Z Transform (Part - II)
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1 Z Transform (Part - II). The Z Transform of the following real exponential sequence x(nt) = a n, nt 0 = 0, nt < 0, a > 0 (a) ; z > (c) for all z z (b) ; z (d) ; z < a > a az az Soln. The given sequence can be written as x(n)a n (n) where a > 0 Z Transform of given sequence is X(z) = Option (b) z z a ; z > a = az, z > a [GATE 990: 2 Marks] 2. A linear discrete-time system has the characteristics equation, z z = 0. The system (a) is stable (b) is marginally stable (c) is unstable (d) stability cannot be assessed from the given information [GATE 992: Marks] Soln. Characteristic equation is given z 3 0.8z = 0 z(z 2 0.8) = 0 z(z 0.9)(z + 0.9) = 0 So, poles are z = 0, 0.9 and 0.9 Note that all three poles are inside the unit circle, so the system is stable Option (a)
2 3. The z transform of a signal is given by z ( z 4 ) ( z. Its final value is ) 2 C(z) = 4 (a) /4 (b) zero Soln. Final value theorem for Z Transform is lim x[n] = lim ( N z z ) X(z) = lim. ( z ) z ( z 4 ) z 4 ( z ) 2 = lim z 4 z ( z 4 ) ( z ) = lim.. (z4 ) z 4 z 4 z (z ) z = lim z 4 (z 2 )(z 2 +) z 4 (z ) = lim z 4 z2 (z+)(z )(z2 +) (z ) (c).0 (d) infinity [GATE 999: 2 Marks] = lim z 4 z2 (z + )(z 2 + ) = = Option (a) 4. If the impulse response of a discrete-time system is h[n] = 5 n u[ n ]. Then the system function H(z) is equal to (a) z and the system is stable z 5 and the system is stable (b) z z 5 (c) z z 5 (d) z z 5 and the system is unstable and the system is unstable [GATE 2002: 2 Marks]
3 Soln. Impulse response h(n) = 5 n u[ n ] Above response is having left handed sequence whose z transform has standard form a n u(n)( n ) z ( a z ) ; ROC z < a Thus 5 n u(n)( n ) ( 5 z ) ; ROC z < 5 or, 5 n u(n)( n ) z z 5 ; ROC z < 5 since ROC is z < 5 so it includes unit circle, system is stable Option (b) 5. A causal LTI system is described by the difference equation 2y[n] = αy[n 2] 2x[n] + βx[n ] The system is stable only if (a) α = 2, β < 2 (b) α > 2, β > 2 (c) α < 2, any value of β (d) β < 2, any value of α [GATE 2004: 2 Marks] Soln. Casual LTI system is described by the difference equation 2 y[n] = α y[n 2] 2 x[n] + β x[n ] Taking z transform 2 Y[z] = α Y(z)z 2 2 X(z) + β X(z)z Y(z) = (β z 2) X(z) (2 α z 2 )
4 Or, H(z) = z(β 2 z) (z 2 α 2) It has poles at ± α 2 and zero at 0 and β 2 For stable system poles must lie inside the unit circle of z plane. But 0 can lie any where in plane. Thus β can be of any value Option (c) 6. The z transform X[z] of a sequence x[n] is given by X[z] = 0.5 2z. It is given that the Region of convergence of X[z] includes the unit circle. The value of x[0] is (a) 0.5 (b) 0 Soln. Given X(z) of a sequence x[n] X(z) = z Above transform is for left handed sequence with ROC z < 2 Corresponding sequence is x(n) = (0.5)2 n u( n ) So, x(0) = 0 (c) 0.25 (d) 0.5 [GATE 2007: 2 Marks] If, the value of x(0) is determined by initial value theorem then it will not be correct, since the sequence x(n) is defined for n < 0 Option (b) 7. A system with transfer function H(z) has impulse response h(n) defined as h(2) =, h(3) = - and h(k) = 0 otherwise. Consider the following statements. S : H(z) is low-pass filter
5 Soln. S 2 : H(z) is an FIR filter Which of the following is correct? (a) Only S 2 is true (b) Both S is and S 2 are false (c) Both S is and S 2 are true, and S 2 is a reason for S (d) Both S is and S 2 are true, and S 2 is not a reason for S [GATE 2009: 2 Marks] Given h(2) = h(3) = h(k) = 0 otherwise The diagram of the response is as shown 3 0 k 2 - Note, that it has the finite magnitude values. So it is having finite impulse response. So, FIR But it is not low pass filter So, S is false Option (a) 8. The transfer function of a discrete time LTI system is given by 2 3 H(z) = 4 z 3 4 z + 8 z 2 Consider the following statements: S : The system is stable and casual tor ROC z > 2
6 S 2 : The system is stable but not causal for ROC z < 4 S 3 : The system is neither stable not causal For ROC 4 < z < 2 Soln. Which one of the following statements is valid? (a) Both S and S 2 are true (b) Both S 2 and S 3 true (i) (ii) We know: (c) Both S and S 3 are true (d) S, S 2 and S 3 are all true [GATE 200: 2 Marks] Causal System : A discrete time LTI system is causal if and only of the ROC of its system function is exterior of a circle, including infinity Stable System : A discrete time LTI system is stable if and only if the ROC of its system function includes the unit circle i.e. z = 2 3 H(z) = 4 z 3 4 z + 8 z 2 = ( 4 z ) + ( 2 z ) ( 4 z ) ( 2 z ) H(z) = ( + 4 z ) ( 2 z ) ROC z > 2 h(n) = ( n 4 ) u(n) + ( n 2 ) u(n) The system is causal, since ROC is exterior of the circle. ROC of H(z) includes unit circle so it is stable So, S is true For ROC : z < 4
7 ROC does not include the unit circle. So system is not stable. So, S 2 is not true For ROC : 4 < z < 2 ROC does not include unit circle. So system is not stable. Also ROC is not exterior of z =. So it not 2 causal So, S 3 is true Both S and S 3 are true 9. Two systems H (z) and H 2 (z) are connected in cascade as shown below. The overall output y(n) is the same as the input x(n) with a one unit delay. The transfer function of the second system H 2 (z) is x(n) H (z) = ( 0.4z ) ( 0.6z ) H 2 (z) y(n) (a) ( 0.6 z ) z ( 0.4 z ) (c) z ( 0.4 z ) ( 0.6 z ) (b) z ( 0.6 z ) ( 0.4 z ) (d) ( 0.4 z ) z ( 0.6 z ) Soln. Given y[n] = x[n ] Taking Z Transfer on both sides Y(z) = z X(z) or Y(z) X(z) = z For cascaded system H(z) = H (z). H 2 (z) [GATE 20: 2 Marks]
8 z = ( 0.4 z ) ( 0.6 z ). H z(z) Or, H 2 (z) = z ( 0.6 z ) ( 0.4 z ) Option (b) 0. Let x[n] = ( 9 )n u(n) ( 3 )n u( n ). The Region of Convergence (ROC) of the Z Transform of x[n] (a) is z > 9 (b) is z < 3 (c) is 3 > z > 9 (d) does not exist [GATE 204: 2 Marks] Soln. Given x[n] = ( 9 )n u(n) ( 3 )n u( n ) Let, x (n) = ( 9 )n u(n) and x 2 (n) = ( 3 )n u( n ) x (n) = ( 9 )n u(n) It is right sided sequence. X (z) = ( 9 ) z ROC : z 9 or z > 9 x 2 (n) = ( 3 )n u( n 3) It is left sided sequence X 2 (z) = ( 3 ) z ROC : z 3
9 When both these transform are added then ROC must be between Option (c) 3 > z > 9. The input-output relationship of a causal stable LTI system is given as y[n] = α y[n ] + β x[n] If the impulse response h[n] of this system satisfies the condition h[n] = 2, n=0 (a) α = β 2 (b) α = + β 2 the relationship between α and β is Soln. Given input output relation for causal, stable, LTI system. y[n] = α y[n ] + β x[n] Taking Z Transfer Also, Y(z) = α z Y(z) + β X(z) Y(z)[ α z ] = β X(z) Y(z) X(z) = H(z) = h(n) = β. α n u[n] β h(n) = 2 n=0 β α z i. e. β α n u[n] = 2 n=0 (c) α = 2β (d) α = 2β [GATE 204: 2 Marks] = 2 or β = 2 2α or α = β α 2 Option (a)
10 . The Z Transform of the sequence X[n] is given by X(z) = ( 2z ) 2, with the region of convergence z > 2. Then, X[2] is [GATE 204: 2 Marks] Soln. Given X(z) = Or, ( 2z ) 2 = z > 2 X(z) = ( 2z ) 2 = + 4 z + 2 z 2 + Expanding by binomial expansion Taking inverse z transform We get, x[n] = {, 4, 2, } So, x[2] = 2
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