# Fourier Transform and Delta Function

I am very new to Fourier analysis, but I understand that through the use of the Fourier transform a signal in the time domain is displayed in the frequency domain, where frequency values are normally displayed along the x-axis, and amplitude is displayed along the y-axis. However, at one point in the textbook I am using, the following is stated:

Let us assume that we have the function $f(t) = \cos(\omega_0 t)$. The spectrum then consists of two delta-functions

$$F(\omega) = \pi \delta(\omega - \omega_0) + \pi \delta(\omega + \omega_0)$$

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This confuses me. When we have $f(t) = \cos(\omega_0 t)$, then I would assume that the Fourier transform should yield an amplitude of $1$ at $\omega = \omega_0$ and $0$ elsewhere. But the delta function is defined as:

$$\delta(\omega - \omega_0) = \left\{ \begin{array}{1 1} \infty & \quad \omega = \omega_0 \\ 0 & \quad \omega \neq \omega_0 \end{array} \right.$$

So wouldn't this give an infinite value at $\omega = \omega_0$?

If anyone can explain the intuition behind the statement in my textbook, then I would be very grateful!

• Instead of assuming that the Fourier transform should yield an amplitude of $1$ at $\omega=\omega_0$, why don't you try to see if this actually happens? – Dilip Sarwate Dec 19 '13 at 18:59
• I don't doubt that the textbook is right. But how do you get from the result above to an accurate representation in the frequency/amplitude domain? – Kristian Dec 19 '13 at 20:24
• Learn about the difference between power and energy. As Hilmar has pointed out to you, a (mathematical) sinusoid has infinite energy although the sinusoids available to us mortals are always of finite duration and thus have finite energy. A mathematical sinusoid has finite power, though, and in this instance, the total power is $\frac{1}{2}$. – Dilip Sarwate Dec 19 '13 at 21:21