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3
The Z Transform
Exercises 3.2.3
1(a)
F(z) =
k=0(1/4)k
zk=
1
1
1/4z
=4z
4z
1
if | z |> 1/4
1(b)
F(z) =
k=0
3k
zk=
1
1 3/z =z
z 3 if | z |> 3
1(c)
F(z) =
k=0(2)k
zk=
1
1
(
2)/z
=z
z + 2if | z |> 2
1(d)
F(z) =
k=0
(2)kzk
= 11 2/z =
z
z 2 if | z |> 2
1(e)
Z{k} = z(z 1)2 if | z |> 1
from (3.6) whence
Z{3k} = 3 z(z 1)2 if | z |> 1
2
uk = e2kT =
e2T
kwhence
U(Z) =z
z
e2T
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Exercises 3.3.6
3
Z{sin kT} = 12
z
z eT 1
2
z
z eT
=z sin T
z2 2z cos T + 1
4
Z{1
2k
} = 2z2z
1
so
Z{yk} = 1z3 2z
2z 1 =2
z2(2z 1)Proceeding directly
Z{yk} =k=3
xk3zk
=r=0
xrzr+3
=1
z3Z {xk} = 2
z2(2z 1)
5(a)
Z1
5
=
r=0
15z
r=
5z
5z + 1| z |> 1
5
5(b)
{cos k} = (1)kso
Z {cos k
}=
z
z + 1 |z
|> 1
6
Z
1
2
k=
2z
2z 1By (3.5)
Z(ak) = zz a
so
Z(kak1) =z
(z a)2c
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9 Final value theorem
(1 z1)X(z) =r=0
xr xr1zr
= x0 +x1 x0
z+
x2 x1z2
+ . . . +xr xr1
zr+ . . .
As z 1 and if limr xr exists, then
limz1
(1 z1)X(z) = limr
xr
10 Multiplication property (3.19): Let Z {xk} =
k=0xkzk
= X(z) then
Zakxk = k=0
akxkzk
= X(z/a)
10 Multiplication property (3.20)
z ddz
X(z) = z ddz
k=0
xkzk
=k=0
kxkzk
= Z {kxk}
The general result follows by induction.
Exercises 3.4.2
11(a)z
z 1 ; from tables uk = 1
11(b)z
z + 1=
z
z (1) ; from tables uk = (1)k
11(c)z
z
1/2
; from tables uk = (1/2)k
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11(d)
z3z + 1
= 13
zz + 1/3
13
(1/3)k
11(e)z
z ; from tables uk = ( )k
11(f)z
z +
2
=z
z (
2)
(
2)k
11(g)1
z 1 =1
z
z
z 1
0; k = 01; k > 0
using first shift property.
11(h)z + 2
z + 1 = 1 +
1
z
z
z + 1 1; k = 0(1)k1; k > 0=
1; k = 0(1)k+1; k > 0
12(a)
Y(z)/z =1
3
1
z 1 1
3
1
z + 2
so
Y(z) =1
3
z
z 1 1
3
z
z + 2 1
3
1 (2)k
12(b)
Y(z) =1
7
z
z 3 z
z + 1/2
1
7
(3)k (1/2)k
12(c)
Y(z) =1
3
z
z
1
+1
6
z
z + 1/2 1
3+
1
6(1/2)k
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12(d)
Y(z) = 23
zz 1/2 23 zz + 1 23 (1/2)
k 23
(1)k
=2
3(1/2)k +
2
3(1)k+1
12(e)
Y(z) =1
2
z
z z
z ( )
=
1
2 z
z e /2 z
z e /2 1
2
(e /2)k (e /2)k
= sin k/2
12(f)
Y(z) =z
z (3 + ) z (3 )=
1
2 z
z
(
3 + ) z
z
(
3
)=
1
2
z
z 2e /6 z
z 2e /6
12
2kek/6 2kek/6
= 2k sin k/6
12(g)
Y(z) =5
2
z
(z 1)2 +1
4
z
z 1 1
4
z
z 3
52
k +1
4
1 3k
12(h)
Y(z)/z =z
(z 1)2(z2 z + 1) =1
(z 1)2 1
z2 z + 1so
Y(z) =z
(z 1)2 13
z
z
1+3
2
zz
13
2
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= z(z 1)2
13
z
z e /3 z
z e /3
k 23
sin k/3 = k +2
3cos(k/3 3/2)
13(a)
X(z) =
k=0xkzk
=1
z+
2
z7
whence x0 = 0 , x1 = 1 , x2 = x3 = . . . = x6 = 0, x7 = 2 and xk = 0, k > 7 .
13(b) Proceed as in Example 13(a).
13(c) Observe that
3z + z2 + 5z5
z5= 5 +
1
z3+
3
z4
and proceed as in Example 13(a).
13(d)
Y(z) =1
z2+
1
z3+
z
z + 1/3
{0, 0, 1, 1}+ {(1/3)k}
13(e)Y(z) = 1 +
3
z+
1
z2 1/2
z + 1/2
{1, 3, 1} 12
0, k = 0(1/2)k, k 1
=
1, k = 05/2, k = 15/4, k = 2
12
(
1/2)k1, k
3
=
1, k = 05/2, k = 15/4, k = 2
18
(
1/2)k3, k
3
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13(f)
Y(z) = 1z 1
2(z 1)2 +
1z 2
0, k = 01 2(k 1) + 2k1, k 1
=
0, k = 03 2k + 2k1, k 1
13(g)
Y(z) =2
z 1 1
z 2
0, k = 02 2k1, k 1
Exercises 3.5.3
14(a) If the signal going into the left D-block is wk and that going into the right
D-block is vk , we have
yk+1 = vk, vk+1 = wk = xk 12
vk
so
yk+2 = vk+1 = xk 12
vk
= xk 1
2 vk = xk 1
2 yk+1
i.e.
yk+2 +1
2yk+1 = xk
14(b) Using the same notation
yk+1 = vk, vk+1 = wk = xk 14
vk 15
yk
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Then
yk+2 = xk 14
yk+1 15
yk
or
yk+2 +1
4yk+1 +
1
5yk = xk
15(a)
z2Y(z) z2y0 zy1 2(zY(z) zy0) + Y(z) = 0
with y0 = 0, y1 = 1
Y(z) =z
(z 1)2so yk = k, k 0.
15(b) Transforming and substituting for y0 and y1
Y(z)/z =2z 15
(z 9)(z + 1)so
Y(z) =3
10
z
z 9 17
10
z
z + 1
thus
yk =3
109k 17
10(1)k, k 0
15(c) Transforming and substituting for y0
and y1
Y(z) =z
(z 2 )(z + 2 )
=1
4
z
z 2e /2 z
z 2e /2
thus
yk =1
42k e
k/2 ek/2 = 2k1 sin k/2, k 0
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15(d) Transforming, substituting for y0 and y1 , and rearranging
Y(z)/z =6z 11
(2z + 1)(z 3)so
Y(z) = 2z
z + 1/2+
z
z 3thus
yk = 2(1/2)k + 3k, k 0
16(a)6yk+2 + yk+1 yk = 3, y0 = y1 = 0
Transforming with y0 = y1 = 0,
(6z2 + z 1)Y(z) = 3zz 1
so
Y(z)/z =3
(z 1)(3z 1)(2z + 1)
andY(z) =
1
2
z
z 1 9
10
z
z 1/3 +2
5
z
z + 1/2
Inverting
yk =1
2 9
10(1/3)k +
2
5(1/2)k
16(b) Transforming with y0 = 0, y1 = 1,
(z2 5z + 6)Y(z) = z + 5 zz
1
whence
Y(z) =5
2
z
z 1 +7
2
z
z 3 6z
z 2so
yk =5
2+
7
2(3)k 6(2)k
16(c) Transforming with y0 = y1 = 0 ,
(z
2
5z + 6)Y(z) =z
z 1/2c
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so
Y(z) = 415
zz 1/2 23 zz 2 + 25 zz 3
whence
yn =4
15(1/2)k 2
3(2)k +
2
5(3)k
16(d) Transforming with y0 = 1, y1 = 0 ,
(z2 3z + 3)Y(z) = z2 3z + zz 1
so
Y(z) =z
z 1 z
z2 3z + 3
=z
z 1 13j
z
z 3+3j
2
zz 3
3j
2
=z
z 1 13j
z
z 3ej/6 z
z 3ej/6
so
yn = 1 23
(3)k ejn/6 ejn/62j
= 1 2(3)n1 sin n/6
16(e) Transforming with y0 = 1, y1 = 2
(2z2 3z 2)Y(z) = 2z2 + z + 6 z(z 1)2 +
z
z 1so
Y(z) =z
z 2+ z z + 5
(z 1)2
(2z + 1)(z 2)=
12
5
z
z 2 2
5
z
z + 1/2 z
z 1 2z
(z 1)2so
yn =12
5(2)n 2
5(1/2)n 1 2n
16(f) Transforming with y0 = y1 = 0,
(z
2
4)Y(z) = 3z
(z 1)2 5z
z 1c
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so
Y(z) =z
z 1 z
(z 1)2 1
2
z
z 2 1
2
z
z + 2
and
yn = 1 n 12
(2)n 12
(2)n
17 Write the transformed equations in the formz 3/20.21
1
z 1/2
c(z)
e(z)
=
zC0zE0
Then c(z)
e(z)
=
1
z2 2z + 0/96
z 1/20.21
1z 3/2
zC0zE0
Solve for c(z) as
c(z) = 1200z
z 1.2 + 4800z
z 0.8and
Ck = 1200(1.2)k + 4800(0.8)k
This shows the 20% growth in Ck in the long term as required.
Then
Ek = 1.5Ck Ck+1= 1800(1.2)k + 7200(0.8)k 1200(1.2)k+1 4800(0.8)k+1
Differentiate wrt k and set to zero giving
0.6 log(1.2) + 5.6x log(0.8) = 0 where x = (0.8/1.2)k
Solving, x = 0.0875 and so
k =log0.0875
log(0.8/1.2)= 6.007
The nearest integer is k = 6, corresponding to the seventh year in view of the
labelling, and C6 = 4841 approx.
18 Transforming and rearranging
Y(z)/z =
z
4
(z 2)(z 3) +1
(z 1)(z 2)(z 3)c
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so
Y(z) = 12
zz 1 + zz 2 12 zz 3
thus
yk =1
2+ 2k 1
23k
19
Ik = Ck + Pk + Gk
= aIk1
+ b(Ck
Ck1
) + Gk
= aIk1 + ba(Ik1 Ik2) + Gkso
Ik+2 a(1 + b)Ik+1 + abIk = Gk+2Thus substituting
Ik+2 Ik+1 + 12
Ik = G
Using lower case for the z transform we obtain
(z2 z + 12
)i(z) = (2z2 + z)G + Gz
z 1whence
i(z)/z = G
1
z2 z + 12+
2
z 1
= G
2
z 1 +1
(z
1+2 )(z
12 )
so
i(z) = G
2
z
z 1 +2
2
z
z 12
e /4 z
z 12
e /4
Thus
Ik = G
2 +
2
2(
12
)k
ek/4 ek/4
= 2G
1 +
1
2
ksin k/4
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20 Elementary rearrangement leads to
in+2 2 cosh in+1 + in = 0
with cosh = 1 + R1/2R2 . Transforming and solving for I(z)/z gives
I(z)/z =zi0 + (i1 2i0 cosh )
(z e)(z e)
=1
2 sinh
i0e
+ (i1 2i0 cosh )z e
i0e + (i1 2i0 cosh )
z e
Thus
ik =(i0e
+ (i1 2i0 cosh ))en (i0e + (i1 2i0 cosh ))en2 sinh
=1
sinh {i1 sinh n i0 sinh(n 1)}
Exercises 3.6.5
21 Transforming in the quiescent state and writing as Y(z) = H(z)U(z) then
21(a)
H(z) =1
z2 3z + 2
21(b)
H(z) =z 1
z2 3z + 2
21(c)
H(z) =1 + 1/z
z3 z2 + 2z + 1
22 For the first system, transforming from a quiescent state, we have
(z2 + 0.5z + 0.25)Y(z) = U(z)
The diagram for this is the standard one for a second order system and is shown
in Figure 3.1 and where Y(z) = P(z), that is yk = pk .
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Figure 3.1: The block diagram for the basic system of Exercise 22.
Transforming the second system in the quiescent state we obtain
(z2 + 0.5z + 0.25)Y(z) = (1 0.6)U(z)
Clearly
(z2 + 0.5z + 0.25)(1 0.6z)P(z) = (1 0.6z)U(z)indicating that we should now set Y(z) = P(z)
0.6zP(z) and this is shown in
Figure 3.2.
Figure 3.2: The block diagram for the second system of Exercise 22
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23(a)
Y(z)/z = 1(4z + 1)(2z + 1)
so
Y(z) =1
2
z
z + 1/4 1
2
z
z + 1/2
yk =1
2(1/4)k 1
2(1/2)k
23(b)
Y(z)/z =z
z2
3z + 3whence
Y(z) =3 +
3
2
3
z
z (3+3 )
2
3
3
2
3
z
z (33 )
2
so
yk =3 +
3
2
3(
3)kek/6 3
3
2
3(
3)kek/6
= 2(
3)k
3
2sin k/6 +
1
2cos k/6
= 2(3)k sin(k + 1)/6
23(c)
Y(z)/z =z
(z 0.4)(z + 0.2)so
Y(z) =2
3
z
z 0.4 +1
3
z
z + 0.2
then
yk =
2
3 (0.4)k
+
1
3 (0.2)k
23(d)
Y(z)/z =5z 12
(z 2)(z 4)so
Y(z) =z
z 2 + 4z
z 4and
yk = (2)k + (4)k+1
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24(a)
Y(z) =1
z2 3z + 2
=1
z 2 1
z 1
yk =
0, k = 0
2k1 1, k > 0
24(b)
Y(z) =1
z 2so
yk =
0, k = 0
2k1, k > 0
25 Examining the poles of the systems, we find
25(a) Poles at z = 1/3 and z = 2/3, both inside | z |= 1 so the system isstable.
25(b) Poles at z = 1/3 and z = 2/3, both inside | z |= 1 so the system isstable.
25(c) Poles at z = 1/2 1/2 , | z |= 1/2, so both inside | z |= 1 and the
system is stable.
25(d) Poles at z = 3/4 17/4, one of which is outside | z |= 1 and so thesystem is unstable.
25(e) Poles at z = 1/4 and z = 1 thus one pole is on | z |= 1 and the other isinside and the system is marginally stable.
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26 To use the convolution result, calculate the impulse response as y,k (1/2)k .Then the step response is
yk =k
j=0
1 (1/2)kj = (1/2)kk
j=0
1 (2)j = (1/2)k 1 (2)k+1
1 2
= (1/2)k(2k+1 1) = 2 (1/2)k
Directly,
Y(z)/z =z
(z
1/2)(z
1)
=2
z
1 1
z
1/2
so
yk = 2 (1/2)k
27 Substituting
yn+1 yn + Kyn1 = K/2n
or
yn+2 yn+1 + Kyn = K/2n+1
Taking z transforms from the quiescent state, the characteristic equation is
z2 z + K = 0
with roots
z1 =1
2+
1
2
1 4Kand z2 = 1
2 1
2
1 4K
For stability, both roots must be inside | z |= 1 so if K < 1/4 then
z1 < 1
1
2
+1
2
1
4K < 1
K > 0
and
z2 > 1 12 1
2
1 4K > 1 k > 2
If K > 1/4 then
| 12
+1
2
4K 1 |2< 1 K < 1
The system is then stable for 0 < K < 1.
When k = 2/9 we have
yn+2 yn+1 +2
9 yn =
1
9
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Transforming with a quiescent initial state
(z2 z + 29
)Y(z) =1
9
z
z 1/2so
Y(z) = z1
9
1
(z 1/2)(z 1/3)(z 2/3)
= 2z
z 1/3 + 2z
z 2/3 4z
z 1/2which inverts to
yn = 2(1/3)n + 2(2/3)n 4(1/2)n
28
z2 + 2z + 2 = (z (1 + ))(z (1 + ))
establishing the pole locations. Then
Y(z) =1
2
z
z
(
1 + )
12
z
z
(
1
)
So since (1 ) = 2e3 /4 etc.,
yk = (
2)k sin3k/4
Exercises 3.9.6
29H(s) =
1
s2 + 3s + 2
Replace s with2
z 1z + 1
to give
H(z) =2(z + 1)2
4(z 1)2 + 6(z2 1) + 22(z + 1)2
=
2(z + 1)2
(4 + 6 + 22)z2 + (42 8)z + (4 6 + 22)c
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This corresponds to the difference equation
(Aq2 + Bq + C)yk = 2(q2 + 2q+ 1)uk
where
A = 4 + 6 + 22 B = 42 8 C = 4 6 + 22
Now put q = 1 + to get
(A22 + (2A + B)+ A + B + C)yk
= 2(22 + 4+ 4)uk
With t = 0.01 in the q form the system poles are at z = 0.9048 and z = 0.8182,
inside | z |= 1. When t = 0.01 these move to z = 0.9900 and z = 0.9802,closer to the stability boundary. Using the form with t = 0.1, the poles are at
= 1.8182 and = 0.9522, inside the circle centre (10, 0) in the -planewith radius 10. When t = 0.01 these move to = 1.9802 and = 0.9950,within the circle centre (100, 0) with radius 100 , and the closest pole to the
boundary has moved slightly further from it.
30 The transfer function is
H(s) =1
s3 + 2s2 + 2s + 1
To discretise using the bi-linear form use s 2T
z 1z + 1
to give
H(z) = T3
(z + 1)3
Az3 + Bz2 + Cz + D
and thus the discrete-time form
(Aq3 + Bq2 + Cq+ D)yk = T3(q3 + 3q2 + 3q+ 1)uk
where
A = T3 + 4T2 + 8T + 8, B = 3T3 + 4T2 8T 3,
C = 3T3 4T2 8T + 3, D = T3 4T2 + 8T 1c
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To obtain the form use s
2
2 +
giving the transfer function as
(2 + )3
A3 + B2 + C+ D
This corresponds to the discrete-time system
(A3 + B2 + C+ D)yk = (33 + 222 + 4+ 8)uk
where
A = 3 + 42 + 8 + 8, B = 62 + 16 + 16,
C = 12 + 16, D = 8
31 Making the given substitution and writing the result in vector-matrix form
we obtain
x(t) =
0
21
3
x(t) +
0
1
u(t)
and
y(t) = [1, 0]x(t)
This is in the general form
x(t) = Ax(t) + bu(t)
y = cTx(t) + d u(t)
The Euler discretisation scheme gives at once
x((k + 1)) = x(k ) + [Ax(k ) + bu(k )]
Using the notation of Exercise 29 write the simplified form equation as
2 +
12 + 8
A+
8
A
yk =
1
A
22 + 4+ 4
uk
Now, as usual, consider the related system
2
+
12 + 8
A +
8
A
pk = uk
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and introduce the state variables x1(k) = pk , x2(k) = pk together with the
redundant variable x3(k) = 2pk . This leads to the representation
x(k) =
0 1 8
A12 + 8
A
x(k) + 0
1
u(k)
yk =
4
A 8
2
A2
,
4
A (12 + 8)
2
A2
x(k) +
2
Au(k)
or
x(k + 1) = x(k) + [A()x(k) + bu(k)]
yk = cT()x(k) + d()uk
Since A(0) = 4 it follows that using A(0), c(0) and d(0) generates the Euler
Scheme when x(k) = x(k) etc.
32(a) In the z form substitution leads directly to
H(z) =12(z2 z)
(12 + 5)z2 + (8 12)z
When = 0.1 this gives
H(z) =12(z2 z)
12.5z2 +11.2z 0.1
(b) The form is given by replacing z by 1 + . Substitution and
rearrangement gives
H() =12(1 + )
2(12 + 5) + (8 12) + 12
when = 0.1 this gives
H() =12(1 + 0.1)
1.252 11.2+ 12
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Glyn James: Advanced Modern Engineering Mathematics, Third edition 181
Review exercises 3.10
1
Z {f(kT)} = Z {kT} = TZ {k} = T z(z 1)2
2
Zak sin k = Z
ak(e k e k)2
=1
2Z(ae )k (ae )k
=1
2
z
z ae z
z ae
=az sin
z2 2az cos + a2
3 Recall that
Zak = z(z a)2
Differentiate twice wrt a then put a = 1 to get the pairs
k z(z 1)2 k(k 1)
2z
(z 1)3
then
Zk2 =
2z
(z 1)3
+z
(z 1)2
=z(z + 1)
(z 1)3
4
H(z) =3z
z 1 +2z
(z 1)2
so inverting, the impulse response is
{3 + 2k}
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5
YSTEP(z) =
z
(z + 1)(z + 2)(z 1)= 1
2
z
z + 1+
1
3
z
z + 2+
1
6
z
z 1Thus
ySTEP,k = 12
(1)k + 13
(2)k + 16
6
F(s) =1
s + 1=
1
s 1
s + 1
which inverts to
f(t) = (1 et)(t)where (t) is the Heaviside step function, and so
F(z) = Z {f(kT)} = zz 1
z
z eTThen
esTF(s) f((t T))
which when sampled becomes f((k 1)T) and
Z {f((k 1)T)} =k=0
f((k 1)T)zk
=1
zF(z)
That is
esTF(s) 1z
F(z)
So the overall transfer function is
z
1
z z
z 1 z
z eT
=1
eT
z eT
7
H(s) =s + 1
(s + 2)(s + 3)=
2
s + 3 1
s + 2
y(t) = 2e3t e2t {2e3kT e2kT}
so
H(z) = 2z
z
e3T z
z
e2T
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Glyn James: Advanced Modern Engineering Mathematics, Third edition 183
8(a) Simple poles at z = a and z = b . The residue at z = a is
limza
(z a)zn1X(z) = limza
(z a) zn
(z a)(z b) =an
a b
The residue at z = b is similarlybn
b a and the inverse transform is the sumof these, that is
an bna b
8(b)(i) There is a only double pole at z = 3 and the residue is
limz3
d
dz(z 3)2 z
n
(z 3)2 =
n3n1
(ii) There are now simple poles at z =1
2
3
2 . The individual residues are
thus given by
limz(1/2
3/2 )
12
32
n
3Adding these and simplifying in the usual way gives the inverse transform
as 2
3sin n/3
9
H(z) =z
z + 1 z
z
2
so
YSTEP(z) =
z
z + 1 z
z 2
z
z 1= 3z
(z 1)(z + 1)(z 2)=
3
2
z
z 1 +1
2
z
z + 1 2 z
z 2so
ySTEP,k =3
2+
1
2(1)k 2k+1
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Glyn James: Advanced Modern Engineering Mathematics, Third edition 185
leading to the difference equation
yn+2 0.9744yn+1 + 0.2231yn = 0.5xn+2 0.4226xn+1
As usual (see Exercise 22), draw the block diagram for
pn+2 0.9744pn+1 + 0.2231pn = xn
then taking yn = 0.5pn+2 0.4226pn+1
yn+2 0.9744yn+1 + 0.2231yn = 0.5pn+4 0.4226pn+3
0.9774(0.5pn+3 0.4226pn+2) + 0.2231(0.5pn+2 0.4226pn+1)
= 0.5xn+2 0.4226xn+1
13yn+1 = yn + avn
vn+1 = vn + bun
= vn + b(k1(xn yn) k2vn)= bk1(xn yn) + (1 bk2)vn
so
yn+2 = yn+1 + a[bk1(xn yn) + (1 bk2)vn]
(a) Substituting the values for k1 and k2 we get
yn+2 = yn+1 + 14
(xn yn)
or
yn+2 yn+1 + 14
yn =1
4xn
Transforming with relaxed initial conditions gives
Y(z) =1
(2z 1)2 X(z)
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(b) When X(z) =A
z 1,
Y(z) =A
4
4
z
z 1 4z
z 1/2 2z
(z 1/2)2
then
yn =A
4
4 4(1/2)n 2n(1/2)n1
14 Substitution leads directly to
yk 2yk1 + yk2T2
+ 3yk yk1
T+ 2yk = 1
Take the z transform under the assumption of a relaxed system to get
[(1 + 3T z + 2T2)z2 (2 + 3T)z + 1]Y(z) = T2 z3
z 1The characteristic equation is thus
(1 + 3T z + 2T2)z2 (2 + 3T)z + 1 = 0
with roots (the poles)
z =1
1 + T, z =
1
1 + 2T
The general solution of the difference equation is a linear combination of these
together with a particular solution. That is
yk = 11 + T
k
+ 11 + 2T
k
+
This can be checked by substitution which also shows that = 1/2 . The
condition y(0) = 0 gives y0 = 0 and since y(t) yk yk1
T, y(0) = 0
implies yk1 = 0. Using these we have
+ +1
2= 0
(1 + T) + (1 + 2T) +1
2= 0
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with solution =
1, = 1/2 so
yk =
1
1 + T
k+
1
2
1
1 + 2T
k+
1
2
The differential equation is simply solved by inverting the Laplace transform
to give
y(t) =1
2(e2t 2et + 1), t 0
Figure 3.3: Response of continuous and discrete systems in Exercise 14 over
10 seconds when T = 0.1
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Figure 3.4: Response of continuous and discrete systems in Exercise 14 over
10 seconds when T = 0.05
15 Substitution for s and simplifying gives
[(4 + 6T + 2T2)z2 + (4T2 8)z + (4 6T + 2T2)]Y(z)= T2(z + 1)2X(x)
The characteristic equation is
(4 + 6T + 2T2)z2 + (4T2 8)z + (4 6T + 2T2) = 0
with roots
z =8 4T2 4T
2(4 + 6T + 2T2)
That is
z =
1
T
1 + T and z =
2
T
2 + T
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Glyn James: Advanced Modern Engineering Mathematics, Third edition 189
The general solution of the difference equation is then
yk =
1 T1 + T
k+
2 T2 + T
k+
This can be checked by substitution which also shows that = 1/2 . The
condition y(0) = 0 gives y0 = 0 and since y(t) yk yk1
T, y(0) = 0
implies yk1 = 0. Using these we have
+ +1
2= 0
1 + T
1 T + 2 + T
2 T +1
2= 0
with solution
=1 T
2 = 2 T
2
Thus
yk =1 T
2
1 T1 + T
k+2 T
2
2 T2 + T
k+
1
2
Figure 3.5: Response of continuous and discrete systems in Exercise 15 over
10 seconds when T = 0.1
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190 Glyn James: Advanced Modern Engineering Mathematics, Third edition
Figure 3.6: Response of continuous and discrete systems in Exercise 14 over 10
seconds when T = 0.05
16
f(t) = t2, {f(k)} = k22 , k 0Now
Z{k2} = z ddz
z
(z 1)2 =z(z + 1)
(z 1)3So
Z{k22} = z(z + 1)2
(z 1)3To get D -transform, put z = 1 + to give
F
() =(1 + )(2 + )2
33
Then the D -transform is
F() = F
() =(1 + )(2 + )
3
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