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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.
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
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) $
