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Lecture 4: Periodic Signals and Representations

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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

From the last lecture we learned how functions can be represented as a series of other functions:

\begin{displaymath}f(t) = \sum_{k=1}^n c_k i_k(t).\end{displaymath}
We discussed how certain classes of things can be built using certain kinds of basis functions. In this lecture we will consider specifically functions that are periodic, and basic functions which are trigonometric. Then the series is said to be a Fourier series.

A signal $f(t)$ is said to be periodic with period $T_0$ if

\begin{displaymath}f(t) = f(t+T_0) \end{displaymath}
for all $t$ . Diagram on board. Note that this must be an everlasting signal. Also note that, if we know one period of the signal we can find the rest of it by periodic extension . The integral over a single period of the function is denoted by
\begin{displaymath}\int_{T_0} f(t) dt.\end{displaymath}
When integrating over one period of a periodic function, it does not matter when we start. Usually it is convenient to start at the beginning of a period.

The building block functions that can be used to build up periodic functions are themselves periodic: we will use the set of sinusoids. If the period of $f(t)$ is $T_0$ , let $\omega_0 = 2\pi/T_0$ . The frequency $\omega_0$ is said to be the fundamental frequency ; the fundamental frequency is related to the period of the function. Furthermore, let $F_0 = 1/T_0$ . We will represent the function $f(t)$ using the set of sinusoids

i_0(t) &=& \cos(0t) = 1 \\
i_1(t) &=& \cos(\omega_0 t + \theta_1) \\
i_2(t) &=& \cos(2\omega_0 t + \theta_2) \\
\vdots \\
\begin{displaymath}f(t) = C_0 + \sum_{n=1}^\infty C_n \cos(n\omega_0 t + \theta_n)
The frequency $n\omega_0$ is said to be the $n$ th harmonic of $\omega_0$ .

Note that for each basis function associated with $f(t)$ there are actually two parameters: the amplitude $C_n$ and the phase $\theta_n$ . It will often turn out to be more useful to represent the function using both sines and cosines. Note that we can write

\begin{displaymath}C_n \cos(n\omega_0 t + \theta_n) = C_n \cos(\theta_n)\cos(n\omega_0
t) - C_n \sin(\theta_n)\sin(n\omega_0 t).
Now let
\begin{displaymath}a_n = C_n \cos \theta_n\quad\quad\quad b_n = -C_n \sin \theta_n
\begin{displaymath}C_n \cos(n\omega_0 t + \theta_n) = a_n \cos(n\omega_0 t) + b_n
\sin(n\omega_0 t)
Then the series representation can be
f(t) &=& C_0 + \sum_{n=1}^\infty C_n \cos(n\omega_0 t + \theta...
... \sum_{n=1}^\infty a_n \cos(n\omega_0 t) + b_n \sin(n\omega_0 t)
The first of these is the compact trigonometric Fourier series . The second is the trigonometric Fourier series. . To go from one to the other use
\begin{displaymath}C_0 = a_0 \end{displaymath}

\begin{displaymath}C_n = \sqrt{a_n^2 + b_n^2} \end{displaymath}

\begin{displaymath}\theta_n = \tan^{-1}(-b_n/a_n). \end{displaymath}

To complete the representation we must be able to compute the coefficients. But this is the same sort of thing we did before. If we can show that the set of functions $\{ \cos(n\omega_0 t),
\sin(n\omega_0 t)\}$ is in fact an orthogonal set, then we can use the same formulas we did before. Check:

\begin{displaymath}\int_{T_0} \cos(n\omega_0 t) \cos(m\omega_0 t)  dt =
...& n \neq m \\
\frac{T_0}{2} & n = m \neq 0
So pairs of cosine functions are orthogonal. Similarly,
\begin{displaymath}\int_{T_0} \sin(n\omega_0 t) \sin(m\omega_0 t)  dt =
...& n \neq m \\
\frac{T_0}{2} & n = m \neq 0
So pairs of sin functions are orthogonal. Also,
\begin{displaymath}\int_{T_0} \sin(n\omega_0 t) \cos(m\omega_0 t) dt = 0 \mbox{ for
all $n$ and $m$.}
So the sines and cosines are orthogonal to each other. This means that we can use our formulas:
\begin{displaymath}a_n = \frac{\la f(t),\cos(n\omega_0 t)\ra}{\la \cos(n\omega_0...
...t) \ra } = \frac{2}{T_0} \int_{T_0} f(t)
\cos(n\omega_0 t) dt \end{displaymath}

\begin{displaymath}b_n = \frac{\la f(t),\sin(n\omega_0 t)\ra}{\la \sin(n\omega_0...
...t) \ra } = \frac{2}{T_0} \int_{T_0} f(t)
\sin(n\omega_0 t) dt \end{displaymath}
The only exception is the coefficient $a_0$ :
\begin{displaymath}a_0 = \frac{\la f(t), 1\ra}{\la 1,1 \ra } = \frac{1}{T_0} \int_{T_0} f(t)
 dt \end{displaymath}

\par Let us start with a square wave:  [1in]
...ldots, \\
0 & \mbox{otherwise}\end{array}\right.

It is interesting to see how the function gets built up at the pieces are added together:

Copyright 2008, by the Contributing Authors. Cite/attribute Resource . admin. (2006, May 23). Lecture 4: Periodic Signals and Representations. Retrieved January 07, 2011, from Free Online Course Materials — USU OpenCourseWare Web site: This work is licensed under a Creative Commons License Creative Commons License