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EECS 451 FREQUENCY RESPONSE
Recall: zon → |H(z)| → H(zo)zon. Eigenfunction of LTI. Now: zon = e+jωon^ → |H(z)| → H(e+jωo^ )e+jωon. and: zon = e−jωon^ → |H(z)| → H(e−jωo^ )e−jωon. cos(ωon) → |H(z)| → |H(ejωo^ )| cos(ωon + arg[H(ejωo^ )]).
Gain: Amplitude increases by factor of |H(ejωo^ )|. Phase: Shift by arg[H(ejωo^ )] = tan−^1 Im[H(e
jωo (^) )] Re[H(ejωo^ )]. DTFT: H(ejωo^ ) =
h(n)e−jωon^ = DT F T [h(n)]. Zero: H(z) has a zero at e±jωo^ → y(n) = 0 in steady-state. Pole: H(z) has a pole at e±jωo^ → y(n) → ∞ blows up.
EX #1: h(n) = ( 12 )nu(n) and x(n) = {... − 1 , 0 , 1 , 0 , − 1.. .} = cos( πn 2 ). H(ejω^ ) = 1/(1 − 12 e−jω^ ) = 1/(1 + j 2 ) = 0. 89 e−j^26.^6
o at ω = π 2 y(n) = 0.89 cos( πn 2 − 26. 6 o) = {... 0. 8 , 0. 4 , − 0. 8 , − 0. 4 , 0. 8 , 0. 4.. .}.
EX #2: h(n) = ( 12 )nu(n) and x(n) = {... − 1 , 1 , − 1 , 1 , − 1.. .} = cos(πn). H(ejω^ ) = 1/(1 + 12 ) = 23 at ω = π → y(n) = 23 cos(πn) = 23 (−1)n. Why 23? y(n) = h(n) ∗ x(n) =
( 12 )i(−1)n−i^ = (−1)n^
(− 12 )i^ = (−1)n^1 −(^1 − 1 2 )^
EX #3a: h(n) = { 12 , + 12 } → H(ejω^ ) = 12 (1 + e−jω^ ) = cos( ω 2 )e−jω/^2. Lowpass.
EX #3b: h(n) = { 12 , − 12 } → H(ejω^ ) = 12 (1 − e−jω^ ) = sin( ω 2 )ej(π−ω)/^2. High.
Notch: H(z) = (z − ejωo^ )(z − e−jωo^ ) (^) z^1 = z − 2 cos(ωo) + z−^1. filter: H(ejω^ ) = 2 cos(ω) − 2 cos(ωo). h(n) = { 1 , −2 cos(ωo), 1 }.
Comb: H(z) = (^2) M^1 +
∑M
k=−M z
−Lk (^) → zeros at z = e L(2j^2 Mπk +1) for k = ± 1... ± LM unless L divides k. See p. 349.
Reson- H(z) = Bz^2 /[(z − rejωo^ )(z − re−jωo^ )] = Bz^2 /[z^2 − 2 r cos(ωo)z + r^2 ]. ator: On unit circle |z| = 1, H(ejω^ ) peaks at ω = ± cos−^1 [ 1+r
2 2 r cos^ ωo]. r → 1 : Resonant freq.' ωo; 3 dB bandwidth' 2(1 − r). See p. 342.
All- H(z) = zD^
∏ (^) z−^1 −z∗ k 1 −zk z−^1 or^
A(1/z) zN^ A(z) →^ H(z)H(1/z) = 1^ → |H(e
jω (^) )| = 1. pass: |zi| < 1 →stable and causal; A(z) = Z{causal signal}.
EECS 451 DTFS AND FREQUENCY RESPONSE
Signal: x(n) = {... 4 , 0 , 1 , 0 , 1 , 0 , 4 , 0 , 1 , 0 , 1 , 0 , 4.. .} has period=N=6. System: y(n) = 12 (x(n) + x(n − 1)) (average the 2 most recent inputs). Goal: Compute Discrete Time Fourier Series (DTFS) of input x(n). Goal: Compute Discrete Time Fourier Series (DTFS) of output y(n).
DTFS: x(n) =
∑N − 1
k=0 Xke
j 2 πnk/N (^) for all n where N =period of x(n).
Compute: Xk = (^) N^1
∑N − 1
n=0 x(n)e
−j 2 πnk/N (^) for k = 0... N − 1. Use fft/N
Note: Like Fourier series, except for finite number N of harmonics.
Compare: x(t) =
−∞ Xke
j 2 πkt/T (^) where Xk = 1 T
∫ T
0 x(t)e
−j 2 πkt/T (^) dt.
Compute: fft([4 0 1 0 1 0],6)/6=[1 .5 .5 1 .5 .5] Note: x(n) real and even → DTFS coefficients real and even (extensions). Note: x(n) = 0 for odd n → DTFS coefficients repeat (the last 3=first 3). DTFS: x(n) = 1 +. 5 ej^2 πn/^6 +. 5 ej^4 πn/^6 + ej^6 πn/^6 +. 5 ej^8 πn/^6 +. 5 ej^10 πn/^6 DTFS: x(n) = 1 + cos( π 3 n) + cos( 23 π n) + cos(πn) after simplifying above. Average 16 (4^2 + 0^2 + 1^2 + 0^2 + 1^2 + 0^2 ) = 3 in time domain agrees with Power: 12 + ( 12 )^2 + ( 12 )^2 + 1^2 + ( 12 )^2 + ( 12 )^2 = 3 by Parseval’s theorem.
DTFT: y(n) = 12 (x(n) + x(n − 1)) → h(n) = { 12 , 12 } → H(ejω^ ) = 12 + 12 e−jω^. DTFT: H(ejω^ ) = 12 + 12 e−jω^ = (cos ω 2 )e−jω/^2 after simplifying. Then have: x(n) : 1 cos( π 3 n) cos( 23 π n) cos(πn) ω : 0 π/ 3 2 π/ 3 π H(ejω^ ) : 1 0. 8666 − π 6 0. 56 − π 3 0 Gain : 1 0. 866 0. 5 0 Phase : 0 −π/ 6 −π/ 3 N A Then: y(n) = 1 + 0.866 cos( π 3 n − π 6 ) + 0.5 cos( 23 π n − π 3 ) + 0 simplifies to y(n) = {... 2 , 2 ,. 5 ,. 5 ,. 5 ,. 5 , 2 , 2 ,. 5 ,. 5 ,. 5 ,. 5 , 2 , 2.. .} Period still=6. Note: Higher frequencies of x(n) reduced in amplitude (attenuated) in y(n). Thus: This system smoothes the input signal (n) (running 2-point average).
