LTI Systems
Chapter 2 Class Notes
Convolution Example
This is an example of convolution done two ways on a fairly simple general signal.
x(t) = u(t)
h(t) = $ {e}^{-\alpha t}u(t) $, \alpha > 0
Now, to convolute them...
1. $ y(t) = x(t)*h(t) = \int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau $
2. $ y(t) = \int_{-\infty}^{\infty}u(\tau){e}^{-\alpha (t-\tau)}u(t-\tau)d\tau $
Since $ u(\tau)*u(t-\tau) $ = 0 when t < 0, also when $ \tau > t $, you can set the limit accordingly. Keep in mind the following steps (4&5) are for t > 0, else the function is equal to 0.
3. $ y(t) = \int_{0}^{t} {e}^{-\alpha (t-\tau)}d\tau = {e}^{-\alpha t} \int_{0}^{t}{e}^{ \alpha \tau}d\tau $
4. $ y(t) = {e}^{-\alpha t}\frac{1}{\alpha}({e}^{\alpha t}-1) = \frac{1}{\alpha}(1-{e}^{-\alpha t}) $
Now you can replace the condition in steps 4&5 with a u(t).
5. $ y(t) = \frac{1}{\alpha}(1-{e}^{-\alpha t})u(t) $
Now, the other way... (by the commutative property)
1. $ y(t) = h(t)*x(t) = \int_{-\infty}^{\infty}h(\tau)x(t-\tau)d\tau $
2. $ y(t) = \int_{-\infty}^{\infty}{e}^{-\alpha (\tau)}u(\tau)u(t-\tau)d\tau $
Since $ u(\tau)*u(t-\tau) $ = 0 when t < 0, also when $ \tau > t $, you can set the limit accordingly. Keep in mind the following step (4) is for t > 0, else the function is equal to 0.
3. $ y(t) = \int_{0}^{t} {e}^{-\alpha \tau}d\tau = \frac{1}{\alpha}(1-{e}^{-\alpha t}) $
Now you can replace the condition in step 4 with a u(t).
4. $ y(t) = \frac{1}{\alpha}(1-{e}^{-\alpha t})u(t) $
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