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==[[Lecture21ECE301S11|Lecture 21]]==
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=== Multiplication Property ===
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<math> \mathcal{F}(x_1(t) x_2(t)) = \frac{1}{2\pi}\mathcal{F}(x_1(t))*\mathcal{F}(x_2(t))</math>
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=== Causal LTI system defined by cst coeff diff equations ===
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<math> \sum_{k=0}^{N}a_k \frac{d^k}{dt^k}y(t) =  \sum_{k=0}^{M}b_k \frac{d^k}{dt^k}x(t)</math>
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What is the frequency response of this system? Recall:
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<math>
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\begin{align}
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\mathcal{H}(\omega) &= \mathcal{F}(h(t)) \\
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\mathcal{Y}(\omega) &= \mathcal{H}(\omega)\mathcal{X}(\omega)
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\end{align}
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</math>
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Steps to solve:
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# Take the F.T. of both sides.
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#: <math>
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\begin{align}
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\mathcal{F}\left(\sum_{k=0}^{N}a_k \frac{d^k}{dt^k}y(t)\right) &= \mathcal{F}\left( \sum_{k=0}^{M}b_k \frac{d^k}{dt^k}x(t) \right) \\
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\sum_{k=0}^{N}a_k \mathcal{F}\left(\frac{d^k}{dt^k}y(t)\right) &= \sum_{k=0}^{M}b_k \mathcal{F}\left(\frac{d^k}{dt^k}x(t)\right) \\
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& \text{recall: }\mathcal{F}\left(\frac{d^n}{dt^n}y(t)\right) = (j\omega)^n Y(\omega) \\
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\sum_{k=0}^{N}a_k \left( j \omega \right)^k \mathcal{Y}(\omega) &= \sum_{k=0}^{M}b_k \left( j \omega \right)^k \mathcal{X}(\omega) \\
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\mathcal{Y}(\omega) &= \frac{\sum_{k=0}^{M}b_k(j\omega)^{k}}{\sum_{k^\prime=0}^{N}a_k(j\omega)^{k^\prime}} \mathcal{X}(\omega) \\
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\mathcal{H}(\omega) &= \frac{\sum_{k=0}^{M}b_k(j\omega)^{k}}{\sum_{k^\prime=0}^{N}a_k(j\omega)^{k^\prime}} \\
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h(t) &= \mathcal{F}^{-1}\left(\mathcal{H}(\omega)\right)
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\end{align}
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</math>
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Here are my lecture notes from ECE301 you can download both files from my dropbox account by Prof. Boutin
 
Here are my lecture notes from ECE301 you can download both files from my dropbox account by Prof. Boutin
  

Revision as of 07:32, 7 March 2011

Lecture 21

Multiplication Property

$ \mathcal{F}(x_1(t) x_2(t)) = \frac{1}{2\pi}\mathcal{F}(x_1(t))*\mathcal{F}(x_2(t)) $

Causal LTI system defined by cst coeff diff equations

$ \sum_{k=0}^{N}a_k \frac{d^k}{dt^k}y(t) = \sum_{k=0}^{M}b_k \frac{d^k}{dt^k}x(t) $

What is the frequency response of this system? Recall:

$ \begin{align} \mathcal{H}(\omega) &= \mathcal{F}(h(t)) \\ \mathcal{Y}(\omega) &= \mathcal{H}(\omega)\mathcal{X}(\omega) \end{align} $

Steps to solve:

  1. Take the F.T. of both sides.
    $ \begin{align} \mathcal{F}\left(\sum_{k=0}^{N}a_k \frac{d^k}{dt^k}y(t)\right) &= \mathcal{F}\left( \sum_{k=0}^{M}b_k \frac{d^k}{dt^k}x(t) \right) \\ \sum_{k=0}^{N}a_k \mathcal{F}\left(\frac{d^k}{dt^k}y(t)\right) &= \sum_{k=0}^{M}b_k \mathcal{F}\left(\frac{d^k}{dt^k}x(t)\right) \\ & \text{recall: }\mathcal{F}\left(\frac{d^n}{dt^n}y(t)\right) = (j\omega)^n Y(\omega) \\ \sum_{k=0}^{N}a_k \left( j \omega \right)^k \mathcal{Y}(\omega) &= \sum_{k=0}^{M}b_k \left( j \omega \right)^k \mathcal{X}(\omega) \\ \mathcal{Y}(\omega) &= \frac{\sum_{k=0}^{M}b_k(j\omega)^{k}}{\sum_{k^\prime=0}^{N}a_k(j\omega)^{k^\prime}} \mathcal{X}(\omega) \\ \mathcal{H}(\omega) &= \frac{\sum_{k=0}^{M}b_k(j\omega)^{k}}{\sum_{k^\prime=0}^{N}a_k(j\omega)^{k^\prime}} \\ h(t) &= \mathcal{F}^{-1}\left(\mathcal{H}(\omega)\right) \end{align} $

Here are my lecture notes from ECE301 you can download both files from my dropbox account by Prof. Boutin

There are bound to be a few errors in the document, if you find them please let me know and I'll fix them ASAP.

Lecture.pdf contains all lectures after lecture 5.

Lecture.pdf

Lecture.tex

Lecture5.pdf

Lectures 1 - 4

Alumni Liaison

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

Dr. Paul Garrett