# CT Convolutions and their bounds

1. Sep 22, 2006

### ktpr2

This question deals with an example found in Signals and Systems, 2ed, page 99, ex 2.7. But I can summarize it here:

EDIT - my latex doesn't seem to work ...

$$x(t) = (1, 0 < t < T) or (0, otherwise)$$

$$h(t) = (t, 0 < t < 2T) or (0, otherwise)$$

I realize there are five bounds to consider: (t<0, 0 <t <T, T < t < 2T, 2T < t <3T, 3T < t)

However, for instance,

$$\\int_T^0 x(\tau) h(t-\tau) d\tau = \\int_T^0 h(t-\tau) d\tau = t\tau - \tau^2/2 (from 0 to T)$$

does not equal the example's answer of
$$t^2/2$$

nor does

(integrate from 2T to 3T)
$$\\int_2*T^3*T x(\tau) h(t-\tau) d\tau = \\int_T^0 h(t-\tau) d\tau = t\tau - \tau^2/2 (from 2T to 3T)$$

does not equal the example's answer of
$$-t^2/2 + Tt + 3/2*T^2$$

So my question is this:

Once I realize the bounds that the convolution must be evaluated for, how do i convert the bounds so that I can integrate them and derive the correct answer?

edit- x(tau) is a square pulse of height 1, from 0 to T (not tau but T)

and

h(t-tau) is a downward sloping right trianlge (90degree corner on the left) with a base from t-2T to t.