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In my Signals and Systems class, we learned that the Fourier Series of a signal $x(t)$ is given by

$$ x(t) = \sum_{k = -\infty}^{\infty} X_k e^{ik\omega_0t} $$

where $\omega_0 = 2\pi/p$ and

$$ X_k = \frac{1}{p} \int_0^p x(t) e^{-ik\omega_0t} \, dt. $$

I have two questions:

1) Why is the summation from $k = -\infty$ to $k = \infty$? That is, why does it include negative values rather than just from $k = 0$ to $k = \infty$?

2) Where does the $e^{-ik\omega_0t}$ in the integral expression for $X_k$ come from?

Michael Davids
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3 Answers3

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Why negative frequencies?

So that you can represent real valued signals. For these, $X_{-k}=\overline{X}_k$.

Why the exponential factor in the coefficient integral?

This has multiple answers.

  • The first technical is that if you insert the series for $x(t)$, then it shifts the position of the constant term to $k$. $$ e^{-ikω_0t}\sum_{m=-\infty}^{\infty}X_me^{imω_0t}=\sum_{m=-\infty}^{\infty}X_me^{i(m-k)ω_0t}=\sum_{m=-\infty}^{\infty}X_{m+k}e^{imω_0t} $$ All oscillating terms integrate to zero, only the constant term where the exponentional has exponent zero gives a non-zero contribution. Thus the exponential factor together with integration sift out the coefficient $X_k$.

  • The more general answer is that the functions $e_k(t)=e^{ikω_0t}$ are an orthogonal basis in the space of periodic functions, and like in cartesian space with euclidean geometry, the coordinates in an orthonormal basis can just be computed with scalar products $X_k=\langle x,e_k\rangle$, where in this case $$ \langle x,y\rangle=\tfrac1p\int_0^p x(t)\overline{y(t)}\,dt. $$ And $\overline{e_k(t)}=e^{-ikω_0t}$.

Lutz Lehmann
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1) Recall Euler's formulae:

$\sin(kw_0t)=1/2(e^{ikw_0t}-e^{-ikw_0t})$

$\cos(kw_0t)=1/2(e^{ikw_0t}+e^{-ikw_0t})$

If your signal is a linear combination of both $\sin$ and $\cos$ then you've got to include $e^{ikw_0t}$ with both positive and negative $k$.

2) Those integral are convolutions, functional analysis analogues of "inner products" that one encounters in analytic geometry. IMO the best way to look at Fourier series by analogy with finite dimensional analytic geometry as follows.

Suppose you have $n$-dimensional space with an orthonormal basis $v_1,...,v_n$, and you want to express a vector $x$ in that basis. What you do is take inner product of $x$ with each of basis vectors and that gives you the coefficients:

$x=\Sigma_{k=1}^nX_kv_k$

where the $k$-th coefficient is given by inner product with $v_k$:

$X_k=x\cdot v_k$

Now, the linear space of (certain) functions is infinite-dimensional. However, if one restricts to "square-integrable" functions the above formulae still hold, with some important clarifications. First, the series become infinite. Second, one needs to properly define what "inner product" is.

It turns out that one can define inner product between square integrable functions as their convolution, and all the usual properties of inner product would hold.

Michael
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I'll answer the second question first.

It can be shown that for every integer $k$ and $l$ $$  \frac{1}{p} \int_0^p \mathrm{e}^{ i k \omega _0 t} \mathrm{e}^{- i l \omega _0 t} \mathrm{d}t = \left\{ \begin{array} {l l} 0 & \quad \text{if $k \neq l$} \\ 1 & \quad \text{if $k = l$} \end{array} \right. $$

Now suppose that the Fourier series is indeed given by $$ x(t) = \sum\limits_{k=-\infty}^\infty X_k \mathrm{e}^{i k \omega _0 t} $$

Plugging this to the definition of the coefficients gives (using $l$ for clarity): $$ \frac{1}{p} \int_0^p x(t) \mathrm{e}^{- i l \omega _0 t} = \frac{1}{p} \int_0^p \left( \sum\limits_{k=-\infty}^\infty X_k \mathrm{e}^{i k \omega _0 t} \right) \mathrm{e}^{- i l \omega _0 t} = \sum\limits_{k=-\infty}^\infty \left( X_k \frac{1}{p} \int_0^p \mathrm{e}^{i k \omega _0 t} \mathrm{e}^{- i l \omega _0 t} \right) = X_l $$

as $k=l$ is the only term in the sum that is non-zero.

To answer the first question consider a function $x(t) = \mathrm{e}^{-i \omega _0 t}$. Using the first equation it is easy to see that $X_k = 0$ for $k \in \{0, 1, 2, ... \} $ so $\sum\limits_{k=0}^\infty X_k \mathrm{e}^{i k \omega _0 t} = 0 \neq \mathrm{e}^{-i \omega _0 t}$. Including the negative values of $k$ in the sum allows us to represent all "nice enough" $p$-periodic functions as a Fourier series.

lamiaani
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