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Seu 2 © Table of Contents
1, Z-Transform
2. Properties of Z-Transform
3. Inverse Z-Transform
4. Solutions of Differential Equation © Z-Transform
Laplace Transform - is used in Analog System as:
1. The Laplace transform definition involves an integral
2. Applying the Laplace transform to certain ordinary differential
equations turns them into simpler algebraic equations
3. Use of the Laplace transform gives rise to the basic concept of the
transfer function of a continuous or analog system.
Z-transform - plays a similar role for discrete systems to that played by
the Laplace transform for systems where the basic variable t is
continuous. Specifically:
1, The z-transform definition involves a summation
2. The z-transform converts certain difference equations to algebraic
equations
3. Use of the z-transform gives rise to the concept of the transfer
function of discrete or digital systems. © Z-Transform
Definition: For a sequence x(n) the z-transform denoted by X(z) is
given by the infinite series:
X(z) = x(0) + x(1)272 + x(2)2-2 +
.
X(z) = > x(n)z-"
n=0
Note:
1. The z-transform only involves the causal terms y(n), n = 0, 1, 2, ...
of the sequence. Noncausal terms are not involved.
2. The infinite series must converge for X(z) to be defined as a precise
function of z.
3. The precise significance of the quantity z need not concern us
except to note that it is complex and, unlike n, is continuous. © Z-Transform
Example 1: Given the sequence below, Find the z-transform of x(n).
x(n) = u(n)
Solution: From the definition, the z-transform is,
X(z) = ¥ u(njz" = > (4) =14(@4)4(21)?+..4+ (21)
n=0 n=0
This is an infinite geometric series that converges to,
1
t-r 1-277
X(Z= >
with a condition |z~*| < 1. Thus, the region of convergence for all
values of z is given as |z| > 1 © Z-Transform
Example 2: Considering the exponential sequence below, find the z-
transform of the sequence x(n).
x(n) = a™u(n)
Solution: From the definition, the z-transform is,
X(z) = > a™u(n)z-" = (az~1)" = 1+ (az~1) + (az“1)* +...
n=0 n=0
Since this is a geometric series which will converge for |az~*| < 1, itis
further expressed as,
a oe
@ = 7 Trart
with a condition |z| > |a| © Z-Transform
1. d(n)
2. au(n)
3. nu(n)
4. a™u(n)
5. na"u(n)
6. asin (bn) u(n)
7. a”cos (bn) u(n)
8. 2|A||P|" cos (nd + #) u(n)
Gay?
zasin (b)
—Dazeos (y+?
z[z—acos (b)]
P—2azcos (a)+a
Az + Az
z—P_| 2—P* © Z-Transform
Example 3: Find the z-transform for each of the following sequences:
a) x(n) = 10u(n)
b) x(n) = 10sin (0.252n)u(n)
c) x(n) = (0.5)"u(n)
d) x(n) = (0.5)"sin (0.257n)u(n)
e) x(n) = e°'"cos (0.257) u(n)
Solution: (a) From the table, we can get the Z-transform as,
X(z) = Z{10u(n)}
(b) From the table, we can get the Z-transform as,
X(z) = 10Z{sin (0.257n)u(n) }
10sin (0.257)z
X(z) = =—_———
(2) = 2 —Fac0s (025m) +1 © Z-Transform
Solution: (c) From the table, we can get the Z-transform as,
X(@) = 1{(0.5)"u(n)}
(d) From the table, we can get the Z-transform as,
X(z) = Z{(0.5)" sin (0.257n)u(n) }
x@ == (0.5)sin (0.252)z .
z° — 2(0.5)cos (0.257) z + 0.5) 1.4142z + 0.25
(e) From the table, we can get the Z-transform as,
X(z) = Z{(e~+) cos (0.257) u(n)}
2(z — e~©1 cos (0.2571) )
z* — 2e-1 cos (0.25) z + e- 9?
X(z) = © Table of Contents
1, Z-Transform
2. Properties of Z-Transform
3. Inverse Z-Transform
4. Solutions of Differential Equation © Properties of Z-Transform
Linearity Property - z-transform is a linear transformation, implies
Zax, (n) + bx2(n)} = aZ{x1(n)} + b2{x2(n)}
where x,(n) and x(n) denote the sampled sequences, while a and b
are the arbitrary constants.
Example 1: Find the z-transform of the sequence defined by,
x(n) = u(n) — (0.5)"u(n)
Solution: Applying the linearity of the z-transform previously discussed,
we have,
X(@) = ZL{x(n)} = Z{u(n)} — Z{(0.5)"u(n)}
Using the table, we have,
X(2) © Properties of Z-Transform
Example 2: Find the z transform of the sequence x(n) =cosh (Bn),
Solution: Using the property of hyperbolic function,
fn + phn
ne
From the linearity property of the Z-transform, we have,
cosh (Bn) =
Bn 4 o-Bn
X(z) = Z{cosh (fn) } = fa
From the Z-transform table, where a” > (e?)”
x0) ~7(=a) +(e)
¥@ <2 2z—(e8 +e#)
@= 2 ls —z(e8 + eF) + ql
1 1
a Bn a —pn
} = 56 }+7Afe 3 © Properties of Z-Transform
Example 3: Find the z transform of the sequence x(n) =cos (wn),
Solution: Using the property of hyperbolic function,
gion 4 g-iwn
2
From the linearity property of the Z-transform, we have,
cos (wn) =
elon + emion 1 ; 1
X(z) = Z{cos(wn)} =Z ae } = Zntee"} + 5 He}
From the Z-transform table, where a” > (e!)"
x) =3( 3a) +7(—s)
@=>5 z—e] 2\z— ele
x)= z[ 2z— (el + el)
@ = 2|22 -— (el +e) 2 41 © Properties of Z-Transform
Right Shift - the general right shift theorem, also called as delays is,
Z{x(n — 1)} = x(-1) + 271 X(z)
Z{x(n — 2)} = x(—2) + x(—1)z71 + 27? X(z)
Z{x(n — m)} = x(—m) + x(—m $+ 12h +... #x(-1) 2-1 + 2 ™X(z)
For causal sequences where x(—1) = x(—2) = ... = 0 results toa
particularly simple z-transform as:
Z{x(n — m)} = z-™X(z)
Note: in a digital system the right shift corresponds to a delay, and is
represented symbolically as:
x(n) x(n 1) x(n ~2) © Properties of Z-Transform
Example 1: Obtain the z-transform of the sequence w(n) = y(n — 2)
using the right shift method.
Solution: the z-transform of w(n) is,
W(z) = w(0) + w(1)z7! + w(2)z7? + w(3)z3 +... F(z
Since w(n) = y(n — 2), we have
W(z) = y(—2) + y(—1)z71 + y(0)z-? + y(1)z3 +. t y(n — 2)z™
W(z) = y(-2) + (-Dz +27 [yO) + yz 7+. + ¥M)z"]
W(z) = y(—2) + y(-1)z-1 + z-?Y(z) © Properties of Z-Transform
Example 2: Determine the z-transform of the following sequence:
y(n) = (0.5) u(n — 5)
where u(n — 5) = 1 forn = 5 and u(n— 5) = 0 forn <5,
Solution: We first use the shift theorem to have,
Y(z) = Z{(0.5)-u(n — 5)} = z-5Z{(0.5)"u(n)}
Using the table, we have, © Properties of Z-Transform
Example 3: Write the following sequence f(n) as a difference of two
unit step sequences and obtain its z-transform.
f(n)
Solution: The sequence can be represented as,
1 n=0,1,2,3,4
fay= { 0 otherwise
In terms of step response, we have,
fm) = u(n) — u(n— 5) © Properties of Z-Transform
Left Shift - the general left shift theorem, also called as advance is,
Z{x(n + 1)} = zX(z) — zx(0)
Z{x(n + 2)} = z?X(z) — 2?x(0) — zx(1)
Z{x(n + m)} = z™X(z) — 2™x(0) — z™-1x(1) — ... — zx(m— 1)
These left shift theorems have simple forms in special cases:
Ifx(0) =0, Z{x(n + 1)} = 2X(2),
Ifx(0) = x(1) = 0, Zf{x(n + 2)} = z?X(z),
Ifx(0) =x(1) =..x(m—1)=0, Z{x(n+2)}=2z™X(z),
Note carefully the occurrence of positive powers of z in the left shift
theorems and of negative powers of z in the right shift theorems. © Properties of Z-Transform
Example 1: Obtain the z-transform of the sequence y(n + 2) using the
left shift method.
Solution:
Z{y(n + 2)} = y(2) + y(3)2z71 + y(4)z-7 +... Hy(n t+ 2)z™
= y(2) + 2?[y(3)2F + yA)e4 + + yz]
= y(2) + 27[¥(z) — y(0) — yz" — y(2)2-7]
= y@)+ 27Y(z) — y(0)z* — y(1)z- ye)
Zy(n + 2)} = 2¥(z) — y(0)z? — y(1)z © Properties of Z-Transform
Multiplication of a sequence by a”: Suppose f(n) is an arbitrary
sequence with z-transform F(z). Consider the sequence v(n) where,
v(n) = a"f(n)
By the z-transform definition,
Z{v(n)} = f(0) + af (Iz + a f(2)z-7 +...
— y a" f(n)z"
n=0
=> re(2)"
n=0
Thus, we have
aa" f(n)} = F (=) © Properties of Z-Transform
Example 1: Write down the z-transform of the sequence v(n) where,
v(n) =e?" sin 3n
Solution: From the z-transform table,
zsin3
Msin3n} = 7a 7c0s3 +1
Using the property, with a = e~?, we replace z by ze* to obtain,
Z{v(n)} = Zf{e~2" sin 3n}
4
_ ze’ sin 3 e
~ (ze*)? — 2ze?cos3 +1 e4 © Properties of Z-Transform
Example 2: Using the property just discussed write down the z-
transform of the sequence w(n) where,
w(n) = e-“"coswn
Solution: Using the z-transform table, we have,
z* — zcoswn
Zz —2zcos3+1
Z{cos wn }
Using the property, with a = e~®, we replace z by ze® to obtain,
Z{v(n)} = Z{e~*" cos wn}
_ Ge)? — ze*cos wn
~ (ze%)? — 2ze*cos3 +1 © Properties of Z-Transform
Multiplication of a sequence by n - the sequence n can be represented
as a unit ramp as,
0 n=-1,-2,...
r= tn n=0,1,3,..
The z-transform is,
Z{r(n)} = 0+ 1271 + 2277 +...
Thus, the z-transform is,
Z{r(n)} = Z{n} = aay with |z|>1
In general,
dF(z)
Unf (n)} =—z— © Properties of Z-Transform
Example 1: By differentiating the z-transform R(z) of the unit ramp
sequence obtain the z-transform of the causal sequence r(n) = nu(n),
Solution: Using the table,
Z _ Zz
{u(n)} =
Using the property,
Z{nu(n)} =— 20)
_ dp zy_ _@-1)@) - 20)
7 ZG a i) “~ 2 (z-1)? © Properties of Z-Transform
Example 2: By differentiating the z-transform R(z) of the unit ramp
sequence obtain the z-transform of the causal sequence nu(n),
Solution: From the previous example,
Z{nu(n)} = eae
Using the property,
dR
Z{n2u(n)} = Z{n-r(n)} =— 2
_ af 2 ]__, @ ata) -2@-1)
“CIE fe mal ~~ 4 (@—1*
o (@-T[z-1- 22] _ -z-1
Z{n2u(n)} =- —G-pr =- 7G © Properties of Z-Transform
Convolution and z-transforms - Consider a discrete system with
transfer function H(z),
—>—__
X(z) Y(z)
In z-domain, at zero initial conditions, we can get the output Y(z), as,
Y(@) = X(2)H(Z) = Z{x(n) * h(n}
In time-domain, we can get the output y(n), as,
y(n) = x(n) * h(n)
Thus, we can get the convolution of x(n) and h(n) as,
y(n) = x(n) * h(n) = Z{X(@)H(@)} © Properties of Z-Transform
Convolution - given two sequences x,(n) and x2(n), their convolution
can be determined as follows:
x(n) = (2) #20 = Yan x2
k=0
where z designates the linear convolution. In z-transform domain,
X(Z) = X1(@)X2(@)
where, X(z) = Z{x(n)}, X1 (2) = Z{xy(n)}, and X2(z) = Z{x2(n)}, © Properties of Z-Transform
Example 1: Given two sequences below.
x(n) = 36(n) + 26(n — 1)
X2(n) = 26(n) — 6(n — 1)
(a) Find the z-transform of their convolution: X(z) = Z{x,(n) * x2(n)},
(b) Determine the convolution sum using the z-transform:
00
x(n) = 24(n) #x2(n) = Y's OOxa(n = 1)
k=0
Solution: Applying z-transform to x,(n) and x,(n), respectively, it
follows that
X,(z) =3 +2271
X2(z) =2-z1 © Properties of Z-Transform
Solution: Using the convolution property, we have
X(2) = X1@)X2@) = B+ 2272-274)
X(z) =6+271-22?
Applying the inverse z-transform and using the shift theorem from the
table, leads to
x(n) = Z-1{6 + 271 — 2277}
x(n) = 66(n) + 6(n — 1) — 26(n — 2) © Properties of Z-Transform
Example 2: Calculate the convolution y(n) of the sequences,
v(n) =a", w(n) = b”
a) using the definition of convolution
b) using z-transform
Solution: From the formula of convolution, we have,
n n
y(n) = > v(k)w(n — k) = ¥ akpn-k
rg
bn [2 + ; + () tat (| © Properties of Z-Transform
Solution: The bracketed sum is a geometric series of common ratio a/b,
ayntt
- G) pe
y(n) = a
b
(b) Using z-transform,
Zz
Viz) =—, W@=
z-a z—b
n+1
y(n) = ZV (Z)W(2)} = 24 (= a - 5}
y(n) =z {saa}
(@—a@—b) © Properties of Z-Transform
Initial Value Theorem - If f(1) is a sequence with z-transform F(z) then
the initial value f (0) is given by,
f (0) = lim F(z)
z0
provided, that this limit exists.
Final Value Theorem - If f(n) is a sequence with z-transform F(z) then
the final value f (0) is given by,
Fo) =lim (1 = 2-F@)
Note: the final value theorem does not hold for z-transform poles
outside the unit circle. © Properties of Z-Transform
Example 1: Obtain the z-transform of the sequence below, and verify
the initial value theorem for the z-transform pair you obtain,
fm =1-a’, O
©, f(n) is,
f(~) =1-a"=1-0 © Table of Contents
1, Z-Transform
2. Properties of Z-Transform
3. Inverse Z-Transform
4. Solutions of Differential Equation © Inverse Z-Transform
Inverse Z-Transform: The z-transform of the sequence x(n) and the
inverse z-transform of the function X(z) are defined as, respectively,
X(z) = Z{x(n)}
x(n) = Z-"{X(z)}
where Z{} is the z-transform operator, while Z~1{} is the inverse z-
transform operator.
The inverse z-transform may be obtained by at least three methods:
1. Partial fraction expansion and look-up table
2. Power series expansion
3. Residue method. © Inverse Z-Transform
Example 1: Find the inverse z-transform for each of the following
functions:
Az Zz
2 X@) = 2+ 77-05
Pane 5z 2z
= Grn G05"
10z
c) X(z)= Forel
4 zw
d) X(z) =a +2%+ Te © Inverse Z-Transform
Solution: (a) From linearity principle,
x(n) = 20-*{1} — 42-7 5) -a & =a)
Using z-transform the table,
x(n) = 26(n) + 4u(n) — (0.5)"u(n)
(b) From linearity principle,
x(n) = 5072 {=a} — 20-1 {=}
(z-1) (z— 0.5)
xy =sa*{—7 Zz 2} - at 0.52 7}
@—-bJ 05° l@—05)
Using z-transform the table,
x(n) = 5nu(n) — 4n(0.5)"u(n) © Inverse Z-Transform
Solution: (c) Since,
X@) = = (22) (St _)
-—zt+1. \sina/\z? — 2zcosat+1
By coefficient matching, we have,
—2cosa=—1
Hence, cos a = 0.5, and a = 60°. Substituting a = 60° into the sine
function leads to:
sina =sin 60° = 0,866
Finally, we have,
10 zsina
x(n) = saat f
0
——sin (60°n)
1
sina z* — 2zcosa + d= 0.866 © Inverse Z-Transform
Solution: (d) By linearity principle,
x(n) = 21 {-5—} FO 6 +0 {e4*— }
Using the z-transform table, we have, © Inverse Z-Transform
Inverse Z-Transform with Partial Fraction Decomposition:
1. Partial fraction with the first-order real pole:
A= R=e-py 2)
zZ—p z=)
2. Partial fraction with the first-order complex poles:
Ae AZ ga PX
Zz-P z—-P* Z |ep
3. Partial fraction with mth-order real poles:
Rn Rm-1 Ry
zp @-pe Gp
_ 1 dk mX(2)
= Gale]
Z=p © Inverse Z-Transform
Residue Method - has its basis in a branch of mathematics called
complex integration. In many cases inversion using residues is easier
than using partial fractions.
+ For First Order Pole:
R = Res[G(2),p] = (@ — p)G@) |
z=p
+ For Nth Order Pole:
1 Kk=1
(k- Diaz real
Ry, = Res|G(2), pli = @-ry" | | © Inverse Z-Transform
Example 1: Find the inverse of the following z-transform:
1
X(z) = ————————
@ = G=-pja ose)
Solution: Eliminating the negative power of z by multiplying the
numerator and denominator by z? yields:
2 2
X@ = 2G=7DG-082) — @=DE-05)
Dividing both sides by z leads to,
X(z) = lea = [A+]
(z-1)(@- 0.5). z-1 2-05.
By residue method, we have,
X(z Zz 1
A= Res © |- “T7057 © Inverse Z-Transform
Solution: Continuation,
B=R x@) 0:5 |= = 05 _ 1
Nes TT Teos 05-1
22 Zz
z-1 2-05
+
x(n) = 2u(n) — (0.5)"u(n) © Inverse Z-Transform
Example 2: Find y(n) if Y(z) is given as,
2(z+1)
’@) = G-pe-2 405)
Solution: Dividing Y(z) by z, we have,
2(z +1)
¥@) = le —D(@—z+t 05)
Applying the partial fraction expansion leads to,
yi [ot A A*
@ =z) 4+ 7-05-j05 'z—-054)05
We first find B as:
Y(z) 1|= 2(z +1)
z’ | 2-2+05
_— 1041)
a1 127-1405
B= Res| © Inverse Z-Transform
Solution: Notice that A and A* form a complex conjugate pair. We
determine A as follows:
Y(z
A= es| @
Zz
z(z+1)
,0.5 + j0.$| = ————__-__
(@-1D@-05 + j05)I_5 ne
(0.5 + j0.5)(0.5 + 0.5 + 1) (0.5 + j0.5)(1.5 + j0.5)
~(5+j05—-105+j05—05+j05) (-05+j05)j
Using the polar form, we get,
_ 0.707/45°)(1.58114/18,43°) _ 5
4=—anoT ssa 90) = 158114/= 161.57
A* = 1.58114/161.57°
Assume that a first-order complex pole has the form,
P=0.5 +0.5j =|P|/@ = 0.707/45° and P* = 0.707/— 45° © Inverse Z-Transform
Solution: Thus, we have,
Az Az A‘z
z—-1 2—-P\z—P*
Using the linearity principle, we have,
= 47-1 Zz oa AZ A‘z
y(n) = {—}+ (otal
Y@) =
Subsequent simplification yields to,
y(n) = 4u(n) + 2|A||P|"cos (nd + ) u(n) © Inverse Z-Transform
Example 3: Find x(n) if X(z) is given as:
Zz
X(z) = ———__——
@=G-pe-05
Solution: Dividing both sides of the previous z-transform by z yields,
X(z) Zz A B Cc
2 = + + 2
z (2-12-05)? z-1 z-05° (z-0.5)
where,
_ X(z) _ Zz
A= res| Zz | ~@=05)",_,
Using the formulas for mth-order real poles in Table 5.3, where m = 2
and p = 0.5, to determine B and C yields
@ o<| - 1
08) aay wale"...
Xx |
z=
B= Ry = Res |= © Inverse Z-Transform
Solution: Continuation...
d Zz -1
=D" =A, a
C=R,=— 0.5)? 2 =-1
Hh = GT sale- ) fe oe
Thus, we have,
—4z —1z
+7 ep
z-1 z-05 (z—0.5)
Taking the inverse z-transform, we have,
Gee tat oe { Z =a } ~ age {=}
x(n) )
X(z) =
u(n) — 4(0.5)"u(n) — 2n(0.5)"u(; © Table of Contents
1, Z-Transform
2. Properties of Z-Transform
3. Inverse Z-Transform
4. Solutions of Difference Equation © Solution of Difference Equations
Solution of Difference Equation Using Z-Transform: Using the shift
theorem of the z-transform, where,
Right Shift Theorem:
Z{x(n — m)} = x(—m) + x(—m + Iz) +... + x(-1) 2-1 + 2 ™X(z)
+ Left Shift Theorem:
Z{x(n + m)} = 2™X(z) — 2™x(0) — 2z™1x(1) — ... — zx(m — 1)
Procedure for the Solution:
1. Apply z-transform to the difference equation.
Substitute the initial conditions.
Solve for the difference equation in z-transform domain.
- Y N
Find the solution in time domain by applying the inverse z-
transform. © Solution of Difference Equations
Example 1: Solve the difference equation,
y(n+1)-y(m) =4d, n=0,1,2,.. y(0)=a
where a and d are constants.
Solution: Applying z-transform for both side of the equation, we have,
dz
2¥(2) ~ zy(0) — ¥(@) ==
Substituting the initial condition y(0) = 0 and solving for Y(z), we got,
(Z-DY@= a +az
Zils
dz az
YO= Groat zai
Then, we take the inverse z-transform to get y(n), © Solution of Difference Equations
Solution: Applying inverse z-transform for both sides,
yen) = IY ay) = dt fh 4 at f=}
y(n) = dnu(n) + au(n)
y(n) = [a + dnju(n)
y(n) =at+dn n=0,1,2,...
Note: This solution give the nth term of an arithmetic sequence with a
constant difference d and initial term a. © Solution of Difference Equations
Example 2: Solve the difference equation,
ym) -ym-1)=d n=0,1,2,... y(-1) =a
where a and d are constants.
Solution: Start by taking the z-transform of the equation as,
dz
¥(@) - e*¥@) + x) =
1
Y(2)Q-z74)-a= =
¥(2) &G = *) = — +a
dz? az
Y(z) =
sy Ss
(2-1)? z-1 © Solution of Difference Equations
Solution: Taking the inverse z-transform, we have,
y(n) = Z{Y(z)} = dz* {z . am} +aZ* { = 1 }
y(m) = d(n + 1)u(n) + au(n)
y(n) = [a+ (r+ 1)dJu(n) © Solution of Difference Equations
Example 3: A digital signal processing (DSP) system is described by the
difference equation below. Determine the solution when the initial
condition is given by y(—1) = 1.
y(n) — 0.5y(n — 1) = 5(0.2)"u(n)
Solution: Applying the z-transform on both sides of the difference
equation, we have
Y¥(z) — 0.5[y(-1) + 2-1¥(z)] = 5Z{0.2"u(n)}
Substituting the initial condition and Z{0.2"u(n)} = z/(z — 0.2), we got,
Y(z) — 0.5[1 + 2-1Y(z)] = 52/(z — 0.2)
Y(z) — 0.5271Y(z) = 0.5 + 5z/(z — 0.2)
Factoring out Y(z) and combining the right-hand side of the equation, it
follows that
5.5z—-0.1
¥(2)(1— 0.5271) = = © Solution of Difference Equations
Solution: Then we obtain,
5.5z—0.1 2(5.5z — 0.1)
Y@) = ooo ooo
(1-052 (¢-0.2) (z—0.5)(z—0.2)
Using the partial fraction expansion method leads to,
Y@) = 5.5z—-0.1 |= f A i B
@=2\G-o5G@-02)~*lz-05 +202
where,
¥(2) 5.5z—0.1
A= Res ,0.5|=-——"| = 8.8333
z—0.2 I-05
Y(z) 5.5z — 0.1
B= Res 0.2) =" | = 3.3333
Z-0.5 |z=0.2
8.8333 —3.3333] 8.8333z —3.3333z
Y(a)= + =
esas 0128 are eee 02 © Solution of Difference Equations
Solution: Taking the inverse z-transform, we have,
y(n) = ZY (2)} = 8.83332-2 {—_} — 3.33332 { ; = 3}
Zz — 0.5.
y(n) = 8.3333(0.5)"u(n) — 3.3333(0.2)"u(n)
y(n) = [8.3333(0.5)" — 3.3333(0.2)"]u(n)
y(n) = 8.3333(0.5)" — 3.3333(0.2)"” n=0,1,2, © Solution of Difference Equations
Example 4: By solving the second order difference equation below,
obtain the general term y(n) of the Fibonacci sequence.
y(n +2) =y(n+ 1) + ym)
y@) =y@)=1
Solution: Taking the z-transform for both equation,
2Y(z) — z?y(0) — zy(1) = z¥(z) — zy(0) + Y@)
PY (2) — 2? 5/4 = 2Y(z) 2+ Y(2)
Y(2)(z* -z-1) = 2?
Y(z) = a
@= z—z-1
Next, solve the quadratic equation z* — z — 1 = 0 and hence factorize
the denominator of Y(z): © Solution of Difference Equations
Solution: Solving for z* — z—1=0,
a=05+V5/2, b=0.5—V5/2
Thus, we have,
2 2
Z—z-1 (@-ae-b)
By partial fraction decomposition,
Y@=
B
De «le-aezpl =? Lats]
Y(z) _ 2 _ a
A= res| z l= o-s5
(z) Zz re)
Be res| | zaalep "ba © Solution of Difference Equations
Solution: By taking the inverse z-transform, we have,
y(n) = £4¥(@)} = az {—_} + Ba i)
Z— a
y(n) = Aa” + Bb” = (+) a+ (-) bn
nti — pnt
WO = 5
Substituting the values of a = 0.5 + v5/2, and b = 0.5 — v5/2, where,
a- b= (0.5 + V5/2) — (0.5 — V5/2) = V5 © Solution of Difference Equations
Example 5: A relaxed (zero initial conditions) DSP system is described
by the difference equation,
y(n) + 0.1y(n — 1) — 0.2y(n — 2) = x(n) + x(n — 1)
a) Determine the impulse response y(n) due to the impulse sequence
x(n) = 6(n)
b) Determine system response y(n) due to the unit step function
excitation, where u(n) = 1 forn > 0.
Solution: Applying the z-transform on both sides of the difference
equations, we yield,
Y(z) + 0.1¥(z)z~4 — 0.2Y(z)z~* = X(z) + X(z)z
Factoring out Y(z) on the left side and substituting X(z) = Z{5(n)} = 1,
Y(z)(1 + 0.1271 — 0.2277) =14+271
1+z1
Y(z) =
@ = Ty0ar* oar © Solution of Difference Equations
Solution: To obtain the impulse response, which is the inverse z-
transform of the transfer function, we multiply the numerator and
denominator by z?,
+z z(z+1)
Y@) == =
z*+01z-0.2 (z—0.4)(z+0.5)
Using the partial fraction expansion method leads to,
Y _ zt+1 | _ A a B
@ =e =p HG +05) ~7lz—-04 77405
where,
Y(z) zt+1
A= Res 0.4) = = 1.5556
Z+05l220.4
Y(z) zt+i1
B= Res 0.5] = =— 0.5556
4 Z+0Al,--0.5 © Solution of Difference Equations
Solution: Thus,
1.5556z —0.5556z
7-04) 2405
which gives the impulse response,
y(n) = 1.5556(0.4)"u(n) — 0.5556(—0.5)"u(n)
(b) To obtain the response due to a unit step function, the input
sequence is set to be x(n) = u(n), and the z-transform is then,
Zz
Y(z) =
X@Z=
z-1
Thus, taking the z-transform,
Y(z)(1 + 0.1274 — 0.2272) = sa +27)
Wee 1+z7 _ 2(z+1)
QT (; +0427 — a=) @—DE—04(z +05) © Solution of Difference Equations
Solution: Using the partial fraction expansion method as before gives,
2.2222z -—1.0370z —0.1852z
"@= 7 +04 + 7405
and the system response is found by using the z-transform table as,
y(n) = 2.2222u(n) — 1.0370(0.4)"u(n) — 0.1852(—0.5)"u(n)
y(n) = [2.2222 — 1.0370(0.4)" — 0.1852(—0.5)"]u(n)