3 Z Transform

download 3 Z Transform

of 12

  • date post

    13-Jul-2016
  • Category

    Documents

  • view

    232
  • download

    5

Embed Size (px)

description

Z Transform

Transcript of 3 Z Transform

  • Signals and Systems

    Z-Transform

    ECE - 2006

    1. If the region of convergence of

    x1[n] +x2[n] is 1/3 < |z| e-0.05

    (D) 5z

    ze5, |z| > e-5

    ECE - 2009

    5. The ROC of Z-Transform of the discrete time sequence ,

    x(z)

    z transform Sampler

    (fs = 10Hz)

    200k +

    5V

    10F

    S x[n]

  • Signals and Systems

    x[n] = (1

    3)

    nu[n] (

    1

    2)

    nu[n 1] is

    (A) 1

    3< |z| 1

    2

    (C) |z| 1

    2

    S2: The system is stable but not causal for ROC: |z| 2. . Then, x [2] is_______.

    EE - 2007

    1. The discrete time signal

    x[n] X (Z) = 3n

    2+nn=0 z

    2n,

    Where denotes a transform pair relationship, is orthogonal to the signal

    (A) y1[ n ] y1 ( z ) = (2

    3)

    nn=0 z

    n

    (B) y2[ n ] y2 ( z )

    = (5n n)n=0 z(2n+1)

    (C) y3[ n ] y3 ( z ) = 2|n|

    n= zn

    (D) y4[ n ] y4 ( z ) =2z4 + 3z2 +1

    EE - 2008

    2. Given X (z) = Z

    (za)2 with |Z|> a, the residue of X(z) zn1 at z = a for n 0 will be

    (A) an-1

    (B) an

    (C) nan

    (D) nan-1

    3. H (z) is a transfer function of a real system. (When a signal x[n] = (1 + j)n is the input to

    such a system, the output is zero. Further, the region of convergence (ROC) of (1

  • Signals and Systems

    1

    2z1)H(z) is the entire Z-plane (Except z = 0). It can then be inferred that H (z) can have a

    minimum of

    (A) one pole and one zero

    (B) one pole and two zero

    (C) two poles and one zero

    (D) two poles and two zeros

    EE - 2014

    4. Let X(z) =1

    1z3 be the Z-transform of a causal signal x[n]. Then, the values of x[2] and

    x[3] are

    (A) 0 and 0

    (B) 0 and 1

    (C) 1 and 0

    (D) 1 and 1

    IN - 2008

    1. The region of convergence of the z-transform of the discrete-time signal

    x[n] = 2nu[n] will be

    (A) |Z|>2

    (B) |Z|< 2

    (C) |Z|>1

    2

    (D) |Z|1

    3

    (1

    2)

    n

    (n 1) z

    z 12

    , |z| 1

    9

    (Right sided sequence, ROC in

    exterior of circle of radius 1

    9

    )

    Thus overall ROC in1

    9< |z| a

    Let Y(z) = zn1 X(z) =zn

    (za)2

    Residue of Y(z)= 1

    1!

    d

    dz[(z a)2Y(z)]|

    z=a

    =d

    dz[zn]|

    z=a

    = n zn1|z=a = n an1 =

    1

    anan

    Note that the Question asked is to find x(n) given

    X(z) = z

    (za)2 with |z|>a , for which the answer is known through the following fundamental

    ZT pairs and properties.

    u(n) z

    z 1, |z| > 1

    n u(n) zd

    dz(

    z

    z 1) =

    z

    (z 1)2,

    |z| > 1

    n an u(n)

    za

    (za 1)

    2 , |z

    a| > 1

    =az

    (z a)2 , |z| > |a|

    n an1u(n) z

    (z a)2 , |z| > |a|

    3. [Ans. A]

    (11

    2z1). H(z)has ROC as entire Z-plane which implies that it is polynomial.

    H(z) has one pole

    4. [Ans. B]

    X(z) =1

    1 z3

    1 + z3 + z6

    1 z3 1

    1 z3

    z3

    z3 z6

    z6

  • Signals and Systems

    z6 z9

    .

    x(z) = 1 + z3 + z6+. .

    x(n) = {1, 0, 0, 1, 0, 0, 1, }

    x(2) = 0 and x(3) = 1

    IN

    1. [Ans. A]

    Given,

    x[n] = 2nu(n)

    x(z) = x[n]zn

    n=

    = 2n u[n]zn

    n=

    = 2n. zn

    n=0

    = (2

    z)

    n

    =1

    1 2z1

    n=0

    ROC |2z1| < 1 |z| > 2

    2. [Ans. B]

    y[n] 1

    3y[n 1] = x[n]

    Y(z)

    X(z)= H(z) =

    1

    1 13 z

    1

    For input x[n] = (1

    2)

    n

    u[n]

    X(z) =z

    (z 12)

    , |z| >1

    2

    Y(z) = H(z)X(z) =1

    (1 13 z

    1)

    z

    (z 12)

    = z

    (1 13) (z

    12)

    Y(z)

    z=

    z

    (1 13) (z

    12)

    =2

    (1 13)

    +3

    (z 12)

    Y(z) =2z

    z 13

    +3z

    z 12

    y(n) = [2 (1

    3)

    n

    + 3 (1

    2)

    n

    ] u(n)

    3. [Ans. C]

  • Signals and Systems

    For a system to be stable, it should contain Z=1 circle in its ROC

    For ROC, (1 1

    4Z1) > 0

    Then, Z >1

    4

    4. [Ans. C]

    Only this option gives poles inside unit circle.