2
$\begingroup$

Let's take a time domain function $x(t) = \cos( 2 \pi f_0 t) $. Its Fourier transform can be represented as

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

as well as

$$X(j\omega) = \pi \left[ \delta(\omega - \omega_0) + \delta(\omega + \omega_0) \right]\tag{2}$$ which was initially wrong, corrected for clarity.

Considering $ \omega = 2 \pi f $, I can interchange between the two expressions. I can tell that the $j$ in the argument of the second expression can be accounted for by the $j$ in the RHS of the first one but I can't seem to get $(1)$ from $(2)$.

Starting with $(2)$:

\begin{align} X(j\omega) &= \pi \left[ \delta(\omega - \omega_0) + \delta(\omega + \omega_0) \right]\\ X(j 2 \pi f) &= \pi \left[ \delta( 2 \pi f - 2 \pi f_0) + \delta( 2 \pi f + 2 \pi f_0) \right] \end{align}

I can't see how to get $(1)$ from here. I know it is very simple but I can't get any good results by Googling it up. I'd appreciate if you do not downvote my post.

$\endgroup$
3
  • 1
    $\begingroup$ There are a lot of mistakes in your question. First, I suppose you mean $x(t)=\cos(2\pi f_0 t)$ instead of $\cos(2\pi\omega_0 t))$. Second, the scale factors in both equations (1) and (2) are wrong. There shouldn't be a $j$ there, and in (2) you're missing a factor of $2\pi$. So both equations you start with are wrong. Please make sure everything is correct, otherwise we don't know where to start. $\endgroup$
    – Matt L.
    Jun 18, 2016 at 21:10
  • $\begingroup$ Hello @MattL. Edited the mistakes. I hope it's fine now? $\endgroup$
    – user22487
    Jun 18, 2016 at 21:26
  • $\begingroup$ Yes, now it looks good, and you got a good answer right away! $\endgroup$
    – Matt L.
    Jun 19, 2016 at 9:58

1 Answer 1

0
$\begingroup$

Your first equation is right but second equation is wrong:

I would proceed in the following way to see their relation. Given $x(t) = \cos(\omega_0 t)$ with $\omega_0 = 2 \pi f_0$, first see that: $$x(t) = \cos(\omega_0 t) = \frac{1}{2}e^{j\omega_0 t} + \frac{1}{2}e^{-j\omega_0 t}$$ whose CTFT is: $$X(\omega) = \pi [\delta(\omega - \omega_0) + \delta(\omega + \omega_0)] ..... (1)$$

which follows from the transform pair: $e^{j\omega_0 t} \leftrightarrow 2\pi \delta(\omega - \omega_0)$

Then use the following property of the impulse $\delta(a x) = \frac{1}{|a|}\delta(x)$ applied with $a=2\pi$ and $x=f$, and then (1) becomes:

\begin{align} X(f) &= \pi [\delta(2\pi f - \omega_0) + \delta(2\pi f + \omega_0)]\\ X(f) &= \pi [\delta(2\pi (f - \omega_0/{2\pi})) + \delta(2\pi (f + \omega_0/{2\pi})]\\ X(f) &= \pi [\frac{1}{2\pi}\delta(f - f_0) + \frac{1}{2\pi}\delta(f + f_0)]\\ X(f) &= \frac{1}{2}\delta(f - f_0) + \frac{1}{2}\delta(f + f_0)\\ \end{align} is the conclusion.

Note that the $j$ in the argument of CTFT $X(j\omega)$ is a notation that is equally valid in both forms as in $X(jω)$ or $X(ω)$, and it can be omitted for simplicity if there is no possibility of confusion, but otherwise remembered when it is necessary to underline the fact that the function is in general complex valued. Note that for the communications engineer the notation $X(f)$ is almost always is the choice. In either case arguments must be interpreted correctly based on the fundamental definition of the corresponding Fourier transform as given by:

$$X(jω) \triangleq \int_{-\infty}^{\infty}x(t)e^{−j\omega t}dt$$ , and $$X(f) \triangleq \int_{-\infty}^{\infty}x(t)e^{−j2\pi f t}dt$$

$\endgroup$
7
  • $\begingroup$ Voted up (+1). However, I still have confusions before I can accept this answer. You showed the proof relating $X(f)$ and $X(j \omega)$. Perhaps a more exact question at this point would be to compare $X(\omega)$ and $X(j \omega)$? How are they related? The original question basically asks regarding the effect of $j$ in the argument. $\endgroup$
    – user22487
    Jun 18, 2016 at 23:08
  • $\begingroup$ I guess I get it. That $j$ also comes from the same property of the impulse (i.e. $\delta(ax) = \frac{1}{|a|} \delta(x)$ where $x=\omega$ and $a=j$. Am I right? $\endgroup$
    – user22487
    Jun 18, 2016 at 23:16
  • $\begingroup$ No. That $j$ in the argument of CTFT is a notataion that is equally valid in both forms such as $X(j\omega)$ and as $X(\omega)$ or $X(jf)$ and $X(f)$ which is usually omitted for simplicity but remembered when it is necessary to underline the fact that the function is in general complex valued as given by the integrals: $X(j\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t}dt$, $X(j f) = \int_{-\infty}^{\infty} x(t) e^{-j2\pi f t}dt$ $\endgroup$
    – Fat32
    Jun 18, 2016 at 23:26
  • $\begingroup$ Perfect. Please edit your answer and place it there, since that's what I have been looking for. Plus, where is the factor of $2 \pi$ as @MattL. stated was missing in my post? $\endgroup$
    – user22487
    Jun 18, 2016 at 23:31
  • $\begingroup$ ok let me put it, but I don't know where your $2\pi$ is... $\endgroup$
    – Fat32
    Jun 18, 2016 at 23:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy