Slecture Fourier transform w f ECE438 - Rhea

Fourier transform as a function of frequency ω versus Fourier transform as a function of frequency f

A slecture by ECE student Dauren Nurmaganbetov

Partly based on the ECE438 Fall 2014 lecture material of Prof. Mireille Boutin.

OUTLINE

Introduction
Theory
Examples
Conclusion
References

Introduction

In my slecture I will explain Fourier transform as a function of frequency ω versus Fourier transform as a function of frequency f (in hertz).

Theory

Review of formulas used in ECE 301

CT Fourier Transform	$\mathcal{X}(\omega)=\mathcal{F}(x(t))=\int_{-\infty}^{\infty} x(t) e^{-i\omega t} dt$
Inverse Fourier Transform	$\, x(t)=\mathcal{F}^{-1}(\mathcal{X}(\omega))=\frac{1}{2\pi} \int_{-\infty}^{\infty}\mathcal{X}(\omega)e^{i\omega t} d \omega\,$

Review of formulas used in ECE 438.

CT Fourier Transform	$X(f)=\mathcal{F}(x(t))=\int_{-\infty}^{\infty} x(t) e^{-i2\pi ft} dt$
Inverse Fourier Transform	$\, x(t)=\mathcal{F}^{-1}(X(f))=\int_{-\infty}^{\infty}X(f)e^{i2\pi ft} df \,$

For more formulas see the table of CT Fourier transform pairs and properties

Examples

1) Let's compute FT of a cosine in two different ways:

First way is by changing FT pair and changing of variable

Let 
 $\, \mathcal\omega={2\pi}f$  ,   $\, \mathcal\omega_0={2\pi}f_0$

Also recall that
 $\displaystyle\delta(\alpha f)=\frac{1}{\alpha}\delta(f)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;for\;\;\alpha>0$

	$x(t) \$	$\longrightarrow$	$\mathcal{X}(\omega)$
	$\cos(\omega_0 t) \$		$\pi \left[\delta (\omega - \omega_0) + \delta (\omega + \omega_0)\right] \$

X(f)=\mathcal{X}({2\pi}f)=\pi \left[\delta ({2\pi}f - {2\pi}f_0) + \delta ( {2\pi}f+ {2\pi}f_0)\right] \

X(f)= \pi \left[\frac{1}{2\pi }\delta (f - f_0) + \frac{1}{2\pi }\delta (f + f_0)\right] \

X(f)= \frac{1}{2}\left[\delta (f - f_0) + \delta (f + f_0)\right] \

Second way is by direct using CTFT formula

X(f)= \frac{1}{2} \left[\delta (f - \frac{\omega_0}{2\pi}) + \delta (f + \frac{\omega_0}{2\pi})\right] \

2) Let's find CTFT of a shifted unit impulse:
  $\delta (t-t_0)\$ 
Keep in mind that:

CT Fourier Transform

X(f)=\mathcal{F}(x(t))=\int_{-\infty}^{\infty} x(t) e^{-i2\pi ft} dt

CT Fourier Transform

X(f)=\mathcal{F}(\delta (t-t_0))=\int_{-\infty}^{\infty} \delta (t-t_0) e^{-i2\pi ft} dt

Thus we get

X(f)=e^{-i2\pi ft} = e^{-i\omega ft}

Conclusion

Observe that the expressions for the FT are different because we used change of variables. 
Also notice that one can transform one expression into  the other using the scaling property of the Dirac delta

References

[1].Mireille Boutin, "ECE438 Digital Signal Processing with Applications," Purdue University August 26,2009

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