# Convolution not commutative for this case?

#### jp1390

Joined Aug 22, 2011
45
Hi all, just needing some clarification on this question, or just another pair of eyes to check if I am doing this correctly. I am wanting to compute the convolution of v(t) and g(t). In doing so I wanted to try to test commutativity, but found that it wasn't working out.   When I tried the first convolution, I yielded the correct result over the desired regions of overlap, but for the second, I did not.

Method 1
$$y(t) = g(t)\ast v(t) = \int_{-\infty}^{\infty}g(\tau)v(\tau-t)d\tau$$

Method 2
$$y(t) = v(t)\ast g(t) = \int_{-\infty}^{\infty}v(\tau)g(\tau-t)d\tau$$ (did not receive correct result)

Region 1

t <= 0;
$$y(t_{1}) = 0$$

Region 2

t >= 0 and (t-2) <= 0
0 <= t <= 2;

$$y(t_{2}) = \int_{0}^{t}(exp{-\tau})(2exp{2(\tau -t)})d\tau$$

$$= 2exp{-2t}\int_{0}^{t}exp{\tau}d\tau$$

$$=2exp{-2t}(exp{t} - 1)$$

The answer for this region should be:

$$y(t_{2}) = 2exp{-t}(1 - exp{-t})$$

I got the correct result when I flipped and shifted v(t) as in Method 1, which is weird because they should be commutative.

Can anyone see where I went wrong?

Thanks,
JP

#### Zazoo

Joined Jul 27, 2011
114
$$=2exp{-2t}(exp{t} - 1)$$

The answer for this region should be:

$$y(t_{2}) = 2exp{-t}(1 - exp{-t})$$
These two results are the same, they are just factored differently. i.e if you distribute the outside term both are equal to:
$$y(t) = 2exp{-t} - 2exp{-2t}$$

$$y(t) = 2exp{-t} - 2exp{-2t}$$