# EECE 301 Signals & Systems Prof. Mark Fowler

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1 EECE 3 Signals & Sysems Prof. Mark Fowler Noe Se #8 C-T Sysems: Frequency-Domain Analysis of Sysems Reading Assignmen: Secion 5.2 of Kamen and Heck /2

2 Course Flow Diagram The arrows here show concepual flow beween ideas. Noe he parallel srucure beween he pink blocks (C-T Freq. Analysis) and he blue blocks (D-T Freq. Analysis). New Signal Models Ch. Inro C-T Signal Model Funcions on Real Line Sysem Properies LTI Causal Ec Ch. 3: CT Fourier Signal Models Fourier Series Periodic Signals Fourier Transform (CTFT) Non-Periodic Signals Ch. 2 Diff Eqs C-T Sysem Model Differenial Equaions D-T Signal Model Difference Equaions Zero-Sae Response Ch. 5: CT Fourier Sysem Models Frequency Response Based on Fourier Transform New Sysem Model Ch. 2 Convoluion C-T Sysem Model Convoluion Inegral Ch. 6 & 8: Laplace Models for CT Signals & Sysems Transfer Funcion New Sysem Model New Sysem Model D-T Signal Model Funcions on Inegers New Signal Model Powerful Analysis Tool Zero-Inpu Response Characerisic Eq. Ch. 4: DT Fourier Signal Models DTFT (for Hand Analysis) DFT & FFT (for Compuer Analysis) D-T Sysem Model Convoluion Sum Ch. 5: DT Fourier Sysem Models Freq. Response for DT Based on DTFT New Sysem Model Ch. 7: Z Trans. Models for DT Signals & Sysems Transfer Funcion New Sysem Model 2/2

3 5.2 Response o Periodic Inpus periodic x() h() y() =? H(ω) Since x() is periodic, wrie i as FS: So, he inpu is a sum of erms k = x jkω x( ) = c k e Linear Sysem: So Oupu = Sum of Individual Responses Bu each individual response is o a complex sinusoid inpu EASY! Indicaes for x() c x e k jkω H(ω) H x jkω ( ω k) c k e (complex: magniude & phase) Sum hese o ge inpu Sum hese o ge oupu k = x jkω x( ) = c k e x jkω y( ) = [ H ( ωk) c k ] e k = FS coefficien of y() 3/2

4 General Insighs from his Analysis. periodic in, periodic ou 2. The sysem s frequency response H(ω) works o modify he inpu FS coefficiens o creae he oupu FS coefficiens: y c k = H ( kω) c x k 4/2

5 Example (Ex. 5.4 wih Some Injeced Realiy) Problem: suppose you have a circui board ha has a digial clock circui on i. I makes he recangular pulse rain shown below: x() (Of course mos digial clock circuis would run much faser) Suppose you need o connec his clock signal o a circui on anoher circui board using a wised pair of wires: x() y() Q: Wha effec does he cable have on he clock signal a he 2 nd board??? Pair of wires can be modeled as an RC circui: x() y() Assume: The circui driving he cable has an infiniesimally small oupu impedance (ha is good!): Thevenin of driver: x() 5/2

6 Assume: The circui being driven by he cable has infinie inpu impedance (ha is good!) i.e. No loading of he RC circui So x() y() (goes o driven circui having infinie inpu impedance) Goal: Perform an analysis o enable you o recommend an accepable value of cable RC ime consan (Analysis Drives Design!) Sep : Analyically find FS of inpu and compue runcaed FS sum: From Ex. 3.4 we ge: Indicaes for x() c x k =, kπ, kπ k = ±, ± 5, ± 9,... k = ± 3, ± 7, ±,..., k = ± 2, ± 4, ± 6,..., k = 2 Sep 2: Find cable s frequency response as a funcion of RC: (See Ex. in secion 5.) x( ) N k = N c x k e jkω Then plo vs. ime H ( ω) = + jωrc o 6/2

7 Sep 3 (opional) (Bu i really helps you see wha is going on!) Look a frequency domain plos of Inpu and Sysem (for various RC values) sem plo of FS coefficiens x c Magniude k coninuous plo of Magniude of sysem s Frequency Resp. H(ω) Sep 4 (opional) (This also really helps you see wha is going on) Compue oupu FS coefficiens: Look a he resul sem plo of N y jkω y( ) c k e k = N c = H kω ) c y k y c k ( Sep 5: Compue runcaed FS sum o see oupu signal Plo vs. ime See plos on nex 3 pages for hree RC ime consan values: x k RC =. s RC =. s RC = s Noe: Shor RC ime consan passes high frequencies beer han long RC ime consan 7/2

8 Inpu Signal x().5.5 RC Circui Analysis w/ Square Wave Inpu Inpu Signal ime (sec) Arifac from summing only a finie # of erms RC =. s Does Decen Job of Passing Mos of he Significan Frequencies of he Inpu Oupu Signal y().5.5 Oupu Signal ime (sec) c x k.4.2 H(f) c y k k f o (Hz) Inpu Signal s FS Coefficiens 5 5 f (Hz) Cable s Freq. Resp. 5 5 k f o (Hz) Oupu Signal s FS Coefficiens 8/2

9 RC Circui Analysis w/ Square Wave Inpu Inpu Signal x().5.5 Inpu Signal ime (sec) RC =. s Does Moderae Job of Passing Mos of he Significan Frequencies of he Inpu Oupu Signal y().5.5 Oupu Signal ime (sec) c x k.4.2 H(f) c y k k f o (Hz) Inpu Signal s FS Coefficiens 5 5 f (Hz) Cable s Freq. Resp. 5 5 k f o (Hz) Oupu Signal s FS Coefficiens 9/2

10 RC Circui Analysis w/ Square Wave Inpu Inpu Signal x().5.5 Inpu Signal ime (sec) RC = s Does Poor Job of Passing Mos of he Significan Frequencies of he Inpu Oupu Signal y() Oupu Signal ime (sec) c x k.4.2 H(f) c y k k f o (Hz) Inpu Signal s FS Coefficiens 5 5 f (Hz) Cable s Freq. Resp. 5 5 k f o (Hz) Oupu Signal s FS Coefficiens /2

11 Insigh from Example: We used a simple model for he cable o make i easy o analyze Bu he mehod would be he same even if we had a more deailed model for he cable The inpu clock signal has nice sharp ransiions due o is significan high frequency componens Cables ha significanly suppressed he inpu s high frequency componens provided a low-qualiy clock signal o he 2 nd board We made assumpions abou he driver circui and he driven circui The driver was assumed o have zero oupu resisance If ha were no rue, is oupu impedance ges added o he resisor and ha would furher degrade he performance (in fac he driver s oupu impedance may be more han he cable resisance in which case i would be he dominan facor The driven circui was assumed o have infinie inpu impedance If ha were no rue we would have o combine i in parallel wih he capacior s impedance his would furher degrade he performance Typically he RC value of a cable increases wih lengh So performance would decrease wih lengh of cable /2

12 funcion [x,]=example_5_4(rc) k=:4:2; K=[k;k+;k+2;k+3]; 4 ypes of indices:,5,9, 2,6,,.. 3,7,, 4,8,2, C_k=./(K*pi); % fill marix wih his form keep firs row C_k(2,:)=zeros(size(k)); % replace 2 nd row w/ zeros C_k(3,:)=-*C_k(3,:); % replace 3 rd row w/ negaives C_k(4,:)=zeros(size(k)); % replace 4 h row w/ zeros c_x_k=c_k(:); % urn ino col. vecor by going down marix columns =-3:(6/8):3; % creae ime vecor wih approp. spacing k=(:max(max(k))); % creae FS erm index [T,K]=meshgrid(,k); % creae ime marix and index marix wo=pi; EXP_pos=exp(j*T.*K*wo); % Each row is a sinusoid erm EXP_neg=exp(-j*T.*K*wo); %% Compue he FS summaion o ge approx. inpu ime signal x=.5+sum(c_x_k(:,ones(,lengh())).*exp_pos) +sum(conj(c_x_k(:,ones(,lengh()))).*exp_neg); % The above cmnd adds up he rows of EXP weighed by he c_x_k subplo(2,3,) plo(,x) xlabel('ime (sec)') ylabel('inpu Signal x()') Malab Code for Example s Plos T = K = 3 subplo(2,3,5) w=:.:max(w_k); % creae finely-spaced frequency H=./(+j*w*RC); % compue Freq hese Freqs plo(w/(2*pi),abs(h)) % plo vs. freq in Hz xlabel('f (Hz)') ylabel(' H(f) ') axis([-.5 5.]) subplo(2,3,6) H_k=./(+j*w_k*RC); % compue Freq Resp a FS freqs H_=./(+j**RC); c_y_k=c_x_k.*(h_k.'); % compue oupu FS coeffs sem([ w_k]/(2*pi),[.5*h_ abs(c_y_k).']) xlabel('k f_o (Hz)') ylabel(' c^y_k ') axis([ ]) subplo(2,3,4) w_k=k*wo; % creae vecor of FS frequencies % In he nex line we have o aach c_=.5 and is freq sem([ w_k]/(2*pi),[.5 abs(c_x_k).']) % plo vs freq in Hz xlabel('k f_o (Hz)') ylabel(' c^x_k ') axis([ ]) subplo(2,3,3) %% FS summaion o ge approx. oupu ime signal y=.5*h_+sum(c_y_k(:,ones(,lengh())).*exp_pos) +sum(conj(c_y_k(:,ones(,lengh()))).*exp_neg); plo(,y) xlabel('ime (sec)') ylabel('oupu Signal y()') 2/2

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