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Hi: After looking around the internet and looking at solutions to similar questions, I was finally able to convince myself of the following mathematically.

If $G(\omega_{0}) = cos(\omega_{0} t)$, then the inverse fourier transform is such that

$ g(t_{0}) = 2 \pi \left[ \frac{\delta(t_{0} - t)}{2} + \frac{\delta( t_{0} + t)}{2} \right] $

Intutively, this means that an impulse at $t = t_{0}$ in the time domain multiplied by $2 \pi$ is equivalent to $cos(\omega_{0}t)$ in the frequency domain. I don't have any intuition for this and I think I'm confused by the fact that a fourier transform which is a representation in the frequency domain can still involve the time $t$.

I thought the "x-axis" in the frequency domain was $\omega$ so where does $t$ come into play ? Thanks for any wisdom here. The background is that I'm trying to learn this on my own by reading various books and articles so any insights are really appreciated. This particular example came from "Fourier Transforms For Pedestrians" which is a very nice textbook except that it sometimes skips steps that aren't always obvious to me. Thanks again. This site is an amazing facility for learning.

mark leeds
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I think the confusion has to do with the notation. Usually when we use $\omega_0$ we mean some fixed frequency, and not an independent variable. So the notation $G(\omega_0)$ for a Fourier transform of some function $g(t)$ is very unfortunate. It actually suggests that it is the value of the function evaluated at some frequency $\omega_0$. For the same reason it is not wise to use the variable $t$ as a parameter in the definition of $G(\omega)$. In your case $t$ is not the independent time variable but some constant determining the shape of $G(\omega)$. A much better and also much more common notation would be

$$G(\omega)=\cos(\omega T)$$

where $T$ is some fixed parameter with the dimension 'time', and $\omega$ is the independent frequency variable.

With the definition of the Fourier transform that I use you have the correspondences

$$\delta(t)\Longleftrightarrow 1\\ \delta(t-T)\Longleftrightarrow e^{-j\omega T}$$

Since

$$\cos(\omega T)=\frac12\left[e^{j\omega T}+e^{-j\omega T}\right]$$

you get

$$g(t)=\frac12[\delta(t+T)+\delta(t-T)]$$

Note that here $t$ is the independent time variable. For both $g(t)$ and $G(\omega)$, $T$ is some fixed parameter, whereas $t$ is the independent time variable (and hence does not occur in $G(\omega)$), and $\omega$ is the independent frequency variable (and hence does not occur in $g(t)$).

Matt L.
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