Unit - 2

Z-transform

Q1) Find the z-transform of the signal x(n) = sin w0n u(n) and find ROC.

A1)

X(z) = 

X(z) = 

= sin = ej - e-j / 2j

= 1/2j     (ejw0n – e –jwon )z-n 

= 1/2j ejwon z-1) n - e-jw0z-1) n

1/2j[ 1/1-ejw0z-1 – 1/1-e-jw0z-1]

= sin w0 z-1/ 1-2cos(wo)z-1 + z-2   ROC |z| >1

Q2) Find the z-transform of the sequencex(n) = a n-1 u(n-1)

A2)

We know that x(n) = an u(n) is

X(z) = 1/1/az-1

By using time shifting property we have Z{ x(n-k)} = z-k X(z)

Therefore

Z{an-1 u(n-1)} = z-1/ 1-az-1  = 1/z-a.            ROC |z| > |a|

Q3) Find the z-transform of the sequencex(n) = an cos nπ/2

A3)

Z{ cos w0n} = 1- (cos w0) z-1/ 1 –(2 cos w0) z-1 + z-2

Since w0=π/2

Z{ cos π/2 n u(n)} = 1/1 + z-2

Using exponential sequence property

Z{ an x(n) } = X(a-1 z)

Z{ an cos nπ/2} = 1/1+(a-1z)-2 = 1/ 1 + a2/z2  = z2/ a2 + z2  

Q4) Find the z transform of the sequence x(n) = n u(n)

A4)

The z-transform of unit step sequence is given by

Z{u(n)} = z/z-1

Z{ n u(n)} = -z d/dz (z/z-1)

= z/(z-1)2    

Q5) If x(n) = x1(n) * x2(n) where x1(n) = (1/3)n u(n) and x2(n) = (1/5)n u(n). Find X(z) by using convolution property.

A5)

X1(z) = 1/ 1- (1/3)z-1         X2(z) = 1/1- (1/5) z-1

X(z) = 1/1-(1/3)z-1 . 1/1-(1/5) z-1  

Using z-transform find the convolution of two sequences.

x1(n) = {1,2,-1,0,3}        x2(n) = { 1,2,-1}

Z{ x1(n) * x2(n)} = X1(z) . X2(z)

X1(z) = 1 + 2z-1 – z-2 + 3 z-4

X2(z) = 1 + 2z-1 – z-2 

(1 + 2z-1 – z-2 + 3 z-4 )  (1 + 2z-1 – z-2  )

= 1 + 4z-1 + 2z-2 – 4 z- 3 + 4 z-4 + 6 z- 5 – 3 z-6 

Q6) Determine the z-transform of the signal x(n ) = rn (sin w0n ) u(n)

A6)

Z{(sin w0n ) u(n)}  = sin w0 z-1/ 1 -2 (cos w0) z-1 + z-2

Z{an x(n)} = X(a-1 z)

Therefore

Z{ rn sin(w0n) u(n) } = (sin w0) (r-1 z)-1/ 1- 2 (cos w0)(r-1z)-1 + (r-1z)-2  

= r(sinw0) z-1/ 1-2r(cos w0) z-1 + r2 z-2

Q7) Determine the signal x(n) whose z-transform is given by  X(z) = log(1- az-1).

A7)

X(z)= log(1-az-1)

Differentiating both side we get

d/dz X(z) = 1/1-az-1 (a z-2) = az-2/1- az-1

-z d/dz { X(z)} = -az-1/1-az-1

= -az-1[ 1/1-az-1]

= -a Z[ a n-1 u(n-1)]                           -------- (1)

From differentiation property

Z{ n x(n)} = -z d/dz [ X(z)]            ------- (2)

Comparing (1) and (2) we get

n  x(n) = -a [ a n-1 u(n-1)]

Or x(n) = -a [a n-1 u(n-1)]/n

Q8) Determine the z-transform of the signal x(n) =1/2 (n2 + n) (1/3) n-1 u(n-1)

A8)

x(n) =1/2 (n2 + n) (1/3) n-1 u(n-1)

=½ n2 (1/3) n-1 u(n-1) +1/2 n (1/3) n-1 u(n-1)

We know that

Z[(1/3) n u(n)] = z/ z-1/3

Using time-shifting property

Z{(1/3) n-1 u(n-1)] = 1/ z- 1/3

Z [ n (1/3) n-1 u(n-1)] = -z d/dz [1/z-(1/3)]

=-z d/dz( 1/z-1/3)= -z [-1/(z-1/3) 2]= z/ (z-1/3)2

Z [ n2 (1/3) u(n-1)] = -z d/dz [ z/(z-1/3)2]

= z(z+1/3)/(z-1/3)2

= -z [ (z-1/3)2 -2z(z-1/3)/(z-1/3)4

= z(z+1/3)/(z-1/3)3 

X(z) = ½[z(z+1/3)/(z-1/3)3 + z/ (z-1/3)2]

= z2/(z-1/3)3

Q9) Find the z-transform for the sequence x[n] = 2 + 3 + 5 + 2

A9)

X(z) = 2 + 3 z-1 + 5 z -2 + 2 z -3

Q10) Find the z-transform and ROC of the signal x(n) = an u(n)

A10) X(z) = 

= an u(n) -----(1)u(n) = 0 for n<0

1 for n≥0

= an ------- (2)

= n   ------- (3)

This is a geometric series of infinite length that is

a + ar + ar2 + ………….. ∞ = a /1-r if |r| <1

Then from equation (3) it converges when |az-1| < 1 or |z| >|a|

Therefore

X(z) = 1/ 1-az-1:  ROC    |z| > |a|

Fig. ROC |z|>|a|

Q11) Find the z-transform of the signal x(n) =-b n u(-n-1). Find ROC

A11)

X(z) = 

X(z) =  bn u(-n-1)u(-n-1) =0 for n ≥0

= 1 for n ≤ -1

= bn   = b-1  =

= b-1z/1- b-1z = z/ z-b = 1/ 1-bz-1                                                               |z| < |b|

Fig. ROC |z|<|b|

Q12) Find the z-transform of x(n) = an u(n) – bn u(-n-1)

A12)

X(z) = 

=  n + b-1

= z/z-a + z/z-bROC |a| < |z| < |b|

Fig. ROC |a| < |z| < |b|

|b|< |a|

Fig. ROC |b|< |a|

|b| >|a|

Q13) Find the partial fraction method of

A13)

By solving A= 1 and B=-1

Therefore,

X(z) =

x(n) = (1/2) n u(n) – (1/4) n u(n)