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:


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.

  • $\begingroup$ do you mean the Discrete-Time Fourier Transform? $\endgroup$ – robert bristow-johnson Feb 20 '16 at 3:06
  • $\begingroup$ See this answer on math.SE to see how to convert Fourier transforms with respect to the Hertzian frequency variable $f$ into Fourier transforms with respect to radian frequency variable $\omega$ in the more general case. Of course, for Dirac deltas, there are additional complications as pointed out in MattL's answer. $\endgroup$ – Dilip Sarwate Feb 20 '16 at 18:15

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:


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


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 $2\pi$:

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

| improve this answer | |

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.