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Continuous-time Fourier transform: from omega to f

In ECE301, you learned that the Fourier transform of a step function $ x(t)=u(t) $ is the following:

$ {\mathcal X} (\omega) = \frac{1}{j \omega} + \pi \delta (\omega ). $

Use this fact to obtain an expression for the Fourier transform $ X(f) $ (in terms of frequency in hertz) of the step function. (Your answer should agree with the one given in this table.) Justify all your steps.


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Answer 1

There are a few things that you need to know to accomplish this problem. The two main formulas that you need are $ \omega = 2 \pi f $ and $ \delta(cx)= \frac{1}{c} \delta(x) $ for c>0.

PROOF

$ \int_{-\infty}^{\infty}\delta(x)dx = 1 $

$ y=cx => \frac{dy}{c}=dx $

$ \int_{-\infty}^\infty \delta(y)\frac{dy}{c}=\frac{1}{c} $

THEREFORE

$ \delta(\omega)=\delta(2\pi f)=\frac{1}{2\pi}\delta(f) $

and

$ {\mathcal X} (\omega) = \frac{1}{j \omega} + \pi \delta (\omega ) = \frac{1}{2}(\frac{1}{j\pi f} + \delta(f)) $

-my

Instructor's comments: In essence, this answer is correct, except that you forgot to state that
$ \delta(cx)= \left\{ \begin{array}{ll} \infty, & x = 0,\\ 0, & \text{ else.} \end{array} \right. $
However the "flow of thoughts" is a bit hard to follow. I would suggest a slight reordering/rearrangement of the arguments. You could save space by giving less details regarding the change of variables when integrating. -pm

Answer 2

I claim that $ c\delta(cx)= \delta(x) $ because of the following two facts:

$ c\delta(cx)= \left\{ \begin{array}{ll} \infty, & x = 0,\\ 0, & \text{ else.} \end{array} \right. $

and

$ \int_{-\infty}^\infty c\delta(cx)dx=\int_{-\infty}^\infty c\delta(y)\frac{dy}{c} = 1 $

Now, using the equation $ \omega=2 \pi f $, we have:

$ X(f)=\frac{1}{j2\pi f} + \pi\delta(2\pi f) = \frac{1}{j2\pi f} + \frac{1}{2}\delta(f) $

Instructor's comments: Pretty good! What do you guys think? -pm


Answer 3

$ c\delta(cx)= \delta(x) $ could be proven this way.

$ \begin{align}\mathcal{F}[x(t)] = \frac{1}{2\pi}\int_{-\infty}^{\infty} x(t)e^{j\omega t} dt \\= \frac{1}{2\pi}X(\omega)\end{align} $

since $ w = 2\pi f $,

$ \begin{align} \frac{1}{2\pi}X(\omega) = \frac{1}{2\pi}X(2\pi f) \end{align} $

$ \begin{align}\mathcal{F}[x(t)] = \int_{-\infty}^{\infty} x(t)e^{j2\pi f t} dt\end{align} $


more to be added


Answer 4

$ {\mathcal X} (\omega) = \frac{1}{j \omega} + \pi \delta (\omega ) $

Since $ \omega $ is equal to $ 2\pi f $

$ {\mathcal X} (2\pi f) = \frac{1}{j2\pi f} + \pi \delta(2\pi f) $

$ c\delta(cx)= \delta(x) $ because,

$ c\delta(cx)= \left\{ \begin{array}{ll} \infty, & x = 0,\\ 0, & \text{ else.} \end{array} \right. $

$ \int_{-\infty}^\infty c\delta(cx)dx=\int_{-\infty}^\infty c\delta(y)\frac{dy}{c} = 1. $

Using the above formula,

$ {\mathcal X} (f) = \frac{1}{j2\pi f} + \frac{1}{2} \delta (f) $

Answer 5

$ \omega=2\pi f, so{\mathcal X}(\omega)= {\mathcal X} (2\pi f) = \frac{1}{j \2\pi f} + 2\pi \delta (2\pi f )/2. Because C\delta(Cf)=\delta(f),{\mathcal X}(f)=\frac{1}{j \2\pi f}+\frac{1}{2}\delta(f) $



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