I try to understand how the Fourier transform changes when I try to compute $X(\omega)$ or $X(f)$. Can someone work me through the maths please for two examples, $x(t) = \exp(j\omega_0 t)$ and $x(t) = \cos(\omega_0t)$. I know what their Fourier transform is in terms of $\omega$ but when I try to compute $X(f)$ from first principles, I am stuck.

  • $\begingroup$ What are the definitions of the Fourier transforms that give you $F(\omega)$ and $F(f)$ from $f(t)$, and how would they differ if the signals were named $g(t)$ instead of $f(t)$? Incidentally, neither $\exp(j\omega_0t)$ nor $\cos(\omega_0)$ have Fourier transforms that exist in the classical sense, and so it might be worth explaining how exactly you "know" what their Fourier transforms are: something from your book? a table of Fourier transforms? And if you calculated the Fourier transforms from first principles all by yourself, including those calculations would help too. $\endgroup$ – Dilip Sarwate Apr 29 at 19:24
  • $\begingroup$ So I calculated FT using F(ω) = $\int_{-\infty} ^\infty \! f(t)exp(-j \omega t) \, \mathrm{d}t$. From this you can work out the maths. (It's a bit hard for me to show it here, since I am not familiar with latex). The result for cos(ω0t) is π[δ(ω-ω0) + δ(ω+ω0)]. In the FT that I used above, If I replace ω with 2πf, then how can I proceed?. I hope this is useful! $\endgroup$ – makala Apr 29 at 19:43
  • $\begingroup$ Great! Now consider $F(f) = \int_{-\infty}^\infty f(t)\exp(-j2\pi f t) dt$ and $G(f) = \int_{-\infty}^\infty g(t)\exp(-j2\pi f t) dt$. Is one of them easier to calculate than the other? $\endgroup$ – Dilip Sarwate Apr 29 at 19:51
  • $\begingroup$ @DilipSarwate, i wouldn't be using $f(\cdot)$ a a function name when $f = \frac{\omega}{2\pi}$ are a variable. I might suggest using a different function name when also using the Fourier Transform with $f$ in Hz. $\endgroup$ – robert bristow-johnson May 1 at 4:21
  • $\begingroup$ So is this about the Duality Theorem of the Fourier Transform? Is that what you need? $\endgroup$ – robert bristow-johnson May 1 at 4:23

You have the obvious relationship $\omega=2\pi f$. So going from a Fourier transform expressed in terms of $\omega$ to a Fourier transform expressed in terms of $f$ is just a matter of substitution.

The problem you've encountered is very likely in the argument of the Dirac delta impulse. Clearly, $\delta(\omega)=\delta(2\pi f)$, but you need to know the following relationship in order to obtain an expression in terms of $\delta(f)$:

$$\delta(ay)=\frac{1}{|a|}\delta(y),\qquad a\neq 0\tag{1}$$

So we have

$$\delta(\omega)=\delta(2\pi f)=\frac{1}{2\pi}\delta(f)\tag{2}$$

E.g., the Fourier transform of $x(t)=e^{j\omega_0t}=e^{j2\pi f_0t}$ is given by

$$\begin{align}\mathcal{F}\big\{e^{j\omega_0t}\big\}&=2\pi\delta(\omega-\omega_0)\\&=2\pi\delta\big(2\pi (f-f_0)\big)\\&=\delta(f-f_0)\tag{3}\end{align}$$

I'm sure that from this you can derive the corresponding expressions for $x(t)=\cos(\omega_0t)$.


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