ECE 308 -13
Frequency Analysis of Signals and Systems
Z. Aliyazicioglu
Electrical and Computer Engineering Department Cal Poly Pomona
Frequency Analysis of Signals and Systems
The Fourier Series for Discrete-Time Periodic Signals
The Fourier representation of signal maps the signal into frequency domain. The Fourier transform provides a different way to interpret signals and systems. It is useful for convolution operation in time domain maps into multiplication in frequency domain It gives us information about system or signal characteristic of behavior
ECE 308-13 2
1 Frequency Analysis of Signals and Systems
The Fourier Series for Discrete-Time Periodic Signals
A given periodic sequence x(n) with period N, that is x(n)=x(n+N)
The Fourier representation of x(n) can be expressed as
N −1 xn()= cejknN2/π , ∑ k k =0
where ck ara the coefficients in the series representation. This equation is called the Discrete-time Fourier Series (DTFS)
The Fourier coefficients { ck k } , = 1,2,..., N − 1 provides the description of x(n) in the frequency domain
N −1 1 − j2/π kn N cxnek = ∑ () N n=0
ECE 308-13 3
Frequency Analysis of Signals and Systems
ck represents the amplitude and phase associated with the frequency component
jknN2/π jnωk sk ==ee
where ωk = 2/πkN
The function Sk are periodic with period N. Hence
skk()nsnN= (+ )
NN−−11 So, that 11−+jkNnN2(ππ )/ − jknN 2 / cxnexneckN+ ===∑∑() () k NNnn==00
Therefore,ck is periodic sequence with fundamental period N.
Instead of focusing in periodic range k=0,1,…, N-1 in frequency 2πk 02<=ω <π domain k N for 0 2 Frequency Analysis of Signals and Systems Example: Determine the spectra of the following signal πn π 1 xn()= cos ω = f = 3 0 3 0 6 Hence x(n) is periodic with fundamental period N=6 N −15 11−−jknN2/ππ jkn 2/6 cxnexnekk ==∑∑() () , = 0,1,...,5 N nn==006 πn 11 xn()== cos ej2/6ππnjn + e− 2/6 32 2 j2ππnjnjn /6 2 (5− 6) /6 2 π 5 /6 ee== e which means that cc−15= 5 1 nπ − jkn2/6π cekk ==∑cos( ) , 0,1,...,5 63n=0 ECE 308-13 5 Frequency Analysis of Signals and Systems Example:(cont) 123π ππ cee=+(1 cos−−jk2/6ππ + cos jk 22/6 + cos e − jk 23/6 π k 63 3 3 45ππ ++coseek−−jk24/6ππ cos jk 25/6 ) , = 0,1,...,5 33 c = 0 1 1 0 c1 = c = 0 2 c = 0 c = 0 c5 = 2 3 4 2 % Find the Fourier Series coefficients of x(n)=cos(pi n/3) for k=1:6 c(k)=0; for n=1:6 c(k)=c(k)+cos(pi*(n-1)/3)*exp(-j*2*pi*(k-1)*(n-1)/6); end c(k)=c(k)/6; end for k=1:6; rats(c(k)) end . ECE 308-13 6 3 Frequency Analysis of Signals and Systems Example: Determine the spectra of the following periodic signal with period N=4 x(n )= {1,1,0,0} N −13 11−−jknN2/ππ jkn 2/4 cxnexnekk ==∑∑() () , = 0,1,2,3 N nn==004 1 ce=+(1− jk2/4π ) , k = 0,1, 2, 3 k 4 1 1 1 c0 = cj=−(1 ) c = 0 cj= (1+ ) 2 1 4 2 3 4 1 c = 2 π 0 c = (c1 =− c2 = 0 2 1 4 4 2 π c3 = (c = 4 3 4 ECE 308-13 7 Frequency Analysis of Signals and Systems Fourier Transform for DT The Fourier transform of a finite-energy discrete time signal x(n) is defined as ∞ Xxne()ω = ∑ ()− jωn n=−∞ X(ω) signal is periodic with period 2π ,,, ∞ Xk(2)ωπ+=∑ xne ()−+jkn(2)ωπ n=−∞ ∞ Xk(2)ωπ+=∑ xnee ()−−jωπnjkn2 n=−∞ ∞ Xk(2)ω +=πω∑ xneX ()− jnω = () n=−∞ The inverse Fourier transform of discrete-time signal is π 1 xn()= X (ω ) ejnω dω 2π ∫ −π ECE 308-13 8 4 Frequency Analysis of Signals and Systems Properties of the Fourier Transform for DT Properties Time Domain Frequency Domain Notation xn(), x12 (), n x () n XX(),ω 12 (),ωω X () Linearity ax12() n+ b x () n aX12()ω + bX ()ω Time Shifting xn()− k eX− jkω ()ω Time Reversal xn()− X()−ω Convolution xn12()* x () n XX12()ω ()ω rlxlxl()= ()* (− ) SXX=−=()ω (ωωω ) XX ()* () Correlation xx12 12 xx12 12 12 jnω Frequency Shifting exn0 () X()ω − ω0 11 XX()()ω ++ωωω − Modulation xn10()cosω n 2200 π 1 Multiplication xnxn() () X ()λ Xd (ωλλ− ) 12 2π ∫ 12 −π dX()ω Differentiation in * j Frequency domain xn() dω Conjugation nx() n X *()−ω ∞ π Parseval’s Theorem **1 xnxn12() ()= X 1 ()ω X 2 ()ωω d ∑ 2π ∫−π n=−∞ ECE 308-13 9 Frequency Analysis of Signals and Systems Example: n xn() 0.8 −∞ ∞∞∞ −−−jnωωω n jn j n Xxnee1()ω ===∑∑∑ () 0.8 (0.8 e ) nnn=−∞ =00 = 1 e jω == 10.8−−ee− jjωω 0.8 ∞−11 − −−−−jωωωnnjnjn Xxnee2(ω )==∑∑ ( ) 0.8 = ∑ (0.8 e ) nn=−∞ =−∞ n =−∞ ∞ 0.8e jω ==(0.8e jnω ) ∑ jω n=1 10.8− e ECE 308-13 10 5 Frequency Analysis of Signals and Systems Example: (cont) 10.810.8e jω − 2 XX()ωωω=+12 () X () = + = 10.8−−−+ee− jjωω 10.8 12(0.8)cosω 0.82 n=-20:20; x=0.8.^abs(n); stem (n,x) xlabel ('n') ylabel ('x(n)') title ('x(n) sequence') figure; w=-2*pi:0.1:2*pi; y=(1-0.8^2)./(1-2*0.8*cos(w)+0.8^2); plot (w,y); xlabel ('\omega') ylabel ('y(\omega)') title ('Fourier transform of x(n) sequence') (Fourier1.m) ECE 308-13 11 Frequency Analysis of Signals and Systems Example: xn()= 0.5n −∞ n=-20:20; b=0.5 x=b.^abs(n); stem (n,x) xlabel ('n') ylabel ('x(n)') title ('x(n) sequence') figure; w=-2*pi:0.1:2*pi; y=(1-b^2)./(1-2*b*cos(w)+b^2); plot (w,y); xlabel ('\omega') ylabel ('y(\omega)') title ('Fourier transform of x(n) sequence') (Fourier2.m) 10.5− 2 X()ω = 1−+ cosω 0.52 ECE 308-13 12 6 Frequency Analysis of Signals and Systems Example: (cont) 10.5− 2 X()ω = 1−+ cosω 0.52 ECE 308-13 13 Frequency Analysis of Signals and Systems Example: xn()= 0.5n 0 < n <∞ ∞∞∞ Xxnee()ω ===∑∑∑ ()−−−jωωωnnjnjn 0.5 (0.5 e ) nnn=−∞ =00 = 1 e jω == 10.5−−ee− jjωω 0.5 ECE 308-13 14 7 Frequency Analysis of Signals and Systems Example: (cont) n=-0:20; b=0.5 x=b.^abs(n); stem (n,x) xlabel ('n') ylabel ('x(n)') title ('x(n) sequence') figure; w=-2*pi:0.1:2*pi; X=exp(j*w)./(exp(j*w)-b); subplot (2,1,1) plot (w,abs(X)); xlabel ('\omega') ylabel ('X(\omega)') title ('Fourier transform of x(n) sequence') subplot (2,1,2); plot (w,phase(X)); xlabel ('\omega') ylabel ('Phase (X(\omega))') title ('Fourier transform of x(n) sequence') Fourier3.m ECE 308-13 15 Frequency Analysis of Signals and Systems n Example: xn()=−() 0.5 un () n=0:20; ∞∞ ∞ b=-0.5 j nnjnjn Xxnee(ω )==−=−∑∑ ()−−−ωωω (0.5) ∑ (0.5 e ) x=b.^n; nn=−∞ =00 n = stem (n,x) 1 e jω xlabel ('n') == − jjωω ylabel ('x(n)') 10.5++ee 0.5 title ('x(n) sequence') figure; w=-2*pi:0.1:2*pi; X=exp(j*w)./(exp(j*w)-b); subplot (2,1,1) wpi=w/pi; plot (wpi,abs(X)); xlabel ('\omega/\pi') ylabel ('X(\omega)') title ('Fourier transform of x(n) sequence') subplot (2,1,2); plot (wpi,phase(X)); xlabel ('\omega/\pi') ylabel ('Phase (X(\omega))') title ('Fourier transform of x(n) sequence') ECE 308-13 16 8 Frequency Analysis of Signals and Systems Example: (cont) ECE 308-13 17 Frequency Analysis of Signals and Systems 20≤ n ≤ 5 Example: xn()= n=0:10; 0 otherwise x=2*(n>=0 & n<=5); stem (n,x) ∞ 5 1− e− j6ω xlabel ('n') Xxnee()ω === ()−−jnωω 2 jn 2 ∑∑1− e− jω ylabel ('x(n)') nn=−∞ =0 title ('x(n) sequence') −−jj33ωω j 3 ω 5 eee− − j ω sin3ω ==22e 2 figure; eee−−jjωω/2 /2− j ω /2 sinω / 2 w=-2*pi:0.01:2*pi; X=2*(1-exp(-j*6*w))./(1-exp(-j*w)); subplot (2,1,1) wpi=w/pi; plot (wpi,abs(X)); xlabel ('\omega/\pi') ylabel ('X(\omega)') title ('Fourier transform of x(n) sequence') subplot (2,1,2); plot (wpi,phase(X)); xlabel ('\omega/\pi') ylabel ('Phase (X(\omega))') title ('Fourier transform of x(n) sequence') ECE 308-13 18 9 Frequency Analysis of Signals and Systems Example: (cont) ECE 308-13 19 10