EECE 31 Signals & Sysems Prof. Mar Fowler Noe Se #1 C-T Signals: Circuis wih Periodic Sources 1/1
Solving Circuis wih Periodic Sources FS maes i easy o find he response of an RLC circui o a periodic source! Use he FS o conver he source ino a sum of sinusoids Do phasor analysis for each of he inpu sinusoids (hin superposiion!) Add up he sinusoidal responses o ge he oupu signal Example: In elecronics you have seen (or will see) how o use diodes and an RC filer circui o creae a DC power supply: 6Hz Sine wave (Period = 16.67 ms) Obviously we can do his for all infiniely many erms bu we can do i for enough and if we do i numerically i is no hard! x() R C Periodic!! Thin Fourier Series!! Periodic Signal x() T = 8.33ms 4 rad/sec /1
Progression of Ideas Periodic Source x() x() R C y()? T = 8.33ms Periodic Source as FS j x( ) c e x() R C y()? c h Phasor Term of FS R @ o Z 1 j C C Need o find he c values numerically or analyically o Z C 1 joc d c c ZC R 1 joc R? = Phasor o 1 c Find 1 jorc 3/1
For his scenario we can find he c analyically The equaion for he FS coefficiens is: c 1 T j T x( ) e d Over his inerval: x() A x() Asin T T T T j T 1 T c Asin e d T T Now apply Calc I ideas o evaluae. Change of variable: T Use a Table of Inegrals and do some algebra & rig o ge: A sin j c e d c A (1 4 ) FS coefficien for full-wave recified sine wave of ampliude A 4/1
So he wo-sided specrum afer he recifier: c c 5 4 A 3 3 5 4 3 A 3 4 5 3 4 5 Now we can us Parseval s Theorem o deermine how many erms we need in our approximaion for he source 1 T A A A P A sin d sin d T T P approx K K K A 4A 1 c K K (1 4 ) K (1 4 ) 5/1
We can loo a he raio of hese wo as a good measure: P K 4A 1 K approx K P A K (1 4 ) 8 1 (1 4 ) Numerically evaluaing his for differen K values shows ha K = 1 reains more han 99.99% of he power. So we can use ha value. So our numerical approach is now his: 1. Numerically evaluae c for = 1 o 1. Numerically conver hem ino he d phasors 3. Conver he phasors ino corresponding FS sinusoidal erms and add hem up We ll do his for: A = 1 vols R = 1 C = 1 F c d A (1 4 1 1 y () 1 ) 1 c jorc j de 6/1
wo=4*pi; % Se fund freq fo=wo/(*pi); % conver o Hz T = *pi/wo; % compue period K=1; % Se number of erms v=(-k):k; % se vecor of indices A=1; % se ampliude of inpu R=1; % se resisance C=1e-6; % se capaciance c=(*a/pi)./(1-4*(v.^)); % compue he inpu FS coefficens d=(1./(1+j*v*wo*r*c)).*c; % compue he oupu FS coefficens Fs = 4*K*fo; % Compue sampling rae (se here o wice he minimum value of Kfo) Ts = 1/Fs; % Compue sample spacing = (-3*T):Ts:(3*T); x_apprx = zeros(size()); % ses up vecor of zeros as firs parial sum for = (-K):K % loop hrough all coefficiens x_apprx = x_apprx + c(+k+1)*exp(j**wo*); % Add curren erm o parial sum end x_apprx = real(x_apprx); y_apprx = zeros(size()); % ses up vecor of zeros as firs parial sum for = (-K):K % loop hrough all coefficiens y_apprx = y_apprx + d(+k+1)*exp(j**wo*); % Add curren erm o parial sum end y_apprx = real(y_apprx); % heory says imaginary pars cancel so enforce his in case % of numerical round-off issues figure(1); plo(,x_apprx,'r',,y_apprx,'g--'); xlabel('ime (seconds)'); ylabel('inpu and Oupu (vols)'); grid figure(); subplo(,1,1); sem(v,abs(c)); subplo(,1,); sem(v,abs(d)) 7/1
Recified sinewave applied o RC circui Oupu of RC circui: DC level wih small wiggle 1 9 Inpu and Oupu (vols) 8 7 6 5 4 3 1 -.5 -. -.15 -.1 -.5.5.1.15..5 ime (seconds) 8/1
8 c 6 4 d 1 1 c jorc Changed via Muliplicaion! d -1-5 5 1 8 6 4 Effec of RC circui: Mae all non-dc FS coefficiens negligible -1-5 5 1 Muliplicaive facor has small magniude here! 9/1
Big Idea: Frequency Response Inpu Oupu j x( ) c e Linear j y () de Circui Inpu s FS Coefficiens d H( ) c o Oupu s FS Coefficiens H() is he Frequency Response of he Circui How o find he Frequency Response of a Circui Assume arbirary phasor X wih frequency Analyze circui o find oupu phasor Y I will always ae his muliplicaive form: Y = H() X All impedances are evaluaed a he arbirary frequency The frequency response funcion H() is he hing ha muliplies X 1/1