Fourier Transform. Find the Fourier series for a periodic waveform Determine the output of a filter when the input is a periodic function

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1 Objectives: Be able to Fourier Transform Find the Fourier series for a periodic waveform Determine the output of a filter when the input is a periodic function Filters with a Single Sinusoidal Input: Suppose you have a filter with an input X and a transfer function G(s): (s)=g(s) X(s) If x(t) is a sinusoid at frequency ω, y(t) will also be a sinusoid at frequency ω. y(t) is related to x(t) by the gain, gain, G, evaluated at s = jω. Example: Find y(t) for the following system: 2sin(3t) X Solution: is the gain for all 's'. The input, X, only exists at, however, so you only care about s = j3 the gain at that frequency. Plugging in s = j3 results in y = 2sin(3t) s=j3 y = ( ) 2sin(3t) y(t)=1.90 sin (3t 87 0 ) Filters with Multiple Sinusoidal Inputs: If x(t) is composed of several sine waves, you can use superposition. The output, y(t) will be the sum of each input times its corresponding gain. Example: Find y(t) for = X x(t)=1 + 2sin(3t)+4sin(5t) Since multiplication and addition are commutative, you can JSG 1 rev November 18, 13

2 Sum the sine waves to produce X, and then filter X to produce (option 1), or ou can filter the sine waves then add to produce (option 2) 1 1 y1 2sin(3t) X 2sin(3t) y2 4sin(5t) 4sin(5t) y3 Option 1 Option 2 The first option is how you would actually implement the filter: a single filter acts on the entire waveform. The second option is easier to analyze: since the filter has only a single sinusoidal input, you can use phasor analysis to find the output as y = y 1 + y 2 + y 3 y1: (s=0) y 1 = y 1 = (2) 1 y 1 = 2 1 s=0 y2: (s=j3) y 2 = 2sin(3t) s=j3 y 2 = ( ) 2sin(3t) y 2 = 1.90 sin (3t 87 0 ) y3: (s=j5) y 3 = 4sin(5t) s=j5 y 3 = ( ) 4sin(5t) y 3 = 2.08 sin (5t ) JSG 2 rev November 18, 13

3 So, y = y 1 + y 2 + y 3 y(t)= sin (3t 87 0 ) sin (5t ) Note that this only works if the input is composed of sinusoids. Filters with Periodic Inputs (Fourier Series) Problem: Determine the output of the following filter with a 1Hz square wave input: x(t) y(t) 1V x(t) 0V 1 second time Solution: if x(t) can be expressed as the sum of a bunch of sine waves, you can Filter x(t) directly (option 1), or Filter each sine wave separately and add up the results (option 2) The second option is simpler in a sense since each filter acts on a single sine wave and phasor analysis applies. a1 sin(b1 t) a1 sin(b1 t) y1 a2 sin(b2 t) x(t) y(t) a2 sin(b2 t) y2 y(t) a3 sin(b3 t) a3 sin(b3 t) y3 Option 1: Option 2: The trick, then, is how to express x(t) in terms of a bunch of sinusoids. This trick is called the Fourier Transform. JSG 3 rev November 18, 13

4 Fourier Transform: Given a function which is periodic in time T: x(t + T)=x(t) Then x(t) can be expressed in terms of sine and cosine terms: x(t)=a 0 + Σ an cos (nω o t) + Σ bn sin (nω o t) where ω 0 = 2π T Each constant, an, bn, can be found using the following equations: a 0 = average(x)= 1 T T x(t)dt a n = T (x(t) cos (nω 0t))dt T cos 2 (nω 0 t) dt b n = T (x(t) sin (nω 0 t))dt T sin 2 (nω 0 t) dt Note: ou can also express x(t) in polar form x(t)=a 0 + Σ cn cos (nω o t +θ n ) or complex exponential form: x(t)=a 0 + Σ cn e j(nωot+θn ) All three forms are equivalent - it's just what you're personal preference is. I personally like the first form. Translation: ω 0 = 2π simply states that the basis is sine and cosine waves which are periodic in time T. T Going right to left, the Fourier transform states that if you add up a bunch of functions which are periodic in time T, the result is periodic in time T. (duh). Going left to right, the Fourier transform states that any periodic function which is not a pure sine wave contains harmonics. This is important, so again... Any periodic function which is not a pure sine wave contains harmonics. For example, the Fourier series expansion of a delta train, square wave, and triangle wave are as follows: JSG 4 rev November 18, 13

5 Delta train x(t)=δ(t) = Σ cos (nω0 t) Square Wave: x(t)= +1 sin(ω 0t)>0 0 otherwise Delta train taken out to harmonics 2 = Σ nπ sin (nω 0 t) n odd Square wave taken out to harmonics Triangle wave 4 x(t)=see below = Σ cos (nω n n odd 2 π 2 0 t) Triangle wave taken out to harmonics JSG 5 rev November 18, 13

6 Sample MATLAB Code: >t = [0:0.001:14]'; >for i=1:11 > n = 2*i-1; > y = y + 2/(n*%pi)*sin(n*t); > end >y = y + 0.5; >plot(t,y); Example: A 1Hz square wave that goes form 0V to +5V is applied to the following filter: Find y(t). = X Solution: Step 1: Convert x(t) to its Fourier series. The DC term is 2.5 (the average of x(t).) The remaining terms are from the above table scaled by 5: x(t)= < t < < t < 1 10 x(t)=2.5 + Σ nπ sin (nω 0 t) n odd 2.5 DC y1 X square wave 3.18sin(6.28t) n=3 1.06sin(18.85t) y2 y3 Option sin(31.4t) n=5 y4 Option 2 JSG 6 rev November 18, 13

7 Step 2: Find the gain of the filter at each frequency contained in x(t). The output will be the gain times the input. n s x(t) G(s) y(t) 0 s = G(0) = s = j2π sin (2πt) G(j2π) = sin (2πt ) 3 s = j6π sin (6πt) G(j6π) = sin (6πt ) 5 s = j10π sin (10πt) G(j10π) = sin (10πt ) 7 s = j14π sin (14πt) G(j14π) = sin (14πt ) etc. Step 3: Add up all the terms notes: y(t)= sin (2πt ) sin (6πt ) sin (10πt ) +... The Fourier transform turned a single problem which is very difficult to solve into an infinite number of problems which are easy to solve. The higher harmonics in y(t) decrease in amplitude quickly. This is a very common phenominum. It results from x(t) having most of its energy near s=0 and G(s) acting as a low pass filter. This lets you express y(t) with just a few terms (rather than an infinite number of terms). A plot of x(t) and y(t) is shown above. y(t) only includes terms out to the 10th harmonics x(t) y(t) JSG 7 rev November 18, 13

8 Example #2: x(t) is a 1Hz square wave like the previous problem. Find the output when passed through the filter: = s s 2 +s+986 X Repeating the above procedure... n s x(t) G(s) y(t) 0 s = G(0) = s = j2π sin (2πt) G(j2π) = sin (2πt ) 3 s = j6π sin (6πt) G(j6π) = sin (6πt ) 5 s = j10π sin (10πt) G(j10π) = sin (10πt 1 0 ) 7 s = j14π sin (14πt) G(j14π) = sin (14πt 87 0 ) etc. so y(t) is y(t)=0.022 sin (2πt ) sin (6πt ) sin (10πt 1 0 ) +... A plot of x(t) and y(t) is shown above. y(t) only includes terms out to the 10th harmonics x(t) y(t) Note that this is very odd: the input is a 1Hz square wave. The output is a 5Hz sine wave (mostly.) Why is the frequency of the output different from the frequency of the input? Actually, it's not. x(t) is not a pure sine wave. As a result, it contains harmnonics. G(s) is a band-pass filter. It passes frequencies near 5Hz and attenuates all other frequencies. The net result is G(s) passes the 5Hz component of x(t). The 5Hz component of a 1Hz square wave has an amplitude of JSG 8 rev November 18, 13

9 Sample MATLAB Code The following function returns the first Fourier coefficients for function y(t). It doesn't do the DC term. function [c] = four(y) t = [1:length(y)]' / length(y); wo = 2*pi; a = zeros(,1); b = zeros(,1); for : a(n) = sum(y.* cos(n*2*pi*t)) / sum(cos(n*2*pi*t).^2); b(n) = sum(y.* sin(n*2*pi*t)) / sum(sin(n*2*pi*t).^2); end c = a - j*b; The following function generates y(t) from its Fourier approximation out to N terms. function plot_four(y, N) t = [1:length(y)]'/length(y); c = four(y); DC = mean(y); f = 0*t + DC; for :N f = f + real(c(n))*cos(2*n*pi*t) - imag(c(n))*sin(2*n*pi*t); end plot(t,y,t,f) JSG 9 rev November 18, 13