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Lecture 4: Exponential Fourier Series

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Schedule :: Periodic Signals :: Fourier Spectrum :: Symmetry :: Fundamental Frequency :: Smoothness of the Function :: Fourier Series :: Exponential Series :: Spectra :: Bandwidth :: Energy of Signals :: Geometric Viewpoint

We have seen how sin and cosine functions can be represented in terms of complex exponentials. It turns out that we can use complex exponentials to represent Fourier series. In many respects, this makes for a simpler representation.

Let's go back to the compact Fourier series representation function, and express it in terms of complex exponentials:

\begin{displaymath}C_n \cos(n \omega_0 t + \theta_n) = D_n e^{jn\omega_0 t} +
D_{-n} e^{-jn\omega_0 t}
\end{displaymath}
where
\begin{displaymath}D_n = \frac{1}{2}e^{j\theta_n}C_n \end{displaymath}

\begin{displaymath}D_{-n} = \frac{1}{2}e^{-j\theta_n}C_{n} \end{displaymath}
We can write the Fourier series:
\begin{eqnarray*}
f(t) &=& C_0 + \sum_{n=1}^{\infty} C_n \cos(n\omega_0 t + \the...
...ega_0 t} \\
&=& \sum_{n=-\infty}^{\infty} D_n e^{j n\omega_0 t}
\end{eqnarray*}
Now, how do we find the coefficients? First, note that if we define the inner product correctly, the exponentials are orthogonal. Up to now, inner products have been defined for real functions. We will extend this now to inner products over complex functions. If $f(t)$ and $g(t)$ are a periodic complex functions with period $T_0$ , define the inner product as
\begin{displaymath}\la f(t),g(t) \ra = \int_{T_0} f(t) \overline{g(t)} dt
\end{displaymath}
where the over-line means complex conjugate. The rules for this inner product are the same as before, except that

\begin{displaymath}\la f,g \ra = \overline{\la g,f \ra} \end{displaymath}
With this inner product, note that
\begin{displaymath}\la e^{j n \omega_0 t}, e^{j m \omega_0 t} \ra =
\int_{T_0} e...
... \omega_0 t} dt =
\frac{1}{j(n-m)\omega_0}[e^{j(n-m)2\pi} - 1]
\end{displaymath}
which is 0 if $n-m \neq 0$ and $T_0$ is $n=m$ . So under this inner product, we have a whole set of orthogonal functions. The geometry of orthogonal functions we talked about before applies, including the orthogonality theorem. We can therefore write
\begin{displaymath}D_n = \proj(f(t),e^{jn\omega_0 t}) = \frac{\la f,e^{j n \omeg...
... t} \ra} = \frac{1}{T_0}
\int_{T_0} f(t) e^{-j n\omega_0 t} dt
\end{displaymath}
(Derive this another way also.)

Note that this is true for all values of $n$ (there are no special cases when $n=0$ ) and there is only one formula (not two, as for sins and cosines). This is my preferred form! In fact, due to its similarity with the Fourier transform to be discussed soon, it is the most common form of the Fourier series. It is, of course, possible to convert from one form to another. For example,

\begin{displaymath}D_n = \frac{1}{2}(a_n - jb_n) \end{displaymath}

Suppose (as is most often the case) that $f(t)$ is a real function. Then

\begin{displaymath}\overline{D_{n}} = \overline{\frac{1}{T_0}\int_{T_0} f(t) e^{...
... \frac{1}{T_0} \int_{T_0} f(t) e^{j n \omega_0 t}
dt = D_{-n}.
\end{displaymath}


\begin{example}Find the exponential F.S. of the square wave function with
period...
...c(n\pi/2) \end{displaymath}Note: $\sinc(0) = 1$. (How do we know?)
\end{example}

Important observation: To compute the F.S. coefficients, multiply the function by an exponential with negative exponent. There are many transforms that electrical engineers use -- Laplace transforms, Z-transforms, Fourier series, Fourier transforms, etc. In all of these, the exponent is negative. Don't forget it!

Copyright 2008, by the Contributing Authors. Cite/attribute Resource . admin. (2006, May 23). Lecture 4: Exponential Fourier Series. Retrieved January 07, 2011, from Free Online Course Materials — USU OpenCourseWare Web site: http://ocw.usu.edu/Electrical_and_Computer_Engineering/Signals_and_Systems/4_8node8.html. This work is licensed under a Creative Commons License Creative Commons License