solucionario parte 3 matemáticas avanzadas para ingeniería - 2da edición - glyn james

7/30/2019 Solucionario parte 3 Matemticas Avanzadas para Ingeniera - 2da Edicin - Glyn James

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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

cPearson Education Limited 2004


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160 Glyn James: Advanced Modern Engineering Mathematics, Third edition

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

Pearson Education Limited 2004


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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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Glyn James: Advanced Modern Engineering Mathematics, Third edition 163

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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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



25/32


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



26/32


27/32


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)



28/32


(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



29/32


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



30/32




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



31/32


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





32/32


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

solucionario parte 3 matemáticas avanzadas para ingeniería - 2da edición - glyn james

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