Solucionario parte 3 Matemáticas Avanzadas para Ingeniería - 2da Edición - Glyn James
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3 The
Transform
Z
Exercises 3.2.3 1(a) F (z) =
∞ (1/4)k k=0
1(b) F (z) =
zk
∞ 3k k=0
1(c) F (z) =
∞ (−2)k k=0
1(d) F (z) =
zk
zk
∞ −(2)k k=0
zk
=
4z 1 = 1 − 1/4z 4z − 1
=
z 1 = 1 − 3/z z−3
=
z 1 = 1 − (−2)/z z+2
if | z |> 2
z 1 =− 1 − 2/z z−2
if | z |> 2
=−
if | z |> 1/4
if | z |> 3
1(e) Z{k} =
z (z − 1)2
if | z |> 1
from (3.6) whence Z{3k} = 3
2
z (z − 1)2
if | z |> 1
k uk = e−2ωkT = e−2ωT
whence U (Z) =
z z − e−2ωT
c Pearson Education Limited 2004
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Exercises 3.3.6 3 Z{sin kωT } = = 4
z z 1 1 − ωT 2 z − e 2 z − e−ωT
z2
z sin ωT − 2z cos ωT + 1
k 2z 1 }= Z{ 2 2z − 1
so Z{yk } =
2 1 2z = 2 × 3 z 2z − 1 z (2z − 1)
Proceeding directly Z{yk } =
∞ xk−3 k=3
5(a)
zk
1 Z − 5 5(b)
=
=
∞ xr 1 2 = × Z {x } = k z r+3 z3 z 2 (2z − 1) r=0
r ∞ −1 r=0
5z
Z {cos kπ} =
By (3.5)
so
5z 5z + 1
| z |>
{cos kπ} = (−1)k
so
6
=
z z+1
| z |> 1
k 2z 1 = Z 2 2z − 1 Z (ak ) = Z (kak−1 ) =
z z−a z (z − a)2
c Pearson Education Limited 2004
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Glyn James: Advanced Modern Engineering Mathematics, Third edition thus
Z (kak ) =
whence
az (z − a)2
k 2z 1 = Z k 2 (2z − 1)2
7(a) sinh kα = so 1 Z {sinh kα} = 2
1 α k 1 −α k (e ) − (e ) 2 2
z z − α z−e z − e−α
=
z2
z sinh α − 2z cosh α + 1
7(b) cosh kα =
1 α k 1 −α k (e ) + (e ) 2 2
then proceed as above.
8(a)
8(b)
k uk = e−4kT = e−4T ;
Z {uk } =
z z − e−4T
1 kT e − e− kT 2 z z sin T 1 z = 2 − Z {uk } = T − T 2 z − e z−e z − 2z cos T + 1 uk =
8(c) uk =
1 2kT e + e− 2kT 2
then proceed as above.
9
Initial value theorem: obvious from definition. c Pearson Education Limited 2004
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162 9
Glyn James: Advanced Modern Engineering Mathematics, Third edition Final value theorem (1 − z
−1
)X(z) =
∞ xr − xr−1
zr
r=0
= x0 + As z → 1 and if
lim r→∞
x2 − x1 x1 − x0 xr − xr−1 + + ... + + ... 2 z z zr xr exists, then lim (1 − z −1 )X(z) = lim xr r→∞
z→1
10
Multiplication property (3.19): Let Z {xk } =
k
Z a xk =
∞ ak xk k=0
10
zk
∞
xk k=0 z k
= X(z) then
= X(z/a)
Multiplication property (3.20) −z
∞ ∞ d xk kxk d X(z) = −z = = Z {kxk } dz dz zk zk k=0
k=0
The general result follows by induction.
Exercises 3.4.2 11(a)
11(b)
11(c)
z ; z−1
from tables uk = 1
z z ; = z+1 z − (−1)
z ; z − 1/2
from tables uk = (−1)k
from tables uk = (1/2)k
c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition 11(d) 1 z 1 z = ←→ (−1/3)k 3z + 1 3 z + 1/3 3 11(e) z ; z− 11(f )
from tables uk = ( )k
√ z z √ = √ ←→ (− 2)k z+ 2 z − (− 2)
11(g) 1 1 z ←→ = z−1 zz−1
0; k = 0 1; k > 0
using first shift property.
11(h)
1 z z+2 1; k=0 =1+ ←→ k−1 (−1) ; k >0 z+1 zz+1 1; k=0 = k+1 ; k>0 (−1)
12(a) Y (z)/z = so Y (z) =
12(b) 1 Y (z) = 7
1 1 1 1 − 3z−1 3z+2
1 z 1 1 z − ←→ 1 − (−2)k 3z−1 3z+2 3
z z − z − 3 z + 1/2
←→
1 k (3) − (−1/2)k 7
12(c) Y (z) =
1 z 1 1 1 z + ←→ + (−1/2)k 3 z − 1 6 z + 1/2 3 6 c Pearson Education Limited 2004
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12(d) Y (z) =
z 2 2 2 z 2 − ←→ (1/2)k − (−1)k 3 z − 1/2 3 z + 1 3 3 =
12(e)
2 2 (1/2)k + (−1)k+1 3 3
z z 1 − Y (z) = 2 z − z − (− ) 1 z z = − 2 z − e π/2 z − e− π/2
1 (e π/2 )k − (e− π/2 )k = sin kπ/2 ←→ 2
12(f ) z √ √ z − ( 3 + ) z − ( 3 − ) z z 1 √ √ − = 2 z − ( 3 + ) z − ( 3 − ) z z 1 − = 2 z − 2e π/6 z − 2e− π/6 1 k kπ/6 k − kπ/6 2 e ←→ = 2k sin kπ/6 −2 e 2 Y (z) =
12(g) Y (z) =
1 z z 1 z 5 − + 2 2 (z − 1) 4z−1 4z−3 ←→
1 5 k+ 1 − 3k 2 4
12(h) Y (z)/z =
z (z −
1)2 (z 2
− z + 1)
so z 1 √ Y (z) = − (z − 1)2 3
=
z
z−
1 1 − 2 2 (z − 1) z −z+1
√ 1+ 3 2
−
z
z−
c Pearson Education Limited 2004
√ 1− 3 2
Glyn James: Advanced Modern Engineering Mathematics, Third edition
1 z −√ = 2 (z − 1) 3
z z − π/3 z−e z − e− π/3
2 2 ←→ k − √ sin kπ/3 = k + √ cos(kπ/3 − 3π/2) 3 3 13(a) X(z) =
∞ xk k=0
zk
=
2 1 + 7 z z
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 1 3 3z + z 2 + 5z 5 =5+ 3 + 4 5 z z z
and proceed as in Example 13(a).
13(d) Y (z) =
1 1 z + 3+ 2 z z z + 1/3
←→ {0, 0, 1, 1} + {(−1/3)k } 13(e) 1 1/2 3 + 2− z z z + 1/2 1 0, k = 0 ←→ {1, 3, 1} − 2 (−1/2)k , k ≥ 1 Y (z) = 1 +
=
1, k = 0 5/2, k = 1 k=2 5/4, 1 − 2 (−1/2)k−1 , k ≥ 3
=
1, k = 0 5/2, k = 1 k=2 5/4, 1 − 8 (−1/2)k−3 , k ≥ 3
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13(f ) Y (z) =
2 1 1 − + z − 1 (z − 1)2 z−2
0, k = 0 1 − 2(k − 1) + 2k−1 , k ≥ 1 0, k = 0 = 3 − 2k + 2k−1 , k ≥ 1
←→
13(g) Y (z) = ←→
2 1 − z−1 z−2
0, k = 0 2 − 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 right D-block is vk , we have yk+1 = vk ,
1 vk+1 = wk = xk − vk 2
so 1 yk+2 = vk+1 = xk − vk 2 1 1 = xk − vk = xk − yk+1 2 2 i.e. 1 yk+2 + yk+1 = xk 2 14(b)
Using the same notation yk+1 = vk ,
1 1 vk+1 = wk = xk − vk − yk 4 5
c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition Then 1 1 yk+2 = xk − yk+1 − yk 4 5 or 1 1 yk+2 + yk+1 + yk = xk 4 5 15(a) z 2 Y (z) − z 2 y0 − 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) =
3 z 17 z − 10 z − 9 10 z + 1
thus yk =
15(c)
3 k 17 9 − (−1)k , k ≥ 0 10 10
Transforming and substituting for y0 and y1 Y (z) = 1 = 4
thus yk =
z (z − 2 )(z + 2 )
z z − z − 2e π/2 z − 2e− π/2
1 k kπ/2 2 e − e− kπ/2 = 2k−1 sin kπ/2, k ≥ 0 4 c Pearson Education Limited 2004
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15(d)
Transforming, substituting for y0 and y1 , and rearranging Y (z)/z =
so Y (z) = 2
6z − 11 (2z + 1)(z − 3)
z z + z + 1/2 z − 3
thus yk = 2(−1/2)k + 3k , k ≥ 0 16(a) 6yk+2 + yk+1 − yk = 3,
y0 = y 1 = 0
Transforming with y0 = y1 = 0 , (6z 2 + z − 1)Y (z) = so Y (z)/z = and Y (z) =
3z z−1
3 (z − 1)(3z − 1)(2z + 1)
9 2 z z 1 z − + 2 z − 1 10 z − 1/3 5 z + 1/2
Inverting yk = 16(b)
1 2 9 − (1/3)k + (−1/2)k 2 10 5
Transforming with y0 = 0, y1 = 1 , (z 2 − 5z + 6)Y (z) = z + 5
whence Y (z) =
7 z z 5 z + −6 2z−1 2z−3 z−2
so yk = 16(c)
z z−1
5 7 k + (3) − 6 (2)k 2 2
Transforming with y0 = y1 = 0 , (z 2 − 5z + 6)Y (z) =
z z − 1/2
c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition so Y (z) =
z 2 z 2 z 4 − + 15 z − 1/2 3 z − 2 5 z − 3
whence yn = 16(d)
4 2 2 (1/2)k − (2)k + (3)k 15 3 5
Transforming with y0 = 1, y1 = 0 , (z 2 − 3z + 3)Y (z) = z 2 − 3z +
so
z z−1
z z − 2 z − 1 z − 3z + 3
1 z z z √ √ −√ − = 3− 3j z−1 3j z − 3+ 3j z − 2 2 1 z z z √ √ −√ − = z−1 3j z − 3ejπ/6 z − 3e−jπ/6 Y (z) =
so
16(e)
√ ejnπ/6 − e−jnπ/6 2 √ = 1 − 2( 3)n−1 sin nπ/6 yn = 1 − √ ( 3)k 2j 3 Transforming with y0 = 1, y1 = 2 (2z 2 − 3z − 2)Y (z) = 2z 2 + z + 6
so
z +z Y (z) = z−2 =
z+5 2 (z − 1) (2z + 1)(z − 2)
12 z 2 z z z − − −2 5 z − 2 5 z + 1/2 z − 1 (z − 1)2
so yn = 16(f )
z z + 2 (z − 1) z−1
12 n 2 (2) − (−1/2)n − 1 − 2n 5 5
Transforming with y0 = y1 = 0 , (z 2 − 4)Y (z) = 3
z z −5 2 (z − 1) z−1
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so
z 1 z 1 z z − − − 2 z − 1 (z − 1) 2z−2 2z+2
Y (z) = and
17
1 1 yn = 1 − n − (2)n − (−2)n 2 2 Write the transformed equations in the form z − 3/2 1 c(z) zC0 = zE0 −0.21 z − 1/2 e(z)
Then
c(z) e(z)
1 = 2 z − 2z + 0/96
Solve for c(z) as c(z) = 1200
z − 1/2 −1 0.21 z − 3/2
zC0 zE0
z z + 4800 z − 1.2 z − 0.8
and 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 k
0.6 log(1.2) + 5.6x log(0.8) = 0 where x = (0.8/1.2) Solving, x = 0.0875 and so k=
log 0.0875 = 6.007 log(0.8/1.2)
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 =
1 z−4 + (z − 2)(z − 3) (z − 1)(z − 2)(z − 3)
c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition so Y (z) =
z 1 z 1 z + − 2z−1 z−2 2z−3
thus yk =
1 1 + 2k − 3k 2 2
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 1 Ik+2 − Ik+1 + Ik = G 2 Using lower case for the z transform we obtain 1 z (z 2 − z + )i(z) = (2z 2 + z)G + G 2 z−1 1 2 i(z)/z = G 2 + z−1 z − z + 12 1 2 + =G 1− z − 1 (z − 1+ 2 )(z − 2 )
whence
so
2 z + i(z) = G 2 z − 1 2 Thus
z z−
√1 e π/4 2
z − 1 z − √2 e− π/4
2 1 k kπ/4 − kπ/4 Ik = G 2 + (√ ) e −e 2 2 k 1 = 2G 1 + √ sin kπ/4 2 c Pearson Education Limited 2004
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Glyn James: Advanced Modern Engineering Mathematics, Third edition Elementary rearrangement leads to in+2 − 2 cosh α in+1 + in = 0
with cosh α = 1 + R1 /2R2 . Transforming and solving for I(z)/z gives I(z)/z =
zi0 + (i1 − 2i0 cosh α) (z − eα )(z − e−α )
α i0 e + (i1 − 2i0 cosh α) i0 e−α + (i1 − 2i0 cosh α) 1 − = 2 sinh α z − eα z − e−α Thus ik =
(i0 eα + (i1 − 2i0 cosh α))enα − (i0 e−α + (i1 − 2i0 cosh α))e−nα 2 sinh α =
1 {i1 sinh nα − i0 sinh(n − 1)α} sinh 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) =
z2
1 − 3z + 2
H(z) =
z2
z−1 − 3z + 2
21(b)
21(c) H(z) = 22
z3
1 + 1/z − z 2 + 2z + 1
For the first system, transforming from a quiescent state, we have (z 2 + 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 . c Pearson Education Limited 2004
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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 (z 2 + 0.5z + 0.25)Y (z) = (1 − 0.6)U (z) Clearly (z 2 + 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 c Pearson Education Limited 2004
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23(a) Yδ (z)/z = so
z 1 z 1 − 2 z + 1/4 2 z + 1/2
Yδ (z) = yk =
1 (4z + 1)(2z + 1)
1 1 (−1/4)k − (1/2)k 2 2
23(b) Yδ (z)/z = whence
so
z2
z − 3z + 3
√ √ 3 + 3 3 − 3 z z √ √ √ − √ Yδ (z) = 2 3 z − (3+ 3 ) 2 3 z − (3− 3 ) 2 2 √ √ 3 + 3 √ k kπ/6 3 − 3 √ k − kπ/6 √ yk = − √ ( 3) e ( 3) e 2 3 2 3 √ √ k 1 3 sin kπ/6 + cos kπ/6 = 2( 3) 2 2 √ = 2( 3)k sin(k + 1)π/6
23(c) Yδ (z)/z = so Yδ (z) = then yk =
z (z − 0.4)(z + 0.2)
1 z 2 z + 3 z − 0.4 3 z + 0.2 2 1 (0.4)k + (−0.2)k 3 3
23(d) Yδ (z)/z = so Yδ (z) =
5z − 12 (z − 2)(z − 4)
z z +4 z−2 z−4
and yk = (2)k + (4)k+1 c Pearson Education Limited 2004
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24(a) Yδ (z) = = yk =
z2
1 − 3z + 2
1 1 − z−2 z−1 0, k = 0 2k−1 − 1, k > 0
24(b) Yδ (z) = so
yk =
25
1 z−2
0, k = 0 2k−1 , k > 0
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 is
stable.
25(b) Poles at z = −1/3 and z = 2/3 , both inside | z |= 1 so the system is stable. √ 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 ± system is unstable.
√
17/4 , one of which is outside | z |= 1 and so the
25(e) Poles at z = −1/4 and z = 1 thus one pole is on | z |= 1 and the other is inside and the system is marginally stable.
c Pearson Education Limited 2004
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Glyn James: Advanced Modern Engineering Mathematics, Third edition To use the convolution result, calculate the impulse response as yδ,k − (1/2)k .
Then the step response is yk =
k
k−j
1 × (1/2)
j=0
k
= (1/2)
k
1 × (2)j = (1/2)k
j=0
1 − (2)k+1 1−2
= (1/2)k (2k+1 − 1) = 2 − (1/2)k Directly, Y (z)/z =
2 1 z = − (z − 1/2)(z − 1) z − 1 z − 1/2
so yk = 2 − (1/2)k 27
Substituting yn+1 − yn + Kyn−1 = 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
1 1√ 1 1√ + 1 − 4K and z2 = − 1 − 4K 2 2 2 2 For stability, both roots must be inside | z |= 1 so if K < 1/4 then z1 =
z1 < 1 ⇒ and z2 > −1 ⇒
1 1√ + 1 − 4K < 1 ⇒ K > 0 2 2 1 1√ − 1 − 4K > −1 ⇒ k > −2 2 2
If K > 1/4 then
1√ 1 + 4K − 1 |2 < 1 ⇒ K < 1 2 2 The system is then stable for 0 < K < 1 . |
When k = 2/9 we have
2 1 yn+2 − yn+1 + yn = 9 9 c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition Transforming with a quiescent initial state 1 z 2 (z 2 − z + )Y (z) = 9 9 z − 1/2 1 1 Y (z) = z 9 (z − 1/2)(z − 1/3)(z − 2/3)
so
=2
z z z +2 −4 z − 1/3 z − 2/3 z − 1/2
which inverts to yn = 2(1/3)n + 2(2/3)n − 4(1/2)n 28 z 2 + 2z + 2 = (z − (−1 + ))(z − (−1 + )) establishing the pole locations. Then Yδ (z) = So since (−1 ± ) =
√
1 z z 1 − 2 z − (−1 + ) 2 z − (−1 − )
2e±3 π/4 etc., √ yk = ( 2)k sin 3kπ/4
Exercises 3.9.6 29 H(s) = Replace s with
2 z−1 to give ∆ z+1 ˜ H(z) =
=
1 s2 + 3s + 2
∆2 (z + 1)2 4(z − 1)2 + 6∆(z 2 − 1) + 2∆2 (z + 1)2
∆2 (z + 1)2 (4 + 6∆ + 2∆2 )z 2 + (4∆2 − 8)z + (4 − 6∆ + 2∆2 ) c Pearson Education Limited 2004
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This corresponds to the difference equation (Aq 2 + Bq + C)yk = ∆2 (q 2 + 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∆2 δ 2 + (2A + B)∆δ + A + B + C)yk = ∆2 (∆2 δ 2 + 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 at ν = −1.8182 and ν = −0.9522 , inside the circle centre (−10, 0) in the ν -plane with 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 the boundary has moved slightly further from it.
30
The transfer function is H(s) =
s3
+
1 + 2s + 1
2s2
To discretise using the bi-linear form use s → ˜ H(z) =
2 z−1 to give T z+1
T 3 (z + 1)3 Az 3 + Bz 2 + Cz + D
and thus the discrete-time form (Aq 3 + Bq 2 + Cq + D)yk = T 3 (q 3 + 3q 2 + 3q + 1)uk where A = T 3 + 4T 2 + 8T + 8, C = 3T 3 − 4T 2 − 8T + 3,
B = 3T 3 + 4T 2 − 8T − 3, D = T 3 − 4T 2 + 8T − 1
c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition To obtain the δ form use s →
179
2δ giving the δ transfer function as 2 + ∆δ (2 + ∆δ)3 Aδ 3 + Bδ 2 + Cδ + D
This corresponds to the discrete-time system (Aδ 3 + Bδ 2 + Cδ + D)yk = (∆3 δ 3 + 2∆2 δ 2 + 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 form we obtain 0 1 0 x(t) ˙ = x(t) + u(t) −2 −3 1 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 8 12 + 8∆ 1 2 2 2 δ+ yk = ∆ δ + 4∆δ + 4 uk δ + A A A Now, as usual, consider the related system 8 12 + 8∆ 2 δ+ pk = u k δ + A A c Pearson Education Limited 2004
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and introduce the state variables x1 (k) = pk , x2 (k) = δpk together with the redundant variable x3 (k) = δ 2 pk . This leads to the representation 1 0 u(k) 12 + 8∆ x(k) + 1 − A
0 δx(k) = 8 − A yk =
8∆2 4 − 2 A A
4∆ (12 + 8∆)∆2 ∆2 − u(k) , x(k) + A A2 A
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(z 2 − z) (12 + 5∆)z 2 + (8∆ − 12)z − ∆
When ∆ = 0.1 this gives H(z) =
12(z 2 − z) 12.5z 2 + −11.2z − 0.1
(b) The γ form is given by replacing z by 1 + ∆γ . rearrangement gives ˜ H(γ) =
γ 2 ∆(12
12γ(1 + ∆γ) + 5∆) + γ(8∆ − 12) + 12
when ∆ = 0.1 this gives ˜ H(γ) =
12γ(1 + 0.1γ) 1.25γ 2 − 11.2γ + 12
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Substitution and
Glyn James: Advanced Modern Engineering Mathematics, Third edition
Review exercises 3.10 1 Z {f (kT )} = Z {kT } = T Z {k} = T
2
z (z − 1)2
ak (e kω − e− kω ) Z a sin kω = Z 2 1 = Z (ae ω )k − (ae− ω )k 2 z 1 z = − 2 z − ae ω z − ae− ω az sin ω = 2 z − 2az cos ω + a2
k
3 Recall that
Z ak =
z (z − a)2
Differentiate twice wrt a then put a = 1 to get the pairs k ←→ then
z (z − 1)2
Z k2 =
k(k − 1) ←→
2z (z − 1)3
2z z z(z + 1) + = 3 2 (z − 1) (z − 1) (z − 1)3
4 H(z) =
3z 2z + z − 1 (z − 1)2
so inverting, the impulse response is {3 + 2k}
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5
z (z + 1)(z + 2)(z − 1) 1 z 1 z 1 z + + =− 2z+1 3z+2 6z−1
YSTEP (z) =
Thus
1 1 1 ySTEP,k = − (−1)k + (−2)k + 2 3 6
6 F (s) =
1 1 1 = − s+1 s s+1
which 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 − z − 1 z − e−T
Then e−sT F (s) ←→ f ((t − T )) which when sampled becomes f ((k − 1)T ) and Z {f ((k − 1)T )} =
∞ f ((k − 1)T ) k=0
That is e−sT F (s) →
zk
=
1˜ F (z) z
1˜ F (z) z
So the overall transfer function is z z 1 − e−T z−1 − = z z − 1 z − e−T z − e−T 7 H(s) =
1 s+1 2 − = (s + 2)(s + 3) s+3 s+2
yδ (t) = 2e−3t − e2t −→ {2e−3kT − e2kT } so ˜ H(z) =2
z z − −3T z−e z − e−2T
c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition 8(a)
183
Simple poles at z = a and z = b . The residue at z = a is lim (z − a)z n−1 X(z) = lim (z − a)
z→a
z→a
The residue at z = b is similarly of these, that is
an zn = (z − a)(z − b) a−b
bn and the inverse transform is the sum b−a n a − bn a−b
8(b) (i) There is a only double pole at z = 3 and the residue is n−1 d zn (z − 3)2 = n3 z→3 dz (z − 3)2 lim
√ 1 3 . The individual residues are (ii) There are now simple poles at z = ± 2 2 thus given by n
√ 1 3 ± 2 2 √ lim√ ± 3 z→(1/2± 3/2 ) Adding these and simplifying in the usual way gives the inverse transform
as
2 √ sin nπ/3 3
9 H(z) = so
z z − z+1 z−2
z z z YSTEP (z) = − z+1 z−2 z−1 3z =− (z − 1)(z + 1)(z − 2) 1 z z 3 z + −2 = 2z−1 2z+1 z−2
so ySTEP,k =
3 1 + (−1)k − 2k+1 2 2
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10
1 z z2 × 1− = Y (z) = (z + 1)(z − 1) z z+1
so yk = (−1)k z2 α+β αβ Y (z) = × 1− + 2 =1 (z − α)(z − β) z z
11
so yk = {δk } = {1, 0, 0, . . .} z The response of the system with H(z) = is clearly given by (z − α)(z − β) Y (z) = 1/z , which transforms to yk = {δk−1 } = {0, 1, 0, 0, . . .}
12
From H(s) =
s the impulse response is calculated as (s + 1)(s + 2) yδ (t) = (2e−2t − e−t ) t ≥ 0
Sampling gives
{yδ (nT )} = 2e−2nT − enT t
with z transform Z {yδ (nT )} = 2
z z − = D(z) −2T z−e z − e−T
Setting Y (z) = T D(z)X(z) gives Y (z) = T 2
z z X(z) − z − e−2T z − e−T
Substituting for T and simplifying gives 1 z − 0.8452 Y (z) = z 2 X(z) 2 z − 0.9744z + 0.2231 so (z 2 − 0.9744z + 0.2231)Y (z) = (0.5z 2 − 0.4226z)X(x) c Pearson Education Limited 2004
Glyn James: Advanced Modern Engineering Mathematics, Third edition 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 13 yn+1 = yn + avn vn+1 = vn + bun = vn + b(k1 (xn − yn ) − k2 vn ) = 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 1 yn+2 = yn+1 + (xn − yn ) 4 or 1 1 yn+2 − yn+1 + yn = xn 4 4 Transforming with relaxed initial conditions gives Y (z) =
1 X(z) (2z − 1)2
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(b) When X(z) =
A , z−1
z A z z 4 Y (z) = −4 −2 4 z−1 z − 1/2 (z − 1/2)2 then yn = 14
A 4 − 4(1/2)n − 2n(1/2)n−1 4
Substitution leads directly to yk − yk−1 yk − 2yk−1 + yk−2 + 2yk = 1 +3 2 T T Take the z transform under the assumption of a relaxed system to get [(1 + 3T z + 2T 2 )z 2 − (2 + 3T )z + 1]Y (z) = T 2
z3 z−1
The characteristic equation is thus (1 + 3T z + 2T 2 )z 2 − (2 + 3T )z + 1 = 0 with roots (the poles) 1 1 , z= 1+T 1 + 2T The general solution of the difference equation is a linear combination of these together with a particular solution. That is z=
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 yk − yk−1 , y (0) = 0 condition y(0) = 0 gives y0 = 0 and since y (t) → T implies yk−1 = 0 . Using these we have 1 =0 2 1 α(1 + T ) + β(1 + 2T ) + = 0 2 α+β+
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with solution α = −1 , β = 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 −2t − 2e−t + 1), t ≥ 0 (e 2
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 + 2T 2 )z 2 + (4T 2 − 8)z + (4 − 6T + 2T 2 )]Y (z) = T 2 (z + 1)2 X(x) The characteristic equation is (4 + 6T + 2T 2 )z 2 + (4T 2 − 8)z + (4 − 6T + 2T 2 ) = 0 with roots z= That is z=
8 − 4T 2 ± 4T 2(4 + 6T + 2T 2 )
2−T 1−T and z = 1+T 2+T
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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 yk − yk−1 condition y(0) = 0 gives y0 = 0 and since y (t) → , y (0) = 0 T implies yk−1 = 0 . Using these we have α+β+ α
2+T 1 1+T +β + =0 1−T 2−T 2
with solution α= Thus
1−T yk = 2
1 =0 2
1−T 2
1−T 1+T
k
β=−
2−T 2
2−T +− 2
2−T 2+T
k +
1 2
Figure 3.5: Response of continuous and discrete systems in Exercise 15 over 10 seconds when T = 0.1 c Pearson Education Limited 2004
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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∆)} = k 2 ∆2 , k ≥ 0
Now Z{k 2 } = −z
z d z(z + 1) = 2 dz (z − 1) (z − 1)3
So Z{k 2 ∆2 } =
z(z + 1)∆2 (z − 1)3
To get D -transform, put z = 1 + ∆γ to give (1 + ∆γ)(2 + ∆γ)∆2 ∆3 γ 3
F∆ (γ) = Then the D -transform is
F∆ (γ) = ∆F∆ (γ) =
(1 + ∆γ)(2 + ∆γ) γ3
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