HW3 - Spring 2006 EE113 HW 3 Solution Due Problem 1 10 d 1 d Z-11 n S = n(1 3 = n(u(n u(n 11)Z n = Z Z dZ 1 Z-1 dZ 1 Z-1 Z =3 n =0 Z =3 n =0 Z

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Unformatted text preview: Spring 2006 EE113 HW 3 Solution ( Due 5/9/06 ) Problem 1: 10 d 1 d Z -11 n S = n(1 3) = n(u (n ) - u (n - 11))Z - n = - Z +Z dZ 1 - Z -1 dZ 1 - Z -1 Z =3 n =0 Z =3 n =0 Z - 11Z -9 + 10Z -10 = 0.7499 = (Z - 1)2 Z =3 Problem 2: a. x(n ) = nu (n - 4 ) d -4 1 Z -4 4 - 3Z -1 , X (Z ) = - Z Z = 2 dZ 1 - Z -1 1 - Z -1 b. x(n ) = n n u (n ) ( ( ) ) R.O.C: Z > 1 n X (Z ) = nZ = , but for any finite Z as n>Z, the sequence become n =0 n =0 Z unbounded. Therefore, the Z-transform does not exist. -1 n ( ) n Problem 3: (4Z )2 + (4Z )3 + ...... = -1 - 4 - n Z -n = -1 4 - n Z -n X (Z ) = log(1 - 4 Z ) = - 4 Z + n = - - n 2 3 n = - n 4 -n By inspection, x(n ) = u (- n - 1) n Problem 4: a. y (n ) + 1 1 y (n - 1) - y (n - 2 ) = 0, y (- 1) = y (- 2 ) = 2 2 4 1 -1 + 1 Y + (Z ) + Z Y (Z ) + 2 Z - Z - 2 Y + (Z ) + 2 Z + 2 Z 2 = 0 2 4 1 -1 1 1 -1 1 -2 + 1 + Z - Z Y (Z ) = -1 + Z + 4 2 2 2 ( ) ( ) Y + (Z ) = - 0.5 + 0.5Z -1 0.31 - 0.81 = + -1 -2 -1 1 + 0.5Z - 0.25Z 1 - 0.31Z 1 + 0.81Z -1 n +1 y (n ) = 0.31n+1 + (- 0.81) u (n ) [ b. 1 1 y (n ) = y (n - 1) + u (n ), y (- 1) = 1 2 4 1 1 Y + (Z ) = Z -1 Y + (Z ) + Z + 2 1 - 0.25Z -1 1 1 - 0.5Z -1 Y + (Z ) = 0.5 + 1 - 0.25Z -1 2.5 -1 + Y + (Z ) = -1 1 - 0.5Z 1 - 0.25Z -1 n n y (n ) = 2.5 (0.5) u (n ) - (0.25) u (n ) n [ ( ) Problem 5: a. R.O.C corresponds to right-sided sequence, so n +1 n -1 Z2 -4 Z 4 Z -1 1 = - x(n ) = X (Z ) = -1 -1 Z - 1 2 1 - 0.5Z 1 - 0.5Z 2 x(0 ) = 1 2 x(1) = - 15 4 x(2) = - 15 8 b. Because R.O.C: 1 < Z < 2 , so 1 u (n + 1) - 4 2 u (n - 1) Z2 1 -1 2 X (Z ) = 2 = = + x(n ) = -u (n ) - 2 n +1 u (- n - 1) -1 -2 -1 -1 Z - 3Z + 2 1 - 3Z + 2 Z 1- Z 1 - 2Z x(0 ) = -1 x(1) = -1 x(2) = -1 Problem 6. (3 4)Z -1 + (1 4)Z -1 Z -1 Z -1 - Z -2 = = 10 10 1 - 3Z -1 1 - (1 3)Z -1 Z 2 - Z + 1 1 - Z -1 + Z - 2 3 3 n -1 n -1 3 n -1 1 1 1 n 1 x(n ) = 3 u (n - 1) + u (n - 1) = 3 + u (n - 1) 4 4 3 4 3 X (Z ) = Problem 7. Z R.O.C: Z > 1 2 , , (Z - 1 2)(Z - 1 3) Z 1 n -1 x(n ) = (Z - 1 2)(Z - 1 3) Z dZ = Residue at Z=1/2 and Z=1/3 2j X (Z ) = 1 1 = 6 u (n ) - 6 u (n ) 2 3 x(0 ) = 0 x(1) = 1 x(2) = 5 6 n n Problem 8. (Sayed 9.3) 1. x(n ) = nu (n - 1) X (Z ) = - Z d dZ Z -1 Z -1 = 1 - Z -1 1 - Z -1 , R.O.C: Z > 1 ( ) 2 2. x(n ) = 1 + n 2 n - 2 u (n - 1) = u (n - 1) + -1 n n n -1u (n - 1) ( ) ( ) X (Z ) = Z -1 d d Z -1 Z -1 Z - 2 + -1 Z -1 - Z = + + -1 - Z 3 dZ dZ 1 - Z -1 1 - Z -1 1 - Z -1 1 - Z -1 ( ) R.O.C: Z > 1 3. x(n ) = n X (Z ) = n = - -n Z -n + n Z -n = n Z n + n Z -n = n =0 n =1 n=0 -1 1 1 - -1 1 - Z 1 - -1 Z -1 R.O.C: < Z < 4. y (n ) - 1 3 1 y (n - 1) + y (n - 2) = x(n - 1) 4 8 3 -1 1 => Y (Z ) - Z Y (Z ) + Z - 2 Y (Z ) = Z -1 X (Z ) 4 8 4 -4 Y (Z ) Z -1 => H (Z ) = = = + 3 -1 1 -2 1 -1 1 X (Z ) 1- Z + Z 1- Z 1 - Z -1 4 8 2 4 For relaxed system, ( i.e. causal ) 1 n -2 1 n-1 h(n ) = - u (n ) , R.O.C: Z > 1 2 2 4 Problem 9. (Sayed 10.1) X (Z ) = (1 3)Z -2 + Z -1 + Z - 1 2 = Z -3 + Z -1 1 1 Z 2 - (1 6 )Z - 1 6 1 - Z -1 1 + Z -1 2 3 n -3 1 1) R.O.C: Z > 1 2 , => x(n ) = 2 1 u (n - 3) + - 3 n -1 u (n - 1) n -1 1 2) R.O.C: Z < 1 3 , => x(n ) = - 2 n -3 1 u (- n + 2 ) - - 3 n -3 u (- n ) n -1 1 3) R.O.C: 1 3 < Z < 1 2 , => x(n ) = - 2 1 u (- n + 2 ) + - 3 u (n - 1) ...
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This note was uploaded on 05/06/2008 for the course EE 113 taught by Professor Walker during the Spring '08 term at UCLA.

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