What is the realtionship between a cosine function and a Dirac delta?

Question

I want to compute the Fourier transformation of a cosine function $\cos(2\pi f_0 t)$, but I don't know if there are any relationships between trigonometric functions and Dirac delta's.

Answer 1

The Fourier transform of a cosine is $\frac{\delta(f-f_0) + \delta(f+f_0)}{2}$. But it is the shifting that characterizes them not the delta functions per se.

While the Fourier transform of $f(t)=1$ is $\delta(f)$. The shifting results from $\cos(2\pi f_0 t) = \frac{e^{i2\pi f_0 t } + e^{-i2 \pi f_0 t}}{2}$. For if $f_1(t) = f(t)e^{i2\pi f_0 t}$. Then

$$\begin{align}\mathcal{F}_1(f) &= \int_{-\infty}^{\infty} (f(t)e^{i2\pi f_0 t})e^{-i2\pi t f}dt\\ &=\int_{-\infty}^{\infty} f(t)e^{-i 2\pi t (f-f_0)}dt\\ &= \mathcal{F}(f-f_0)\end{align}$$

Answer 2

The Fourier transform of a function is another function that tells you the frequency content of the original function. A sine or a cosine (a horizontal shift does not change the frequency content) is a wave with a pure frequency, as opposed to a general sum of sines and cosines with different frequencies.

The Fourier transform of a pure wave of frequency $f_0$ has two peaks, one at $f_0$ and another at $-f_0$. Essentially the reason for this is that you cannot tell the difference between a real wave with a given frequency or the negative of that frequency, they look the same. Accordingly, the transforms of $\sin$ and $\cos$ are:

$\mathcal F [\sin(f_0t)](f) = \frac{\delta(f-f_0)-\delta(f+f_0)}{2i}$

$\mathcal F [\cos(f_0t)](f) = \frac{\delta(f-f_0)+\delta(f+f_0)}{2}$