Difference between revisions of "HW 7 Allen Humphreys ECE301Fall2008mboutin" - Rhea

Revision as of 13:33, 24 October 2008

4.1

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}} $