(New page: 4.1 ==Continuous Time Fourier Transform Pair for Aperiodic and Periodic Signals== ::<math> x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} X(j \omega) e^{j\omega t} \, d\omega </math> (4.8...) |
|||
| (4 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
| − | + | 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== | ||
| Line 21: | Line 21: | ||
==Properties of CT Fourier Transforms== | ==Properties of CT Fourier Transforms== | ||
| − | + | [[Table_of_Formulas_and_Properties_ECE301Fall2008mboutin|Are Here]] | |
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ==System's Characterized by Linear Constant-Coefficient Differential Equations== | |
| − | + | System's in sigma notation: | |
| − | + | ||
| − | + | ::<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> | |
| − | + | ||
| − | + | Example: | |
| − | + | ||
| − | + | --What is the frequency response of the general form system described above. | |
| − | + | ||
| − | + | <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> | |
| + | |||
| + | The sums will be constants so linearity applies | ||
| + | <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> | ||
| + | |||
| + | If you apply the differentiation property of a Fourier transform k times: | ||
| + | |||
| + | <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> | ||
| + | |||
| + | Output signal: | ||
| + | |||
| + | |||
| + | <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> | ||
| + | |||
| + | 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: | ||
| + | |||
| + | <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 14:38, 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.
Contents
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
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}} $
