# Linear System: Symmetric Under Time Reversal?

In class, my professor mentioned that "Linear systems must be symmetric under time reversal" in an off-handed way and did not clarify further. I assume this is true, but I'm not sure how one would show this based off the definition of linear that I once learned.

I'm assuming symmetric under time reversal means: $$x(n) = x(-n)$$

How is this shown by the definition of linearity? $$a x_1(n) + b x_2(n) \rightarrow a y_1(n) + b y_2(n)$$

• You need to ask your prof. It could be that they were teaching about some specific subject and they meant that the linear systems that they were talking about that day needed to be time-symmetric. Otherwise I don't think the statement makes much sense as it stands. Commented Jan 27 at 19:11

I can't know what your professor meant, but what can be shown quite easily for linear time-invariant systems is that time-reversing the input and the system's impulse response will result in a time-reversed output. Let $$x[n]$$ and $$h[n]$$ be the input signal and the impulse response, respectively. Then the output is given by the convolution sum

$$y[n]=\sum_{k=-\infty}^{\infty}x[k]h[n-k]\tag{1}$$

Now we define the time-reversed input and impulse response by

$$x^-[n]=x[-n],\qquad h^-[n]=h[-n]$$

The convolution sum of these two sequences is the time-reversed output $$y[-n]$$:

\begin{align*} \sum_{k=-\infty}^{\infty}x^-[k]h^-[n-k] &= \sum_{k=-\infty}^{\infty}x[-k]h[-n+k] \\ &= \sum_{k=-\infty}^{\infty}x[k]h[-n-k] \\ &= y[-n] \end{align*}

Note that both, $$x[n]$$ and $$h[n]$$ need to be time-reversed in order to obtain a time-reversed output.

• Wouldn’t the system need to be time invariant? The OP only mentions “linear”
– Jdip
Commented Jan 24 at 17:07
• @Jdip: Yes, I assumed time-invariance. I believe that this is what must be meant, even though you're right that in the question it only says "linear". Commented Jan 24 at 17:13