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ENGR 3324: Signals and Systems Ch6 Continuous-Time Signal Analysis Engineering and Physics University of Central Oklahoma Dr. Mohamed Bingabr Outline • Introduction • Fourier Series (FS) representation of Periodic Signals. • Trigonometric and Exponential Form of FS. • Gibbs Phenomenon. • Parseval’s Theorem. • Simplifications Through Signal Symmetry. • LTIC System Response to Periodic Inputs. Sinusoidal Wave and phase x(t) = Asin(t) = Asin(250t) x(t) A t T0 = 20 msec x(t-0.0025)= Asin(250[t-0.0025]) = Asin(250t-0.25)= Asin(250t-45o) A t td = 2.5 msec Time delay td = 25 msec correspond to phase shift =45o Representation of Quantity using Basis • Any number can be represented as a linear sum of the basis number {1, 10, 100, 1000} Ex: 10437 =10(1000) + 4(100) + 3(10) +7(1) • Any 3-D vector can be represented as a linear sum of the basis vectors {[1 0 0], [0 1 0], [0 0 1]} Ex: [2 4 5]= 2 [1 0 0] + 4[0 1 0]+ 5[0 0 1] Basis Functions for Time Signal • Any periodic signal x(t) with fundamental frequency 0 can be represented by a linear sum of the basis functions {1, cos(0t), cos(20t),…, cos(n0t), sin(0t), sin(20t),…, sin(n0t)} Ex: x(t) =1+ cos(2t)+ 2cos(2 2t)+ 0.5sin(23t)+ 3sin(2t) x(t) =1+ cos(2t)+ 2cos(2 2t)+ 3sin(2t)+ 0.5sin(23t) + + + = Purpose of the Fourier Series (FS) FS is used to find the frequency components and their strengths for a given periodic signal x(t). The Three forms of Fourier Series • Trigonometric Form • Compact Trigonometric (Polar) Form. • Complex Exponential Form. Trigonometric Form • It is simply a linear combination of sines and cosines at multiples of its fundamental frequency, f0=1/T. n 1 n 1 xt a0 an cos2f 0 nt bn sin 2f 0 nt • a0 counts for any dc offset in x(t). • a0, an, and bn are called the trigonometric Fourier Series Coefficients. • The nth harmonic frequency is nf0. Trigonometric Form • How to evaluate the Fourier Series Coefficients (FSC) of x(t)? n 1 n 1 xt a0 an cos2nf 0t bn sin 2nf 0t To find a0 integrate both side of the equation over a full period 1 a0 xt dt T0 T0 Trigonometric Form n 1 n 1 xt a0 an cos2nf 0t bn sin 2nf 0t To find an multiply both side by cos(2mf0t) and then integrate over a full period, m =1,2,…,n,… 2 an xt cos2nf 0t dt T0 T0 To find bn multiply both side by sin(2mf0t) and then integrate over a full period, m =1,2,…,n,… 2 bn xt sin 2nf 0t dt T0 T0 Example f t a0 an cos2nt bn sin 2nt f(t) 1 n 1 e-t/2 0 • Fundamental period T0 = • Fundamental frequency f0 = 1/T0 = 1/ Hz 0 = 2/T0 = 2 rad/s a0 an 1 2 0 0 2 2 e dt e 1 0.504 e t 2 t 2 2 cos2nt dt 0.504 2 1 16n 8n bn e sin 2nt dt 0.504 2 0 1 16n an and bn decrease in amplitude as n . 2 t 2 2 cos2nt 4n sin 2nt f t 0.5041 2 n 1 1 16n To what value does the FS converge at the point of discontinuity? Dirichlet Conditions • A periodic signal x(t), has a Fourier series if it satisfies the following conditions: 1. x(t) is absolutely integrable over any period, namely x(t ) dt T0 2. x(t) has only a finite number of maxima and minima over any period 3. x(t) has only a finite number of discontinuities over any period Compact Trigonometric Form • Using single sinusoid, xt C0 dc component Cn cos2nf 0t n n 1 nth harmonic C0 a0 • Cn , and n are related to the trigonometric coefficients an and bn as: Cn an bn 2 2 and bn n tan an 1 The above relationships are obtained from the trigonometric identity a cos(x) + b sin(x) = c cos(x + ) Role of Amplitude in Shaping Waveform xt C0 Cn cos2nf 0t n n 1 Role of the Phase in Shaping a Periodic Signal xt C0 Cn cos2nf 0t n n 1 Compact Trigonometric f t C0 Cn cos2nt n f(t) 1 n 1 e-t/2 0 a0 0.504 • Fundamental period T0 = • Fundamental frequency f0 = 1/T0 = 1/ Hz 0 = 2/T0 = 2 rad/s f t 0.504 0.504 n 1 2 1 16n 2 2 an 0.504 2 1 16n 8n bn 0.504 2 1 16n C0 ao 0.504 2 Cn a b 0.504 2 1 16 n 1 bn tan 1 4n n tan an 2 n 2 n cos 2nt tan 1 4n Line Spectra of x(t) • The amplitude spectrum of x(t) is defined as the plot of the magnitudes |Cn| versus • The phase spectrum of x(t) is defined as the plot of the angles Cn phase(Cn ) versus • This results in line spectra • Bandwidth the difference between the highest and lowest frequencies of the spectral components of a signal. Line Spectra f(t) C0 0.504 1 e-t/2 n tan 1 4n 0 2 Cn 0.504 2 1 16 n f t 0.504 0.504 n 1 2 1 16n 2 cos 2nt tan 1 4n f(t)=0.504 + 0.244 cos(2t-75.96o) + 0.125 cos(4t-82.87o) + 0.084 cos(6t-85.24o) + 0.063 cos(8t-86.24o) + … Cn n 0.504 0.244 0.125 0.084 0 2 4 6 -/2 0.063 8 10 HW8_Ch6: 6.1-1 (a,d), 6.1-3, 6.1-7(a, b, c) Exponential Form • x(t) can be expressed as xt j 2f 0 nt D e n n j 2f 0 nt To find Dn multiply both side by e over a full period, m =1,2,…,n,… and then integrate 1 Dn xt e j 2f 0 nt dt , n 0, 1, 2,.... To To Dn is a complex quantity in general Dn=|Dn|ej D-n = Dn* |Dn|=|D-n| Even Dn = - D-n Odd D0 is called the constant or dc component of x(t) Line Spectra of x(t) in the Exponential Form • The line spectra for the exponential form has negative frequencies because of the mathematical nature of the complex exponent. x(t ) ... | D 2 | e j 2 e j 20t | D1 | e j1 e j0t D0 | D1 | e j1 e j0t | D2 | e j 2 e j 20t ... x(t ) C0 C1 cos(0t 1 ) C2 cos( 20t 2 ) ... C0 = D0 Cn = 2|Dn| Cn = Dn Example Find the exponential Fourier Series for the squarepulse periodic signal. f(t) /2 1 1 jnt Dn e dt 2 / 2 sin n / 2 0.5sinc( n / 2) n 1 D0 2 n even 0 Dn 1 / n n odd 0 n for all n 3,7,11,15, n 3,7,11,15, 2 /2 /2 2 • Fundamental period T0 = 2 • Fundamental frequency f0 = 1/T0 = 1/2 Hz 0 = 2/T0 = 1 rad/s Exponential Line Spectra |Dn| 1 1 Dn 1 1 Example The compact trigonometric Fourier Series coefficients for the square-pulse periodic signal. 1 C0 2 0 n even Cn 2 n odd n 0 for all n 3,7,11,15, n n 3,7,11,15, f(t) 1 2 /2 /2 1 2 ( n 1) / 2 x(t ) cos nt (1) 1 2 n 1 n 2 odd Does the Fourier series converge to x(t) at every point? 2 Gibbs Phenomenon x3 (t ) x9 (t ) Gibbs Phenomenon – Cont’d x21 (t ) x45 (t ) overshoot: about 9 % of the signal magnitude (present even if N ) Example Find the exponential Fourier Series and sketch the corresponding spectra for the impulse train shown below. From this result sketch the trigonometric spectrum and write the trigonometric Fourier Series. T (t ) Solution 0 Dn 1 / T0 1 T0 (t ) T0 jn0t e n Cn 2 | Dn | 2 / T0 C0 | D0 | 1 / T0 1 T0 (t ) T0 1 2 cos( n0t ) n 1 -2T0 -T0 T0 2T0 Relationships between the Coefficients of the Different Forms Dn 0.5an jbn D n D n 0.5an jbn Dn 0.5Cn n 0.5Cn e j n D0 a0 C0 an Dn D n 2 ReDn bk j Dn D n 2 ImDn an Cn cos n bn Cn sin n a0 D0 c0 Cn an bn 2 2 bn n tan an Cn 2 Dn 1 n Dn C0 a0 D0 Parseval’s Theorem • Let x(t) be a periodic signal with period T • The average power P of the signal is defined as 1 P T T /2 T / 2 2 x(t ) dt • Expressing the signal as xt C0 Cn cos( n0t n ) n 1 it is also P C0 0.5Cn 2 n 1 2 P D 2 Dn 2 0 n 1 2 Simplifications Through Signal Symmetry • If x (t) is EVEN: It must contain DC and Cosine Terms. Hence bn = 0, and Dn = an/2. • If x(t) is ODD: It must contain ONLY Sines Terms. Hence a0 = an = 0, and Dn=-jbn/2. LTIC System Response to Periodic Inputs e j 0 t H(s) H(j) H ( j0 )e j0t A periodic signal x(t) with period T0 can be expressed as x(t ) jn0 t D e n n For a linear system x(t ) jn0 t D e n n H(s) H(j) y (t ) jn0t D H ( jn ) e n 0 n HW9_Ch6: 6.3-1(a,d), 6.3-5, 6.3-7, 6.3-11, 6.4-1, 6.4-3 Fourier Series Analysis of DC Power Supply A full-wave rectifier is used to obtain a dc signal from a sinusoid sin(t). The rectified signal x(t) is applied to the input of a lowpass RC filter, which suppress the timevarying component and yields a dc component with some residual ripple. Find the filter output y(t). Find also the dc output and the rms value of the ripple voltage. R=15 sint Full-wave rectifier x(t) C=1/5 F y(t) Fourier Series Analysis of DC Power Supply 2 Dn (1 4n 2 ) D0 2 / 2 j 2 nt x(t ) e 2 ( 1 4 n ) n 1 H ( j ) 3 j 1 y (t ) D H ( jn )e n n 0 jn0t PDC 4 / 2 Pripple 2 | Dn |2 n 1 Dn 2 (1 4n ) 36n 1 2 2 Pripple 0.0025 ripple rms Pripple 0.05 2 j 2 nt y (t ) e 2 Ripple rms is only 5% ( 1 4 n )( j 6n 1) n of the input amplitude Fourier Series Analysis of DC Power Supply clear all t=0:1/1000:3*pi; for i=1:100 n=i; yp=(2*exp(j*2*n*t))/(pi*(1-4*n^2)*(j*6*n+1)); This Matlab code will n=-i; plot y(t) for -100 n yn=(2*exp(j*2*n*t))/(pi*(1-4*n^2)*(j*6*n+1)); 100 and find the ripple y(i,:)=yp+yn; power according to the end equations below yf = 2/pi + sum(y); 2 j 2 nt y (t ) e plot(t,yf, t, (2/pi)*ones(1,length(yf))) 2 ( 1 4 n )( j 6n 1) n axis([0 3*pi 0 1]); Pripple 2 | Dn |2 0.0025 Power=0; for n=1:50 Power(n) = abs(2/(pi*(1-4*n^2)*(j*6*n+1))); end TotalPower = 2*sum((Power.^2)); figure; stem( Power(1,1:20)); n 1