Convert Fourier transform of a signal to $\omega$ form?

Question

Suppose we have this signal:

$$x(t)=5 \sin(2\pi 1000 t) \cos(2\pi 10000 t)$$

I calculated the Fourier transform as below:

$$X(f)=\left(j\frac{5}{4}\right)\left[\delta(f-9000)+\delta(f+11000)-\delta(f-11000)-\delta(f+9000)\right]$$

and then I would like to convert it to $\omega$ form. I know that the result is

$$X(\omega )=\left(j\frac{5}{2}\right)\left[\delta(\omega -9000)+\delta(\omega +11000)-\delta(\omega -11000)-\delta(\omega +9000)\right]$$

I know that $\omega=2\pi f T$ where $T$ is sampling rate.

What I cannot understand is how we convert the Fourier transform of our signal to $\omega$ form using above formula.

Answer 1

In your example there is no sampling involved, so you simply have $\omega=2\pi f$, where $f$ is the frequency in Hertz, and $\omega$ is the frequency in radians.

Your first result for $X(f)$ looks correct. However, your result for $X(\omega)$ is wrong. What you need to know to be able to convert an expression with Dirac deltas from $f$ to $\omega$ is the following relation:

$$\delta(af)=\frac{1}{|a|}\delta(f)\tag{1}$$

With $a=2\pi$ and $\omega=2\pi f$ you get from $(1)$

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

You get the two following basic Fourier transform correspondences, depending on whether you use $f$ or $\omega=2\pi f$ as the frequency domain variable:

$$\begin{align}e^{j2\pi f_0t}&\longleftrightarrow \delta(f-f_0)\\ e^{j\omega_0t}&\longleftrightarrow 2\pi\delta(\omega-\omega_0)\end{align}$$

So your expression for $X(\omega)$ misses a factor of $\pi$, and the sums and differences of the two frequencies in the arguments of the Dirac deltas need to be multiplied by $2\pi$:

$$X(\omega)=\frac{5\pi j}{2}[\delta(\omega-2\pi9000)-\delta(\omega+2\pi9000)+\delta(\omega+2\pi11000)-\delta(\omega-2\pi11000)]$$