Chapter 3

Z Transform

CHAPTER HIGHLIGHTS

☞ Sampling ☞ Analysis of Discrete-time LTI Systems

☞ Sampling Theorem ☞ Characterization of Discrete Time LTI Systems
☞ The z-transform ☞ The Unilateral z-transform
☞ Region of Convergence ☞ Relationship between z and Laplace Transform
☞ z-transform Properties ☞ Structures for Realization of IIR Systems
☞ Inverse z-transform

Sampling To prevent high-frequency noise components having fre-

quency higher than ws /2 an analogue filter is used before
Let x(t) be any continuous signal periodically sampled at
sampler to attenuate signals with frequency higher than
equal sample time period Ts to get discrete-time signal rep-
ws/2. Such filters are called antialiasing filters.
resented by x(n)

x(t) Sampling x(n) Z Transform

A discrete-time LTI system with impulse response h(n), the
x(n) = x(nTs) output y(n) of the system to the complex exponential input
of the form zn is y(n) = T{zn}
n = 0, ±1, ±2, ±3 ∞
= h(n) × zn
Ts – sampling time period ∞
= ∑ h( k ) z n - k
Sampling Theorem k = -∞
∞
Let x(t) is strictly band-limited signal with Fourier spectrum.
X(jw) = 0 for |w| > wm
= zn ∑ h( k ) z - k ,
k = -∞
Then, x(t) can be recovered without loss of information
= H(z) zn
from its sample signal x(n) = x(nTs), if the sampling fre-
∞
quency ws is greater than twice of the maximum frequency
wm (of x(t))
where H(z) = ∑ h(n) z - n
-∞
ws > 2wm The variable z is generally complex valued.
Minimum sampling frequency ws = 2wm is called When |z| is not restricted to unity, the summation is
Nyquist sampling rate. referred to as the z transform of h(n).
In case of sampling frequency ws < 2wm
Xp(jw) which is spectrum of xp(t) = x(nTs) no longer Definition
contains all information of X(jw); hence, x(t) is not fully For general discrete-time signal x(n), the z transfer function
recoverable from sampler xp(t), and there is definite loss of X(z) is defined as follows:
∞
original signal x(t).
The loss of information in such sampling case (ws < 2wm) X(z) = ∑ x( n) z - n
n = -∞
is due to super imposition of high-frequency components
onto the less-frequency component. This phenomenon is The variable is generally complex valued and is expressed
called frequency folding or aliasing. in polar form as z = rejW.
Due to aliasing, high-frequency noise components get Where ‘r’ is the magnitude of z and ‘W’ is the angle of ‘z’
mixed with signal frequencies between 0 to ws/2. x(n) and X(z) are said to form a z transform pair as
x(n) ↔ X(z)
3.192  |  Part III  •  Unit 2  •  Signals and Systems

Region of Convergence The pole zero plots are

The range of values of the complex variable z for which the Im{z}
z transform converges is called the region of convergence
Consider the sequence x(n) = anu(n)
∞
Z transform of x(n), X(z) = ∑ a n u( n) z-n
a 1
Re{z}
Im{z}
n = -∞

∞
= ∑ (az -1 )n
n= 0

For the convergence, this summation should be finite. 0<a<1 Re{z}

This ROC is the range of values of z for which |az−1| < 1, 1
a
or |z| > |a| then
∞
1 z
X(z) = ∑ (az -1 )n = 1 - az -1 , |z| > |a| = z - a , |z| > |a|. a>1
n=0

There is one zero, at z = 0, and one pole at z = a, ROC and Properties of ROC
the pole zero plot for the examples are shown in figure. 1. The ROC of X(z) consists of a ring in the z plane
centred about the origin. In some cases, the inner
Re(z) Re(z)
boundary can extend inward to the origin, in which
case the ROC becomes a disc. In other case, the outer
boundary can extend outward to infinity.
x Im(z) x Im(z)
a 1 1a a1 Im{z}
z-plane

0<a<1 a >1

Re(z) Re{z}
Re(z)

x Im(z) x Im(z)
1 a a− 1

2. ROC does not contain any poles; At poles, X(z) is

infinite, and therefore, by definition, it does not converge
−1 < a < 0 a < −1
3. If x(n) is a finite sequence, then X(z) converges for
Figure 1  ROC of the form |z| > |a|. every value of z, then ROC is the entire z plane except
possibly z = 0 and/ or z = ∞
Now let x(n) = −an u(−n -1), then z transform X(z) 4. If x(n) is a right-handed sequence, (x(n) = 0, for n < N1
+∞ -1 < ∞), then ROC is of the form |z| > rmax or ∞ > |z| > rmax
=- ∑ a n u ( - n - 1) z - n = - ∑ ( az -1 ) n where rmax is the largest magnitude of any of the poles
n = -∞ n = -∞ of X(z). Thus, ROC is the exterior of the circle |z| = rmax
∞ 5. If x(n) is a left-handed sequence (x(n) = 0, for n >
= - ∑ ( a -1 z ) n . By taking n = −n, N2 > -∞), then ROC is of the form |z| < rmin or 0 < |z| <
n =1 rmin where rmin is the smallest magnitude of any of the
∞ poles of X(z), and thus, ROC is interior of the circle |z|
= 1 - ∑ ( a -1 z ) n = rmin
n= 0
6. If x[n] is two-sided sequence, (x(n) is infinite duration
If|a-1 z| < 1, then the above summation converges, i.e., for sequence). Then, the ROC is of the form r1 < |z| < r2
|z| < |a| where r1, r2 are the magnitudes of the two poles of
X(z). It forms like a ring.
1 - a -1 z z
X(z) = 1 - = = , |z| < |a| 7. Relationship between the ROC and the time extent of
1 - a z 1 - a -1 z z - a
-1
a signal
 Chapter 3  •  Z Transform  |  3.193

x(n) Im{z} -u(–n - 1) |z| < 1

1
1 − z −1
d(n - m) z-m All z except 0
(m > 0) or
Re {z}
r+ ∞(m < 0)
n anu(n) 1 |z| > |a|
1 − az −1
-anu(-n -1) 1 |z| < |a|
A right-sided signal has an ROC of form |z| > r+ 1 − az −1

x(n) nanu(n) az −1 |z| > |a|

Im{z}
(1 − az −1 )2
-nanu(-n -1) az −1 |z| < |a|
(1 − az −1 )2
Re{z}
n r_ (n + 1)anu(n) 1 |z| > |a|
(1 − az −1 )2
(cos W0n) u(n) z 2 − (cos Ω0 ) z |z| > 1
z2 − (2cos Ω0 ) z + 1
A left-sided signal has an ROC of form |z| < r-
(sin W0n) u(n) (sin Ω0 ) z |z| > 1
x(n) Im{z} z 2 − (2cos Ω0 ) z + 1

(rncos W0n) u(n) z 2 − ( r cos Ω0 ) z |z| > r

z 2 − (2r cos Ω0 ) z + r 2
r+ Re{z}
r− (rnsin W0n)u(n) ( r sin Ω0 ) z |z| > r
n
z 2 − (2r cos Ω0 ) z + r 2

an 0 < n < N −1 1 − aN z − N |z| > 0

0 otherwise 1 − az −1
A two-sided signal has an ROC of form r+ < |z| < r-

Z Transforms of the Some Solved Examples

Common Sequences
For unit impulse signal d [n], its z transform is given by d (n) Example 1
∞ What are the ROC and z transforms of the signal x(n) = 5
←⎯
→
z
∑ d (n) z - n =1
⎛ 1⎞
n
⎛ 1⎞
n
n = -∞
ROC is entire z plane, including z = 0, and z = ∞ ⎜⎝ ⎟⎠ u(n) - 4 ⎜⎝ ⎟⎠ u(n)?
4 3
For delayed unit impulse
∞ Solution
d (n - n0) ←⎯→
z
∑ d ( n - n0 ) z - n = z - n0
∞ ⎧⎪ ⎛ 1 ⎞ n n ⎫⎪
n = -∞ ⎛ 1⎞
ROC is entire z plane, expect z = 0
X(z) = ∑ ⎜⎝ 4 ⎟⎠
⎨5 u ( n) - 4 ⎜⎝ ⎟⎠ u( n)⎬ z
3
-n

n = -∞ ⎩⎪ ⎭⎪
For advanced unit impulse
∞ n ∞ n
⎛ 1⎞ ⎛ 1⎞
∞ = 5 ∑ ⎜ ⎟ z -n - 4 ∑ ⎜ ⎟ z -n
⎝ ⎠ ⎝ ⎠
d (n + n0) ←⎯→
z
∑ d (n + n0 ] z - n = z n0 n= 0 4 n= 0 3
n = -∞
∞ n
∞ n
ROC is entire z plane, except z = ∞ ⎛1 ⎞ ⎛1 ⎞
= 5 ∑ ⎜ z -1 ⎟ - 4 ∑ ⎜ z -1 ⎟ (1)
n= 0
⎝ 4 ⎠ ⎝
n= 0 3
⎠
x(n) X(z) ROC
d(n) 1 All z 5
1 - z -1 + z -1
5 4 3
u(n) 1 |z| > 1 = - =
1 − z −1
1 -1 1 -1 ⎛ 1 -1 ⎞ ⎛ 1 -1 ⎞
1- z 1- z ⎜⎝1 - z ⎟⎠ ⎜⎝1 - z ⎟⎠
4 3 4 3
(Continued)
3.194  |  Part III  •  Unit 2  •  Signals and Systems

2 ⎛ 2⎞ Im{z}
1 - z -1 z⎜z- ⎟
3 ⎝ 3⎠
= =
⎛ 1 -1 ⎞ ⎛ 1 -1 ⎞ ⎛ 1 ⎞ ⎛ 1⎞
⎜⎝1 - z ⎟⎠ ⎜⎝1 - z ⎟⎠ ⎜⎝ z - ⎟⎠ ⎜⎝ z - ⎟⎠
4 2 4 3
Re{z}
For the convergence of X(z), both sums in the equation (1) b
1
1 1 b
should converge, which requires z -1 < 1, and z -1 < 1
4 3
1 1 Roc for o<b<1
or equivalently |z| > and |z| >
4 3
1
By combining these two regions, the ROC will be |z| >
3 Z Transform Properties
1. Linearity
Example 2 x1(n) ↔ X1(z), ROC = R1
What is the z transform and ROC of the signal x2(n) ↔ X2(z), ROC = R2
⎧a n 0 ≤ n ≤ N - 1, a > 0 a1x1(n) + a2x2(n) ↔ a1X1(z) + a2X2(z)
x(n) = ⎨ R1 ⊃ R1 ∩ R2
⎩0 otherwise 2. Time shifting
Solution x1(n) ↔ X(z), ROC = R
x(n - n0) ↔ z–n0 X(z), R1 = R ∩ {0 < |z| < ∞}
N -1 N -1
X(z) = ∑ an z - n = ∑ (az -1 )n 3. Multiplication by z0n
n=0 n= 0
⎛ z⎞
z0nx(n) ↔ X ⎜ ⎟ , R1 = |z0|R
1- ( az -1 ) N
1 z N - aN ⎝ z0 ⎠
= -
= N -1 .
1 - ( az )
1 z z-a anx(n) ↔ X(a-1 z), R1 = |a| R
N -1
1 ( z - a) ( a z N -2
+ K1 + .... + a N -1 )
= e + jW0 n x( n) ↔ X (e - jW0 n ), R1 = R
z N -1 z-a
1 4. Time reversal
= ( z N -1 + K1 z N - 2 + ..... + a N -1 )
z N -1 ⎛ 1⎞ 1
x(-n) ↔ X ⎜ ⎟ , R1 =
⎝ z⎠ R
There is a pole of order N - 1 at z = 0, and ROC is all values
of z except z = 0 5. Multiplication by ‘n’
dX ( z ) 1
nx(n) ↔ -z ,R =R
Example 3 dz
What is the z transform of x(n) = b|n| b > 0 6. Accumulation
n
Solution 1
∑ x( k ) ↔
1 - z -1
X ( z ) , R1 ⊃ R ∩ {|z| > 1}
x(n) = bnu(n) + b-n u(-n - 1) k = -∞
7. Convolution
1
z
bn u(n) ←⎯
→ , |z| > b, x1(n) × x2(n) ↔ X1(z) X2 (z), R1 ⊃ R1 ∩ R2
1 - bz -1
n 8. Conjugation
⎛ 1⎞ z -1 1
⎜⎝ ⎟⎠ u( - n - 1) ←⎯ → - -
, |z| < x*(n) ↔ X*(z*) with ROC = R
b 1- b z 1 1 b
9. First difference
When b > 1, there is no common ROC for the above functions, x(n) - x(n - 1) ↔ (1 - z-1)X(z),
But for 0 < b < 1, the ROCs of above two equations will R1 ∩ R ⊃ {|z| > 0}
overlap, and thus, the z transform for composite sequence is 10. Initial value theorem
if x(n) = 0, for n < 0, then x(0) = Lt X ( z )
1 1 1 z→∞
X(z) = -
- - -
, b < |z| < 11. Final value theorem
1- b z 1 1- b z1 1 b
1 Lt x( n) = Lt ( z - 1) X ( z ), Poles of (z-1) X(z) must
z z n→∞ z →1
= - , b < |z| <
z - b z - b -1 b lie inside the unit circle.
12. Time expansion
b2 -1 z 1 n
= , b < |z| < xk(n) = x(n/k) ↔ X(zk), R1 = R1/k, for n, k, and are
-
b ( z - b)( z - b )
1 b integers k
 Chapter 3  •  Z Transform  |  3.195

Inverse Z Transform For the above two functions, ROC can be combined
1
Inversion of the z transform to find the sequence x(n) from as |z| >
its z transform X(z) is called inverse z transform. 3
n n
1 ⎛ 1⎞ ⎛ 1⎞
So, x(n) = 2 ⎜ ⎟ u( n) + ⎜ ⎟ u( n)
x(n) = z-1 {X(z)}. x(n) = ∫ X ( z ) z n-1dz
2p j  ⎝ 3⎠ ⎝ 5⎠
C
(ii) When ROC is
Where C is a counter clockwise contour of integration 1 1
enclosing the origin <| z |<
5 3
Example 4 1 1
ROC is |z| < and |z| >
Consider the z transform X(z) 3 5
So the functions
11
3 - z -1 n
15 ⎛ 1⎞ z 1 1
= ⎜⎝ ⎟⎠ u( n) ←⎯ →
1 -1
| z |>
⎛ 1 -1 ⎞ ⎛ 1 -1 ⎞ 5
1- z
5
⎜⎝1 - z ⎟⎠ ⎜⎝1 - z ⎟⎠ 5
3 5
n
Find the inverse z transform for different ROCs ⎛ 1⎞ z 1 1
- ⎜ ⎟ u[ - n - 1] ←⎯ → | z |<
1 1 1 1 ⎝ 3⎠ 1 -1 3
(i) |z| > (ii)  < | z | <   (iii) |z| < 1- z
3 5 3 5 3
By combining above two functions
Solution Re{z }

11 -1
3- z
X(z) = 15
⎛ 1 -1 ⎞ ⎛ 1 -1 ⎞
⎜⎝1 - z ⎟⎠ ⎜⎝1 - z ⎟⎠ Im{z}
3 5
1 1
⎛ 11⎞ ⎡ ⎤ 5 3
z -1 ⎜ 3 z - ⎟
⎝ 15 ⎠ 1 ⎢ 2 1 ⎥
X(z) = = ⎢ + ⎥
⎛ 1⎞ ⎛ 1 ⎞ z -1 ⎢ 1 1
z -2 . ⎜ z - ⎟ ⎜ z - ⎟ z- z- ⎥
⎝ 3⎠ ⎝ 5⎠ ⎣ 3 5⎦ n n
⎛ 1⎞ ⎛ 1⎞
2 1 x(n) = ⎜ ⎟ u( n) - 2 ⎜ ⎟ u( - n - 1)
X(z) = + ⎝ 5⎠ ⎝ 3⎠
1 -1 1 -1 1
1- z 1- z (iii) When ROC is |z| < ; in this case, the ROC is inside
3 5 5
the both poles
(i) When ROC is |z| > 1/3 n
There are two poles z = 1/3 and 1/5, and ROC lies out- ⎛ 1⎞ z 1 1
- ⎜ ⎟ u( - n - 1) ←⎯ → | z |<
side outermost pole, so the inverse z transform is a ⎝ 5⎠ 1 -1 5
n
1- z
⎛ 1⎞ z 1 1 5
right-sided sequence ⎜ ⎟ u( n) ←⎯ → ,| z | > n
⎝ 5⎠ 1 5 ⎛ 1⎞ 1 1
1 - z -1 z
- ⎜ ⎟ u( - n - 1) ←⎯ → | z |<
5 ⎝ 3⎠ 1 -1 3
n 1- z
⎛ 1⎞ z 1 1 3
⎜⎝ ⎟⎠ u( n) ←⎯ →
1 -1
,| z | > By combining the above two equations
3 3
1- z
3 Re{z}

x x Im{z}
x x Im{z} 1 1
1 1 5 3
5 3

n n
⎛ 1⎞ ⎛ 1⎞
x(n) = -2 ⎜ ⎟ u( - n - 1) - ⎜ ⎟ u( - n - 1)
⎝ 3⎠ ⎝ 5⎠
3.196  |  Part III  •  Unit 2  •  Signals and Systems

Example 5 Example 7
What is the inverse z transform of X(z) = log(z + a) - log z -1
Inverse z transform for X(z) = , |z| > |a|
z, |z| > |a| (1 - az -1 ) 2
Solution Solution
X(z) = log (z + a) - log z, |z| > |a| Z 1
anu(n) ←⎯ → |z| > |a|
1 - az -1
⎛ z + a⎞
= log ⎜ , |z| > |a|
⎝ z ⎟⎠ Consider x(n) = an u(n)

X(z) = log (1 + az-1), |z| > |a| Z d

nx(n) ←⎯→ -z { X ( z )}
dz
By using multiplication by n property,
d ⎧ 1 ⎫ - az -2 + az -1
d X ( z) −z ⎨ ⎬ = -z =
n x(n) ←⎯
z
→ -z (1) dz ⎩1 - az -1 ⎭ (1 - az -1 ) 2 (1 - az -1 ) 2
dz
dX ( z ) a n z z -1
= .( - z -2 ) x( n) ←⎯ →
dz 1 + az -1 a (1 - az -1 ) 2

dX ( z ) az -1 z -1 n
-z  = |z| > |a|  (2) So, the inverse z transform of is x( n) =
-
dz 1 + az -1 n n (1 - az ) 1 2 a
. a u( n) = n.an-1u(n)
1
z a
(-a)n u(n) ←⎯
→ , |z| > |a|
1 + az -1
a
Analysis of Discrete-Time Lti
z
a(-a)n u(n) ←⎯ → , |z| > |a| Systems
1 + az -1
By using time-shifting property Function of a Discrete-Time LTI Systems
z a The output y(n) of a discrete-time LTI system equals the
→ z-1
a(-a)n–1 u[n - 1] ←⎯ ,
1 + az -1 convolution of the input x(n) with the impulse response h(n)
that is x(n) ∗ h(n)
|z| > |a|(3)
By applying z transform Y(z) = X(z) H(z)
So, from equations (1), (2), and (3) Y ( z)
H(z) =
nx(n) = a(-a)n-1 u(n - 1) X ( z)
The transform H(z) of h(n) is referred to as the system func-
-( - a) n u( n - 1)
x(n) = tion (or transfer function) of the system
n
h[n] x[n] ∗ h[n]
Example 6 x[n] y[n]

What is the inverse z transform of X(z) = 5 z4 + 3 z2 + 2 +

z-1 + 4 z-3 0 < |z| < ∞ X[z] Y[z]
H[z] . X[z]
H[z]
Solution
z
d(n) ←⎯→ 1 Figure 2  Impulse response and system function.
for any x(n)
Characterization of Discrete-Time LTI
x(n - n0) ↔ z-n0 X(z) Systems
So, d (n + 4) ↔ z4 (i) Causality
d (n + 2) ↔ z2 For causal discrete-time LTI system, we have h(n) =
0, n < 0.
d (n - 1) ↔ z-1 Since h(n) is right-sided signal, the corresponding
d (n - 3) ↔ z-3 requirement on H(z) is that the ROC of H(z) must
be of the form |z| > rmax, exterior of circle, including
X(z) = 5z4 + 3z2 + 2 + z-1 + 4z-3 infinity.
Then, x(n) = 5 d (n + 4) + 3d (n + 2) + 2 d (n) +d (n - 1) + 4 Similarly, if system is anticausal, (h(n) = 0, n ≥ 0), the
d (n - 3) ROC of H(z) must be of the form |z| < rmin
 Chapter 3  •  Z Transform  |  3.197

(ii) Stability M
A discrete-time LTI system is BIBO stable if and
Y ( z)
∑ bk z - k
∞ k =0
H(z) = =
only if ∑ | h( n) | < ∞ corresponding requirement X ( z) N
n= -∞ ∑ ak z - k
on H(z) is that ROC of H(z) contains the unit circle k =0

(i.e., |z| = 1) So, H(z)) is always rational

The difference equation by itself does not provide infor-
mation about which ROC to associate with the algebraic
Causal and Stable Systems
expression H(z). An additional constraint, such as the cau-
If the system is both causal and stable, then all of the poles sality or stability of the system, however, serves to specify
of H(z) must lie inside the unit circle of the z-plane because the region of convergence. For example, if the system is
the ROC is of the form |z| > rmax and since unit circle is causal, the ROC will be outside the outermost pole. If the
included in ROC, we must have rmax < 1 system is stable, the ROC must include the unit circle.
(iii) Invertibility
If H(z) is the transfer function of a system S, the Si, its Systems Interconnections
1 For two LTI system (h1(n), h2(n)) in cascade, the overall
inverse system has a transfer function Hi(z) = .
H ( z) impulse response h(n) is given by h(n) = h1(n) × h2(n)
The inverse system Si undoes the operation of S, ­corresponding system function H(z) = H1(z).H2(z)
Hence if H(z) is placed in cascade with Hi(z), the x[n]
transfer function of the composite system is unity. h2[n]
y[n] x[n] y[n]
h1[n] = h1[n] ∗ h2[n]
h[n] hi[n]
y[n] H1[z] H2[z] H1[z]H2[z]
x[n ] Hi [z] x[n]
H [z]
The impulse response of a parallel combination of two LTI
S Si systems is given by

H(z). H1(z) = 1, that is, h(n) × hi(n) = d(n). The zeros h(n) = h1(n) + h2(n)
of H(z) are the poles of Hi(z), and the poles of H(z) and H(z) = H1(z) + H2(z)
1
are the zeros of Hi(z) as Hi(z) = . If H(z) is x[n]
H ( z) y[n]
h1[n] x[n] y[n]
both stable and causal, then all poles of H(z) will h1[n] + h2[n]
lie inside the unit circle, If Hi(z) is to be both stable
and causal, then all poles of Hi(z) have to lie inside H1[z] H1[z] + H2[z]
the unit circle, the poles of Hi(z) = zeros of H(z). h2[n]
So, for the systems H(z) and Hi(z) to be stable and
causal, all the poles and zero of H(z) must lie inside H2[z]
the unit circle. A system with all poles and zeros
When two systems are connected in feedback interconnec-
inside the unit circle is termed as minimum phase
system. tion, as shown in figure, then the overall system function for
the feedback system
System Function of LTI System Described Y ( z) H1 ( z )
by Linear Constant Coefficient Difference H(z) = =
X ( z ) 1 + H1 ( z ) H 2 ( z )
Equations
x[n]+
A distinct-time LTI system for which input x(n) and output + h1[n] y [n] x[n] H1( z)
y(n) satisfy the general linear constant coefficient difference − 1+ H1( z) H2 ( z) y [n]
equation is of the form H1[z]

N M
∑ ak y(n - k ) = ∑ bk x(n - k ) h2[n]
k =0 k =0
H2[z]
By applying z transform using time shift property
Example 8
N M
∑ ak z -k
Y ( z) = ∑ bk z -k X ( z) Consider the an LTI system for which the input x(n) and
k =0 k =0 output y(n) satisfy the linear constant coefficient difference
equations
3.198  |  Part III  •  Unit 2  •  Signals and Systems

1 1 ROC of XI(z) is always outside a circle in the z plane.

y(n) - y(n - 1) = x(n) + x(n - 1), Time-shifting property of unilateral z transform if x(n) ↔
3 5
XI(z), then for m ≥ 0
then what is the impulse response of the system when
(i) system is stable, (ii) system is anticausal? x(n - m) ↔ z-m XI(z) + z-m+1 x[-1]
+ z-m+2 x[-2] + ...... + x[-m]
Solution
x(n + m) ↔ zmXI(z) - zmx[0] - zm-1x[1] ....... -z x[m - 1]
Given difference equations is
1 1 The unilateral z transform, that is particularly useful in
y(n) - y(n - 1) = x(n) + x(n - 1) analysing causal systems specified by linear constant coef-
3 5
ficient difference equations with non-zero initial conditions
By considering z transform
( unilateral z )
1 1 Time delay x(n - 1) ←⎯⎯⎯⎯→ z-1 X(z) + x(-1)
Y(z) - z -1 Y ( z ) = X ( z ) + z -1 X ( z )
3 5 ( unilateral z )
Time delay x(n + 1) ←⎯⎯⎯⎯→ z X(z) - z x[0]
⎛ 1 ⎞ ⎛ 1 ⎞
Y(z) ⎜1 - z -1 ⎟ = X(z) ⎜1 + z -1 ⎟
⎝ 3 ⎠ ⎝ 5 ⎠ Relationship Between Z and
⎛ 1 -1 ⎞
1+ z Laplace Transform
Y ( z ) ⎜⎝ 5 ⎟⎠ 1
H(z) = = For a general signal, x(t) is sampled at sampling rate to
X ( z ) ⎛ 1 -1 ⎞ get discrete value x(kT) which has z transform T
⎜⎝1 - z ⎟⎠
3 ∞

For ROC, we need to consider the system conditions X(z) = ∑ x( kT ) z - k

k = -∞
(i) When the system is stable, it includes unit circle
1 The same general signal x(t) can be considered as the impulse
|z| > (and the system is causal too) 1
3 sampled at the same rate and may be represented as
T
1 1 z -1 1
H(z) = + |z| > ∞
1
1 - z -1
5 1 -1
1- z
3 x(t) = ∑ x( kT )d (t - kT )
3 3 k = -∞
By considering inverse z transform Laplace transform of above equation is
n n -1 ∞
⎛ 1⎞ 1 ⎛ 1⎞
h(n) = ⎜ ⎟ u(n) + ⎜ ⎟
⎝ 3⎠ 5 ⎝ 3⎠
u( n - 1) is the impulse X(s) = ∑ x( kT )e - ksT
k = -∞
response, when the system is causal (or) stable if esT = z, we can write as
1 ∞
(ii) When the system is anticausal, |z| <
3 X(s) = ∑ x( k ) z - k = X(z)
k = -∞
1 1 z -1 1
H(z) = + , |z| < Thus, z = esT
1 5 1 -1 3 ln z = ST.
1 - z -1 1- z
3 3 1
s= ln z
by considering inverse z transform T
n n -1 X(s) = X ( z ) z = eTs
⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞
h(n) = - ⎜ ⎟ u( - n - 1) - ⎜ ⎟ ⎜ ⎟ u[-(n-1)-1]
⎝ 3⎠ ⎝ 5 ⎠ ⎝ 3⎠
n n -1 Structures for the Realization
⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞
h(n) = - ⎜ ⎟ u( - n - 1) - ⎜ ⎟ ⎜ ⎟
⎝ 3⎠ ⎝ 5 ⎠ ⎝ 3⎠
u(-n) of IIR Systems
In time domain, the representation Nth order IIR system is
is the impulse response, when the system is anti causal
N M
   y(n) = – ∑ a k y (n - k ) + ∑ b k x (n - k ) (i)
The Unilateral Z Transform k =1 k =0

The unilateral (or one-sided) z transform XI(z) of a sequence and the z-domain representation of an Nth order IIR system is,
∞ -1 -2 -M
Y ( z ) b 0 + b 1 z + b 2 z + ............. + b M z
x(n) is defined as XI(z) = ∑ x(n) z - n and differs from the H(z) =
X ( z)
=
1 + a 1 z -1 + .........+ a N z - N
n= 0
bilateral transform in that the summation is carried over only
In the above representation, the value of M gives the number
n ≥ 0.
of zeros and N gives the number of poles of the system.
 Chapter 3  •  Z Transform  |  3.199

Types of structures for realizing the IIR systems are as On cross multiplying equation (2) and (3), we get
follows:   W(z) = X(z) – a1z–1W(z) – a2z–2W(z) …– aNz–NW(z)(4)
Direct form – I Structure and
Consider the difference equation governing an IIR system.   Y(z) = b0W(z) + b1z–1W(z) + ….. + bMz–MW(z)(5)
Consider the equation (1) The equations (4) and (5) represent the IIR system in
y(n) + a1y(n − 1) + a2y(n − 2) + …….. aNy(n − N) z-domain and it can be realized by a direct structure called
= b0x(n) + b1x(n − 1) + b2x(n − 2) + …. bMx(n − M) direct form-II.
In direct form-II, the number of delays is equal to order
on taking z transform both sides
of the system.
we get,
X(z) Y(z) X(z) Y(z)
Y(z) = –a1z–1Y(z) – a2z–2Y(z) – ……… – aNz–NY(z) + b0X(z) H H1 H2
+ b1z–1X(z) + ………. + bMz–MX(z)
Direct form -I
X(z) Y(z)
b0 + +
X(z) Y(z)
z−1 z−1 H2 H1
b1 – + + −a1
Direct form -II
z−1 z−1

− a2
Figure 4  Conversion of Direct form-I to Direct form-II.
b2 + +

X(z) Y(z)
W(z)
bN−1 + − aN−1 + b0 +
+

z−1 z−1 −1
bN −aN
z
−1
− a1 z W(z) −1
−1
− b1 z W(z)
z W(z)
+ − a1 b1
H2 +
H1

−1
Figure 3  Direct form-I structure of IIR system. z
−2 −2
− a2 z W(z) −2 − b2 z W(z)
This provides a direct relation between time-domain and + − a2
z W(z)
b2 +
z-domain equations.
⇒ This required more memory elements (z–1).

Direct form-II structure of IIR system −1

z
This structure can be realized which uses less number of delay − aN−1 z W(z)
−1
−(N −1)W(z)
−1
− bN−1 z W(z)
z
elements or memory elements than the direct form-I. Consider + − aN−1 − bN−1 +
the general difference equation governing an IIR system.
N M −1

∑ a k y (n - k ) + ∑ b k x (n - k )
z
Y(n) = – − aN z
−N
W(z) −N
−N
− bN z W(z)
k =1 k =0 z W(z)
− aN bN
on taking z transform of the above equation.
We can obtained Figure 5  Direct form-II structure of IIR system.
Y(z) [1 + a1z–1 + a2z–2 + ……… + aNz–N] = X(z) [b0 + b1z–1
+ b2z–2 ………. + bMz–M]
Solved Examples
-1 -2 -M
Y ( z ) b 0 + b 1 z + b 2 z + ............. + b M z
= Example 9
X ( z) 1 + a 1 z -1 + .........+ a N z - N
What is the z transform of the signal shown in figure?
Y ( z) W ( z) Y ( z)
Let =
X ( z) X ( z) W ( z) x(n)

W ( z) 1 2
Where, = (2)
X ( z ) 1 + a 1z -1 + a 2z -2 + ......... a N z - N
Y ( z) -1 -2 -M
   W ( z ) = b 0 + b 1z + b 2z + ........... b M z  (3) 0
n
1 2 3 4 5
3.200  |  Part III  •  Unit 2  •  Signals and Systems

Solution Example 11
Here, x(0) = x(1) = ..... = x(5) = 2, and x(6) = x(7) =.....= 0 A discrete LTI system has the difference equation y(n + 1) −
∞ 2y(n) = x(n + 1), if the initial conditions are y(−1) = −1, and
X(z) = ∑ x( n) z - n
⎛ 1⎞
n
n = -∞ the input x(n) = ⎜ ⎟ u(n) then find output y(n).
2 2 2 2 2 ⎝ 3⎠
=2+ + + + +
z z 2 z3 z 4 z5 Solution
2 Given y(n + 1) − 2y(n) = x(n + 1)
= 5 ( z 5 + z 4 + z 3 + z 2 + z 4 + 1) for all z ≠ 0
z
y(n) − 2y(n − 1) = x(n)
⎛ 1 1 1 1 1⎞
(or) X(z) = 2 ⎜1 + + 2 + 3 + 4 + 5 ⎟
⎝ z z z z z ⎠ y(n) u(n) ↔ Y(z)
This expression in geometric progression, the summation of 1 1
Y(z) + y[−1] = Y(z) + (-1)
y(n − 1) u(n) ↔
a( r n - 1) z z
n terms with initial value a is Sn =
r -1 x(−1) = x(−2)…… = x(−n) = 0, x(n) is causal input
⎛ ⎛ ⎛ 1⎞ 6 ⎞ ⎞ n
⎛ 1⎞
⎜ 1 ⎜ ⎜ ⎟ - 1⎟ ⎟ x(n) = ⎜ ⎟ u( n)
⎜ ⎝⎝ z⎠ ⎠⎟ 2z ⎝ 3⎠
1
So, X(z) = 2 ⎜ = (1 - z -6 )
⎛1 ⎞ ⎟ z - 1 ⇒ X(z) =
1 -1
⎜ ⎜ - 1⎟ ⎟ 1- z
⎜ ⎝z ⎠ ⎟ 3
⎝ ⎠
By taking the z transform for the difference equation
Example 10
⎡1 ⎤
Find the z transform of the signal x(n) depicted in the figure Y(z) − 2 ⎢ Y ( z ) - 1⎥ = X(z)
⎣ 2 ⎦
x(n)
5 ⎡ 2⎤
Y(z) ⎢1 - ⎥ + 2 = X(z)
4 ⎣ z⎦
3
2 1
-2
1 1 -1
1- z
n X ( z) - 2 3
1 2 3 4 5 6 Y(z) = =
1 - 2 z -1 1 - 2 z -1
Solution 2 2
1 - 2 + z -1 -1 + z -1
The signal x(n) can be expressed as = 3 = 3
- ⎛ 1 -1 ⎞ - ⎛ 1 -1 ⎞
x(n) = n for 0 ≤ n ≤ 5 (1 - 2 z ) ⎜1 - z ⎟ (1 - 2 z ) ⎜1 - z ⎟
1 1
⎝ 3 ⎠ ⎝ 3 ⎠
x(n) = n{u(n) − u(n − 6)}
= n u(n) − nu(n − 6) z ( -3 z + 2) ⎡ A B ⎤
Y(z) = =z⎢ + ⎥
( z - 2) (3 z - 1) ⎣ z - 2 3 z - 1⎦
= n u(n) − (n − 6 + 6) u(n − 6)
= nu(n) − (n − 6) u(n − 6) − 6 u(n − 6) ⎡ ⎤
⎡ -4 1 3 1 ⎤ -1 ⎢ 4 1 ⎥
u(n) ↔
1 =z⎢ - ⎥ = 5 ⎢1 - 2 z -1 + ⎥
1 - Z -1 ⎣ 5 z - 2 5 (3 z - 1) ⎦ ⎢ 1
1 - z -1 ⎥
⎣ 3 ⎦
z -1
nu(n) ↔
-1 ⎡ n ⎤
n
(1 - z -1 ) 2 ⎛ 1⎞
⇒ y(n) = ⎢ 4.2 u( n) + ⎜ ⎟ u( n) ⎥
x(n − k) ↔ z−k X(z) 5 ⎢⎣ ⎝ 3⎠ ⎥⎦
So applying z transform ⎡ ⎛ 1⎞ ⎤
n
+
= − 0.2 ⎢ 2 + ⎜ ⎟ ⎥ u( n)
n 2
z -1 1 z -1 1 6 ⎢⎣ ⎝ 3⎠ ⎥
X(z) = - - 6 , |z| > 1 ⎦
- - z (1 - z -1 )
(1 - z )1 2 z (1 - z )
6 1 2
Example 12
z -1 - z -7 - 6 z -6 (1 - z -1 ) 5 z -7 - 6 z -6 + z -1 For a causal system specified by the transfer function H(z) =
X(z) = = z
(1 - z -1 ) 2 (1 - z -1 ) 2 , find the zero - state response to input
z - 0.4
z - 6z + 5
6
= x(n) = (0.5)n u(n) + 2 2n u(-n - 1)
z 5 ( z - 1) 2
 Chapter 3  •  Z Transform  |  3.201

Solution -1 ⎛ -1 -2 ⎞
W(z) = -z ⎜ z ⎟⎠
Zero-state response means y(n) with zero initial conditions ⎛ 1 -1 ⎞ ⎝ 3
2

1 2 z ( z - 2 - 2 z + 1) ⎜⎝1 + z ⎟⎠
X(z) = - = 3
1 - 0.5 z -1 1- 2 z -1 ( z - 0.5) ( z - 2)
-1 z -1 1
- z ( z + 1) = 2
with ROC |z| >
= 3 ⎛ 1 ⎞ 3
-1
( z - 0.5) ( z - 2) ⎜⎝1 + z ⎟⎠
3
The ROC of X(z) is -n
For the first term |z| > 0.5 and For the second term |z| < 2 ⎛ 1⎞
y(n) = (5)n u(−n) = ⎜ ⎟ u[ - n]
The ROC for X(z) is common region given by 0.5 < |z| < 2 ⎝ 5⎠
n
- z ( z + 1) ⎛ 1⎞ z1
and X(z) = z
⎜⎝ ⎟⎠ u(n) ←⎯ → , ROC |z| >
( z - 0.5) ( z - 2) 5 1 5
z-
- z 2 ( z + 1) 5
Then Y(z) = X(z). H(z) =
( z - 0.4) ( z - 0.5) ( z - 2) z ⎛ 1⎞ 1
x(-n) ←⎯ → X ⎜ ⎟ with ROC
as H(z) is causal system ROC is |z| > 0.4 ⎝ z⎠ Rx
The common region of convergence of X(z) and H(z) is 0.5 1
< |z| < 2, so z 1 1
So, Y(z) = ROC >
- z 2 ( z + 1) 1 1 |z| 5
Y(z) = -
( z - 0.4) ( z - 0.5) ( z - 2) z 5
Y ( z) - z ( z + 1) - z -1
= = |z| < 5
z ( z - 0.4) ( z - 0.5) ( z - 2) 1 -1
-z
-7 1 5 5 1 5
= + - By applying convolution y(n) × w(n), and z transform, we
2 ( z - 0.4) ( z - 0.5) 2 z - 2
have X(z) = Y(z) W(z)
-7 z z 5 z
Y(z) = + 5. - , - z -1 -1 z -1
2 z - 0.4 z - 0.5 2 z - 2 = .
1 -1 3 ⎛ 1 ⎞ 2
-z -1
0.5 < |z| < 2 5 ⎜⎝1 + z ⎟⎠
3
⎡ -7 ⎤ 5
y(n) = ⎢ (0.4) n + 5(0.5) n ⎥ u(n) + 2n u(-n -1) 1
⎣ 2 ⎦ 2 the common ROC is |z| > , and |z| < 5 and X(z) =
5 1 z 3
Example 13 .
z - 5 3 ⎛ 1⎞ 2
Find the z transform of the signal x(n) ⎜⎝ z + ⎟⎠
3
⎧⎪ ⎛ -1⎞ n ⎫⎪ 5 z 1
= ⎨n ⎜ ⎟ u( n)⎬ * (5) n u( - n) = with ROC < |z| < 5
⎝ ⎠
⎩⎪ 3 3 2 3
⎭⎪ ⎛ 1⎞
( z - 5) ⎜ z + ⎟
⎝ 3⎠
Solution
We can write x(n) = w(n) × y(n) Example 14
n
⎛ -1⎞ Find the transfer function and impulse response of a causal
w(n) = n ⎜ ⎟ u(n)
⎝ 3⎠ LTI system, if the input to the system is x(n) = u(n) and
n output is y(n) = d(n)
⎛ -1⎞ z 1
⎜⎝ ⎟⎠ u(n) ←⎯ →
1 -1
with ROC Solution
3
1+ z
3 The z transform of inputs and outputs
1
|z| > 1
3 X(z) = , ROC |z| > 1,
n 1 - z -1
⎛ -1⎞ z
w(n) = n ⎜ ⎟ u(n) ←⎯ → W(z) Y(z) = 1, ROC for all z
⎝ 3⎠
⎧ ⎫ Then transfer function H(z)
d ⎪ 1 ⎪ Y ( z) 1
= -z ⎨ ⎬ = =
dz ⎪ 1 -1 ⎪ X ( z) 1
1+ z
⎩ 3 ⎭ (1 - z -1 )
3.202  |  Part III  •  Unit 2  •  Signals and Systems

= 1 − z−1, with ROC all z except at z = 0 ⎛ 1⎞ ⎛ 1⎞

By taking inverse z transform,
1 ⎜⎝ z - ⎟⎠ ⎜⎝ z - ⎟⎠
3 6
h(n) = d(n) − d(n − 1) is the impulse response. Hi(z) = =
H ( z) ⎛ 1⎞ ⎛ 1⎞
⎜⎝ z + ⎟⎠ ⎜⎝ z - ⎟⎠
Example 15 4 4
An LTI system is described by the difference equation is the transfer function of inverse system. All the poles and
1 1 1 1 zeros of H(z) lie inside the unit circle, so stable and causal
y(n) − y(n − 1) + y(n − 2) = x(n) + x(n − 1) − inverse system exists.
2 18 20 20
x(n − 2) And this is a minimum phase system.
Does a stable and causal LTI inverse system exists?
Example 16
Solution A discrete time LTI model for two path communication
1 1 1 1 channel is y(n) = x(n) + a x(n − 1), Find the difference
y(n) − y(n − 1) + y(n − 2) = x(n) + x(n − 1) −
2 18 20 20 equation description of the inverse system.
x(n − 2) Solution
By considering z transform
y(n) = x(n) + ax(n − 1), By taking z transform Y(z) = X(z)
1 −1 1 −2 + az−1 X(z)
Y(z) − z Y(z) + z Y(z) = X(z) +
2 18 Y ( z)
Transfer function H(z) = = 1 + a z-1
X ( z)
z -1 1
X ( z ) - z -2 X ( z )
20 20 The inverse system has transfer function
1 -1 1 -2 1 1 1 1 Y ( z)
1+ z - z z2 + z- Hi(z) = = = i
Y ( z) 20 20 20 20 -
H(z) = = = H ( z ) 1 + az 1 X i ( z)
X ( z) 1 1 1 1
1 - z -1 + z -2 z2 - z +
2 18 2 18 Yi(z) + az−1 Yi(z) = Xi(z),
⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ By taking inverse z transform
⎜⎝ z + ⎟⎠ ⎜⎝ z - ⎟⎠ ⎜⎝ z + ⎟⎠ ⎜⎝ z - ⎟⎠
4 5 4 5 y(n) + a y(n − 1) = x(n)
= =
⎛ 1 ⎞⎛ 1 ⎞ ⎛ 1 ⎞⎛ 1 ⎞
⎜⎝ z - ⎟⎠ ⎜⎝ z - ⎟⎠ ⎜⎝ z - ⎟⎠ ⎜⎝ z - ⎟⎠ The inverse system is both stable and causal if |a| < 1.
3 6 3 6

Exercises
Practice Problems 1
Direction for questions 1 to 22:  Select the correct alterna- 3. Z transform of nanu(n) is
tive from the given choices. - az az
(A) , |z| > |a| (B) , |z| > |a|
1. x(n) = anu(n) where a is real. Z transform of x(n) is ( z - a) 2 ( z - a)
az az
z z (C) , |z| > |a| (D) , |z| < |a|
(A) , |z| < |a| (B) , |z| > |a| ( z - a) 2 ( z - a) 2
z-a z-a
4. Inverse z transform of log (1 − az−1), |z| > |a| is
z z
(C) , |z| > |a| (D) , |z| < |a| 1 -1 n
z+a z+a (A) anu(n − 1) (B) a u(n − 1)
n n
2. Z transform of an+1u(n + 1) is -1 1 n
(C) anu(n + 1) (D) a u(n + 1)
n n
z2 z2
(A) , |a| < |z| < ∞ (B) , |a| < |z| < ∞ 5. Find the initial and final values of z-domain signal
z-a z+a
2 z -1
z z 2 , |a| > |z| < ∞
(C) , |a| < |z| < ∞ (D) 1 - 1.8 z -1 + 0.8 z -2
z-a z-a (A) 10, 1 (B) 1, 10 (C) 0, 10 (D) 0, ∞
 Chapter 3  •  Z Transform  |  3.203

6. The transfer function of a system is H(z) = 1 − 3z−1, the z2 1

system response for input x(n) is (C) , |z| >
⎛ 1⎞ ⎛ 1⎞ 2
(A) x(n) − 0.3 x(n − 1) ⎜⎝ z + ⎟⎠ ⎜⎝ z - ⎟⎠
(B) x(n) + 0.3 x(n − 1) 2 4
z2 1
(C) x(n) − 3 x(n + 1) (D) , |z| >
(D) x(n) − 3 x(n − 1) ⎛ 1⎞ ⎛ 1⎞ 2
⎜⎝ z - ⎟⎜ z - ⎟
n n 2⎠ ⎝ 4⎠
⎛ 1⎞ ⎛ 1⎞
7. Unilateral z transform of ⎜ ⎟ u( n) + ⎜ ⎟ u( n) and
⎝ 3⎠ ⎝ 5⎠ z ( 2 z - 5/ 6)
ROC are respectively? 12. For a Z transform X(z) = . Find out the
⎛ 1 ⎞ ⎛ 1⎞
z z 1 ⎜⎝ z - ⎟⎠ ⎜⎝ z - ⎟⎠
(A) + ; |z| > 2 3
( z - 1/ 3) ( z - 1/ 5) 5 sequences when the region of convergence is |z| > 1/2
z z 1
(B) + ; |z| > ⎡⎛ 1 ⎞ n ⎛ 1 ⎞ n ⎤
( z - 1/ 3) ( z - 1/ 5) 3 ⎢⎜ ⎟ - ⎜ ⎟ u( n) ⎥
(A)
⎢⎣⎝ 2 ⎠ ⎝ 3⎠ ⎥⎦
z z 1 1
(C) + ; > |z| > n n
( z - 1/ 3) ( z - 1/ 3) 5 3 ⎛ 1⎞ ⎛ 1⎞
⎜⎝ ⎟⎠ u( n) - ⎜⎝ ⎟⎠ u( - n - 1)
(B)
2 3
z z 1 1
(D) + ; < |z| <
( z - 1/ 3) ( z - 1/ 3) 5 3 ⎡⎛ 1 ⎞ n ⎛ 1 ⎞ n ⎤
⎢⎜ ⎟ + ⎜ ⎟ ⎥ u( n)
(C)
⎢⎣⎝ 2 ⎠ ⎝ 3⎠ ⎥
⎦
8. Find f(0), f(1), f(2), and f(3), if n n
⎛ 1⎞ ⎛ 1⎞
F(z) = 3
z2 + 2 (D) ⎜⎝ ⎟⎠ u( n) + ⎜⎝ ⎟⎠ u( - n - 1)
2 3
z - 3z 2 + 3z - 1
13. Find the sequence when region of convergence
(A) 0, 9, 31, 16 (B) 0, 1, 3, 8 1 1
< z < for above X(z),
(C) 1, 3, 3, 1 (D) 0, 1, 2, 4 3 2
sin t n n
9. The minimum Nyquist sampling rate is ⎛ 1⎞ ⎛ 1⎞
pt - ⎜ ⎟ u( - n - 1) + ⎜ ⎟ u( n)
(A)
(A) 2/p (B) 1/p (C) p (D) 2p ⎝ 2⎠ ⎝ 3⎠
10. Match the following: ⎛ 1⎞
n
⎛ 1⎞
n

List-I List-II ⎜⎝ ⎟⎠ u( - n - 1) + ⎜⎝ ⎟⎠ u( n)
(B)
2 3
  x(n)  ROC n n
⎛ 1⎞ ⎛ 1⎞
d(n)
 (1)  (a)  All z ⎜⎝ ⎟⎠ u( n) - ⎜⎝ ⎟⎠ u( - n - 1)
(C)
2 3
  (2)  u(n) (b) |z| > 1
n n
  (3)  −u(−n − 1) (c)  All z except 0 ⎛ 1⎞ ⎛ 1⎞
d(n − m)
 (4)  (d)  |z| < 1 ⎜⎝ ⎟⎠ u( n) + ⎜⎝ ⎟⎠ u( - n - 1)
(D)
2 3
  1 2 3 4  1 2 3 4 14. Impulse response for the cascade of LTI systems hav-
(A) a b c d (B) b a c d n
⎛ 1⎞
(C) a b d c (D) b a d c ing impulse response h1(n) = ⎜ ⎟ u(n) and h2 (n) =
n
⎝ 2⎠
⎛ 1⎞
3 1 ⎜⎝ ⎟⎠ u(n) is?
11. y(n) − y( n - 1) + y( n - 2) = x(n) where x(n) and 4
4 8
y(n) are the input and output of the systems, then deter- n n
⎛ 1⎞ ⎛ ⎞ 1
mine system function. (A)
⎜⎝ ⎟⎠ (2n - 1) u( n) (B)
⎜⎝ ⎟⎠ ( 2n +1 - 1) u( n)
4 4
z2 1 n n
(A) |z| > ⎛ 1⎞ ⎛ 1⎞
⎛ 1⎞ ⎛ 1⎞ 2 n +1
⎜⎝ ⎟⎠ ( 2 + 1) u( n) (D)
(C) ⎜⎝ ⎟⎠ ( 2 + 1) u( n)
n
⎜⎝ z + ⎟⎠ ⎜⎝ z - ⎟⎠ 4 4
2 4
15. What is the Nyquist rate for the signal x(t) = cos2000pt
z2 1
(B) ,| z |< + 6sin 9000pt
⎛ 1⎞ ⎛ 1⎞ 2 (A) 9 kHz (B) 4 kHz
⎜⎝ z + ⎟⎠ ⎜⎝ z - ⎟⎠
2 4 (C) 2 kHz (D) 11 kHz
3.204  |  Part III  •  Unit 2  •  Signals and Systems

16. Which of the following is the inverse Z transform of 1

z S1: System is stable and causal for ROC |z| >
X(z) = |z| < 2 2
( z - 2)( z - 3) 1
S2: System is stable but not causal for ROC: |z| <
(A) [2n − 3n] u(−n − 1) 4
(B) [2n − 3n] u(n + 1) S3: System is neither stable nor causal for ROC:
(C) [3n − 2n] u(−n − 1)
1 1
(D) [2n - 3n] u(n) <| z |<
4 2
Which one of the following statements is valid?
17. The impulse response of a relaxed linear time invari-
(A) Both S1 and S2 are true
ant system is h(n) = anu(n) with |a| < 1. Determine the
(B) Both S2 and S3 are true
value of the step response as n → ∞
(C) Both S1 and S3 are true
⎛ 1 - a n +1 ⎞ ⎡1 - a n -1 ⎤ (D) All are true
(A)
⎜⎝ 1 - a ⎟⎠ u ( n) (B)
⎢ ⎥ u(n)
⎣ 1-a ⎦ 20. ROC of z transform of the discrete time sequence x(n)
n n
⎛ 1⎞ ⎛ 1⎞
⎡1 + a n +1 ⎤ ⎡1 + a n +1 ⎤ = ⎜ ⎟ u(n) − ⎜ ⎟ u[−n − 1] is
(C)
⎢ ⎥ u(n) (D)
⎢ ⎥ u(n) ⎝ 3⎠ ⎝ 2⎠
⎣ 1- a ⎦ ⎣ 1- a ⎦ 1 1
(A) |z| > (B) |z| <
18. Consider Z transform X(z) = 5z2 + 4z−1 + 3;0 < |z| < ∞, 3 2
the inverse z transform x(n) is? 1 1
(A) 5d(n + 1) (C) |z| for all (D) < |z| <
3 2
(B) 3d(n)
21. A causal LTI system is described by the difference
(C) 4d(n + 2)
equation 2y(n) = ay(n − 1) − 2x(n) + bx(n − 1).The
(D) 5d(n + 2) + 4d(n − 1) + 3d(n)
system is stable for which values of a, b?
19. The transfer function of D.T, LTI system is given by (A) |a| > 2, |b| < 1 (B) |a| < 2, any b
3 (C) |a| < 3, |b| > 1 (D) |a| > 2, any b
2 - z -1
4 2 2. A sequence x(n) with z transform X(z) = z4 + z2 − 2z +
H(z) =
3 1 2 − 3z−4 is applied as an input to a linear, time-invariant
1 - z -1 + z -2
4 8 system with impulse response h(n) = 2 d(n − 3), the
output at n = 4 is
Consider the following statements
(A) 2 (B) 0 (C) –2 (D) –6

Practice Problems 2
Direction for questions 1 to 21:  Select the correct alterna- 3. Z transform X(Z) of a sequence x(n) is given by X(Z) =
tive from the given choices. 0.5
, it is given that the ROC of X(Z) includes unit
1. Two D.T systems with impulse response h1(n) = d(n − 1 - 2 z -1
1), h2(n) = d(n − 2) are connected in cascade, the overall circle, the value of x(0) is
(A) 0 (B) 0.5 (C) 1 (D) ∞
Impulse response of the cascaded system is?
(A) d(n − 1) (B) d(n − 2) 1 2
4. If ROC of x1(n) + x2(n) is < z < , then the ROC of
(C) d(n − 3) (D) d(n − 4) 3 3
x1(n) − x2(n) includes
2. A system with transfer function H(z) has impulse 1 2
response h(n), defined as h(2) = 1, h(3) = −1, and h(n) (A) < |z| < 3 (B) < |z| < 3
3 3
= 0 otherwise, consider following statements
3 1 2
S1: H(z) is a low-pass filter (C) < |z| < 3 (D) < z <
2 3 3
S2: H(z) is an FIR filter
n
Which one of the following is correct?
(A) Only S2 is true
5. y(n) = ∑ x( k )
k =0
(B) Both S1 and S2 are false Which one of the following correctly relate the z trans-
(C) Both S1 and S2 are true, and S1 is reason for S2 form of the input and output denoted by X(z) and Y(z),
(D) Both S1 and S2 are true, and S2 is a reason for S1 respectively?
 Chapter 3  •  Z Transform  |  3.205

d X ( z) 3z 2
Y(z) =
(A) Y(z) =
X ( z ) (B) 14. The initial value of x(n), if X(z) =
dz 1 - z -1 ( z + 3)( z - 3)
(C) −1
Y(z) = X(z)z (D) Y(z) = (1 − z−1)X(z) (A) 3 (B) 6
n (C) ∞ (D) 0
1 ⎛ 1⎞ 6 n
6. x(n) = ⎜ ⎟ u( n) - (3) u( - n - 1) . Find out ROC of 15. An LTI system defined as y(n) + 0.5y(n − 1) = x(n) +
7 ⎝ 2⎠ 7 0.3x(n − 1). The transfer function in z plane is
Z transform?
z + 0.3 z - 0.3
(A) 1/2 > |z| < 3 (B) |z| > 1/2 (A) (B)
(C) |z| < 3 (D) 1/2 < |z| < 3 z + 0.5 z - 0.5
7. The range of values a and b for which the z transform with z + 0.3 z - 0.3
(C) (D)
impulse response h(n) = an, n ≥ 0 = bn, n < 0 will be stable z - 0.5 z + 0.5
(A) |a| < 1, |b| < 1 (B) |a| < 1, |b| > 1 16. A signal x(n) = {1, −1, 2, 1, 3, 2}. The ROC
(C) |a| > 1, |b| > 1 (D) |a| > 1, |b| < 1 z transform X(z) is
8. The Z transform of a-n u (−n −1) is (A) |z| > 3
az 1 az 1 (B) −1 < |z| < 3
(A) , z < (B) , z < (C) |z| < 3
1 + az a 1 - az a (D) the entire z plane except at z = 0
- az 1 az 1 17. The minimum sampling frequency of x(t) = 5 cos 50pt
(C) , z < (D) , z >
1 - az a 1 - az a + 2 cos 200pt − 10cos 100pt is
9. Z transform of x(n) = {3, −2, 4, 1, 0, −3} is (A) 100 Hz (B) 200 Hz
↑ (C) 50 Hz (D) Cannot be sampled
(A) 3z2 + 2z + 4 + z−1 + 3z−2,   2 < |z| < 4 18. The minimum number of samples required to exactly
(B) Does not exist describe x(t) = 10 cos 4 pt + 4 sin 8pt is
(C) 3z2 − 2z + 4 + z−1 − 3z−3, 0 < |z| < ∞ (A) 4 samples per second.
(D) 3z2 − 2z + 4 + z−1 − 3z−3, |z| < 0, |z| < ∞ (B) 2 samples per second.
10. Z transform of u(n + 1) is (C) 8 samples per second.
z z2 (D) 16 samples per second.
(A) (B)
z -1 z -1 19. Aliasing occurs when sampling frequency ws is (wm –
z -1 1 band limited signal frequency)
(C) -1 (D) (A) ws = 2wm (B) ws < 2wm
1- z z + z -2
-1
2 (C) ws > 2wm (D) 0
11. Inverse z transform of , (|z| > 2) is
z-2 20. Interpolation is the process of
(A) 2n-1 u(n + 1) (B) 2n+1 u(n − 1) (A) inserting N - 1 unity sequence values to x(n)
(C) 2 u(n + 1)
n (D) 2n u(n − 1) (B) inserting N - 1 zero sequence values to x(n)
12. Z transform and Laplace transform are related by (C) deleting N - 1 unity sequence values from x(n)
ln z (D) deleting N - 1 zero sequence values from x(n)
(A) s = ln z (B) s =
T 21. Given signal m(t) = sin 5pt + 5sin 10pt is sampled instan-
(C) s = T ln z (D) Not related taneously, the maximum interval between samples is
13. The final value of x(n) is (A) 0.1 seconds. (B) 0.2 seconds.
(A) Lt zX(z) (B) Lt zX(z) (C) 10 seconds. (D) none.
z→∞ z→0
(C) Lt (1 − z−1) X(z) (D)
Lt (1 − z−1) X(z)
z→∞ z→1

Previous Years’ Questions

1. The z transform of a system is 2. A 1 kHz sinusoidal signal is ideally sampled at 1,500
z samples/sec and the sampled signal is passed through
H(z) =
z - 0.2 an ideal low-pass filter with cut-off frequency 800 Hz.
If the ROC is z < 0.2, then the impulse response of The output signal has the frequency? [2004]
the system is  [2004] (A) Zero Hz (B) 0.75 kHz
(0.2) u( n) (B)
(A)
n
(0.2) n u( - n - 1) (C) 0.5 kHz (D) 0.25 kHz
3. A casual LTI system is described by the difference
- (0.2) u( n) (D)
- (0.2) u( - n - 1)
n n
(C) equation
3.206  |  Part III  •  Unit 2  •  Signals and Systems

2 y( n) = a y( n - 2) - 2 x( n) + b x( n - 1) . 8. The samples x(n), (n = 0, 1, 2, ….) are given by

 [2008]
The system is stable only if [2004]
(A) 5(1–e–0.05n) (B) 5e–0.05n
(A) a = 2, b < 2 (C) 5(1–e–5n) (D) 5e–5n
(B) a > 2, b >2 9. The expression and the region of convergence of the z
transform of the sampled signal are [2008]
(C) a < 2, any value of b
(D) b < 2 , any value of a 5z 5z
(A) -5 ,| z | < e -5 (B)  -0.05 ,| z | < e -0.5
4. 
The region of convergence of z transform of the z-e z-e
n n 5z 5z
⎛ 5⎞ ⎛ 6⎞
sequence ⎜ ⎟ u ( n) - ⎜ ⎟ u( - n - 1) must be [2005] (C) -0.05 ,| z | > e -0.05 (D)  ,| z | > e -5
⎝ 6⎠ ⎝ 5⎠ z-e z - e -5
5 6 10. The ROC of Z transform of the discrete time sequence
(A) Z < (B) Z >
6 5 ⎛ 1⎞
n
⎛ 1⎞
n
5 6 6 x( n) = ⎜ ⎟ u( n) - ⎜ ⎟ u( - n - 1) is [2009]
(C) < Z < (D) < Z <∞ ⎝ 3⎠ ⎝ 2⎠
6 5 5
1 1 1
5. If the region of convergence of X1(n)+X2(n) (A) < | z | < (B) | z |>
1 2 3 2 2
X1 ( n) + X 2 ( n) is < Z < , then the region of con- 1
3 3 (C) | z |< (D) 2 < |z| < 3
vergence of X1 ( n) - X 2 ( n) includes 3
[2006] 11. A system with transfer function H(z) has impulse
1 2 response h(n → n) defined as h(2) = 1, h(3) = −1 and
(A) < Z < 3 (B) < Z <3
3 3 h(k) = 0 otherwise. Consider the following statements
3 1 2 S1: H(z) is a low-pass filter
(C) < Z < 3 (D) < z < S2: H(z) is a FIR filter
2 3 3 Which of the following is correct? [2009]
6. The minimum sampling frequency (in samples/sec) (A) Only S2 is true
required to reconstruct the following signal from its (B) Both S1 and S2 are false
samples without distortion (C) Both S1 and S2 are true, and S2 is a reason for S1
3 2 (D) Both S1 and S2 are true, but S2 is not a reason for S1
⎛ sin 2p1000t ⎞ ⎛ sin 2p1000t ⎞
x (t ) = 5 ⎜ ⎟ + 7⎜ ⎟⎠
⎝ pt ⎠ ⎝ pt 12. Consider the z transform X(z) = 5z2 + 4z−1 + 3; 0 < |z|
3 2 < ∞. The inverse z transform x(n) is [2010]
⎛ sin 2p1000t ⎞ ⎛ sin 2p1000t ⎞ (A) 5d(n + 2) + 3d (n) + 4d (n − 1)
x (t ) = 5 ⎜ ⎟⎠ + 7 ⎜⎝ ⎟⎠ would
⎝ pt pt (B) 5d(n − 2) + 3d(n) + 4d(n + 1)
be: [2006]
(C) 5u(n + 2) + 3u(n) + 4u(n − 1)
(A) 2 × 103 (B) 4 × 103 (D) 5u(n − 2) + 3u(n) + 4u(n + 1)
(C) 6 × 10 3 (D) 8 × 103 13. The transfer function of a discrete time LTI system is
7. The z transform X(z) of a sequence x(n) is given given by
0.5 3
X(z) = . It is given that the region of conver- 2 - z -1
1 - 2 Z -1 H ( z ) = 4
gence of X(z) includes the unit circle. The value of 3 1
1 - z -1 + z -2
x(0) is:  [2007] 4 8
(A) –0.5 (B) 0 (C) 0.25 (D) 0.5 Consider the following statements:
S1: The system is stable and causal for ROC:|z|>½
Direction for questions 8 and 9: S2: The system is stable but not causal for ROC:|z|<¼
In the following network, the switch is closed at t = 0– and S3: The system is neither stable nor causal for ROC:
the sampling starts from t = 0. The sampling frequency is ¼ < |z| < ½
10 Hz. Which one of the following statements is valid?
 [2010]
S 10 µF (A) Both S1 and S2 are true
X(z) (B) Both S2 and S3 are true
X(n)
+ Sampler (C) Both S1 and S3 are true
5V z-transform
− 200K (fs = 10 Hz)
(D) S1, S2, and S3 are all true
 Chapter 3  •  Z Transform  |  3.207

14. The Nyquist sampling rate for the signal S(t) 21. An FIR system is described by the system function
sin(500pt ) sin(700pt ) 7 3
s( t ) = × is given by [2010] H ( z ) = 1 + z -1 + z -2 The system is  [2014]
pt pt 2 2
(A) maximum phase
(A) 400 Hz (B) 600 Hz
(B) minimum phase
(C) 1,200 Hz (D) 1,400 Hz
(C) mixed phase
15. Two systems H1(z) and H2(z) are connected in cascade (D) zero phase
as shown below. The overall output y(n) is the same as
22. Let x(n) = x(−n). Let X(z) be the z transform of x(n). If
the input x(n) with a one unit delay. The transfer func-
0.5 + j0.25 is a zero of X(z), which one of the follow-
tion of the second H2(z) is [2011]
ing must also be a zero of X(z).[2014]
(A) 0.5 − j0.25 (B) 1/(0.5 + j0.25)
(1 − 0.4z 1 )
x(n) H1(z)= H2(z) y(n) (C) 1/(0.5 − j0.25) (D) 2 + j4
(1 − 0.6z 1 )
23. The input/output relationship of a causal stable LTI
(1 - 0.6 z -1 ) z -1 (1 - 0.6 z -1 ) system is given as y(n) = a y(n − 1) + β x(n). If the
(A) (B)
z -1 (1 - 0.4 z -1 ) (1 - 0.4 z -1 ) impulse response h(n) of this system satisfies the con-
∞
z -1 (1 - 0.4 z -1 ) (1 - 0.4 z -1 ) dition ∑ n= 0 h(n) = 2, the relationship between a and
(C) (D)
-
(1 - 0.6 z )1 z 1 (1 - 0.6 z -1 )
- β is [2014]
(A) a = 1 − β/2 (B)
a = 1 + β/2
16. If x(n) = (1/ 3) n - (1/ 2) n u( n) , then the region of con- (C) a = 2β (D) a = −2β
vergence (ROC) of its Z transform in the Z plane will 24. Let x(t) = cos(10 πt) + cos(30 πt) be sampled at 20 Hz
be [2012] and reconstructed using an ideal low-pass filter with
1 1 1 cut-off frequency of 20  Hz. The frequency/frequen-
(A) < z < 3 (B) < z <
3 3 2 cies present in the reconstructed signal is/are  [2014]
1 1 (A) 5 Hz and 15 Hz only
(C) < Z <3 (D) < z (B) 10 Hz and 15 Hz only
2 3
(C) 5 Hz, 10 Hz, and 15 Hz only
17. Let y(n) denote the convolution of h(n) and g(n), (D) 5 Hz only
n
⎛ 1⎞ 25. For an all pass system
where h(n) = ⎜ ⎟ u(n) and g(n) is a causal sequence.
⎝ 2⎠
If y(0) = 1 and y(1) = ½, then g(1) equals [2012] -1
H ( z ) = ( z - b) ,
(A) 0 (B) ½ (C) 1 (D) 3/2 (1 - az -1 )
18. A band-limited signal with a maximum frequency of where |H(e–jω)| = 1, for all ω. If Re(a) ≠ 0, Im(a) ≠ 0,
5  kHz is to be sampled. According to the sampling then b equals [2014]
theorem, the sampling frequency which is not valid is (A) a (B) a*
[2013] (C) 1/a* (D) 1/a
(A) 5 kHz (B) 12 kHz
(C) 15 kHz (D) 20 kHz 26. Let H1(z) = (1 − pz−1)−1, H2(z) = (1 – qz−1)−1, H(z) =
H1(z) + rH2(z). The quantities p, q, r are real numbers.
19. Consider two real-valued signals, x(t) band-limited to
[−500 Hz, 500 Hz] and y(t) band-limited to [−1 kHz, 1 1
Consider p = , q = - , |r| < 1. If the zero of H(z)
1  kHz]. For z(t) = x(t) • y(t), the Nyquist sampling 2 4
frequency (in kHz) is _______. [2014] lies on the unit circle, then r = _____ [2014]
n n 27. The z transform of the sequence x(n) is given by X(z)
⎛ 1⎞ ⎛ 1⎞ 1
20. Let x( n) = ⎜ - ⎟ u( n) - ⎜ - ⎟ u( - n - 1). The Region = , with the region of convergence |z| > 2.
⎝ 9⎠ ⎝ 3⎠ (1 - 2 z -1 ) 2
of Convergence (ROC) of the z transform of x(n)
[2014] Then, x(2) is _________. [2014]
1 1
(A) is |z| > (B) is |z| < 28. The sequence x(n) = 0.5n u(n), where u(n) is the unit
9 3
step sequence, is convolved with itself to obtain y(n).
1 1 +∞
(C) is > |z| >
3 9
(D) does not exist Then, ∑ n=-∞ y(n) is _______ [2014]
3.208  |  Part III  •  Unit 2  •  Signals and Systems

29. Two causal discrete-time signals x[n] and y[n] are (B) Im

related as y[n] = ∑ m = 0 x[m] . If the z-transform of

n
Unit circle z-plane
2
y[n] is , the value of x[2] is _____. [2015]
z ( z - 1) 2
30. Consider the sequence x[n] = an u[n] + bn u[1], where Re
u[n] denote the unit step sequence and 0 < |a| < |b| < 1. 0.5
The region of convergence (ROC) of the z transform
of x[n] is [2016]
(A) |z| > |a| (B) |z| > |b|
(C) |z| < |a| (D) |a| < |z| < |b|  
31. A discrete-time signal x[n] = d[n - 3] + 2d[n - 5] has
z-transform X(z). If Y(z) = X(-z) is the z-transform of (C) Im
another signal y[n], then [2016]
Unit circle z-plane
(A) y[n] = x[n] (B) y[n] = x[-n]
(C) y[n] = -x[n] (D) y[n] = -x[-n]
32. The ROC (region of convergence) of the z-transform
of a discrete time signal is represented by the shaded
Re
region in the z-plane. If the signal x[n] = (2.0)|n|, 0.5 2
-∞ < n < + ∞, then the ROC of its z-transform is rep-
resented by [2016]
(A) Im

Unit circle z-plane

 

(D) Im

Unit circle z-plane

Re
0.5 2

Re
0.5 2
 

(ROC does not exit)

 
 Chapter 3  •  Z Transform  |  3.209

Answer Keys
Exercises
Practice Problems 1
1. B 2. A 3. C 4. B 5. C 6. D 7. A 8. B 9. B 10. C
11. D 12. C 13. A 14. B 15. A 16. A 17. A 18. D 19. C 20. D
21. B 22. B

Practice Problems 2
1. C 2. A 3. B 4. D 5. B 6. D 7. B 8. B 9. C 10. B
11. D 12. B 13. D 14. A 15. A 16. D 17. B 18. C 19. B 20. B
21. A

Previous Years’ Questions

1. D 2. C 3. C 4. C 5. D 6. C 7. B 8. B 9. C 10. A
11. A 12. A 13. C 14. C 15. B 16. C 17. A 18. A 19.  2.99 to 3.01
20. C 21. C 22. B 23. A 24. A 25. B 26. –0.5 27. 12 28. 4 29. 0
30. B 31. C 32. D