| Line 9: | Line 9: | ||
x(t) = u(t) | x(t) = u(t) | ||
| − | h(t) = {e}^{-\alpha t}u(t), \alpha > 0 | + | h(t) = <math>{e}^{-\alpha t}u(t)</math>, \alpha > 0 |
Now, to convolute them... | Now, to convolute them... | ||
| Line 17: | Line 17: | ||
2. <math>y(t) = \int_{-\infty}^{\infty}u(\tau){e}^{-\alpha (t-\tau)}u(t-\tau)d\tau</math> | 2. <math>y(t) = \int_{-\infty}^{\infty}u(\tau){e}^{-\alpha (t-\tau)}u(t-\tau)d\tau</math> | ||
| − | Since <math>u(\tau)*u(t-\tau) = 0 when t < 0, also when \tau > t</math>, you can set the limit accordingly. | + | Since <math>u(\tau)*u(t-\tau)</math> = 0 when t < 0, also when <math>\tau > t</math>, 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. | Keep in mind the following steps (4&5) are for t > 0, else the function is equal to 0. | ||
| Line 34: | Line 34: | ||
2. <math>y(t) = \int_{-\infty}^{\infty}{e}^{-\alpha (\tau)}u(\tau)u(t-\tau)d\tau</math> | 2. <math>y(t) = \int_{-\infty}^{\infty}{e}^{-\alpha (\tau)}u(\tau)u(t-\tau)d\tau</math> | ||
| − | Since <math>u(\tau)*u(t-\tau) = 0 when t < 0, also when \tau > t</math>, you can set the limit accordingly. | + | Since <math>u(\tau)*u(t-\tau)</math> = 0 when t < 0, also when <math>\tau > t</math>, you can set the limit accordingly. |
Keep in mind the following step (4) is for t > 0, else the function is equal to 0. | Keep in mind the following step (4) is for t > 0, else the function is equal to 0. | ||
Revision as of 19:21, 16 March 2008
LTI Systems
Chapter 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) $
End
