Revision as of 10:59, 12 December 2008 by Mcwalker (Talk | contribs)


Convolution Example

Example of CT convolution

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 $
  3. 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.

  4. $ y(t) = \int_{0}^{t} {e}^{-\alpha (t-\tau)}d\tau = {e}^{-\alpha t} \int_{0}^{t}{e}^{ \alpha \tau}d\tau $
  5. $ y(t) = {e}^{-\alpha t}\frac{1}{\alpha}({e}^{\alpha t}-1) = \frac{1}{\alpha}(1-{e}^{-\alpha t}) $
  6. Now you can replace the condition in steps 4&5 with a u(t).

  7. $ 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 $
  3. 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.

  4. $ y(t) = \int_{0}^{t} {e}^{-\alpha \tau}d\tau = \frac{1}{\alpha}(1-{e}^{-\alpha t}) $
  5. Now you can replace the condition in step 4 with a u(t).

  6. $ y(t) = \frac{1}{\alpha}(1-{e}^{-\alpha t})u(t) $

End

Name --dennis.m.snell.1, Sun, 30 Sep 2007 22:25:27

Michael, why did you put your name at the end of this page?

Name --michael.a.mitchell.2, Mon, 01 Oct 2007 15:54:00

Wasn't Sure if the authorship issue had been solved yet. (in class it was said that only the last person to make a change to a page would be credited with it's authorship)

Name --dennis.m.snell.1, Mon, 01 Oct 2007 16:51:27

The authorship issue was not an issue. It was mentioned in class, but by a student asking about it. There is a log of every action and every edit on this kiwi that can be reviewed each week. You are safe in leaving out your name. Sometime soon the editing will be reworked; however, and you might add your name to some other special page, but it will just get lost at the bottom of a topic. I removed your name here, but worry not, you are not forgotten.

... --john.w.fawcett.1, Mon, 15 Oct 2007 11:15:56

why is this under "Exams" as it's parent? Wouldn't Chapter 3 be better?

... --john.w.fawcett.1, Mon, 15 Oct 2007 11:18:06

Sorry, meant Chapter 2. I'll go ahead and add a backlink to chapter 2, but leave this one to Exams up for now.

Frequency and Impulse Response Example

Frequency and Impulse Response of a causal LTI system defined by a difference equation

For the discrete time L.T.I. system described by

$ y[n]-\frac{1}{2}y[n-1]=x[n]+\frac{1}{2}x[n-1] $

Find the frequency response H($ \omega\ $) and the impulse response h[n] of the system.

Frequency Response:

1: Take the Fourier transform of the equation,

$ Y(\omega)-\frac{1}{2}e^{-j\omega}Y(\omega)=X(\omega)+\frac{1}{2}e^{-j\omega}X(\omega) $

2: Solve for Y($ \omega\ $)/X($ \omega\ $), which is the frequency response H($ \omega\ $),

$ H(\omega)=\frac{Y(\omega)}{X(\omega)}=\frac{1+\frac{1}{2}e^{-j\omega}}{1-\frac{1}{2}e^{-j\omega}} $

Impulse Response:

1: Expand into two terms using partial fraction expansion (Guide to Partial Fraction Expansion) to facilitate use of inverse Fourier transform,

$ H(\omega)=\frac{1}{1-\frac{1}{2}e^{-j\omega}}+\frac{1}{2}\frac{e^{-j\omega}}{1-\frac{1}{2}e^{-j\omega}} $

2: Take the inverse Fourier transform of H($ \omega\ $) (Fourier Transform Table),

$ h[n]={\left(\frac{1}{2} \right)}^{n}u[n]+\frac{1}{2}{\left(\frac{1}{2} \right)}^{n-1}u[n-1] $

3: Simplify if so inclined,

for n = 0
$ h[n] = 1\ $
for n > 0
$ h[n] = {\left(\frac{1}{2} \right)}^{n-1} $

Back to ECE301

Difference Equation in Class Example

Issue with LTI system defined by Difference Equation

I do not understand this difference equation, so could someone please explain it...

The Fourier Transform gives one answer, but it seems as though there are really more answers. One in paticular, c=0, appears to give better results for the step response...please see attached for explanation of what I am talking about...

$ y[n] - y[n-1] = x[n]\ $
$ \Rightarrow H(\omega) = \frac{1}{1 - e^{-j \omega}} $

From table in book:

$ \begin{align} u[n] &\overset {\mathfrak{F}}{\longleftrightarrow} \frac{1}{1 - e^{-j \omega}} + \sum_{k=-\infty}^{\infty} \pi \delta (\omega - 2 \pi k) \\ 1 &\overset {\mathfrak{F}}{\longleftrightarrow} 2 \pi \sum_{l=-\infty}^{\infty} \delta (\omega - 2 \pi l) \end{align} $
$ \Rightarrow H(\omega) = \frac{1}{1 - e^{-j \omega}} + \sum_{k=-\infty}^{\infty} \pi \delta (\omega - 2 \pi k) - \frac{1}{2} \cdot 2 \pi \sum_{l=-\infty}^{\infty} \delta (\omega - 2 \pi l) $
$ \Rightarrow h[n] = \mathfrak{F}^{-1}(H(\omega)) = u[n] - \frac{1}{2}\ $

Find step response:

$ \begin{align} y[n] &= x[n] * h[n] = u[n] * h[n] = \sum_{k=-\infty}^{\infty} u[k]h[n-k] \\ &= \sum_{k=-\infty}^{\infty} u[k] \left ( u[n-k] - \frac {1}{2} \right ) = \sum_{k=-\infty}^{\infty} u[k]u[n-k] - \frac{1}{2} \sum_{k=-\infty}^{\infty} u[k] \\ &= \sum_{k=0}^{\infty} u[n-k] - \frac{1}{2} \sum_{k=0}^{\infty} 1 = -\infty \end{align} $
(this seems unreasonable!)

Note: The original eq. $ y[n] - y[n-1] = x[n]\ $ can be expressed as:

$ h[n] - h[n-1] = \delta [n]\ $

By observation, $ h[n] = u[n] + c\ $ for any constant c. And further: the step response is divergent for $ c \ne 0\ $.

So, what is c equal to?


once again --mireille.boutin.1, Fri, 26 Oct 2007 14:30:38

Dividing by zero is not recommended.

Please clarify --ross.a.howard.1, Fri, 26 Oct 2007 18:16:00

Can you please clarify where I divided by zero? I still do not know what to do with the arbitrary constant that comes up in the solution...

Back to ECE301

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett