1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

ANSWERS 613

Section 9.3. Digital Signal F ilters: page 390

la. Substitute x(n] = cos(23 n) and get y[n] = x(n] + x(n - l] + x[n -2] =

cos(23n)- 4 cos(^2 ~") + 4 v'3sin(^2 ~")-~ cos(2~")-4v'3sin(2~") = 0.

Substitute x[n] = sin(^2 :i n) and get y[n] = x [n] + x[n - l ] + x[n - 2] =

sin(^2 :in) - ~J3 cos(^2 ~") - ~ sin(23") + ~J3 cos(2~") - ~ sin(23") = 0.

le. Substitute x(n] = cos(^3 ; n) and get y[n] = x[n] + v'2x(n - l] + x[n -2] =

cos(3; n) + \1'2(-Y,/-cos(^3 ~") + Y,}-sin(^3 ~" )) - sin(3~") = 0.

Substitute x[n] = sin(^3 ; n) and get y(n] = xlnJ + v'2x[n - l] + x[n - 2] =

sin(3; n) + VZ(-1cos(^3 ~") - Y,}-sin(^3 ~")) +cos(^3 ~") = 0.

3a. A(/J) = ll + v'2e-i^8 + e-^2 i^8 I, and
A(0.10) = ll + J2e-O lOi + e-0-^2 0il = 13. 3872150 - i 0.33985511 = 3.4042219,
A(~)= jl + v'2e=fl + e- "il = ll + VZ(-i) + (-l)j = v'2,
A(3;) = ll + v'2e -~·· + e - ~·'I= jl + v'2(--7!) + ii= 0,
A(2.40) = ll + v'2e-^2 ·^4 0i + e-^4 soil = j0.044666 8 + i0.04091541 = 0. 0605739.

5a. A[8J = 1/ 11 - £e-i^8 + ~e-^2 i^8 j, and

A(O) = l/j1-1e-i^0 + ~e-i^0 j = 1/ 11 - ~+~I=~= 1.2857143,

Arn)= 1/ 1 1-~e-~" + ie-~··1 =1; 11 - ~(~ - if)+ i(-~ -if)I =

l/l^4 *fil = *9 = 2. 0647416 ,

A(^2 :i) = 1/11- ~e~ + ie~I = 1/ 1 1-~(-4 - ~)+§(-~+~)I=

l/l^10 +~J3il =^9 fl = 0.6803360,

A(ir) = 1/11-~e-i" + ~e-i^2 " 1 =1/ 11 - ~(-1) +~I= : 9 = 0.4736842.

7a. y[n] - h fn - 1) = !x[n] and bo =! and ao = -~- The transfer function

I

is H(z) = 1 _f,-•.

A(B) =! / 11 - ~e-i^111 , and

A(O) = 1; 11 - ~e-^101 = !/ 11 - ~I = 1,

A({)= i/11-ie-i1</^4 I = i/ 11 - i 1;;1=0.3529047,

A(~)= 1;11 - ~e-;.-12 1 = ifll - i(-i)I = 0.2,

A(ir) = f; 1 1-fe-;"I =!/II-i(-1) 1 =0.1428571.

The higher frequencies are attenuated and A(/J) <! when/}> 1.186.

7 c. y[n]-~y[n- 1 ] = 116 x[n] and bo = ( 6 and%=-~. The transfer function

1$ · H(z ) = l-lAh z-1.

16

A(B) =^16 / ll - ..2..e-i^11 1, and

A(O) -- i/11 16 - 16 rne-^101 = l/11 16 - llj 16 = (^1) ,
A(!!:)= 4 1-/ 16 ll - 16 15 e- i"/^4 1=1-/ 16 ll - !.§I 16 v'2 - i i = 0. (^0840399) '

AG)= \ 6 /ll - J;e-i"/^2 1= 116 /ll-~~(-i)I = 0.0455960,

A(ir) =will - 16 e- i"I = ( 6 /ll - (~(-1) 1 =0.0 32258 1.

The higher frequencies are attenuated and A( /J) < 116 when /} > 1.083.