If i take the Fast Fourier Transform (FFT) of a cosine function, what has turned this cosine function into its complex exponential form which consists of $e^{i \omega t} + e^{-i \omega t}$ ?

Because on its own, cosine has a single frequency at omega, but the FFT must be interpreting cosine as a sum of its complex exponential because its output is amplitudes at frequencies $-\omega$ and $\omega$. So whats going on here?


3 Answers 3


I will try to explain it in an intuitive, not rigorous way.

The main idea behind the Fourier Transform is to "project" one signal $s(t)$ on another $\phi(t)$. If the projection is not zero, then $\phi(t)$ is "included" in $s(t)$, in the sense that you can write $s(t)$ as $$s(t) = A\phi(t) + \text{other signals},$$ with $A \neq 0$.

The projection is defined as $$A = \int_{-\infty}^\infty s(t) \phi^*(t)\,dt.$$

You can of course project $s(t)$ on multiple signals; you can find the projection on $\theta(t)$ $$B = \int_{-\infty}^\infty s(t) \theta^*(t)\,dt$$ and then you can say that $$s(t) = B\theta(t) + \text{other signals}.$$

In the case when the projection of $\phi(t)$ on $\theta(t)$ is zero (in this case they are said to be orthogonal), you can go further and say that $$s(t) = A\phi(t) + B\theta(t) + \text{other signals}.$$ Since you found two components of $s(t)$, you can expect the "other signals" portion of this last equation to be "smaller" (have less energy) than when you project on $\phi(t)$ or $\theta(t)$ alone.

Now, what Fourier says is that if you project your signal on the set $\lbrace e^{j\omega t} \rbrace$, the "other signals" part of the equations above will be zero. So, this set is special: it's said to be a "complete" basis for all signals (it's not the only complete basis, though).

So, if you look at the Fourier Transform integral, you'll notice that it is projecting your time-domain signal on this complete set. It turns out that, when you project the cosine $\cos(\omega_0 t)$ on this set of exponentials, you get exactly two projections that are not zero. This means that the cosine can be written as the sum of these two projections.

  • $\begingroup$ Intuitive, that’s very good! $\endgroup$
    – Gilles
    Oct 25, 2018 at 18:13

The function $x(t)=\cos(\omega_0t)$ has a Fourier transform that doesn't consist of only one impulse at $\omega=\omega_0$. In fact, it consists of two impulses: that one and another one at $\omega = -\omega_0$.

This can be seen from Euler's formula:

$$\cos(\omega_0t) = \frac{e^{j\omega_0t}+e^{-j\omega_0t}}{2}$$

There it is clear where the two impulses come from.


What you seem to be missing in your two questions is an understanding of Euler's Equation:

$$ e^{i\theta} = \cos(\theta) + i \cdot \sin(\theta) $$

Loosely, conceptually, it says that the trigonometric functions are actually exponential in nature. Specifically, it describes a conversion of a distance along the circumference of the unit circle in the complex plane $(\theta)$ to the underlying complex value in the form of $a+bi$. My first blog article, The Exponential Nature of the Complex Unit Circle tries to intuitively explain how this equation defines a point on the unit circle.

Now go the same distance along the circumference in the other direction and you get to the complex conjugate point.

$$ e^{-i\theta} = \cos(\theta) - i \cdot \sin(\theta) $$

Add the two values and you get a real number.

$$ e^{i\theta} + e^{-i\theta} = 2\cos(\theta) $$

From there, the value of $\cos$ can be solved for and you get the cosine equation.

$$ \cos(\theta) = \frac{ e^{i\theta} + e^{-i\theta} }{ 2 } $$

What the complex value in the DFT bin means is the topic of my second blog article.

You other question doesn't specify the continuous vs. discrete cases. There is a Fourier Transform is each. The DFT is the one for the latter. In the discrete case, only the sinusoidal signals with a whole number of cycles within the sample frame will have only the bins corresponding to the frequency have non-zero values. If there are not a whole number of cycles the bins neighboring the closest frequency bin will have tapering values. This is called leakage.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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