FOURIER ANALYSIS Lucas Illing 2008 Contents 1 Fourier Series 2 1.1 General Introduction . 2 1.2 Discontinuous Functions . 5 1.3 Complex Fourier Series . 7 2 Fourier Transform 8 2.1 Definition . 8 2.2 The issue of convention . 11 2.3 Convolution Theorem . 12 2.4 Spectral Leakage . 13 3 Discrete Time 17 3.1 Discrete Time Fourier Transform . 17 3.2 Discrete Fourier Transform (and FFT) . 19 4 Executive Summary 20 1 1. Fourier Series 1 Fourier Series 1.1 General Introduction Consider a function f(τ) that is periodic with period T . f(τ + T ) = f(τ) (1) We may always rescale τ to make the function 2π periodic. To do so, define 2π a new independent variable t = T τ, so that f(t + 2π) = f(t) (2) So let us consider the set of all sufficiently nice functions f(t) of a real variable t that are periodic, with period 2π. Since the function is periodic we only need to consider its behavior on one interval of length 2π, e.g. on the interval (−π; π). The idea is to decompose any such function f(t) into an infinite sum, or series, of simpler functions. Following Joseph Fourier (1768-1830) consider the infinite sum of sine and cosine functions 1 a0 X f(t) = + [a cos(nt) + b sin(nt)] (3) 2 n n n=1 where the constant coefficients an and bn are called the Fourier coefficients of f. The first question one would like to answer is how to find those coefficients. To do so we utilize the orthogonality of sine and cosine functions: Z π Z π 1 cos(nt) cos(mt) dt = [cos((m − n)t) + cos((m + n)t)] dt −π −π 2 8 2π; m = n = 0 <> = π; m = n 6= 0 :>0; m 6= n ( 2π; m = n = 0 = (4) πδmn; m 6= 0 2 General Introduction Similarly, Z π Z π 1 sin(nt) sin(mt) dt = [cos((m − n)t) − cos((m + n)t)] dt −π −π 2 ( 0 m = 0 = (5) πδmn m 6= 0 and Z π Z π 1 sin(nt) cos(mt) dt = [sin((m − n)t) + sin((m + n)t)] dt −π −π 2 = 0 (6) Using the orthogonality and the assumed expression for the infinite series given in Eq. (3), it follows that the Fourier coefficients are 1 Z π an = f(t) cos(nt) dt (7) π −π 1 Z π bn = f(t) sin(nt) dt (8) π −π This initial insight by Fourier was followed by centuries of a work on the second obvious question: Are the RHS and LHS in Eq. (3) actually the same? Clearly one needs to determine for which class of functions f the infinite series on the right hand side of Eq. (3) will converge. That is, what is a sufficiently nice function f? The precise answer is not of concern here, it suffices to know that the Fourier series exists and converges for periodic functions of the type you are used to, e.g. functions for which first and second order derivatives exists almost everywhere, that are finite and have at most a finite number of discontinuities and zero crossings in the interval (−π; π). When determining a the Fourier series of a periodic function f(t) with period T , any interval (t0; t0 + T ) can be used, with the choice being one of conve- nience or personal preference. For example, in the rescaled time coordinates considering the interval (0; 2π) works just as well as considering (−π; π) as we have done. If a function is even so that f(t) = f(−t), then f(t) sin(nt) is odd. (This follows since sin(nt) is odd and an even function times an odd function is an odd function.) Therefore, bn = 0 for all n. Similarly, if a function is odd so that f(t) = −f(−t), then f(t) cos(nt) is odd. (This follows since cos(nt) is even and an even function times an odd function is an odd function.) Therefore, an = 0 for all n. 3 1. Fourier Series ' $%& $%! $%" 123.4/51 $%# $ !! !" !# $ # " ! ()*+,-./0/ Figure 1: A full-wave-rectifier converts a sinusoidal input, sin(!t), to j sin(!t)j. Example - Rectified sine wave: A first step in converting AC-power from the power-grid to the DC-power that most devices need is to utilize a full-wave rectifier, such as the diode bridge shown in Fig. 1, which converts a sinusoidal input to an output that is the absolute value of the input sine-wave. One notes immediately that for a sinusoidal input, the output of the rectifier is periodic with half of the period of the input. The fundamental frequency of the output is twice the input frequency. How can that be? The reason is that the circuit is not a linear circuit. The presence of diodes makes this circuit nonlinear and allows the circuit to shift power from the fundamental frequency to twice its value. One might wonder whether that is all that is happening. Does the output have contributions (power) at other frequencies? To answer this we look at the Fourier series of the output. Since the output f = j sin(!t)j is even, i.e. f(t) = f(−t), no terms of the form sin(n!t) will appear in the answer. It suffices to determine the an coefficients. For a0 one obtains 1 Z 0 1 Z π a0 = − sin(!t) d(!t) + sin(!t) d(!t) π −π π 0 2 Z π 4 = sin(!t) d(!t) = (9) π 0 π 4 Discontinuous Functions and for the remaining an one gets 2 Z π an = sin(!t) cos(n!t) d(!t) π 0 2 Z π 1 = [− sin((n − 1)!t) + sin((n + 1)!t)] d(!t) π 0 2 1 1 −1 = fcos(nπ − π) − 1g + fcos(nπ + π) − 1g π n − 1 n + 1 ( − 4 1 ; n even = π n2−1 (10) 0; n odd: Note, that the sine and cosine functions are orthogonal on the interval (−π; π). They are not orthogonal on the interval (0; π) and we do get a nonzero contribution for even n. To summarize the result, 1 2 4 X cos(n!t) j sin(!t)j = − : (11) π π n2 − 1 n=2;4;6;::: For an input with frequency f0, the output has a DC-offset, the part that we really care about when building a DC-voltage supply. It has no contribution at f = f0. It does have contributions at frequencies 2f0; 4f0; 6f0;:::. 1.2 Discontinuous Functions In the above example, Eq. (11), the nth coefficient decreases as 1=n2. This decay of the coefficients is in contrast to the Fourier series of a square wave 1 4 X 1 f (t) = sin(n!t) (12) sw π n n=1;3;5;::: where the nth coefficient falls off as 1=n. This is true in general 1 1. If f(t) has discontinuities, the nth coefficient decreases as 1=n. The convergences is slow and many terms need to be kept to approximate such a function well. 1G. Raisbeck, Order of magnitude of Fourier coefficients. Am. Math. Mon. 62, 149-155 (1955). 5 1. Fourier Series Figure 2: The Gibbs phenomenon is an overshoot (or "ringing") of Fourier series and other eigenfunction series occurring at simple discontinuities. Shown in color are the first few partial sums of the square-wave Fourier series. (Source math- world.wolfram.com/GibbsPhenomenon.html) This means that a function generator that generates square waves through the addition of sinusoidal waveforms needs to have a bandwidth (max. freq. it can generate) that is large compared to the frequency of the square-wave that is generated. 2. If f(t) is continuous (although possibly with discontinuous derivatives) the nth coefficient decreases as 1=n2. There is another consequence of a discontinuity in f(t) that can cause trouble in practical applications, where one necessarily only adds a finite number of sinusoidal terms. The nth partial sum of the Fourier series of a piecewise continuously differentiable periodic function f behaves at a jump discontinuity in a peculiar manner. It has large oscillations near the jump, which might increase the maximum of the partial sum above that of the function itself. It turns out that the Fourier series exceeds the height of a square wave by about 9 percent. This is the so-called Gibbs phenomenon, shown in Fig. 2. Increasing the number of terms in the partial sum does not decrease the magnitude of the overshoot but moves the overshoot extremum point closer and closer to the jump discontinuity. You will have the opportunity to explore this in the lab. 6 Complex Fourier Series 1.3 Complex Fourier Series At this stage in your physics career you are all well acquainted with complex numbers and functions. Let us then generalize the Fourier series to complex functions. To motivate this, return to the Fourier series, Eq. (3): 1 a0 X f(t) = + [a cos(nt) + b sin(nt)] 2 n n n=1 1 int −int int −int a0 X e + e e − e = + a + b 2 n 2 n 2i n=1 1 −∞ a0 X an − ibn X a−m + ib−m = + eint + eimt (13) 2 2 2 n=1 m=−1 where we substituted m = −n in the last term on the last line. Equation (13) clearly suggests the much simpler complex form of the Fourier series +1 X in(2πf0)t x(t) = Xn e : (14) n=−∞ with the coefficients given by 1 Z T=2 −in(2πf0)t Xn = x(t) e dt (15) T −T=2 Here, the Fourier series is written for a complex periodic function x(t) with arbitrary period T = 1=f0.