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| − | == | + | [[Category:problem solving]] |
| + | [[Category:ECE301]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:Fourier transform]] | ||
| + | [[Category:signals and systems]] | ||
| + | == Example of Computation of Fourier transform of a CT SIGNAL == | ||
| + | A [[CT_Fourier_transform_practice_problems_list|practice problem on CT Fourier transform]] | ||
| + | ---- | ||
Compute the Fourier Transform of the signal | Compute the Fourier Transform of the signal | ||
| − | <math>\ x(t)= t \sin(2 \pi | + | <math>\ x(t)= \int_{-\infty}^{t} \tau \sin(2 \pi \tau+ \pi/4) d\tau </math> |
By definition the Fourier Transform of a signal is defined as: | By definition the Fourier Transform of a signal is defined as: | ||
| − | <math>X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt</math> | + | <math> F[x(t)] = X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt</math> |
| − | First | + | First expressing the signal in as a Fourier series: |
| − | <math>\ x(t)=\sin( | + | However before finding the transform we note that integration in the time domain is just division in the frequency domain. So the game plan is to find the Fourier series of x'(t) then divide by the frequency in the frequency space. |
| + | |||
| + | <math>\ x'(t)=\sin(2\pi t+ \pi/4) = \frac{e^{2 \pi jt + \pi/4}}{2j} - \frac{e^{-2 \pi jt -j \pi/4}}{2j}</math> | ||
| + | |||
| + | <math> X'(\omega)=\int_{-\infty}^{\infty} \frac{e^{j \pi/4}}{2j} e^{j2 \pi} e^{-j\omega t}dt - \int_{-\infty}^{\infty} \frac{e^{-j \pi/4}}{2j} e^{-j2 \pi} e^{-j\omega t}dt</math> | ||
| + | |||
| + | Using some foresight we see that a straight up integration of the expression above will yield something infinite or indeterminate, we take advantage of the known Fourier transform of a complex exponential. | ||
| + | |||
| + | <math> \int_{-\infty}^{\infty} x(t) dt = \frac{X(\omega)}{\omega} - X(0) \pi \delta(\omega)</math> | ||
| + | |||
| + | <math> X'(\omega)= \frac{e^{j \pi/4}}{2j} F[e^{j2 \pi}] - \frac{e^{-j \pi/4}}{2j} F[e^{-j2 \pi}] </math> | ||
| + | |||
| + | Noting that <math>\ F[e^{j\omega_0}] = 2 \pi \delta(\omega - \omega_0) </math> | ||
| + | |||
| + | <math>\ X'(\omega) = j \pi \delta(\omega + 2\pi) e^{-j \pi /4}- j \pi \delta(\omega + 2\pi) e^{j \pi /4}</math> | ||
| + | |||
| + | Since <math>\ X(\omega) = 0 </math> | ||
| + | |||
| + | <math> X(\omega) =\frac{j \pi}{\omega} \delta(\omega + 2\pi) e^{-j \pi /4}- \frac{j \pi}{w} \delta(\omega + 2\pi) e^{j \pi /4} | ||
| + | ---- | ||
| + | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | ||
Latest revision as of 11:23, 16 September 2013
Example of Computation of Fourier transform of a CT SIGNAL
A practice problem on CT Fourier transform
Compute the Fourier Transform of the signal
$ \ x(t)= \int_{-\infty}^{t} \tau \sin(2 \pi \tau+ \pi/4) d\tau $
By definition the Fourier Transform of a signal is defined as:
$ F[x(t)] = X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt $
First expressing the signal in as a Fourier series:
However before finding the transform we note that integration in the time domain is just division in the frequency domain. So the game plan is to find the Fourier series of x'(t) then divide by the frequency in the frequency space.
$ \ x'(t)=\sin(2\pi t+ \pi/4) = \frac{e^{2 \pi jt + \pi/4}}{2j} - \frac{e^{-2 \pi jt -j \pi/4}}{2j} $
$ X'(\omega)=\int_{-\infty}^{\infty} \frac{e^{j \pi/4}}{2j} e^{j2 \pi} e^{-j\omega t}dt - \int_{-\infty}^{\infty} \frac{e^{-j \pi/4}}{2j} e^{-j2 \pi} e^{-j\omega t}dt $
Using some foresight we see that a straight up integration of the expression above will yield something infinite or indeterminate, we take advantage of the known Fourier transform of a complex exponential.
$ \int_{-\infty}^{\infty} x(t) dt = \frac{X(\omega)}{\omega} - X(0) \pi \delta(\omega) $
$ X'(\omega)= \frac{e^{j \pi/4}}{2j} F[e^{j2 \pi}] - \frac{e^{-j \pi/4}}{2j} F[e^{-j2 \pi}] $
Noting that $ \ F[e^{j\omega_0}] = 2 \pi \delta(\omega - \omega_0) $
$ \ X'(\omega) = j \pi \delta(\omega + 2\pi) e^{-j \pi /4}- j \pi \delta(\omega + 2\pi) e^{j \pi /4} $
Since $ \ X(\omega) = 0 $
$ X(\omega) =\frac{j \pi}{\omega} \delta(\omega + 2\pi) e^{-j \pi /4}- \frac{j \pi}{w} \delta(\omega + 2\pi) e^{j \pi /4} ---- [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] $
