# Convolution of even functions is even

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) ?

• 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. – Fat32 Oct 12 '17 at 20:45

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 $t$.

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

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.
• 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 $y(-t)=y(t)$. – robert bristow-johnson Oct 13 '17 at 2:02
• +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. – Dilip Sarwate Oct 13 '17 at 4:13

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.