Difference between revisions of "Practice CTFT from omega to f step function" - Rhea

Latest revision as of 09:44, 11 November 2013

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.

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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}\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) $

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

$ \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) $

Answer 6

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) $

Answer 7

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} $

Answer 8

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) $

Answer 9

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) $

Back to ECE438 Fall 2011 Prof. Boutin