Solving impulse response y[n] = x[n]− y[n − 2] x[n] = δ[n] z-transform iteration € −2 y[0]= x[0]− y[−2]=1− 0 =1 Y (z) = X(z) − z Y (z) € y[1]= x[1]− y[−1]= 0 − 0 = 0 1+ z−2 Y (z) = X(z) y[2]= x[2]− y[0]= 0 −1= −1 ( ) Y (z) 1 system y[3]= x[3]− y[1]= 0 − 0 = 0 H (z) = = € −2 function € y[4]= x[4]− y[2]= 0 − (−1) =1 X(z) 1+ z € ( ) € M € 1 Y (z) = H (z)X(z) = X(z) € (1+ z−2) € € € Remember: 1 z 2 z 2 − cos(ωˆ )z Y (z) = = X(z) = −2 2 2 1+ z z +1 z − 2 cos(ωˆ )€z + 1 ( ) c ⇓ Inverse z-transform (lookup) x[n] = cos(ωˆ n)u[n] € 2π π 2π y[n]= h[n]= cos nu [n] € ωˆ = = € 2 4 € 4 y[n]= {1,0,−1,0,1,K} € € € € The signal has same z-transform y[n] = x[n]− y[n − 2] as system. The signal is the impulse response of the system ⇓ z-transform 2 € Y (z) 1 z system zeros = roots(z 2) = 0,0 H (z) = = = 2 € 2 2 function poles = roots(z +1) = ± j X(z) (1+ z− ) z +1 n € n Poles: values of z for input x [ n ] = z where output€ y[n] = H(z)z → ∞ Zeros: values of z for input x [ n ] = z n where output y[n] = H(z)z n → 0 € € 2π € signal 1 y[n]= h[n]= cos€ nu [n] € 4 0.8 ⇓ z-transform 0.6 2 1 z 0.4 Y (z) = = 1 z−2 z 2 +1 0.2 € + 2 € ( ) 0 Imaginary part -0.2 TextEnd Poles: z locations of y[n] = z n -0.4 -0.6 Zeros: related to the magnitude and phase of y[n] = z n € -0.8 -1 The closer a zero is to a pole, the smaller € -1 -0.5 0 0.5 1 the effect the pole. Real part € Solve difference equation 2π y[n] = x[n]− y[n − 2] h[n]= cos nu [n] impulse response 4 x[n] = u[n] step input iteration € y[0]= x[0]− y[−2]=1− 0 =1 € € y[1]= x[1]− y[−1]=1− 0 =1 y[2]= x[2]− y[0]=1−1= 0 y[3]= x[3]− y[1]=1−1= 0 € y[4]= x[4]− y[2]=1− 0 =1 € € convolution M € x[n] 1 1 1 1 1 1 1 … € h[n] 1 0 -1 0 1 0 - 1 … € =============================================== 1 1 1 1 1 1 1 0 0 0 0 0 0 -1 -1 -1 -1 -1 … 0 0 0 0 1 1 1 M =============================================== 1 1 0 0 1 … € z-transforms y[n] = x[n]− y[n − 2] x[n] = u[n] ⇓ z ⇓z 1 Y (z) = X(z) − z−2Y (z) X(z) = € € 1− z−1 1 Y (z) = X(z) = H (z)X(z) multiplication y[n] Y(z) € −2 € € (1+ z ) € € 2π 1 1 1 € cos nu [n] ⇔ −2 Y (z) = 4 (1+ z ) −2 −1 1 1+ z 1− z n ( ) ( ) € a€ u[n] ⇔ −1 (1− z ) € ⇓ partial fraction expansion (distinct complex roots) € 1 2 1 2 1 z−1 € n y[n]€ Y (z) = + + € € €(1 − z−1) (1+ z−2) 2 (1+ z−2 ) 0€ 1/2+1/2+0=1€ 1 1/2+0+1/2=1 ⇓ Inverse z-transform (lookup) € €2 1/2-1/2+0=0 1 1 2π 1 2π 3 1/2+0-1/2=0 € y[n] = €+ cos n + cos (n −1) ⇒ 2 2 4 2 4 4 1/2+1/2+0=1 5 1/2+0+1/2=1 € € Partial fraction expansion 1 A B Y(z) = = −1 + −1 We know: 1 2z−1 1 3 z−1 1+ 2z 1− 3 z ( + )( − 4 ) ( ) ( 4 ) right sided sequence 1 ⇔ anu[n] 1− az−1 ⇓ cross multiply z > a € € left sided sequence 3 −1 −1 A(1− 4 z ) + B(1+ 2z ) Y(z) € 1 n = −1 −1 € ⇔ −a u[−n −1] 3 € −1 (1+ 2z )(1− 4 z ) 1− az z < a ⇓ collect terms € € 3 € −1 € (− 4 A + 2B)z + (A + B) Y(z) = −1 3 −1 (1+ 2z )(1− 4 z ) € 1 Y(z) = −1 3 −1 (1+ 2z )(1− 4 z ) A B = + We know: −1 3 −1 (1+ 2z ) (1− 4 z ) right sided sequence € 1 n −1 ⇔ a u[n] 3 −1 1− az (− 4 A + 2B)z + (A + B) = z > a € 1+ 2z−1 1− z−1 ( )( ) left sided sequence 1 n − 3 A + 2B = 0 A + B = 1 match coefficients€ ⇔ −a u[−n −1] 4 1−€az −1 € 8 3 z < a A = B = 11 11 € € € € 8 11 3 11 1 Y(z) = + = € € −1 3 −1 1+ 2z−1 1− 3 z−1 (1+ 2z ) (1− 4 z ) ( )( 4 ) € € Inverse z-transform 1 Y(z) = −1 3 −1 (1+ 2z )(1− 4 z ) 8 11 3 11 = + We know: −1 3 −1 right sided sequence (1+ 2z ) (1− 4 z ) poles:3/4,-2 € 1 ⇔ anu[n] ⇓ Inverse z-transform −1 n 1− az 8 n 3 3 z > a y[n]= (−2) u[n]+ u[n] z > 2 € 11 11 4 € causal, unstable left sided sequence 1 € ⇔ −anu[−n −1] 1−€az −1 ROC € € z < a 1 € 200 RSS RSS € . 100 0 10 0 10 Inverse z-transform 1 Y(z) = −1 3 −1 (1+ 2z )(1− 4 z ) 8 11 3 11 = + We know: −1 3 −1 right sided sequence (1+ 2z ) (1− 4 z ) poles:3/4,-2 € 1 ⇔ anu[n] ⇓ Inverse z-transform −1 n 1− az 8 n 3 3 3 z > a y[n]= − (−2) u[−n −1]− u[−n −1] z < 4 € 11 11 4 € left sided sequence anticausal, unstable 1 € ⇔ −anu[−n −1] 1−€az −1 € € z < a 5 € . 1 € LSS LSS 0 10 0 10 Inverse z-transform 1 Y(z) = −1 3 −1 (1+ 2z )(1− 4 z ) 8 11 3 11 = + We know: −1 3 −1 right sided sequence (1+ 2z ) (1− 4 z ) poles:3/4,-2 € 1 ⇔ anu[n] ⇓ Inverse z-transform −1 n 1− az 8 n 3 3 z > a y[n]= − (−2) u[−n −1]+ u[n] € 11 11 4 3 z 2 € 4 < < left sided sequence two-sided, stable 1 € ⇔ −anu[−n −1] 1−€az −1 € € z < a 1 € 1 . 0 LSS RSS € 1 10 0 10 Inverse z-transform 1 Y(z) = −1 3 −1 (1+ 2z )(1− 4 z ) 8 11 3 11 = + We know: −1 3 −1 right sided sequence (1+ 2z ) (1− 4 z ) poles:3/4,-2 € 1 ⇔ anu[n] ⇓ Inverse z-transform 1− az−1 n z a 8 n 3 3 > € y[n]= (−2) u[n]− u[−n −1] 3 < z ∩ z > 2 11€ 11 4 4 left sided sequence not possible 1 € ⇔ −anu[−n −1] 1−€az −1 z < a € € € € Inverse z-transform 1 Y(z) = −1 3 −1 (1+ 2z )(1− 4 z ) 8 11 3 11 = + We know: −1 3 −1 right sided sequence (1+ 2z ) (1− 4 z ) poles:3/4,-2 € 1 ⇔ anu[n] ⇓ Inverse z-transform −1 n 1− az 8 n 3 3 z a y[n]= (−2) u[n]+ u[n] z > 2 > € 11 11 4 € causal, unstable left sided sequence n 1 n 8 n 3 3 z <€ 3 ⇔ −a u[−n −1] y[n]= − (−2) u[−n −1]− u[−n −1] 4 1−€az −1 11 11€ 4 € anticausal, unstable z < a n 8 n 3 3 3 y[n]= − (−2) u[−n −1]+ €u [n] 4 < z < 2 11 11 4 € € € two-sided, stable n 8 n 3 3 y[n]= (−2) u[n]− u[−€n −1] 3 z z 2 € 11 11 4 4 < ∩ > not possible € € Partial fraction expansion II 4 + 7.6z−1 4 + 7.6z−1 Y(z) = = −6z−2 + 5z−1 + 4 1+ 2z−1 4 − 3z−1 ( )( ) We know: 1+1.9z−1 factor into form right sided sequence = 1 n −1 3 −1 1 p z−1 1 p z−1 ⇔ a u[n] (1+ 2z )(1− 4 z ) ( + 1 )( − 2 ) −1 € 1− az z > a A B = + −1 3 −1 left sided sequence (1+ 2z ) (1− 4 z ) € € € 1 n −1 ⇔ −a u[−n −1] −1 1−€az −1 1+1.9z A = Y(z) 1+ 2z = = 0.036 z < a ( ) z=−2 3 −1 1− 4 z € ( ) z=−2 −1 3 −1 1+1.9z € B = Y(z) 1− 4 z = = 0.964 ( ) z= 3 −1 4 1+ 2z € ( ) z= 3 € 4 0.036 0.964 Y(z) = + −1 3 −1 (1+ 2z ) (1− 4 z ) € € Effects of a zero zeros: 0,-1.9 zeros: 0,0 1+1.9z−1 Y(z) 1 = −1 −1 Y(z) 3 = −1 −1 1+ 2z 1− 4 z poles:3/4,-2 3 ( )( ) (1+ 2z )(1− 4 z ) 0.036 0.964 Y(z) 8 11 3 11 = −1 + −1 = + 3 −1 3 −1 1+ 2z 1− 4 z 1+ 2z 1− z € ( ) ( ) ( ) ( 4 ) € n n 3 n 8 n 3 3 y[n]= −0.036(−2) u[−n −1]+ 0.964( 4 ) u[n] y[n]= − (−2) u[−n −1]+ u[n] 11 11 4 3 € 4 < z < 2 3 € 4 < z < 2 1 € zero at z=-1.9 1 0.5 close€ to pole at z=-2, € . pole’s effect reduced 0 0 € (0.036 vs. 0.727) 0.5 1 10 0 10 10 0 10 two sided sequences Partial fraction expansion III 1+1.9z−1 Y(z) = We know: −1 3 −1 (1+ 2z )(1− 4 z ) right sided sequence 1 z = ⇔ anu[n] 4 + 7.6z−1 1− az−1 z − a = −6z−2 + 5z−1 + 4 z > a € 4z2 + 7.6z left sided sequence = 2 € € 1 z n 4z + 5z − 6 1 = ⇔ −a u[−n −1] € 1− az− €z − a z2 +1.9z = factor into form (z-p1)(z-p2) z < a z + 2 z − 3 ( )( 4 ) Hint: use matlab’s root command € A B € = + 3 € (z + 2) (z − 4 ) € € Partial fraction expansion III 1+1.9z−1 Y(z) = 1+ 2z−1 1− 3 z−1 We know: ( )( 4 ) right sided sequence 2 1 z n 4z2 + 7.6z z +1.9z = ⇔ a u[n] = = 1− az−1 z − a 2 z + 2 z − 3 4z + 5z − 6 ( )( 4 ) z > a € Az Bz = + + C left sided sequence z + 2 z − 3 1 z ( ) ( 4 ) € € = ⇔ −anu[−n −1] € € 1− az−1 €z − a 4z2 + 7.6z C = = 0 z < a 4z2 5z 6 + − z=0 € Y(z)(z − 3 ) z2 +1.9z B = 4 = =€0.964 z z= 3 z(z + 2) 3 € 4 z= 4 € Y(z)(z + 2) z2 +1.9z A = = = 0.036 z z(z 3 ) z=−2 − 4 z 2 € =− € Partial fraction expansion III 1+1.9z−1 Y(z) = 1+ 2z−1 1− 3 z−1 We know: ( )( 4 ) right sided sequence 2 1 z n 4z + 7.6z −1 = ⇔ a u[n] = 2 1− az z − a 4z + 5z − 6 z > a € 0.036z 0.964z = + left sided sequence z + 2 z − 3 1 z ( ) ( 4 ) € € anu[ n 1] € −1 = ⇔ − − − n n 1− az €z − a y[n]= −0.036 −2 u[−n −1]+ 0.964 3 u[n] ( ) ( 4 ) z < a 3 < z < 2 € 4 1 € € € .