chapter 03 solucionario glyn james

3

The Z Transform

Exercises 3.2.3

1(a)

F (z) =∞∑

k=0

(1/4)k

zk=

11− 1/4z

=4z

4z − 1if | z |> 1/4

1(b)

F (z) =∞∑

k=0

3k

zk=

11− 3/z

=z

z − 3if | z |> 3

1(c)

F (z) =∞∑

k=0

(−2)k

zk=

11− (−2)/z

=z

z + 2if | z |> 2

1(d)

F (z) =∞∑

k=0

−(2)k

zk= − 1

1− 2/z= − z

z − 2if | z |> 2

1(e)

Z{k} =z

(z − 1)2if | z |> 1

from (3.6) whence

Z{3k} = 3z

(z − 1)2if | z |> 1

2

uk = e−2ωkT =(e−2ωT

)k

whence

U(Z) =z

z − e−2ωT

c©Pearson Education Limited 2004

160 Glyn James: Advanced Modern Engineering Mathematics, Third edition

Exercises 3.3.6

3

Z{sin kωT} =12

z

z − eωT− 1

2

z

z − e−ωT

=z sinωT

z2 − 2z cos ωT + 1

4

Z{(

12

)k

} =2z

2z − 1so

Z{yk} =1z3 ×

2z

2z − 1=

2z2(2z − 1)

Proceeding directly

Z{yk} =∞∑

k=3

xk−3

zk=

∞∑r=0

xr

zr+3 =1z3 ×Z {xk} =

2z2(2z − 1)

5(a)

Z{−1

5

}=

∞∑r=0

(−15z

)r

=5z

5z + 1| z |> 1

5

5(b)

{cos kπ} ={(−1)k

}so

Z {cos kπ} =z

z + 1| z |> 1

6

Z{(

12

)k}

=2z

2z − 1

By (3.5)

Z {(ak)}

=z

z − a

so

Z {(kak−1)}

=z

(z − a)2


Glyn James: Advanced Modern Engineering Mathematics, Third edition 161

thus

Z {(kak)}

=az

(z − a)2

whence

Z{

k

(12

)k}

=2z

(2z − 1)2

7(a)

sinh kα =12(eα)k − 1

2(e−α)k

so

Z {sinh kα} =12

(z

z − eα− z

z − e−α

)=

z sinhα

z2 − 2z cosh α + 1

7(b)

cosh kα =12(eα)k +

12(e−α)k

then proceed as above.

8(a)

uk =(e−4kT

)=(e−4T

)k; Z {uk} =

z

z − e−4T

8(b)

uk =12

(e kT − e− kT

)

Z {uk} =12

(z

z − e T− z

z − e− T

)=

z sinT

z2 − 2z cos T + 1

8(c)

uk =12(e 2kT + e− 2kT

)then proceed as above.

9 Initial value theorem: obvious from definition.



9 Final value theorem

(1− z−1)X(z) =∞∑

r=0

xr − xr−1

zr

= x0 +x1 − x0

z+

x2 − x1

z2 + . . . +xr − xr−1

zr+ . . .

As z → 1 and if limr→∞ xr exists, then

limz→1

(1− z−1)X(z) = limr→∞ xr

10 Multiplication property (3.19): Let Z {xk} =∑∞

k=0xk

zk = X(z) then

Z {akxk

}=

∞∑k=0

akxk

zk= X(z/a)

10 Multiplication property (3.20)

−zd

dzX(z) = −z

d

dz

∞∑k=0

xk

zk=

∞∑k=0

kxk

zk= 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



11(d)z

3z + 1=

13

z

z + 1/3←→ 1

3(−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=

1z

z

z − 1←→

{0; k = 01; k > 0

using first shift property.

11(h)z + 2z + 1

= 1 +1z

z

z + 1←→

{1; k = 0(−1)k−1; k > 0

={

1; k = 0(−1)k+1; k > 0

12(a)

Y (z)/z =13

1z − 1

− 13

1z + 2

so

Y (z) =13

z

z − 1− 1

3z

z + 2←→ 1

3(1− (−2)k

)

12(b)

Y (z) =17

(z

z − 3− z

z + 1/2

)←→ 1

7((3)k − (−1/2)k

)

12(c)

Y (z) =13

z

z − 1+

16

z

z + 1/2←→ 1

3+

16(−1/2)k



12(d)

Y (z) =23

z

z − 1/2− 2

3z

z + 1←→ 2

3(1/2)k − 2

3(−1)k

=23(1/2)k +

23(−1)k+1

12(e)

Y (z) =12

(z

z − − z

z − (− )

)

=12

(z

z − e π/2 −z

z − e− π/2

)

←→ 12

((e π/2)k − (e− π/2)k

)= sin kπ/2

12(f)

Y (z) =z(

z − (√

3 + )) (

z − (√

3− ))

=12

(z

z − (√

3 + )− z

z − (√

3− )

)

=12

(z

z − 2e π/6 −z

z − 2e− π/6

)

←→ 12

(2ke kπ/6 − 2ke− kπ/6

)= 2k sin kπ/6

12(g)

Y (z) =52

z

(z − 1)2+

14

z

z − 1− 1

4z

z − 3

←→ 52k +

14(1− 3k

)

12(h)

Y (z)/z =z

(z − 1)2(z2 − z + 1)=

1(z − 1)2

− 1z2 − z + 1

so

Y (z) =z

(z − 1)2− 1√

3

(z

z − 1+√

32

− z

z − 1−√3

2

)



=z

(z − 1)2− 1√

3

(z

z − e π/3 −z

z − e− π/3

)

←→ k − 2√3

sin kπ/3 = k +2√3

cos(kπ/3− 3π/2)

13(a)

X(z) =∞∑

k=0

xk

zk=

1z

+2z7

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 that3z + z2 + 5z5

z5 = 5 +1z3 +

3z4

and proceed as in Example 13(a).

13(d)

Y (z) =1z2 +

1z3 +

z

z + 1/3

←→ {0, 0, 1, 1}+ {(−1/3)k}

13(e)

Y (z) = 1 +3z

+1z2 −

1/2z + 1/2

←→ {1, 3, 1} − 12

{0, k = 0(−1/2)k, k ≥ 1

=

1, k = 05/2, k = 15/4, k = 2− 1

2 (−1/2)k−1, k ≥ 3

=

1, k = 05/2, k = 15/4, k = 2−1

8 (−1/2)k−3, k ≥ 3



13(f)

Y (z) =1

z − 1− 2

(z − 1)2+

1z − 2

←→{

0, k = 01− 2(k − 1) + 2k−1, k ≥ 1

={

0, k = 03− 2k + 2k−1, k ≥ 1

13(g)

Y (z) =2

z − 1− 1

z − 2

←→{

0, k = 02− 2k−1, k ≥ 1

Exercises 3.5.3

14(a) If the signal going into the left D-block is wk and that going into the rightD-block is vk , we have

yk+1 = vk, vk+1 = wk = xk − 12vk

so

yk+2 = vk+1 = xk − 12vk

= xk − 12vk = xk − 1

2yk+1

i.e.

yk+2 +12yk+1 = xk

14(b) Using the same notation

yk+1 = vk, vk+1 = wk = xk − 14vk − 1

5yk



Then

yk+2 = xk − 14yk+1 − 1

5yk

or

yk+2 +14yk+1 +

15yk = xk

15(a)

z2Y (z)− z2y0 − zy1 − 2(zY (z)− zy0) + Y (z) = 0

with y0 = 0, y1 = 1

Y (z) =z

(z − 1)2

so 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) =310

z

z − 9− 17

10z

z + 1

thus

yk =310

9k − 1710

(−1)k, k ≥ 0

15(c) Transforming and substituting for y0 and y1

Y (z) =z

(z − 2 )(z + 2 )

=14

(z

z − 2e π/2 −z

z − 2e− π/2

)

thus

yk =14

2k(e kπ/2 − e− kπ/2

)= 2k−1 sin kπ/2, k ≥ 0



15(d) Transforming, substituting for y0 and y1 , and rearranging

Y (z)/z =6z − 11

(2z + 1)(z − 3)

soY (z) = 2

z

z + 1/2+

z

z − 3

thusyk = 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) =3z

z − 1

soY (z)/z =

3(z − 1)(3z − 1)(2z + 1)

andY (z) =

12

z

z − 1− 9

10z

z − 1/3+

25

z

z + 1/2

Inverting

yk =12− 9

10(1/3)k +

25(−1/2)k

16(b) Transforming with y0 = 0, y1 = 1,

(z2 − 5z + 6)Y (z) = z + 5z

z − 1

whenceY (z) =

52

z

z − 1+

72

z

z − 3− 6

z

z − 2so

yk =52

+72(3)k − 6 (2)k

16(c) Transforming with y0 = y1 = 0,

(z2 − 5z + 6)Y (z) =z

z − 1/2



so

Y (z) =415

z

z − 1/2− 2

3z

z − 2+

25

z

z − 3

whence

yn =415

(1/2)k − 23(2)k +

25(3)k

16(d) Transforming with y0 = 1, y1 = 0,

(z2 − 3z + 3)Y (z) = z2 − 3z +z

z − 1

so

Y (z) =z

z − 1− z

z2 − 3z + 3

=z

z − 1− 1√

3j

{z

z − 3+√

3j2

− z

z − 3−√3j

2

}

=z

z − 1− 1√

3j

{z

z −√3ejπ/6− z

z −√3e−jπ/6

}so

yn = 1− 2√3(√

3)k ejnπ/6 − e−jnπ/6

2j= 1− 2(

√3)n−1 sinnπ/6

16(e) Transforming with y0 = 1, y1 = 2

(2z2 − 3z − 2)Y (z) = 2z2 + z + 6z

(z − 1)2+

z

z − 1

so

Y (z) =z

z − 2+ z

{z + 5

(z − 1)2(2z + 1)(z − 2)

}

=125

z

z − 2− 2

5z

z + 1/2− z

z − 1− 2

z

(z − 1)2

so

yn =125

(2)n − 25(−1/2)n − 1− 2n

16(f) Transforming with y0 = y1 = 0,

(z2 − 4)Y (z) = 3z

(z − 1)2− 5

z

z − 1



soY (z) =

z

z − 1− z

(z − 1)2− 1

2z

z − 2− 1

2z

z + 2

andyn = 1− n− 1

2(2)n − 1

2(−2)n

17 Write the transformed equations in the form(z − 3/2−0.21

1z − 1/2

)(c(z)e(z)

)=(

zC0

zE0

)

Then (c(z)e(z)

)=

1z2 − 2z + 0/96

(z − 1/2

0.21−1

z − 3/2

)(zC0

zE0

)

Solve for c(z) asc(z) = 1200

z

z − 1.2+ 4800

z

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 =log 0.0875

log(0.8/1.2)= 6.007

The nearest integer is k = 6, corresponding to the seventh year in view of thelabelling, and C6 = 4841 approx.

18 Transforming and rearranging

Y (z)/z =z − 4

(z − 2)(z − 3)+

1(z − 1)(z − 2)(z − 3)



so

Y (z) =12

z

z − 1+

z

z − 2− 1

2z

z − 3

thus

yk =12

+ 2k − 123k

19

Ik = Ck + Pk + Gk

= aIk−1 + b(Ck − Ck−1) + Gk

= aIk−1 + ba(Ik−1 − Ik−2) + Gk

so

Ik+2 − a(1 + b)Ik+1 + abIk = Gk+2

Thus substituting

Ik+2 − Ik+1 +12Ik = G

Using lower case for the z transform we obtain

(z2 − z +12)i(z) = (2z2 + z)G + G

z

z − 1

whence

i(z)/z = G

[1

z2 − z + 12

+2

z − 1

]

= G

[2

z − 1+

1(z − 1+

2 )(z − 1−2 )

]

so

i(z) = G

[2

z

z − 1+

22

{z

z − 1√2e π/4

− z

z − 1√2e− π/4

}]

Thus

Ik = G

[2 +

22

(1√2)k{

e kπ/4 − e− kπ/4}]

= 2G

[1 +

(1√2

)k

sin kπ/4

]



20 Elementary rearrangement leads to

in+2 − 2 cosh α in+1 + in = 0

with coshα = 1 + R1/2R2 . Transforming and solving for I(z)/z gives

I(z)/z =zi0 + (i1 − 2i0 cosh α)

(z − eα)(z − e−α)

=1

2 sinhα

[i0e

α + (i1 − 2i0 cosh α)z − eα

− i0e−α + (i1 − 2i0 cosh α)

z − e−α

]

Thus

ik =(i0eα + (i1 − 2i0 cosh α))enα − (i0e−α + (i1 − 2i0 cosh α))e−nα

2 sinhα

=1

sinhα{i1 sinhnα− i0 sinh(n− 1)α}

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 shownin Figure 3.1 and where Y (z) = P (z), that is yk = pk .



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 inFigure 3.2.

Figure 3.2: The block diagram for the second system of Exercise 22



23(a)

Yδ(z)/z =1

(4z + 1)(2z + 1)so

Yδ(z) =12

z

z + 1/4− 1

2z

z + 1/2

yk =12(−1/4)k − 1

2(1/2)k

23(b)

Yδ(z)/z =z

z2 − 3z + 3whence

Yδ(z) =3 +√

3

2√

3

z

z − (3+√

3 )2

− 3−√3

2√

3

z

z − (3−√3 )

2

so

yk =3 +√

3

2√

3(√

3)ke kπ/6 − 3−√3

2√

3(√

3)ke− kπ/6

= 2(√

3)k

[√3

2sin kπ/6 +

12

cos kπ/6

]

= 2(√

3)k sin(k + 1)π/6

23(c)

Yδ(z)/z =z

(z − 0.4)(z + 0.2)so

Yδ(z) =23

z

z − 0.4+

13

z

z + 0.2then

yk =23(0.4)k +

13(−0.2)k

23(d)

Yδ(z)/z =5z − 12

(z − 2)(z − 4)so

Yδ(z) =z

z − 2+ 4

z

z − 4and

yk = (2)k + (4)k+1



24(a)

Yδ(z) =1

z2 − 3z + 2

=1

z − 2− 1

z − 1

yk ={

0, k = 02k−1 − 1, k > 0

24(b)

Yδ(z) =1

z − 2

so

yk ={

0, k = 02k−1, 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 thesystem 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.



26 To use the convolution result, calculate the impulse response as yδ,k− (1/2)k .Then the step response is

yk =k∑

j=0

1× (1/2)k−j = (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)=

2z − 1

− 1z − 1/2

soyk = 2− (1/2)k

27 Substitutingyn+1 − yn + Kyn−1 = K/2n

oryn+2 − yn+1 + Kyn = K/2n+1

Taking z transforms from the quiescent state, the characteristic equation is

z2 − z + K = 0

with rootsz1 =

12

+12

√1− 4K and z2 =

12− 1

2

√1− 4K

For stability, both roots must be inside | z |= 1 so if K < 1/4 then

z1 < 1⇒ 12

+12

√1− 4K < 1⇒ K > 0

andz2 > −1⇒ 1

2− 1

2

√1− 4K > −1⇒ k > −2

If K > 1/4 then

| 12

+ 12

√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 +29yn =

19



Transforming with a quiescent initial state

(z2 − z +29)Y (z) =

19

z

z − 1/2

so

Y (z) = z19

[1

(z − 1/2)(z − 1/3)(z − 2/3)

]

= 2z

z − 1/3+ 2

z

z − 2/3− 4

z

z − 1/2

which 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

Yδ(z) =12

z

z − (−1 + )− 1

2

z

z − (−1− )

So since (−1± ) =√

2e±3 π/4 etc.,

yk = (√

2)k sin 3kπ/4

Exercises 3.9.6

29

H(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) + 2∆2(z + 1)2

=∆2(z + 1)2

(4 + 6∆ + 2∆2)z2 + (4∆2 − 8)z + (4− 6∆ + 2∆2)



This corresponds to the difference equation

(Aq2 + Bq + C)yk = ∆2(q2 + 2q + 1)uk

where

A = 4 + 6∆ + 2∆2 B = 4∆2 − 8 C = 4− 6∆ + 2∆2

Now put q = 1 + ∆δ to get

(A∆2δ2 + (2A + B)∆δ + A + B + C)yk

= ∆2(∆2δ2 + 4∆δ + 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 δ form with t = 0.1, the poles are atν = −1.8182 and ν = −0.9522, inside the circle centre (−10, 0) in the ν -planewith radius 10. When t = 0.01 these move to ν = −1.9802 and ν = −0.9950,within the circle centre (−100, 0) with radius 100, and the closest pole to theboundary 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) =T 3(z + 1)3

Az3 + Bz2 + Cz + D

and thus the discrete-time form

(Aq3 + Bq2 + Cq + D)yk = T 3(q3 + 3q2 + 3q + 1)uk

where

A = T 3 + 4T 2 + 8T + 8, B = 3T 3 + 4T 2 − 8T − 3,

C = 3T 3 − 4T 2 − 8T + 3, D = T 3 − 4T 2 + 8T − 1



To obtain the δ form use s→ 2δ

2 + ∆δgiving the δ transfer function as

(2 + ∆δ)3

Aδ3 + Bδ2 + Cδ + D

This corresponds to the discrete-time system

(Aδ3 + Bδ2 + Cδ + D)yk = (∆3δ3 + 2∆2δ2 + 4∆δ + 8)uk

where

A = ∆3 + 4∆2 + 8∆ + 8, B = 6∆2 + 16∆ + 16,

C = 12∆ + 16, D = 8

31 Making the given substitution and writing the result in vector-matrix formwe obtain

x(t) =[

0−2

1−3

]x(t) +

[01

]u(t)

and

y(t) = [1, 0]x(t)

This is in the general form

x(t) = Ax(t) + bu(t)

y = cT x(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 δ form equation as

[δ2 +

12 + 8∆A

δ +8A

]yk =

1A

[∆2δ2 + 4∆δ + 4

]uk

Now, as usual, consider the related system

[δ2 +

12 + 8∆A

δ +8A

]pk = uk



and introduce the state variables x1(k) = pk , x2(k) = δpk together with theredundant variable x3(k) = δ2pk . This leads to the representation

δx(k) =

0 1

− 8A−12 + 8∆

A

x(k) +

[01

]u(k)

yk =[(

4A− 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 EulerScheme 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 γ form is given by replacing z by 1 + ∆γ . Substitution andrearrangement gives

H(γ) =12γ(1 + ∆γ)

γ2∆(12 + 5∆) + γ(8∆− 12) + 12

when ∆ = 0.1 this gives

H(γ) =12γ(1 + 0.1γ)

1.25γ2 − 11.2γ + 12



Review exercises 3.10

1

Z {f(kT )} = Z {kT} = TZ {k} = Tz

(z − 1)2

2

Z {ak sin kω}

= Z{

ak(e kω − e− kω)2

}

=12Z {(ae ω)k − (ae− ω)k

}=

12

(z

z − ae ω− z

z − ae− ω

)

=az sinω

z2 − 2az cos ω + a2

3 Recall that

Z {ak}

=z

(z − a)2

Differentiate twice wrt a then put a = 1 to get the pairs

k ←→ z

(z − 1)2k(k − 1)←→ 2z

(z − 1)3

then

Z {k2} =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}



5YSTEP(z) =

z

(z + 1)(z + 2)(z − 1)

= −12

z

z + 1+

13

z

z + 2+

16

z

z − 1Thus

ySTEP,k = −12(−1)k +

13(−2)k +

16

6

F (s) =1

s + 1=

1s− 1

s + 1which inverts to

f(t) = (1− e−t)ζ(t)

where ζ(t) is the Heaviside step function, and so

F (z) = Z {f(kT )} =z

z − 1− z

z − e−T

Thene−sT F (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

=1zF (z)

That ise−sT F (s)→ 1

zF (z)

So the overall transfer function is

z − 1z

(z

z − 1− z

z − e−T

)=

1− e−T

z − e−T

7

H(s) =s + 1

(s + 2)(s + 3)=

2s + 3

− 1s + 2

yδ(t) = 2e−3t − e2t −→ {2e−3kT − e2kT }so

H(z) = 2z

z − e−3T− z

z − e−2T



8(a) Simple poles at z = a and z = b . The residue at z = a is

limz→a

(z − a)zn−1X(z) = limz→a

(z − a)zn

(z − a)(z − b)=

an

a− b

The residue at z = b is similarlybn

b− aand the inverse transform is the sum

of these, that is {an − bn

a− b

}

8(b)

(i) There is a only double pole at z = 3 and the residue is

limz→3

d

dz(z − 3)2

zn

(z − 3)2={n3n−1}

(ii) There are now simple poles at z =12±√

32

. The individual residues arethus given by

limz→(1/2±√

3/2 )±(

12 ±

√3

2 )n

√3

Adding these and simplifying in the usual way gives the inverse transformas {

2√3

sinnπ/3}

9

H(z) =z

z + 1− z

z − 2so

YSTEP(z) =(

z

z + 1− z

z − 2

)z

z − 1

= − 3z

(z − 1)(z + 1)(z − 2)

=32

z

z − 1+

12

z

z + 1− 2

z

z − 2so

ySTEP,k =32

+12(−1)k − 2k+1



10

Y (z) =z2

(z + 1)(z − 1)×(

1− 1z

)=

z

z + 1so

yk = (−1)k

11

Y (z) =z2

(z − α)(z − β)×(

1− α + β

z+

αβ

z2

)= 1

soyk = {δk} = {1, 0, 0, . . .}

The response of the system with H(z) =z

(z − α)(z − β)is clearly given by

Y (z) = 1/z , which transforms to

yk = {δk−1} = {0, 1, 0, 0, . . .}

12 From H(s) =s

(s + 1)(s + 2)the impulse response is calculated as

yδ(t) = (2e−2t − e−t) t ≥ 0

Sampling gives{yδ(nT )} =

{2e−2nT − enTt

}with z transform

Z {yδ(nT )} = 2z

z − e−2T− z

z − e−T= D(z)

Setting Y (z) = TD(z)X(z) gives

Y (z) = T

[2

z

z − e−2T− z

z − e−T

]X(z)

Substituting for T and simplifying gives

Y (z) =12z

[z − 0.8452

z2 − 0.9744z + 0.2231

]X(z)

so(z2 − 0.9744z + 0.2231)Y (z) = (0.5z2 − 0.4226z)X(x)



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 +14yn =

14xn

Transforming with relaxed initial conditions gives

Y (z) =1

(2z − 1)2X(z)



(b) When X(z) =A

z − 1,

Y (z) =A

4

[4

z

z − 1− 4

z

z − 1/2− 2

z

(z − 1/2)2

]

then

yn =A

4[4− 4(1/2)n − 2n(1/2)n−1]

14 Substitution leads directly to

yk − 2yk−1 + yk−2

T 2 + 3yk − yk−1

T+ 2yk = 1

Take the z transform under the assumption of a relaxed system to get

[(1 + 3Tz + 2T 2)z2 − (2 + 3T )z + 1]Y (z) = T 2 z3

z − 1

The characteristic equation is thus

(1 + 3Tz + 2T 2)z2 − (2 + 3T )z + 1 = 0

with roots (the poles)

z =1

1 + T, z =

11 + 2T

The general solution of the difference equation is a linear combination of thesetogether with a particular solution. That is

yk = α

(1

1 + T

)k

+ β

(1

1 + 2T

)k

+ γ

This can be checked by substitution which also shows that γ = 1/2. The

condition y(0) = 0 gives y0 = 0 and since y′(t) → yk − yk−1

T, y′(0) = 0

implies yk−1 = 0. Using these we have

α + β +12

= 0

α(1 + T ) + β(1 + 2T ) +12

= 0



with solution α = −1, β = 1/2 so

yk = −(

11 + T

)k

+12

(1

1 + 2T

)k

+12

The differential equation is simply solved by inverting the Laplace transformto give

y(t) =12(e−2t − 2e−t + 1), t ≥ 0

Figure 3.3: Response of continuous and discrete systems in Exercise 14 over10 seconds when T = 0.1




15 Substitution for s and simplifying gives

[(4 + 6T + 2T 2)z2 + (4T 2 − 8)z + (4− 6T + 2T 2)]Y (z)

= T 2(z + 1)2X(x)

The characteristic equation is

(4 + 6T + 2T 2)z2 + (4T 2 − 8)z + (4− 6T + 2T 2) = 0

with roots

z =8− 4T 2 ± 4T

2(4 + 6T + 2T 2)

That is

z =1− T

1 + Tand z =

2− T

2 + T



The general solution of the difference equation is then

yk = α

(1− T

1 + T

)k

+ β

(2− T

2 + T

)k

+ γ

This can be checked by substitution which also shows that γ = 1/2. The

condition y(0) = 0 gives y0 = 0 and since y′(t) → yk − yk−1

T, y′(0) = 0

implies yk−1 = 0. Using these we have

α + β +12

= 0

α1 + T

1− T+ β

2 + T

2− T+

12

= 0

with solutionα =

1− T

2β = −2− T

2Thus

yk =1− T

2

(1− T

1 + T

)k

+−2− T

2

(2− T

2 + T

)k

+12




Figure 3.6: Response of continuous and discrete systems in Exercise 14 over 10seconds when T = 0.05

16

f(t) = t2, {f(k∆)} ={k2∆2} , k ≥ 0

NowZ{k2} = −z

ddz

z

(z − 1)2=

z(z + 1)(z − 1)3

SoZ{k2∆2} =

z(z + 1)∆2

(z − 1)3

To get D -transform, put z = 1 + ∆γ to give

F′∆(γ) =

(1 + ∆γ)(2 + ∆γ)∆2

∆3γ3

Then the D -transform is

F∆(γ) = ∆F′∆(γ) =

(1 + ∆γ)(2 + ∆γ)γ3


chapter 03 solucionario glyn james

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