in the time domain is equivalent to convolution in the frequency domain
  3. Multiplication of two real functions is a real function
  4. Inverse transform of a real function is even.
$\endgroup$
  • 2
    $\begingroup$ of course it's easier in the frequency domain, given the theorem that the Fourier Transform of a convolution is the product of the Fourier Transforms. if the OP wants to do it directly with the integral definition all Under needs to do is use the fact that $h(-t)=h(t)$ and $x(-t)=x(t)$ by definition of what even symmetry means, and a change in variable of integration in the integral, to prove that $y(-t)=y(t)$. $\endgroup$ – robert bristow-johnson Oct 13 '17 at 2:02
  • 2
    $\begingroup$ +1 but 4 is not correct: the inverse transform of a real even function is a real even function. The inverse transform of a real function of frequency is a complex function of time. not a real function, and not necessarily even. $\endgroup$ – Dilip Sarwate Oct 13 '17 at 4:13
2
$\begingroup$

Given that the input $x(t)$ to and the impulse response $h(t)$ of a linear time-invariant system are both even functions, the output $y(t)$ is given by $$y(t)=x(t)*h(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau$$ To show that the output is even, it must be shown that $y(t)=y(-t)$. $$y(-t)=\int_{-\infty}^{\infty}x(\tau)h(-t-\tau)d\tau=\int_{-\infty}^{\infty}x(\tau)h(-(t+\tau))d\tau$$ Since $h(t)$ is even $$y(-t)=\int_{-\infty}^{\infty}x(\tau)h(\tau+t)d\tau$$ This output corresponds to an input of $x(-t)$ since by linear time-invariance of the system it would be $$\int_{-\infty}^{\infty}x(\tau)\delta(\tau+t)d\tau=\int_{-\infty}^{\infty}x(-t)\delta(\tau+t)d\tau=x(-t)\int_{-\infty}^{\infty}\delta(\tau+t)d\tau=x(-t)$$

So the output $y(-t)$ is given by the input $x(-t)$, but $x(-t)=x(t)$. This mean $y(-t)=y(t)$. The output of an even input to a linear time-invariant system having an even impulse response is also even.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.