Very simple question regarding $X(f)$ vs. $X(j\omega)$

Question

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.

Answer 1

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