Topic: Continuous-time Fourier transform: from omega to f

## Question

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.

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

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

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

\begin{align}\int_{-\infty}^{\infty} \delta(\frac{x}{a})dx = \int_{-\infty}^{\infty} \delta(\frac{x}{|a|})dx \end{align}

Changing integration variable $y = \frac{x}{|a|} , dy = \frac{dx}{|a|}$ ,

\begin{align}\int_{-\infty}^{\infty} \delta(\frac{x}{a})dx = |a|\int_{-\infty}^{\infty} \delta(y)dy \end{align}

Changing integration variable again from y to x $x = y , dx = dy$ ,

\begin{align}\int_{-\infty}^{\infty} \delta(\frac{x}{a})dx = |a|\int_{-\infty}^{\infty} \delta(x)dx \end{align}

Therefore, $\delta(\frac{t}{c}) = c\delta(t)$

So, using the equation $\omega=2 \pi f, \frac{\omega}{2\pi} = f,$, we have:

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

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

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,

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

$\int_{-\infty}^\infty c\delta(cx)dx=\int_{-\infty}^\infty c\delta(y)\frac{dy}{c} = 1$ $\omega=2 \pi f$

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

The definition of the delta function is

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

and

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

Therefore

$c\delta(cx)= \delta(x)$ by the definition and substitution of $cx$ for $x$ in the integral.

Then change what $X(\omega)$ is a function of to $X(2\pi f)$:

$X(\omega) -> X(f) = \frac{1}{jf} + \pi \delta (f)$

But $\omega = 2\pi f$ so

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

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

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

since

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

so

$\delta(cx)= \frac{1}{c} \delta(x) ; c>0$

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

therefore

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

\begin{align} {\mathcal X} (\omega) &= \frac{1}{j \omega} + \pi \delta (\omega ) \\ &= \frac{1}{2}(\frac{1}{j\pi f} + \delta(f)) \end{align}

Because

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

We can show:

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

Which shows that $c \delta [cx] = \delta [x]$

Then, by comparing with the formula given in the problem statement:

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

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

And knowing that, $\int\limits_{-\infty}^{\infty}\delta(x)dx=1$

It can be shown that, $\int\limits_{-\infty}^{\infty}\delta(ax)dx \overset{\underset{\mathrm{y=ax}}{}}{=} \int\limits_{-\infty}^{\infty}\frac{1}{a}\delta(y)dy = \frac{1}{a}$

So, $\delta(ax) = \frac{1}{a}\delta(x)$

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

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