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4.1
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In retrospect, we didn't really cover that much material, but the amount of knowledge we should've garnered is massive. A lot of the problems are tricky because the first step is to intuitively decide the method of solving them.
  
 
==Continuous Time Fourier Transform Pair for Aperiodic and Periodic Signals==
 
==Continuous Time Fourier Transform Pair for Aperiodic and Periodic Signals==
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==Properties of CT Fourier Transforms==
 
==Properties of CT Fourier Transforms==
===Linearity===
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[[Table_of_Formulas_and_Properties_ECE301Fall2008mboutin|Are Here]]
===Time Shifting===
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===Conjugation and Conjugate Symmetry===
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===Differentiation and Integration===
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===Time and Frequency Scaling===
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Quiz?
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===Duality===
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===Parseval's Relation===
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===Convolution===
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===Multiplication===
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==System's Characterized by Linear Constant-Coefficient Differential Equations==
 
==System's Characterized by Linear Constant-Coefficient Differential Equations==
Lecture 15: Sections 4.2-4.7.
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System's in sigma notation:
Lecture 16:
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Lecture 17:
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::<math> \sum_{k=0}^{n} a_k \frac{d^{k} y(t)}{dt^{k}} = \sum_{k=0}^{m} b_k \frac{d^{k} x(t)}{dt^{k}} </math>
Lecture 18:
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October break-no classes
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Example:
Lecture 11: Sections 4.0-4.1
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Lecture 12: Sections 4.2-4.4. 4.7
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--What is the frequency response of the general form system described above.
Lecture 13: Sections 4.5-4.7. 
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Lecture 14: Sections 5.0-5.3. 
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<math> \mathcal{F} \Bigg (\sum_{k=0}^{n} a_k \frac{d^{k} y(t)}{dt^{k}}\Bigg ) = \mathcal{F} \Bigg ( \sum_{k=0}^{m} b_k \frac{d^{k} x(t)}{dt^{k}} \Bigg ) </math>
Lecture 15: Sections 5.4-5.5, 5.8
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The sums will be constants so linearity applies
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<math>= \sum_{k=0}^{n} a_k \mathcal{F} \Bigg (\frac{d^{k} y(t)}{dt^{k}}\Bigg ) = \sum_{k=0}^{m} b_k \mathcal{F} \Bigg (  \frac{d^{k} x(t)}{dt^{k}} \Bigg )</math>
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If you apply the differentiation property of a Fourier transform k times:
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<math>= \sum_{k=0}^{n} a_k \times (j\omega)^{k} \times \mathcal{Y}(\omega) = \sum_{k=0}^{n} b_k \times (j\omega)^{k} \times \mathcal{X}(\omega)</math>
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Output signal:
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<math>=  \mathcal{Y}(\omega) = \frac{\sum_{k=0}^{n} b_k \times (j\omega)^{k}}{\sum_{k=0}^{n} a_k \times (j\omega)^{k}} \times \mathcal{X}(\omega)</math>
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Since this equation is the output signal of a system with input X in the frequency domain the output signal is the frequency (a constant) times the input signal.
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Thus:
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<math> H(j\omega ) = \frac{\sum_{k=0}^{n} b_k \times (j\omega)^{k}}{\sum_{k=0}^{n} a_k \times (j\omega)^{k}} </math>

Latest revision as of 13:38, 24 October 2008

In retrospect, we didn't really cover that much material, but the amount of knowledge we should've garnered is massive. A lot of the problems are tricky because the first step is to intuitively decide the method of solving them.

Continuous Time Fourier Transform Pair for Aperiodic and Periodic Signals

$ x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} X(j \omega) e^{j\omega t} \, d\omega $ (4.8)
$ X(j\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} \, dt $ (4.9)

The Fourier transform exists if the signal is absolutely integrable or if the signal has a finite number of discontinuities within any finite interval. (See Page 290)

Fourier Transform from the Fourier Series

This is useful for signals that fail to satisfy the previous properties of a signal that is guaranteed a Fourier Transform.

A signal represented as the sum of complex exponentials:

$ x(t) = \sum_{k = -\infty}^{+\infty} a_ke^{jk\omega_0t} $

with ak's:

$ a_k = \frac{1}{T}\int_{T} x(t)e^{-jk\omega_0 t} \, dt $

$ \xrightarrow{\mathcal{F}} $

$ X(j\omega) = \sum_{k = -\infty}^{+\infty} 2\pi a_k \delta(\omega - k\omega_0) $


Properties of CT Fourier Transforms

Are Here

System's Characterized by Linear Constant-Coefficient Differential Equations

System's in sigma notation:

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

Example:

--What is the frequency response of the general form system described above.

$ \mathcal{F} \Bigg (\sum_{k=0}^{n} a_k \frac{d^{k} y(t)}{dt^{k}}\Bigg ) = \mathcal{F} \Bigg ( \sum_{k=0}^{m} b_k \frac{d^{k} x(t)}{dt^{k}} \Bigg ) $

The sums will be constants so linearity applies $ = \sum_{k=0}^{n} a_k \mathcal{F} \Bigg (\frac{d^{k} y(t)}{dt^{k}}\Bigg ) = \sum_{k=0}^{m} b_k \mathcal{F} \Bigg ( \frac{d^{k} x(t)}{dt^{k}} \Bigg ) $

If you apply the differentiation property of a Fourier transform k times:

$ = \sum_{k=0}^{n} a_k \times (j\omega)^{k} \times \mathcal{Y}(\omega) = \sum_{k=0}^{n} b_k \times (j\omega)^{k} \times \mathcal{X}(\omega) $

Output signal:


$ = \mathcal{Y}(\omega) = \frac{\sum_{k=0}^{n} b_k \times (j\omega)^{k}}{\sum_{k=0}^{n} a_k \times (j\omega)^{k}} \times \mathcal{X}(\omega) $

Since this equation is the output signal of a system with input X in the frequency domain the output signal is the frequency (a constant) times the input signal.

Thus:

$ H(j\omega ) = \frac{\sum_{k=0}^{n} b_k \times (j\omega)^{k}}{\sum_{k=0}^{n} a_k \times (j\omega)^{k}} $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood