CT Convolutions and their bounds

[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.

Thank you for your time.

 

[mighty2000]

it is important to understand and accurately interpret mathematical concepts such as convolution. In this case, the example from Signals and Systems, 2ed, page 99, ex 2.7 introduces the concept of convolution with a specific example of two signals, x(t) and h(t). It is important to note that the bounds given in the problem are crucial in correctly evaluating the convolution integral.

First, let's clarify the given signals x(t) and h(t). x(t) is a square pulse of height 1, from 0 to T, which means that x(t) is equal to 1 for t values between 0 and T, and equal to 0 otherwise. Similarly, h(t) is a downward sloping right triangle with a base from t-2T to t, which means that h(t) is equal to t for t values between 0 and 2T, and equal to 0 otherwise.

Now, let's look at the bounds given in the problem. We have t < 0, 0 < t < T, T < t < 2T, 2T < t < 3T, and 3T < t. These bounds correspond to the different intervals where the convolution integral needs to be evaluated. For example, in the first interval t < 0, both x(t) and h(t) are equal to 0, so the integral will also be equal to 0.

In the second interval, 0 < t < T, we have x(t) = 1 and h(t) = t. Substituting these values into the integral, we get:

\\int_0^T x(\tau) h(t-\tau) d\tau = \\int_0^T 1 \cdot t d\tau = t\tau (from 0 to T) = Tt

This matches the example's answer of t^2/2. Similarly, in the third interval T < t < 2T, we have x(t) = 1 and h(t) = t. Substituting these values into the integral, we get:

\\int_T^0 x(\tau) h(t-\tau) d\tau = \\int_T^0 1 \cdot t d\tau = t\tau (from T to 2T) = Tt - T^2

This matches the example's answer of -