Signals & Systems Lecture 4 Fourier Series Properties & Discrete-Time Fourier Series Alp Ertürk alp.erturk@kocaeli.edu.tr
Fourier Series Representation of Continuous-Time Periodic Signals Synthesis equation: x t = a k e jkω 0t = a k e jk 2π/T 0 t Analysis equation: a k = 1 T T x(t)e jkω 0t dt = 1 T T x(t)e jk 2π/T 0 t dt
Fourier Series 1) Linearity x t FS a k y t FS b k z t = Ax t + By t FS c k = Aa k + Bb k
Fourier Series c k = 1 T T z(t)e jkω 0t dt = 1 T T Ax t + By t e jkω 0t dt = A 1 T T x(t)e jkω 0t dt + B 1 T T y(t)e jkω 0t dt = Aa k + Bb k
Fourier Series 2) Time Shifting x t FS a k y t = x t t 0 FS b k = e jkω 0t 0 a k
Fourier Series b k = 1 T T y(t)e jkω 0t dt = 1 T T x(t t 0 )e jkω 0t dt = 1 T T x(τ)e jkω 0 τ+t 0 dτ = e jkω 0t 0 1 T T x(τ)e jkω 0τ dτ = e jkω 0t 0 a k
Fourier Series 3) Time Reversal x t FS a k y t = x t FS b k = a k
Fourier Series b k = 1 T T y(t)e jkω 0t dt = 1 T T x( t)e jkω 0t dt = 1 T T x(τ)e jkω 0 τ dτ = 1 T T x(τ)e j( k)ω 0τ dτ = a k
Fourier Series 4) Time Scaling x t FS a k y t = x(αt) FS b k = a k But the Fourier Series representation changes!!!
Fourier Series b k = 1 T T y(t)e jk αω 0 t dt = 1 T T x(αt)e jk αω 0 t dt = 1 T T x(τ)e jk αω 0 τ/α dτ = 1 T T x(τ)e jkω 0τ dτ = a k
Fourier Series x t = a k e jkω 0t y t = x αt = b k e jkω 1t = a k e jkαω 0t
Fourier Series 5) Multiplication x t FS a k y t FS b k x t y t FS h k = l= a l b k l
Fourier Series h k = 1 T T x t y t e jkω 0t dt = 1 T T n= a n e jnω 0t l= b l e jlω 0t e jkω 0t dt = 1 T T n= l= a n b l e j k n+l ω 0t dt = n l a n b l δ k n + l
Fourier Series h k = 1 T T x t y t e jkω 0t dt = n l a n b l δ k n + l = n a n b k n
Fourier Series 6) Conjugation and Conjugate Symmetry x t FS a k x t FS a k
Fourier Series x(t) = x t = k= k= a k e jkω 0t a k e jkω 0t = k= a k e jkω 0t = l= a l e jlω 0t
Fourier Series 7) Parseval s Relation 1 T T x(t) 2 dt = k= a k 2 1 T T k= a k e jkω 0t 2 dt = 1 T T k= a k 2 dt = k= a k 2 Total average power in a periodic signal equals the sum of average powers in all of its harmonic components
Fourier Series: Example - 1 Find the Fourier Series coefficients for the periodic square wave with period T given by: x t = 1 t < T 1 0 T 1 < t < T/2
Fourier Series: Example - 1 T 1 dt = 2T 1 a 0 = 1 T T1 T a k = 1 T 1 e jkω 0 t dt = 1 T T1 jkω 0 T e jkω 0T 1 e jkω 0( T 1 ) = 2 kω 0 T e jkω 0T 1 e jkω 0T 1 2j = sin kω 0T 1 kπ, k 0
Fourier Series: Example - 1 For example, for T = 4T 1 : a 0 = 1 2, a k = sin kπ/2 kπ, k 0 a 1 = a 1 = 1 π a 3 = a 3 = 1 3π a 5 = a 5 = 1 5π...
Fourier Series: Example - 1 T = 4T1 T = 8T1 T = 16T1
Fourier Series: Example - 2 A period of the signal g(t) with a fundamental period of 4 is given below. Find the Fourier series coefficients of g(t).
Fourier Series: Example - 2 Notice that for T = 4 and T 1 = 1, this signal is related to the signal from the previous example by: g t = x t 1 1/2
Fourier Series: Example - 2 x t FS a k y t = x t 1 FS b k = e jkω 01 a k = e jkπ/2 a k c t = 1 2 FS ck = 0 for k 0 1/2 for k = 0
Fourier Series: Example - 2 g t = x t 1 1 2 FS d k = b k + c k d k = a ke jkπ/2 for k 0 a 0 1 2 for k = 0 From the previous example, we have: a 0 = 1 2 a k = sin kπ/2 kπ, k 0
Fourier Series: Example - 2 Therefore: d k = a ke jkπ/2 for k 0 a 0 1 2 for k = 0 = sin kπ/2 kπ e jkπ/2 for k 0 0 for k = 0
Fourier Series: Example - 3 Consider the triangular wave signal z(t) with period T = 4 and fundamental frequency of ω 0 = π/2
Fourier Series: Example - 3 The derivative of this signal is the signal in the previous example Therefore, using the relation: dx(t) dt FS jkω0 a k d k = jk π/2 e k e k = 2d k jkπ = 2 sin πk/2 j kπ 2 e kπ/2, k 0 e 0 = 1 2 (Use the area under one period)
Fourier Series: Example - 4 Find the Fourier series coefficients of the impulse train: x t = k= δ(t kt)
Fourier Series: Example - 4 a k = 1 T T x(t)e jkω 0t dt = 1 T T k= δ(t kt) e jkω 0t dt T/2 = 1 δ(t)e jkω0t dt T T/2 = 1 T
Fourier Series: Example - 4 We can also solve this problem using Fourier series properties by noting that the derivative of the g(t) from the example before is q(t) and q t = x t + T 1 x(t T 1 ) Note: Not the old x(t) of the first example!!
Fourier Series: Example - 4
Fourier Series: Example - 4 q t = x t + T 1 x(t T 1 ) Fourier series coefficients b k of q t are related to a k of x t : b k = e jkω 0T 1 a k e jkω 0T 1 a k = 2j sin kω 0T 1 T
Fourier Series: Example - 4 b k = 2j sin kω 0T 1 T q t is the derivative of g t, therefore: b k = jkω 0 c k c k = b k = 2j sin kω 0T 1 jkω 0 jkω 0 T = sin kω 0T 1 kπ, k 0 Using the area under one period: c 0 = 2T 1 T
Approximation by Fourier Series clear all; close all; clc; N = 100; x = linspace(-1,1,n); f = sign(x); for M = 1:2:15 figure; sum = 0.*x; for j = 1:2:M sum = sum + 4/pi*sin(j*pi*x)/j; end plot(x, sum, 'r'); hold on; plot(x,f,'linewidth',2); title([num2str(m) ' coefficents']) end
Approximation by Fourier Series clear all; close all; clc; N = 100; x = linspace(0,2*pi,100); f = 0.5 * (pi-x); for M = 1:2:15 figure; sum = 0.*x; for j = 1:M sum = sum + ((1/j)*sin(j*x)); end plot(x, sum, 'r'); hold on; plot(x,f,'linewidth',2); title([num2str(m) ' coefficents']) end
Fourier Series clear all; close all; clc; syms t; xt = heaviside(t + 1) - heaviside(t - 1); T0 = 4; coef = 15; cnt = 0; a = zeros(1,2*coef+1); for k = -coef:1:coef cnt = cnt + 1; if k == 0 a(cnt) = (1/T0) * int(xt, t, -T0/2, T0/2); else a(cnt) = (1/T0) * int(xt*exp(-j*2*pi*k*t/t0), t, -T0/2, T0/2); end end figure; subplot(1,2,1); stem(real(a)); subplot(1,2,2); stem(imag(a));
Fourier Series syms t; yt = heaviside(t) - heaviside(t - 2) - 1/2; T0 = 4; coef = 15; b = zeros(1,2*coef+1); cnt = 0; for k = -coef:1:coef cnt = cnt + 1; if k == 0 b(cnt) = (1/T0) * int(yt, t, -T0/2, T0/2); else b(cnt) = (1/T0) * int(yt*exp(-j*2*pi*k*t/t0), t, -T0/2, T0/2); end end figure; subplot(1,2,1); stem(real(b)); subplot(1,2,2); stem(imag(b));
Fourier Series cnt = 0; for k = -coef:1:coef cnt = cnt + 1; if k == 0 c(cnt) = a(cnt) - 1/2; else c(cnt) = a(cnt) * exp(-j*k*pi/2); end end figure; subplot(1,2,1); stem(real(c)); subplot(1,2,2); stem(imag(c)); title('fourier Series coefficients (using the properties)');
Discrete-Time Fourier Series
Discrete-Time Fourier Series A discrete-time signal x[n] is periodic with period N if: x n = x[n + N] Discrete-time complex exponential signals that are periodic with period N are given by: φ k n = e jkω 0n = e jk 2π/N n, k = 0, ±1, ±2, φ k n = φ k+rn n
Discrete-Time Fourier Series x n = a k φ k n = a k e jkω 0n = a k ejk 2π/N n k k k Since k has N successive integer values, we can write: x n = a k φ k n = a k e jkω 0n = a k ejk 2π/N n k=<n> k=<n> k=<n>
Discrete-Time Fourier Series x 0 = k=<n> a k x 1 = k=<n> a k e j2πk/n x N 1 = k=<n> j2πk N 1 /N a k e
Discrete-Time Fourier Series Synthesis equation: x n = k=<n> a k e jkω 0n = k=<n> jk 2π/N n a k e Analysis equation: a k = 1 x[n]e jkω0n = 1 jk 2π/N n x[n]e N N n=<n> n=<n>
Discrete-Time Fourier Series Example - 1 Find Fourier series coefficients of the signal: x n = sin ω 0 n For ω 0 = 2π N : x n = 1 2j ej 2π/N n 1 2j e j 2π/N n
Discrete-Time Fourier Series Example - 1 a k = 1 x[n]e jk 2π/N n N n=<n> a k = 1 1 N 2j ej 2π/N n 1 e j 2π/N n e 2j n=<n> jk 2π/N n x n = k=<n> a k ejk 2π/N n a 1 = 1 2j, a 1 = 1 2j
Discrete-Time Fourier Series Example - 1 If N = 5:
Discrete-Time Fourier Series Example - 2 Find Fourier series coefficients of the signal: x n = 1 + sin 2π N n + 3 cos 2π N n + cos 4π N n + π 2 x n = 1 + 1 2j ej 2π/N n e j 2π/N n + 3 2 ej 2π/N n e j 2π/N n + 1 2 4πn ej N + π 2 4πn j e N + π 2
Discrete-Time Fourier Series Example - 2 x n = 1 + 1 2j ej 2π/N n e j 2π/N n + 3 2 ej 2π/N n j 2π/N n e + 1 2 4πn ej N + π 2 4πn j e N + π 2 Collecting terms: x n = 1 + 3 2 + 1 2j e j 2π/N n + 3 2 1 2j e j 2π/N n + 1 2 ejπ/2 e j2 2π/N n + 1 2 e jπ/2 e j2 2π/N n
Discrete-Time Fourier Series Example - 2 x n = 1 + 3 2 + 1 2j e j 2π/N n + 3 2 1 2j e j 2π/N n + 1 2 ejπ/2 j2 2π/N n e + 1 2 e jπ/2 j2 2π/N n e a 0 = 1 a 1 = 3 2 + 1 2j = 3 2 1 2 j, a 1 = 3 2 1 2j = 3 2 + 1 2 j a 2 = 1 2 j, a 2 = 1 2 j
Discrete-Time Fourier Series Example - 2
Discrete-Time Fourier Series Example - 2
Properties of Discrete-Time Fourier Series
Properties of Discrete-Time Fourier Series