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CHAPTER 2 2.21. The rectangular rules for numerical integration are illustrated in Fig. P2.21. The leftside rule is depicted in Fig. P2.21(a), and the rightside rule is depicted in Fig. P2.21(b). The integral of x (t ) is approximated by the sum of the rectangular areas shown for each rule. Let y ( kT ) be the numerical integral of x (t ), 0 ≤ t ≤ kT. x(t)
x(k)
kT
x(k + 1)
(k + 1)T
t
(a) x(t)
x(k + 1)
kT
(k + 1)T
t
(b)
FIGURE P2.21 Rectangular rules for integration: (a) left side; (b) right side.
(a)
Write the difference equation relating y ( k + 1) , y ( k ) , and x ( k ) for the leftside rule. (b) Find the transfer function Y ( z ) /X ( z ) for part (a). (c) Write the difference equation relating y ( k + 1) , y ( k ) , and x ( k + 1) for the rightside rule. (d) Find the transfer function Y ( z ) /X ( z ) for part (c). (e) Express y ( k ) as a summation on x ( k ) for the leftside rule. (f) Express y ( k ) as a summation on x ( k ) for the rightside rule.
17
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Solution: (a) y(k + 1) = y(k ) + Tx(k ) (b) zY ( z ) = Y ( z ) + TX ( z ) ⇒
Y ( z) T = X ( z) z − 1
(c) y(k + 1) = y(k ) + Tx(k + 1) (d) zY ( z ) = Y ( z ) + TzX ( z ) ⇒
Y ( z) Tz = X ( z) z − 1
(e) y(1) = y(0) + Tx(0) y(2) = y(1) + Tx(1) = y(0) + T ( x(0) + x(1) ) y(3) = y(2) + Tx(2) = y(0) + T [ x(0) + x(1) + x(2)] k −1
∴ y(k ) = y(0) + T ∑ x(n) n =0
(f) y(1) = y(0) + Tx(1) y(2) = y(1) + Tx(2) = y(0) + T [ x(1) + x(2)] k
∴ y(k ) = y(0) + T ∑ x(n) n =1
2.22. The trapezoidal rule (modified Euler method) for numerical integration approximates the integral of a function x (t ) by summing trapezoid areas as shown in Fig. P2.22. Let y (t ) be the integral of x (t ) . x(t)
x(k) x(k + 1) kT
(k + 1)T
t
FIGURE P2.22 Trapezoidal rule for numerical integration.
(a) Write the difference equation relating y ⎡⎢⎣( k + 1)T ⎤⎦⎥ , y ( kT ), x ⎡⎣⎢( k + 1)T ⎤⎦⎥ , and x ( kT ) for this rule. (b) Show that the transfer function for this integrator is given by 18
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Y ( z)
X ( z)
=
(T 2)( z + 1) z −1
Solution: (a) y(k + 1) = y(k ) + T (b) zY ( z) = Y ( z) +
x(k ) + x(k + 1) 2
T T z +1 X ( z) [ X ( z) + zX ( z)] ⇒ Y ( z) = 2 2 z −1
2.23. (a) The transfer function for the rightside rectangularrule integrator was found in Problem 2.21 to be Y ( z ) /X ( z ) = Tz/ ( z − 1) . We would suspect that the reciprocal of this transfer function should yield an approximation to a differentiator. That is, if w( kT ) is a numerical derivative of x (t ) at t = kT ,
W ( z) X ( z)
=
z −1 Tz
Write the difference equation describing this differentiator. (b) Draw a figure similar to those in Fig. P2.21 illustrating the approximate differentiation. (c) Repeat part (a) for the leftside rule, where W ( z ) /X ( z ) = T / ( z − 1) . (d) Repeat part (b) for the differentiator of part (c).
Solution: (a) Tz W ( z) = zX ( z) − X ( z) w(k + 1) =
1 [ x(k + 1) − x(k )] T
(b) x calculated slope kT (k + 1)T t
(c) TW ( z) = zX ( z) − X ( z)
19
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x
calculated slope
kT (k + 1)T t
w(k ) =
1 [ x(k + 1) − x(k ) ] T
2.31. Find the ztransform of the number sequence generated by sampling the time function e(t ) = t every T seconds, beginning at t = 0 . Can you express this transform in closed form?
e(t) = t; E(z) = 0 + Tz −1 + 2Tz −2 +! =
Solution:
Tz (z − 1)2
2.32. (a) Write, as a series, the ztransform of the number sequence generated by sampling the time function e(t ) = ε−t every T seconds, beginning at t = 0 . Can you express this transform in closed form? (b) Evaluate the coefficients in the series of part (a) for the case that T = 0.05 s . (c) The exponential e(t ) = ε−bt is sampled every T = 0.2 s , yielding the ztransform ⎛ 1⎞ ⎛ 1 ⎞2 ⎛ 1 ⎞3 E(z) = 1 + ⎜⎜ ⎟⎟⎟ z−1 + ⎜⎜ ⎟⎟⎟ z−2 + ⎜⎜ ⎟⎟⎟ z−3 + ⎜⎝ 2 ⎠ ⎜⎝ 2 ⎠ ⎜⎝ 2 ⎠
Evaluate b.
Solution: (a) E( z) = 1 + ε−T z −1 + ε−2T z −2 +L = 1 + (ε −T z −1 ) 1+ (ε −T z −1 )2 +! =
1 z = −T −1 1− ε z z − ε −T
(b) E(z) = 1+ (0.9512z −1 )1 + (0.9512z −1 )2 +! = (c) ε−bT
T = 0.2
z z − 0.9512
= ε−0.2b = 0.5
20
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∴−0.2b = ln (0.5) = −0.6931 ⇒ b = −3.466
2.33. Find the ztransforms of the number sequences generated by sampling the following time functions every T seconds, beginning at t = 0 . Express these transforms in closed form. (a) e(t ) = ε−at − t−T (b) e(t ) = ε ( )u (t − T ) − t−5T (c) e(t ) = ε ( )u (t − 5T )
Solution: (a) e(t) = ε−at ⇒ E(z) = 1+ ε−aT z −1 + ε−2aT z −2 +! =
z 23. z − ε−aT
(b) e(t ) = ε−(t −T )u(t − T ) # z & 1 E(z) = z −1 + ε−T z −2 + ε−2T z −3 +! = z −1 % (= $ z − ε−T ' z − ε−T
(c) e(t ) = ε−(t −5T )u(t − 5T ) # z & 1 E(z) = z −5 + ε−T z −6 + ε−2T z −7 +! = z −5 % (= $ z − ε−T ' z 4 (z − ε−T )
2.41. A function e(t ) is sampled, and the resultant sequence has the ztransform E ( z) =
z 3 − 2z z 4 − 0.9z 2 + 0.8
Solve this problem using E ( z ) and the properties of the ztransform. (a) Find the ztransform of e(t − 2T )u (t − 2T ) . (b) Find the ztransform of e(t + 2)u (t ) . (c) Find the ztransform of e(t − T )u (t − 2T ) .
21
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Solution: (a) [e(t − 2T )u(t − 2T )] =
(z 3 − 2z)z −2 z 4 − 0.9z 2 + 0.8
(b) e(0) = 0, e(1) = 1 ∴ [e(t + T )u(t)] = z[E(z) − e(0) − e(1)z −1 ]
⎡ z3 − 2z 1⎤ −1.1z 2 + 0.8 = z⎢ 4 − = ⎥ 2 4 2 ⎣ z − 0.9 z + 0.8 z ⎦ z − 0.9 z + 0.8
(c) [e(t − T )u(t − 2T )] = e(T )z −2 + e(2T )z −3 +! = z −1[E( z) − e(0)] = z −1E( z), since e(0) = 0 =
z2 − z z 4 − 0.9 z 2 + 0.8
2.42. A function e(t ) is sampled, and the resultant sequence has the ztransform E ( z) =
z−b z − cz 2 + d 2
Find the ztransform of ε akT e( kT ) . Solve this problem using E ( z ) and the properties of the ztransform. Solution: By complex translation zε−aT − b ⎡ ε akT e( kT )⎤ = E(zε−aT ) = ⎢⎣ ⎥⎦ 2 −2aT − cz 2ε−2aT + d z ε
2.51. From Table 23, z ( z − cos aT ) ⎡⎢cos akT ⎤⎥ = ⎦ ⎣ 2 z − 2z cos aT + 1
22
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(a) Find the conditions on the parameter a such that ⇥ ⎡⎢⎣cos akT ⎤⎥⎦ is first order (polezero cancellation occurs). (b) Give the firstorder transfer function in part (a). (c) Find a such that ⇥ ⎡⎢⎣cos akT ⎤⎦⎥ = ⇥ ⎡⎢⎣u ( kT )⎤⎥⎦ , where u ( kT ) is the unit step function.
Solution:
(a) poles: z =
z cos a ± 4cos 2 a − 4 = cos(a) ± j sin(a) 2
∴ pole = cos a, provided sin a = 0 ⇒ a = 0, ± π, ± 2π, K , ± nπ Then cos a = (−1)n ∴ poles = cos a (b) E ( z ) =
z ( z − cos a) z , a = ± nπ, n = 0, 1, … = ( z − cos a)( z − cos a) z − cos a
(c) E ( z ) =
z z , ∴ cos a = 1, a = 0, ± 2π, ± 4π, … = z − cos a z − 1
2.52. Find the ztransform, in closed form, of the number sequence generated by sampling the time function e(t ) every T seconds beginning at t = 0 . The function e(t ) is specified by its Laplace transform, E ( s) =
(
2 1 − ε−5s s(s + 2)
),
T = 1s
Solution: E1 (s) =
2 1 −1 = + s(s + 2) s s + 2
∴ e1 (t ) = (1 − ε−2t )u(t ) ⇒ e1 (kT ) = (1 − ε−2kT )u(kT ) ∴ E1 (z) = (1+ z −1 + z −2 +!) − (1− ε−2T z −1 + ε−4T z −2 +!)
23
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=
1 1 z z (1 − ε−2 ) z − = − = , T =1 1 − z −1 1 − ε−2 z −1 z − 1 z − ε−2T ( z − 1)( z − ε−2 )
E ( z ) = E1 ( z ) − z −5 E1 ( z ) =
(1 − ε−2 )( z 5 − 1) 0.8647( z 5 − 1) = z 4 ( z − 1)( z − ε−2 ) z 4 ( z − 1)( z − 0.1353)
2.61. Solve the given difference equation for x ( k ) using: ⎧1, k = 0, 1 x(k ) − 3x(k − 1) + 2 x(k − 2) = e(k ), e(k ) = ⎨ ⎩0, k ≥ 2 x (−2) = x (−1) = 0
(a) The sequential technique. (b) The ztransform. (c) Will the finalvalue theorem give the correct value of x ( k ) as k → ∞ ?
Solution: (a) x(0) = e(0) = 1 x(1) = e(1) + 3x(0) = 4 x(2) = e(2) + 3x(1) − 2 x(0) = 10 x(3) = 0 + 3(10) − 2(4) = 22 x(4) = 0 + 3(22) − 2(10) = 46
(b) [1 − 3z −1 + 2 z −2 ] X ( z) = E( z) = 1 + z −1 = X ( z) =
z +1 z
z2 z +1 z ( z + 1) 3 ⎤ ⎡ −2 × = = z⎢ + ⎥ z ( z − 1)( z − 2) ( z − 1)( z − 2) ⎣ z −1 z − 2⎦
∴ x(k ) = −2 + 3(2)k
(c) No, since the final value does not exist.
24
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2.62. Given the difference equation y ( k + 2) −
3 1 y ( k + 1) + y ( k ) = e( k ) 4 8
where y (0) = y (1) = 0, e(0) = 0 , and e( k ) = 1, k = 1, 2,… . (a) Solve for y ( k ) as a function of k, and give the numerical values of y ( k ), 0 ≤k ≤ 4. (b) Solve the difference equation directly for y ( k ), 0 ≤ k ≤ 4, to verify the results of part (a). (c) Repeat parts (a) and (b) for e( k ) = 0 for all k, and y (0) = 1, y (1) = −2 .
Solution: (a)
⎡ z ⎤ 1 = E(z) = [u(k − 1)] = z −1 ⎢ ⎥ ⎣ z − 1⎦ z − 1
1⎤ ⎡ 2 3 ⎢ z − 4 z + 8 ⎥ Y ( z) = E( z) ⎣ ⎦
Y ( z) 1 1 −8 8 3 −16 64 3 = · = + + + 1 1 ⎞⎛ 1 ⎞ z −1 z z −1 z − z z −1 4 ⎛ z ⎜ z − ⎟⎜ z − ⎟ 2 2 ⎠⎝ 4⎠ ⎝ k
64 ⎛ 1 ⎞ ⎛1⎞ ∴ y (k ) = −8δ(0) + 8 − 16 ⎜ ⎟ + ⎜ ⎟ 3 3 ⎝4⎠ ⎝2⎠
k
∴ y(0) = 0; y(1) = 0; y(2) = 0; y(3) = 1; y(4) =
(b) y(k + 2) = e(k ) +
7 4
3 1 y (k + 1) − y(k ) 4 8
3 1 y(2) = 0 + (0) − (0) = 0 4 8 3 1 y (3) = 1 + (0) − (0) = 1 4 8 3 1 y(4) = 1 + (1) − (0) = 7 4 4 8
25
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3 1 (c) (a) y(k + 2) − y(k + 1) + y(k ) = 0 4 8 3 1 ∴ z 2 [Y ( z ) − y(0) − y(1) z −1 ] − z[Y ( z ) − y(0)] + Y ( z ) = 0 4 8 3 1⎤ 3 ⎡ ∴ ⎢ z 2 − z + ⎥ Y ( z) = z 2 − 2z − z 4 8⎦ 4 ⎣ ⎡ ⎤ z −1 4 ⎥= ∴ Y (z) = z ⎢ ⎢ ⎥ 1 1 ⎢⎣ z − 2 z − 4 ⎥⎦
(
)(
)
k k ⎡ ⎤ ⎛ 1⎞ ⎛ 1⎞ 10 ⎥ −9 + ⇒ y(k) = −9 ⎜ ⎟ + 10 ⎜ ⎟ z⎢ ⎢z − 1 ⎝ 2⎠ ⎝ 4⎠ z− 1 ⎥ ⎣ 4⎦ 2
y(0) = 1, y(1) = −2, y(2) = −13 8, y(3) = − 31 32, y(4) = − 67 128
(b) y(k + 2) =
3 1 y(k + 1) − y(k ) 4 8
y(2) =
3 1 (−2) − (1) = −13 8 4 8
y (3) =
3 ⎛ 13 ⎞ 1 ⎜ − ⎟ − (−2) = −31 32 4⎝ 8 ⎠ 8
y (4) =
3 ⎛ 31 ⎞ 1 ⎛ 13 ⎞ 67 ⎜− ⎟ − ⎜− ⎟ = − 4 ⎝ 32 ⎠ 8 ⎝ 8 ⎠ 128
2.63. Given the difference equation x ( k ) − x ( k − 1) + x ( k − 2) = e( k )
where e( k ) = 1 for k ≥ 0. (a) Solve for x ( k ) as a function of k, using the ztransform. Give the values of x (0), x (1) , and x (2) .
(b) Verify the values x (0) , x (1) , and x (2) , using the powerseries method. (c) Verify the values x (0) , x (1) , and x (2) by solving the difference equation directly. (d) Will the finalvalue property give the correct value for x (∞)?
26
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Solution: (a) [1 − z −1 + z −2 ] X ( z) = E( z) =
X ( z) =
z3 , ( z − 1)( z 2 − z + 1)
z z −1 poles: z =
1 3 ± j = 1∠ ± 60° 2 2
k k* 1 X ( z) = + 1 + 1 * with p = 1∠60° z z − 1 z − p1 z − p1
k1 =
=
z2 ( z − 1)( z − 1∠ − 60°)
= z =1∠60°
1∠120° (.5 + j.866 − 1)(.5 + j.866 − .5 + j.866)
1∠120° = 0.5774 ∠ − 90° 1∠120° [ j 2(0.866)]
∴ aT = ln ( p1 ) = 0; bT = arg p1 =
π 3
A = 2 k1 = 1.155; θ = arg k1 = −90° ⎛π ⎞ ⎛π ⎞ ∴ x(k ) = 1 + 1.155 cos ⎜ k − 90° ⎟ = 1 + 1.155 sin ⎜ k ⎟ 3 ⎝ ⎠ ⎝3 ⎠
x(0) = 1, x(1) = 2, x(2) = 2 1+ 2z −1 + 2z −2 +! (b) z 3 − 2z 2 + 2z − 1 z 3 3
2
z − 2z + 2z − 1
∴ x(0) = 1 x(1) = 2 x(2) = 2
2z 2 − 2z + 1 2z 2 − 4z + 4 − 2z −1 2z +!
(c) x(k ) = 1 + x(k − 1) − x(k − 2) x(0) = 1 + 0 − 0 = 1 x(1) = 1 + 1 − 0 = 2 x(2) = 1 + 2 − 1 = 2
(d) No, 3 poles for X(z) on the unit circle.
27
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2.64. Given the difference equation x ( k + 2) + 3x ( k + 1) + 2x ( k ) = e( k )
where ⎧ 1, ⎪ k=0 ⎪ e( k ) = ⎨ ⎪ 0, otherwise ⎪ ⎪ ⎩ x (0) = 1 x (1) = −1
(a) Solve for x ( k ) as a function of k. (b) Evaluate x (0) , x (1) , x (2) , and x (3) in part (a). (c) Verify the results in part (b) using the powerseries method. (d) Verify the results in part (b) by solving the difference equation directly.
Solution: (a) z 2 [ X ( z) − x(0) − x(1) z −1 ] + 3z[ X ( z) − x(0)] + 2 X ( z) = E( z) = 1 ∴ X ( z) =
1 + z 2 − z + 3z z 2 + 2 z + 1 z + 1 = 2 = z 2 − 3z + 2 z + 3z + 2 z + 2
1 ⎡ z +1 ⎤ ⎡1 ⎤ ∴ X ( z) = z ⎢ = z⎢ 2 + 2 ⎥ ⎥ + + z z z z ( 2) 2 ⎣ ⎦ ⎣ ⎦
1 1 ∴ x(k ) = δ(k ) + (−2)k 2 2
(b) x(0) = 1, x(1) = −1, x(2) = 2, x(3) = −4
28
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1− z −1 + 2z −2 − 4z −3 +… (c) z + 2 z + 1 z+2 −1 −1− 2z −1 2z −1 2z −1 + 4z −2 −4z −2 !
(d) x(k + 2) = e(k ) − 3x(k + 1) − 2x(k ) x(2) = 1 − 3(−1) − 2(1) = 2 x(3) = 0 − 3(2) − 2(−1) = −4
2.65. Given the difference equation x ( k + 3) − 2.2x ( k + 2) + 1.57x ( k + 1) − 0.36x ( k ) = e( k )
where e( k ) = 1 for all k ≥ 0, and x (0) = x (1) = x (2) = 0 . (a) Write a digital computer program that will calculate x ( k ) . Run this program solving for x (3) , x (4) , . . . , x (25) .
(b) Using the sequential technique, check the values of x ( k ), 0 ≤ k ≤ 5. (c) Use the ztransform and the powerseries method to verify the values x ( k ), 0 ≤ k ≤ 5.
Solution: (a) x0 = 0; x1 = 0; x2 = 0; for k = 0:5; x3 = 2.2*x2 – 1.57*x1 + 0.36*x0 + 1 x0 = x1; 29
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x1 = x2; x2 = x3; end (b) x(k + 3) = e(k ) + 2.2x(k + 2) − 1.57 x(k + 1) + 0.36x(k ) x(3) = 1 + 0 − 0 + 0 = 1 x(4) = 1 + 2.2(1) − 0 + 0 = 3.2 x(5) = 1 + 2.2(3.2) − 1.57(1) = 6.47
(c) [ z 3 − 2.2 z 2 + 1.57 z − 0.36] X ( z) = E( z) = X ( z) =
z z −1
z ( z − 1)( z − 2.2 z 2 + 1.57 z − 0.36) 3
z −3 + 3.2z −4 + 6.47z −5 +! z 4 − 3.2z 3 + 3.77z 2 − 1.93z + 0.36z z − 3.2 + 3.77z −1 −! 3.2 − 3.77z −1 3.2 − 10.24z −1 +! 6.47z −1 +! !
∴ x(3) = 1 x(4) = 3.2 x(5) = 6.47
2.71. (a) Find e(0) , e(1) , and e(10) for E ( z) =
0.1 z ( z − 0.9)
using the inversion formula. (b) Check the value of e(0) using the initialvalue property. (c) Check the values calculated in part (a) using partial fractions. 30
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(d) Find e( k ) for k = 0, 1, 2, 3, and 4 if ⇥ ⎡⎣⎢ e( k )⎤⎦⎥ is given by E ( z) =
(z
2
)
1.98z
(
− 0.9z + 0.9 ( z − 0.8) z 2 − 1.2z + 0.27
)
(e) Find a function e(t ) which, when sampled at a rate of 10 Hz (T = 0.1s) , results in the transform E ( z ) = 2z/ ( z − 0.8) . (f) Repeat part (e) for E ( z ) = 2z/ ( z + 0.8) . (g) From parts (e) and (f), what is the effect on the inverse ztransform of changing the sign on a real pole? Solution: (a) e(k ) =
0.1z k −1
0.1z k − 2
= ∑ ∑ residues z ( z − 0.9) residues z − 0.9
k = 0 : fcn =
0.1 0.1 = 0.1235 , ∴ residue z =0.9 = z ( z − 0.9) (0.9)2
residue z =0 =
d ⎡ 0.1 ⎤ −0.1(1) = ⎢ ⎥ dz ⎣ z − 0.9 ⎦ z = 0 ( z − 0.9)2
2
= z =0
−0.1 = −0.1235 (0.9)2
∴ e(0) = 0 k = 1: e(1) =
0.1 0.1 + =0 z − 0.9 z =0 z z =0.9
k = 10 : e(10) = 0.1(0.9)8
(b) e(0) = lim E ( z) = lim z →∞
(c)
z →∞
0.1 =0 z ( z − 0.9)
k3 k k 0.1 E( z) = 2 = 12 + 2 + z z z − 0.9 z ( z − 0.9) z k1 =
1 0.1 1 −0.1 = − ; k3 = 0.9 9 (0.9)2 8.1
k2 =
d ⎡ 0.1 ⎤ −1 = , from (a) ⎢ ⎥ dz ⎣ z − 0.9 ⎦ z =0 8.1
31
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∴ e(k ) =
−1 1 1 δ(k ) − δ(k − 1) + (0.9)k 8.1 9 8.1
x(0) = −
1 1 1 0.9 +0+ = 0; x(1) = − 0 − + =0 8.1 8.1 9 8.1
x(10) = − 0 − 0 +
(d) E(z) =
0.1 (0.9)10 = 0.1(0.9)8 (0.9)2
1.98z = 1.98z −4 + (⋅)z −5 + (⋅)z −6 +! z 5 +!
∴ e(0) = e(1) = e(2) = e(3) = 0; e(4) = 1.98
(e) E ( z ) = ∴a =
2z 2z = z − 0.8 z − ε− aT
∴ε− aT = 0.8 ⇒ aT = 0.2231
0.2231 = 2.231, ∴ e(t ) = 2ε−2.231t u (t ) 0.1
(f) E ( z ) =
2z ; ∴ε− aT ε jπ = −0.8 ⇒ aT = 2.231 z − (−0.8)
∴ e(t ) = 2e−2.231t cos10πt where
(g) (e) e(k ) = (0.8) k ;
ωs = 10π 2
(f ) e(k ) = (−0.8) k
∴ sign alternates on e(k ).
2.72. For the number sequence {e( k )} , E ( z) =
z 2
( z + 1)
(a) Apply the finalvalue theorem to E ( z ) . (b) Check your result in part (a) by finding the inverse ztransform of E ( z ) . 2
(c) Repeat parts (a) and (b) with E ( z ) = z/ ( z − 1) . 2
(d) Repeat parts (a) and (b) with E ( z ) = z/ ( z − 0.9) .
32
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2
(e) Repeat parts (a) and (b) with E ( z ) = z/ ( z − 1.1) .
Solution: (a) e(∞) = lim ( z − 1) E ( z ) = z →1
z ( z − 1) ( z + 1) 2
=0 z =1
⎡ z ⎤ (b) e(k ) = z −1 ⎢ = k (−1)k , ∴ e(∞)unbounded 2⎥ ⎣ ( z − 1) ⎦
(c) (a) e(∞) = lim ( z − 1) z →1
z , ∴ unbounded ( z − 1)2
(b) e(k ) = k , ∴ unbounded (d) (a) e(∞) = lim ( z − 1) z →1
z =0 ( z − 0.9)2
(b) e(k ) = k (0.9)k ; ∴ e(∞) → 0 (e) (a) e(∞) = lim ( z − 1) z →1
z =0 ( z − 1.1)2
(b) e(k ) = k (1.1)k ; ∴ e(∞) is unbounded.
2.73. Find the inverse ztransform of each E ( z ) below by the four methods given in the text. Compare the values of e( z ) , for k = 0, 1, 2, and 3, obtained by the four methods. (a) E ( z ) =
(c) E ( z ) =
0.5z ( z − 1)( z − 0.6)
0.5( z + 1)
( z − 1)( z − 0.6)
(b)
E ( z) =
(d)
E ( z) =
0.5 ( z − 1)( z − 0.6)
z ( z − 0.7)
( z − 1)( z − 0.6)
(e) Use MATLAB to verify the partialfraction expansions.
Solution:
33
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0.5z −1 + 0.8z −2 + 0.98z −3 +! (a) (i) z 2 − 1.6z + 0.6 0.5z 0.5z − 0.8 + 0.3z −1 0.8 − 0.3z −1 0.8 − 1.28z −1 +! 0.98z −1 +!
(ii)
0.5 1.25 −1.25 1.25 z 1.25 z E( z) ; ∴ E( z) = = = + − ( z − 1)( z − 0.6) z − 1 z − 0.6 z z − 1 z − 0.6
∴ e(k ) = 1.25(1 − 0.6k )u(k )
(iii) z k −1 E ( z ) =
e( k ) =
0.5 z k ( z − 1)( z − 0.6)
0.5(1) k 0.5(0.6) k + = 1.25(1 − 0.6k )u ( k ) 1 − 0.6 0.6 − 1
(iv) E1 ( z ) = E2 ( z) =
0.5 z ⇒ e1 (k ) = 0.5(0.6)k z − 0.6
1 ⇒ e2 (0) = 0; e2 (k ) = 1, k ≥ 1 z −1
e(0) = e1 (0)e2 (0) = (0.5)(0) = 0 e(1) = e1 (0)e2 (1) + e1 (1)e2 (0) = (0.5)(1) + (0.3)(0) = 0.5 e(2) = e1 (0)e2 (2) + e1 (1)e2 (1) + e1 (2)e2 (0) = 0.5 × 1 + 0.3 × 1 + 0.18 × 0 = 0.8
e(3) = 0.5 × 1 + 0.3 × 1 + 0.18 × 1 + 0.108 × 0 = 0.98
(b) e(0) = 0 e(k ) = 1.25 − 2.083(0.6)k , k ≥ 1 E(z) = 0.5z −2 + 0.8z −3 + 0.98z −4 + 1.088z −5 +!
(c) e(0) = 0; e(k ) = 2.5 − 3.33(0.6)k , k ≥ 1 E(z) = 0.5z −1 + 1.30z −2 + 1.78z −3 + 2.068z −4 + 2.2408z −5 +!
(d) e(k ) = 0.75 + 0.25(0.6)k 34
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E(z) = 1+ 0.9z −1 + 0.84z −2 + 0.804z −3 +!
(e) num=[0 0 0.5]; den=[1 –1.6 0.6]; [r, p, k] = residue (num, den)
2.81. Given in Fig. P2.81 are two digitalfilter structures, or realizations, for secondorder filters. e(k)
T
T d2
d1
+
d0
+
+
+
y(k)
y(k)
T


T
c1 c0 (a) b2
b1
e(k)
+
f(k) 

T
T
b0
+
+
y(k)
+
a1
a0 (b)
FIGURE P2.81 Digitalfilter structures: (a) 3D; (b) 1D. 35
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(a) Write the difference equation for the 3D structure of Fig. P2.81(a), expressing y ( k ) as a function of y ( k − i) and e( k − i) . (b) Derive the filter transfer function Y ( z ) /E ( z ) for the 3D structure by taking the ztransform of the equation in part (a). (c) Write the difference equation for the 1D structure of Fig. P2.81(b). Two equations are required, with one for f ( k ) and one for y ( k ) . (d) Derive the filter transfer function Y ( z ) /E ( z ) for the 1D structure by taking the ztransform of the equations in part (c) and eliminating F ( z ) . (e) From parts (b) and (d), relate the coefficients α i , βi to ai , bi such that the two filters realize the same transfer function. (f) Write a computerprogram segment that realizes the 3D structure. This program should be of the form used in Example 2.10. (g) Write a MATLABprogram segment that realizes the 1D structure. This program should be of the form used in Example 2.10.
Solution: (a) y(k ) = β2e(k ) + β1e(k − 1) + β0e(k − 2) − α1 y(k − 1) − α0 y(k − 2) (b) ⎣⎡1 + α1 z −1 + α0 z −2 ⎤⎦ Y ( z ) = ⎡⎣β2 + β1 z −1 + β0 z −2 ⎤⎦ E ( z ) Y ( z ) β2 z 2 + β1 z + β0 = 2 E( z) z + α1 z + α0
(c) f (k ) = e(k ) − a1 f (k − 1) − a0 f (k − 2) y(k ) = b2 f (k ) + b1 f (k − 1) + b0 f (k − 2)
(d) F ( z ) = E ( z ) − (a1 z −1 + a0 z −2 ) F ( z ) ⇒ F ( z ) =
Y ( z ) = (b2 + b1 z −1 + b0 z −2 ) F ( z ) =
(e) αi = ai
E( z) 1 + a1 z −1 + a0 z −2
b2 z 2 + b1 z + b0 z 2 + a1 z + a0
E( z)
and βi = bi , i = 1, 2
36
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(f) ykminus2 = 0; ykminus1 = 0; ekminus2 = 0; ekminus1 = 0; ek = 1; for k = 0:5 yk=b2*ek+b1*ekminus1+b0*ekminus2–a1*ykminus1a0*ykminus2; [k, ek, yk] ekminus2 = ekminus1; ekminus1 = ek; ykminus2 = ykminus1; ykminus1 = yk; end (g) fkminus2 = 0; fkminus1 = 0; ek = 1; for k = 0:5 fk=eka1*fkminus1–a0*fkminus2; yk = b2*fk+b1*fkminus1+b0*fkminus2; [k, ek, yk] fkminus2 = fkminus1; fkminus1 = fk; end
37
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2.82. Shown in Fig. P2.82 is the secondorder digitalfilter structure 1X. b2
g1 e(k)
g3
+
+
f1(k)
+ T

y(k)
+ g2
g2
g4
+
+
f2(k)
T
+ g1
FIGURE P2.82 Digitalfilter structure 1X.
This structure realizes the filter transfer function D ( z ) = b2 +
A A∗ + z− p z− p∗
where p and p ∗ (conjugate of p) are complex. The relationships between the filter coefficients and the coefficients in Fig. P2.82 are given by g1 = Re( p) g3 = −2 Im ( A) g2 = Im ( p) g4 = 2 Re( A)
(a) To realize this filter, difference equations are required for f1 ( k ), f2 ( k ), and y ( k ) . Write these equations. (b) Find the filter transfer function Y ( z ) /E ( z ) by taking the ztransform of the equations of part (a) and eliminating F1 ( z ) and F2 ( z ) . (c) Verify the results in part (b) using Mason’s gain formula. (d) Write a MATLABprogram segment that realizes the 1X structure. This program should be of the form of that is used in Example 2.10. 38
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Solution: (a) f1 (k ) = g1 f1 (k − 1) − g2 f2 (k − 1) + g3e(k ) f2 (k ) = g1 f2 (k − 1) + g1 f1 (k − 1) + g4e(k ) y(k ) = b2e(k ) + f 2 (k − 1)
(b) (1) F1 ( z) = g1 z −1F1 ( z) − g2 z −1F2 ( z) + g3 E( z) (2) F2 ( z) = g1 z −1F2 ( z) + g2 z −1F1 ( z) + g4 E( z) (3) Y ( z) = b0 E( z) + z −1F2 ( z) ∴ (1) ( z − g1 ) F1 ( z) + g2 F2 ( z) = g3 zE( z) (2) − g2 F1 ( z) + ( z − g1 ) F2 ( z) = g4 zE( z) z − g1 g3 zE ( z ) − g 2 g 4 zE ( z ) ( g 4 z 2 − g1 g 4 z + g 2 g3 z ) ∴ F2 ( s) = = E( z) z − g1 g2 ( z − g1 )2 + g 22 − g2
∴
z − g1
g z + g 2 g3 − g1 g 4 Y ( z) = b2 + 4 E( z) ( z − g1 )2 + g 22
also, D( z ) = b2 +
= b2 +
= b2 +
(c) D( z ) = b0 +
= b0 +
Re( A) + j Im( A) Re( A) − j Im( A) + z − Re( p) − j Im( p) z − Re( p) + j Im( p) 1 ( g4 2
− jg3 )
z − g1 − jg 2
+
1 ( g4 2
+ jg3 )
z − g1 + jg 2
g 4 z − g1 g 4 + g 2 g3 ( z − g1 ) 2 + g 22
g2 g3 z −2 + g 4 (1 − g1 z −1 ) 1 − g1 z −1 − g1 z −1 + g12 z −2 + g 22 z −2 g 4 z + g 2 g3 − g1 g 4 z 2 − 2 g1 z + g12 + g 22
(d) f1kminus1 = 0; f2kminus1 = 0; 39
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ek = 1; for k = 0:5 yk = b0*ek+f2kminus1; [k, ek, yk] f1k = g1*f1kminus1 – g2*f2kminus1 + g3*ek; f2k = g1*f2kminus1 + g2*f1kminus1 + g3*ek; f1kminus1 = f1k; f2kminus1 = f2k; end
2.83. Given the secondorder digitalfilter transfer function D( z) =
2z 2 − 2.4z + 0.72 z 2 − 1.4z + 0.98
(a) Find the coefficients of the 3D structure of Fig. P2.81 such that D ( z ) is realized. (b) Find the coefficients of the ID structure of Fig. P2.81 such that D ( z ) is realized. (c) Find the coefficients of the IX structure of Fig. P2.82 such that D ( z ) is realized. The coefficients are identified in Problem 2.82. (d) Use MATLAB to verify the partialfraction expansions in part (c). (e) Verify the results in part (c) using Mason’s gain formula.
Solution: (a) β2 = 2, β1 = −2.4, β0 = 0.72, α1 = −1.4, α0 = 0.98 (b) b2 = 2, b1 = −2.4, b0 = 0.72, a1 = −1.4, a0 = 0.98
(c) poles: z =
(
1.4 ± 1.42 − 4(0.98) 2
)
1
2
= 0.7 ± j 0.7 = 0.99 ∠ ± 45°
40
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D( z ) = 2 +
∴A=
A A* + z − 0.7 − j 0.7 z − 0.7 + j 0.7
2 z 2 − 2.4 z + 0.72 j1.96 − (1.68 + j1.68) + 0.72 = z − 0.7 + j 0.7 z =0.99∠ 45° j1.4
= 0.2 + j 0.6857
∴ g1 = 0.7 g2 = 0.7
g3 = 1.371 g4 = 0.4
(d) num =[2 2.4 .72]; den =[1 1.4 0.98];
[r,p,k,]=residue(num, den) (e) Δ = 1 − (0.7 z −1 + 0.7 z −1 + 0.4z −2 ) + 0.49z −2 = 1 − 1.4 z −1 + 0.98z −2 D( z) = 2 + = 2+
1 [1.371 (0.7) z −2 + 0.4 z −1 (1 + 0.7 z −1 )] Δ 0.4 z − 1.24 2 z 2 − 2.4 z + 0.72 = 2 z − 1.4 z + 0.98 z − 1.4 z + 0.98 2
2.91. Find two different statevariable formulations that model the system whose difference equation is given by: (a) y ( k + 2) + 6 y ( k + 1) + 5y ( k ) = 2e( k ) (b) y ( k + 2) + 6 y ( k + 1) + 5y ( k ) = e( k + 1) + 2e( k ) (c) y ( k + 2) + 6 y ( k + 1) + 5y ( k ) = 3e( k + 2) + e( k + 1) + 2e( k )
Solution: (a)
Y ( z) 2 = 2 U ( z) z + 6z + 5
(1)
control canonical: 41
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b2
O b1
u(k)
+
T
x2(k)
O
x1(k)
T
y(k)
+
b0 2
a1 5
6
a0
⎡0 1⎤ ⎡0 ⎤ x(k + 1) = ⎢ x( k ) + ⎢ ⎥ u ( k ) ⎥ ⎣ −5 −6 ⎦ ⎣1 ⎦
y(k ) = [2 0] x(k )
(2)
observer canonical: u(k)
b1
b0 Z +
a0
b2 O
T
5
x2(k)
+
a1
O T
x1(k)
+
y(k)
6
⎡ −6 1 ⎤ ⎡0⎤ x(k + 1) = ⎢ x( k ) + ⎢ ⎥ u ( k ) ⎥ ⎣ −5 0 ⎦ ⎣ 2⎦
y(k ) = [1 0] x(k )
(b)
x(k + 1) = same as (a) Y ( z) z+2 (1) control canonical: = 2 y (k ) = [2 1]x(k ) U ( z) z + 6 z + 5
(2) observer canonical: ⎡ −6 1 ⎤ ⎡1 ⎤ x(k + 1) = ⎢ x( k ) + ⎢ ⎥ u ( k ) ⎥ ⎣ −5 0 ⎦ ⎣ 2⎦
y(k ) = [1 0] x(k )
(c)
x(k + 1) = same as (a) Y ( z ) 3z 2 + z + 2 (1) control canonical: = 2 y (k ) = [−13 −17]x( k ) + 3u (k ) U ( z) z + 6 z + 5
42
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(2) observer canonical: ⎡ −6 1 ⎤ ⎡1 ⎤ x(k + 1) = ⎢ x( k ) + ⎢ ⎥ u ( k ) ⎥ ⎣ −5 0 ⎦ ⎣ 2⎦
y(k ) = [1 0] x(k ) + 3u(k )
2.92. Write the state equations for the observer canonical form of a system, shown in Fig. 210, which has the transfer function given in (251) and (261) G ( z) =
bn−1z n−1 + + b1z + b0 z n + an−1z n−1 + + a1z + a0
Solution: " a $ n−1 $ x(k + 1) = $ an−2 $ " $# 0
" b 1 0 ! 0 % $ n−1 ' 0 1 ! 0 ' x(k) + $ bn−2 $ ' $ " " " " ' $ b ' 0 0 ! 0 & # 0
% ' ' ' u(k) ' ' &
y(k) = !" 1 0 0 ! 0 #$ x(k)
2.101. Find a statevariable formulation for the system described by the coupled secondorder difference equations given. The system output is y ( k ) , and e1 ( k ) and e2 ( k ) are the system inputs. Hint: Draw a simulation diagram first. x ( k + 2) + v ( k + 1) = 4e1 ( k ) + e2 ( k ) v ( k + 2) − v ( k ) + x ( k ) = 2e1 ( k ) y ( k ) = v ( k + 2) − x ( k + 1) + e1 ( k )
Solution:
43
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e1(k)
4
+
T
+

x2(k)
T
x1(k) x(k) +
2 
+
e2(k)
T +
⎡0 ⎢0 x(k + 1) = ⎢ ⎢0 ⎢ ⎣ −1
y(k)
0⎤ ⎡0 ⎥ ⎢4 0 0 −1⎥ x( k ) + ⎢ ⎢0 0 0 1⎥ ⎥ ⎢ 0 1 0⎦ ⎣2
1 0
x4(k)
T
x3(k) n(k)
0⎤ 1 ⎥⎥ e( k ) 0⎥ ⎥ 0⎦
y(k ) = x4 (k + 1) − x2 (k ) + e1 (k ) = − x1 (k ) + x3 (k ) − x2 (k ) + e1 (k ) ∴ y(k ) = [−1 −1 1 0] x(k ) + [1 0] e(k )
2.102. Consider the system described by ⎡ ⎤ ⎡ ⎤ x ( k + 1) = ⎢⎢ 0 1 ⎥⎥ x ( k ) + ⎢⎢ 1 ⎥⎥ u ( k ) ⎢⎣ 0 3 ⎦⎥ ⎢⎣ 1 ⎥⎦
y ( k ) = ⎡⎢ −2 1 ⎤⎥ x ( k ) ⎣ ⎦
(a) Find the transfer function Y ( z ) /U ( z ) . (b) Using any similarity transformation, find a different state model for this system. (c) Find the transfer function of the system from the transformed state equations. (d) Verify that A given and A w derived in part (b) satisfy the first three properties of similarity transformations. The fourth property was verified in part (c).
Solution: ⎡ z −1 ⎤ ⎥ ; Δ = zI − A = z ( z − 3) = Δ ⎣ 0 z − 3⎦
(a) zI − A = ⎢
44
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⎡ z − 3 1 ⎤ ⎡1⎤ Y ( z) 1 = C[ zI − A]−1 B = [ −2 1] ⎢ z ⎥⎦ ⎢⎣1⎥⎦ U ( z) Δ ⎣ 0 =
⎡1⎤ − z + 4 1 [ −2 z + 6 z − 2] ⎢ ⎥ = Δ ⎣1⎦ z ( z − 3)
⎡ 1 ⎡1 −1⎤ ; P −1 = ⎢ 2 ⎥ ⎣1 1 ⎦ ⎣− 12
(b) P = ⎢
1
2⎤
1
2⎦
⎥
⎡0 1⎤ ⎡1 −1⎤ ⎡ 1 2 ⎥ ⎢ ⎥ ⎢ 1 ⎥ = ⎢− 1 1 ⎦ ⎣ 2 2 ⎦ ⎣ 0 3⎦ ⎣1
⎡ 1 A w = P −1AP = ⎢ 2 ⎣− 12
1
2⎤
2⎤
⎡1 1⎤ ⎥ ⎢ ⎥ 1 2 ⎦ ⎣ 3 3⎦ 1
⎡2 2⎤ =⎢ ⎥ ⎣1 1 ⎦ ⎡1 B w = P −1B = ⎢ 2 ⎣ −1 2
1 1
2⎤
⎡1⎤ ⎡1 ⎤ ⎥ ⎢1⎥ = ⎢0 ⎥ ⎣ ⎦ 2⎦ ⎣ ⎦
⎡1 −1⎤ Cw = CP = [ −2 1] ⎢ ⎥ = [ −1 3] ⎣1 1 ⎦ ⎡2 2⎤ ⎡1 ⎤ ∴ w (k + 1) = ⎢ w ( k ) + ⎢ ⎥ u( k ) ⎥ ⎣1 1 ⎦ ⎣0 ⎦
y(k ) = [−1 3] w(k ) ⎡z − 2
(c) zI − A w = ⎢ ⎣ −1
−2 ⎤ ; Δ = zI − A w = z 2 − 3z + 2 − 2 = z ( z − 3) z − 1⎥⎦
2 ⎤ ⎡1 ⎤ ⎡z −1 Y ( z) 1 = Cw [ zI − A w ]−1 B w = [ −1 3] ⎢ z − 2 ⎥⎦ ⎢⎣0 ⎥⎦ U ( z) Δ ⎣ 1 =
⎡ z − 1⎤ − z + 4 1 [ −1 3] ⎢ ⎥ = Δ ⎣ 1 ⎦ z ( z − 3)
(d) zI − A =
z
1
0 z −3
= z 2 − 3z; zI − A w =
z−2
−2
−1
z −1
= z ( z − 3)
∴ z1 = 0, z2 = 3 A =
0 1 0 3
= 0 = z1 z2 ; A w =
2 2 1 1
=0
tr A = 3 = z1 + z2 ; tr Aw = 3
45
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2.103. Consider the system of Problem 2.102. A similarity transformation on these equations yields ⎡ d ⎤ ⎢ 1 0 ⎥ w( k + 1) = ⎢ ⎥ w( k ) + B wu ( k ) ⎢ 0 d2 ⎥ ⎢⎣ ⎦⎥ y ( k ) = C wx ( k )
(a) Find d1 and d2 . (b) Find a similarity transformation that results in the A w matrix given. Note that this matrix is diagonal. (c) Find B w and C w . (d) Find the transfer functions of both sets of state equations to verify the results of this problem.
Solution: (a) Let z1 , z2 be the characteristic value of A. d1 = z1 ,
d2 = z2
⎡ z −1 ⎤ zI − A = ⎢ ⎥ , ∴ zI − A = z ( z − 3); ∴ z1 = 0, z2 = 3 ⎣ 0 z − 3⎦ ⎡ 0 −1⎤ ⎡ m ⎤
⎡0 ⎤
−m = 0
11 21 (b) ( z1I − A)m1 = ⎢ ⎥ ⎢ m ⎥ = ⎢ 0 ⎥ ⇒ −3m = 0 0 − 3 ⎣ ⎦ ⎣ 21 ⎦ ⎣ ⎦ 21
⎡1 ⎤ ∴ m21 = 0, let m11 = 1, ∴ m1 = ⎢ ⎥ ⎣0⎦ ⎡ 3 −1⎤ ⎡ m12 ⎤ ⎡ 0 ⎤ ( z2 I − A)m2 = ⎢ ⎥ = ⎢ ⎥ ⇒ 3m12 − m22 = 0 ⎥⎢ ⎣ 0 0 ⎦ ⎣ m22 ⎦ ⎣ 0 ⎦ ⎡1⎤ ∴ let m12 = 1, m22 = 3, ∴ m2 = ⎢ ⎥ ⎣ 3⎦ ⎡1 1⎤ ⎡1 −1 3⎤ ∴M = ⎢ , M = 3, M −1 = ⎢ ⎥ ⎥ ⎣ 0 3⎦ ⎣0 1 3 ⎦ ⎡1 −1 3⎤ ⎡0 1⎤ ⎡1 1⎤ ⎡1 −1 3⎤ ⎡0 3⎤ ⎡0 0 ⎤ M −1AM = ⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥=⎢ ⎥ ⎣ 0 1 3 ⎦ ⎣ 0 3⎦ ⎣ 0 3⎦ ⎣ 0 1 3 ⎦ ⎣ 0 9 ⎦ ⎣ 0 3 ⎦
46
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⎡1 −1 3⎤ ⎡ 2 3⎤ ⎥=⎢ ⎥ ⎣0 1 3 ⎦ ⎣1 3 ⎦
(c) B w = M −1B = ⎢
⎡1 1 ⎤ Cw = CM = [ −2 1] ⎢ ⎥ = [ −2 1] ⎣ 0 3⎦ ⎡0 0⎤ ⎡ 2 3⎤ ∴ w(k + 1) = ⎢ ⎥ w ( k ) + ⎢ 1 3 ⎥ u( k ) 0 3 ⎣ ⎦ ⎣ ⎦
y(k ) = [−2 1] w(k )
(d) See Problem 2.102(a) for the first transfer function. 0 ⎤ ⎡z zI − A w = ⎢ ⎥ ; zI − A w = z ( z − 3) = Δ ⎣ 0 z − 3⎦ ⎡ z − 3 0 ⎤ ⎡ 2 3⎤ Y ( z) 1 = Cw [ zI − A w ]−1 B w = [ −2 1] ⎢ z ⎥⎦ ⎢⎣ 1 3 ⎥⎦ U ( z) Δ ⎣ 0 =
⎡ 2 3⎤ − 4 z + 4 + 13 z −z + 4 1 = [ −2 z + 6 z ] ⎢ ⎥ = 3 Δ Δ z ( z − 3) ⎣1 3 ⎦
2.104. Repeat Problem 2.102 for the system described by ⎡ ⎤ ⎡ ⎤ x ( k + 1) = ⎢⎢ 1 0 ⎥⎥ x ( k ) + ⎢⎢ 2 ⎥⎥ u ( k ) ⎢⎣ 0 0.5 ⎦⎥ ⎢⎣ 1 ⎥⎦
y ( k ) = ⎡⎢ 1 2 ⎤⎥ x ( k ) ⎣ ⎦
(a) Find the transfer function Y ( z ) /U ( z ) . (b) Using any similarity transformation, find a different state model for this system. (c) Find the transfer function of the system from the transformed state equations. (d) Verify that A given and A w derived in part (b) satisfy the first three properties of similarity transformations. The fourth property was verified in part (c).
Solution: (a)
47
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⎡ 1 ⎢ z −1 Y ( z) = C[ zI − A]−1 B = [1 2 ] ⎢ U ( z) ⎢ 0 ⎢⎣
⎤ ⎥ ⎥ 1 ⎥ z − 0.5 ⎥⎦ 0
⎡2⎤ ⎢1 ⎥ ⎣ ⎦
⎡ 2 ⎤ ⎢ z −1 ⎥ 2 2 4z − 3 = [1 2 ] ⎢ + = ⎥= ⎢ 1 ⎥ z − 1 z − 0.5 ( z − 1)( z − 0.5) ⎣⎢ z − 0.5 ⎦⎥
⎡1 ⎡1 −1⎤ , P −1 = ⎢ 2 ⎥ ⎣1 1 ⎦ ⎣ −1 2
(b) P = ⎢
⎡1 ∴ A w = P −1AP = ⎢ 2 ⎣ −1 2 ⎡1 B w = P −1B = ⎢ 2 ⎣ −1 2
2⎤
1
2⎦
⎥
⎡1 ⎥ ⎢ 1 2 ⎦ ⎣0 1
2⎤
1
0⎤ 1 ⎥ 2⎦
⎡1 −1⎤ ⎡ 1 2 ⎢1 1 ⎥ = ⎢ −1 ⎣ ⎦ ⎣ 2
⎡1 −1⎤ ⎡ 3 4 ⎥ ⎢ 1 ⎥ = ⎢ −1 1 ⎦ ⎣ 4 4 ⎦ ⎣1 1
4⎤
−1 3
4⎤
4
⎥ ⎦
⎡ 2⎤ ⎡ 3 2 ⎤ =⎢ ⎥ 1 ⎥ ⎢1 ⎥ 2⎦ ⎣ ⎦ ⎣ −1 2 ⎦
1
2⎤
⎡1 −1⎤ Cw = CP = [1 2] ⎢ ⎥ = [3 1] ⎣1 1 ⎦ − 14 ⎤ ⎡ 3 ⎡ 3 ⎤ ∴ w (k + 1) = ⎢ 4 w ( k ) + ⎢ 2 ⎥ u( k ) ⎥ 3 4 ⎦ ⎣− 14 ⎣− 12 ⎦
y(k ) = [3 1] x(k ) ⎡z − 34
(c) zI − A w = ⎢
1
⎣
4
9 1 4 ⎤ , zI − A w = z 2 − 1.5 z + − = z 2 − 1.5 z + 0.5 = Δ z − 3 4 ⎥⎦ 16 16 1
Y ( z) 1 ⎡z − 34 = Cw [ zI − A w ]−1 B w = [3 1] ⎢ U ( z) Δ ⎣ − 14 =
− 14 ⎤ ⎡ 3 2 ⎤ z − 3 4 ⎥⎦ ⎢⎣ − 1 2 ⎥⎦
⎡ 3 ⎤ 1 4z − 3 [3z −2.5 z − 1.5] ⎢ 2 ⎥ = 1 Δ ⎣ − 2 ⎦ ( z − 1)( z − 0.5)
(d) zI − A =
z −1
0
0
z − 0.5
= z 2 − 1.5 z + 0.5; zI − A w = z 2 − 1.5 z + 0.5
∴ z1 = 1, z2 = 0.5 A =
1
0
0 0.5
= 0.5 = z1 z2 ; A w =
3
4
− 14
− 14 3
4
=
9 1 − = 0.5 16 16
tr A = 1.5 = z1 + z2 ; tr A w = 1.5
48
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2.111. Consider a system with the transfer function G ( z) =
Y ( z)
U ( z)
=
2 z ( z − 1)
(a) Find three different statevariable models of this system. (b) Verify the transfer function of each state model in part (a), using (284).
Solution: (a) G ( z ) = G1 ( z )G2 ( z ) =
2 2 z −2 = z 2 − z 1 − z −1
(1) x1(k)
+
Z
T
x2
x1 y(k)
T
change Z to 2 ⎡ 0 1⎤ ⎡0⎤ x(k + 1) = ⎢ x( k ) + ⎢ ⎥ u ( k ) ⎥ ⎣ 0 1⎦ ⎣2⎦
y(k ) = [1 0] x(k )
(2)
G( z ) =
2 2 −2 = + = G1 ( z ) + G2 ( z ) z ( z − 1) z z −1
T
x1(k)
Z 
x1(k)
+
T
x2(k)
y(k)
Z

change Z to 2 ⎡0 0⎤ ⎡1⎤ x(k + 1) = ⎢ x( k ) + ⎢ ⎥ u ( k ) ⎥ ⎣0 1 ⎦ ⎣1⎦
y(k ) = [−2 2] x(k )
49
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(3) u(k)
T
Z
x2 +
T
x1 y(k)
+
change Z to 2 ⎡1 1 ⎤ ⎡0⎤ x(k + 1) = ⎢ x( k ) + ⎢ ⎥ u ( k ) ⎥ ⎣0 0⎦ ⎣ 2⎦
y(k ) = [1 0] x(k )
(b) (1)
⎡ z −1 ⎤ 2 zI − A = ⎢ ⎥ ; zI − A = z − z = Δ ⎣ 0 z − 1⎦ G( z ) = C[ zI − A]−1 B =
⎡ z − 1 1⎤ ⎡0⎤ 1 ⎡0⎤ 1 2 = [ z − 1 1] ⎢ ⎥ = [1 0] ⎢ ⎥ ⎢ ⎥ z ⎦ ⎣ 2⎦ Δ Δ ⎣ 0 ⎣ 2 ⎦ z ( z − 1)
0 ⎤ ⎡z 2 (2) zI − A = ⎢ ⎥ ; zI − A = Δ = z − z 0 z − 1 ⎣ ⎦ G ( z ) = C[ zI − A]−1 B =
⎡ z − 1 0⎤ ⎡2⎤ 1 ⎡2 z − 2⎤ 1 2 = [ −1 1] ⎢ = [−1 1] ⎢ ⎥ ⎢ ⎥ ⎥ z ⎦ ⎣ 2⎦ Δ Δ ⎣ 0 ⎣ 2 z ⎦ z ( z − 1)
⎡ z − 1 −1⎤ (3) zI − A = ⎢ ; zI − A = z 2 − z = Δ z ⎥⎦ ⎣ 0 G ( z ) = C[ zI − A]−1 B =
1 ⎤ ⎡z 1 [1 0] ⎢ 0 z − 1⎦⎥ Δ ⎣
⎡0⎤ 1 ⎡0⎤ 2 ⎢ 2 ⎥ = Δ [ z 1] ⎢ 2 ⎥ = z ( z − 1) ⎣ ⎦ ⎣ ⎦
2.112. Consider a system described by the coupled difference equation y ( k + 2) − v ( k ) = 0 v ( k + 1) + y ( k + 1) = u ( k )
where u ( k ) is the system input. (a) Find a statevariable formulation for this system. Consider the outputs to be y ( k + 1) and v ( k ) . Hint: Draw a simulation diagram first. (b) Repeat part (a) with y ( k ) and v ( k ) as the outputs.
50
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(c) Repeat part (a) with the single output v ( k ) . (d) Use (284) to calculate the system transfer function with v ( k ) as the system output, as in part (c); that is, find V ( z ) /U ( z ) . (e) Verify the transfer function V ( z ) /U ( z ) in part (d) by taking the ztransform of the given system difference equations and eliminating Y ( z ) . (f) Verify the transfer function V ( z ) /U ( z ) in part (d) by using Mason’s gain formula on the simulation diagram of part (a).
Solution: (a) y(k + 2)
u(k)
u(k + 1)
+ 
T
T
y(k + 1) x2(k)
T
y(k) x1(k)
u(k1) x3(k)
⎡0 1 0⎤ ⎡0 ⎤ x(k + 1) = ⎢⎢ 0 0 1 ⎥⎥ x(k ) + ⎢⎢ 0 ⎥⎥ u (k ) ⎢⎣ 0 −1 0 ⎥⎦ ⎢⎣1 ⎥⎦
⎡ x (k ) ⎤ ⎡ 0 1 0 ⎤ y0 (k ) = ⎢ 2 ⎥ = ⎢ ⎥ x(k ); y0 (k ) = output ⎣ x3 (k ) ⎦ ⎣ 0 0 1 ⎦
(b) x(k + 1) = same as (a) ⎡ x (k ) ⎤ ⎡1 0 0 ⎤ y0 (k ) = ⎢ 1 ⎥ = ⎢ ⎥ x( k ) ⎣ x3 (k ) ⎦ ⎣0 0 1 ⎦
(c) x(k + 1) = same as (a) y0 (k ) = x3 (k ) = [0 0 1] x(k ) ⎡ z −1 0 ⎤ z −1⎥⎥ ; zI − A = z 3 − (− z ) = z 3 + z = Δ ⎢⎣ 0 1 z ⎥⎦
(d) zI − A = ⎢⎢ 0
51
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⎡ z2 + 1 z2 ⎢ Cof[ zI − A] = ⎢ z z2 ⎢ z ⎣⎢ 1
⎡ z2 + 1 z 0⎤ ⎥ ⎢ 1 z ⎥ ; [ zI − A]−1 = ⎢ z 2 z2 Δ ⎥ ⎢ z 2 ⎦⎥ z ⎢⎣ 0
⎡ z2 + 1 z ⎢ Y ( z) 1 ∴ 0 = C[ zI − A ]−1 B = [0 0 1] ⎢ z 2 z2 Δ U ( z) ⎢ z ⎣⎢ 0
1 = ⎡⎣ 0 z Δ
1⎤ ⎥ z⎥ ⎥ z 2 ⎦⎥
1⎤ ⎥ z⎥ ⎥ z 2 ⎦⎥
⎡0⎤ ⎢ ⎥ ⎢0 ⎥ ⎢⎣ 1 ⎥⎦
⎡0⎤ z2 z ⎤ z ⎦ ⎢⎢ 0 ⎥⎥ = 3 = 2 z − z z +1 ⎢⎣1 ⎥⎦ 2
(e) z 2Y ( z ) − V ( z ) = 0 ⇒ Y ( z ) =
1 V ( z) z2
1 zV ( z) + zY ( z) = zV ( z) + V ( z) = U ( z) z ∴
V ( z ) Y0 ( z ) 1 z = = = 2 1 U ( z) U ( z) z + z +1 z
(f) From (a):
1 u(z) 1
z1
y0(z)
z1 1
∴
z1
make u and y capital letters
Y0 ( z ) z −1 z = = 2 −2 U ( z) 1 + z z +1
2.113. Given the system described by the state equations ⎡ 1 0 0 ⎢ x ( k + 1) = ⎢⎢ 1 1 0 ⎢ 0 1 0 ⎣
⎤ ⎡ 1 ⎤ ⎥ ⎥ ⎢ ⎥ x (k ) + ⎢ 0 ⎥ u(k ) ⎥ ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎦ ⎦ ⎣
y ( k ) = ⎡⎢ 0 0 1 ⎤⎥ x ( k ) ⎣ ⎦
52
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(a) Calculate the transfer function Y ( z ) /U ( z ) , using (284). (b) Draw a simulation diagram for this system, from the state equations given. (c) Use Mason’s gain formula and the simulation diagram to verify the transfer function found in part (a).
Solution: ⎡z −1
0⎤ z − 1 0 ⎥⎥ ; Δ = z 3 − 2 z 2 + z = z ( z − 1) 2 −1 z ⎥⎦ 0
(a) zI − A = ⎢⎢ −1 ⎢⎣ 0
⎡ z ( z − 1) 1 ⎤ z ⎢ ⎥ Cof ( zI − A) = ⎢ 0 z ( z − 1) z − 1 ⎥ , ( zI − A) −1 ⎢ 0 0 ( z − 1) 2 ⎥⎦ ⎣
⎡ 1 ⎢ −1 z ⎢ ⎢ 1 =⎢ 2 ⎢ ( z − 1) ⎢ 1 ⎢ 2 ⎣⎢ z ( z − 1)
0 1 z −1 1 z ( z − 1)
⎤ 0⎥ ⎥ ⎥ 0⎥ ⎥ 1⎥ ⎥ z ⎦⎥
⎡1 ⎤ G ( z ) = C[ zI − A] B = [0 0 1][ zI − A] ⎢⎢ 0 ⎥⎥ ⎢⎣ 0 ⎥⎦ −1
−1
⎡1 ⎤ ⎡ 1 1 1⎤ ⎢ ⎥ 1 1 =⎢ = 3 ⎥ ⎢0⎥ = 2 2 z z z ( − 1) z z z z z z2 + z ( 1) ( 1) 2 − − − ⎣ ⎦ ⎢0⎥ ⎣ ⎦
(b) e(k)
+
T
x1(k)
+
+ T
x2(k)
+
T
x3(k) y(k)
(c) Δ = 1 − z −1 − z −1 + z −2 = 1 − 2 z −1 + z −2
53
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∴ G( z) =
1 z −3 = 3 Δ z − 2z2 + z
2.114. Section 2.9 gives some standard forms for state equations (simulation diagrams for the control canonical and observer canonical forms). The MATLAB statement ⎡⎢ A,B,C,D⎤⎥ = tf 2ss( num,den) ⎦ ⎣
generates a standard set of state equations for the transfer function whose numerator coefficients are given in the vector num and denominator coefficients in the vector den. (a) Use the MATLAB statement given to generate a set of state equations for the transfer function G ( z) =
3z + 4 z + 5z + 6 2
(b) Draw a simulation diagram for the state equations in part (a). (c) Determine if the simulation diagram in part (b) is one of the standard forms in Section 2.9.
Solution: (a) n = [0 3 4]; d = [1 5 6]; [A,B,C,D] = tf2ss(n, d) ⎡ −5 −6 ⎤ ⎡1 ⎤ x(k + 1) = ⎢ x( k ) + ⎢ ⎥ u ( k ) ⎥ ⎣1 0⎦ ⎣0⎦
y(k ) = [3 4] x(k )
(b) 3
u(k)
+
T
x1(b)
T
x2(b)
4
+
+
y(k)
5
6
54
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(c) Yes, it is the control canonical form with the states renumbered.
2.121. Consider the system described in Problem 2.102. ⎡ ⎤ ⎡ ⎤ x ( k + 1) = ⎢⎢ 0 1 ⎥⎥ x ( k ) + ⎢⎢ 1 ⎥⎥ u ( k ) ⎢⎣ 0 3 ⎥⎦ ⎢⎣ 1 ⎥⎦
y ( k ) = ⎡⎢ −2 1 ⎤⎥ x ( k ) ⎣ ⎦
(a) Find the transfer function of this system. (b) Let u ( k ) = 1, k ≥ 0 (a unit step function) and x (0) = 0 . Use the transfer function of part (a) to find the system response. (c) Find the state transition matrix Φ( k ) for this system. (d) Use (290) to verify the step response calculated in part (b). This calculation results in the response expressed as a summation. Then check the values y (0) , y (1) , and y (2) . (e) Verify the results of part (d) by the iterative solution of the state equations.
Solution: ⎡ z −1 ⎤ ⎥ ; Δ = zI − A = z ( z − 3) = Δ ⎣ 0 z − 3⎦
(a) zI − A = ⎢
⎡ z − 3 1 ⎤ ⎡1⎤ Y ( z) 1 = C[ zI − A]−1 B = [ −2 1] ⎢ z ⎥⎦ ⎢⎣1⎥⎦ U ( z) Δ ⎣ 0 =
(b) Y ( z ) =
⎡1⎤ − z + 4 1 [ −2 z + 6 z − 2] ⎢ ⎥ = Δ ⎣1⎦ z ( z − 3)
(− z + 4) z z ( z − 3)( z − 1)
4 1 − 32 Y ( z) −z + 4 = = 3+ + 6 z z ( z − 1)( z − 3) z z − 1 z − 3
55
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∴ y (0) = 0
⎧⎪ 4 3 − 3 2 + 1 6 = 0, k = 0 ∴ y (k ) = ⎨ k k ≥1 ⎪⎩ − 3 2 + 1 6 (3)
3 1 + = −1 2 2 3 3 y (2) = − + = 0 2 2 y (1) = −
(c)
Φ( z ) = z ( zI − A) −1
⎡ z −3 ⎢ z ( z − 3) = z⎢ ⎢ ⎢ 0 ⎣
1 ⎤ ⎡1 ⎢ z ( z − 3) ⎥ ⎥ = z⎢z z ⎥ ⎢0 ⎢⎣ z ( z − 3) ⎥⎦
⎡ 1 1 k ⎢ δ(k) − δ(k) + (3) ∴ Φ(k) = ⎢ 3 3 k ⎢ 0 (3) ⎣ k −1
1 − 13 ⎤ + 3 ⎥ z z −3 ⎥ 1 ⎥ z − 3 ⎦⎥
⎤ ⎥ ⎥ ⎥ ⎦ ⎡
k −1
⎤
(d) y(k) = ∑ CΦ(k − 1 − j)Bu( j) = ∑ ⎡⎣ −2 1 ⎤⎦ Φ(k − 1 − j) ⎢ 1 ⎥ j=0 j=0 ⎣ 1 ⎦ ⎡ 2 1 k − j−1 ⎢ δ(k − 1 − j) + (3) ⎡ ⎤ = ∑ ⎣ −2 1 ⎦ ⎢ 3 3 j=0 k − j−1 ⎢ (3) ⎣ k −1
⎤ ⎥ k −1 ⎡ −4 1 k − j−1 ⎤ ⎥ = ∑ ⎢ 3 δ(k − j − 1) + 3 (3) ⎥ ⎦ ⎥ j=0 ⎣ ⎦
k −1 ⎤ ⎡ −4 1 = ∑ ⎢ δ(k − 1 − j) + (3) k −1− j ⎥ 3 ⎦ j=0 ⎣ 3
4 1 4 1 y(0) = 0; y(1) = − δ(0) + (3)0 = − + = −1 3 3 3 3 4 1 4 1 4 1 y(2) = − δ(1) + (3)1 − δ(0) + (3)0 = 1 − + = 0 3 3 3 3 3 3
⎡ 0 1⎤ ⎡0 ⎤ ⎡1⎤ ⎡1⎤ ⎡1⎤ + ⎢ ⎥ = ⎢ ⎥ ; y (1) = [ −2 1] ⎢ ⎥ = −1 ⎥ ⎢ ⎥ ⎣ 0 3⎦ ⎣ 0 ⎦ ⎣1⎦ ⎣1⎦ ⎣1⎦
(e) x(1) = ⎢
⎡ 0 1⎤ ⎡1⎤ ⎡1⎤ ⎡ 2 ⎤ ⎡ 2⎤ x(2) = ⎢ + ⎢ ⎥ = ⎢ ⎥ ; y(2) = [ −2 1] ⎢ ⎥ = 0 ⎥ ⎢ ⎥ ⎣ 0 3⎦ ⎣1⎦ ⎣1⎦ ⎣ 4 ⎦ ⎣ 4⎦
2.122. The system described by the equations ⎡ ⎤ ⎡ ⎤ x ( k + 1) = ⎢⎢ 1 0 ⎥⎥ x ( k ) + ⎢⎢ 2 ⎥⎥ u ( k ) ⎢⎣ 0 0.5 ⎥⎦ ⎢⎣ 1 ⎥⎦
56
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y ( k ) = ⎡⎢⎣1 2⎤⎥⎦ x ( k ) T is excited by the initial conditions x (0) = ⎡⎢⎣ −1 2 ⎤⎥⎦ with u ( k ) = 0 for all k.
(a) Use (289) to solve for x ( k ), k ≥ 0. (b) Find the output y ( z ) . (c) Show that Φ( k ) in (a) satisfies the property Φ(0) = I. (d) Show that the solution in part (a) satisfies the given initial conditions. (e) Use an iterative solution of the state equations to show that the values y ( k ) , for k = 0, 1, 2, and 3, in part (b) are correct. (f) Verify the results in part (e) using MATLAB.
Solution: (a)
0 ⎤ ⎡z −1 zI − A = ⎢ ; zI − A = Δ = ( z − 1)( z − 0.5) 0 z − 0.5⎥⎦ ⎣ ⎡ 1 ⎢ z −1 z 0.5 0 − ⎡ ⎤ 1 =⎢ ( zI − A −1 ) = ⎢ ⎥ z − 1⎦ ⎢ Δ⎣ 0 0 ⎣⎢ ⎡ z ⎢ z −1 ∴ Φ(k ) = / −1 ⎢ ⎢ 0 ⎣⎢
⎤ ⎥ ⎥ 1 ⎥ z − 0.5 ⎦⎥ 0
⎤ ⎥ ⎡1 0 ⎤ ⎥=⎢ ⎥ z ⎥ ⎣ 0 0.5k ⎦ z − 0.5 ⎦⎥ 0
⎡1 0 ⎤ ∴ x(k ) = Φ(k )x(0) = ⎢ k⎥ ⎣0 0.5 ⎦
⎡1 ⎤ ⎡ 1 ⎤ ⎢ 2⎥ = ⎢ k⎥ ⎣ ⎦ ⎣ 2(0.5) ⎦
⎡ 1 ⎤ (b) y(k ) = Cx(k ) = [1 2] ⎢ = 1 + 4(0.5) k k⎥ ⎣ 2(0.5) ⎦ ⎡1 0 ⎤ ⎡1 0 ⎤ (c) Φ(0) = ⎢ =⎢ ⎥=I 0⎥ ⎣0 0.5 ⎦ ⎣0 1 ⎦
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⎡ 1 ⎤ ⎡1 ⎤ (d) x(k ) k =0 = ⎢ =⎢ ⎥ k⎥ ⎣ 2(0.5) ⎦ ⎣ 2 ⎦
(e) From (b),
y (0) = 5 y (2) = 2 y (1) = 3 y (3) = 1.5
⎡1 ⎤ y (0) = Cx(0) = [1 2 ] ⎢ ⎥ = 5 ⎣2⎦ ⎡1 0 ⎤ x(1) = ⎢ ⎥ ⎣ 0 0.5⎦
⎡ 1 ⎤ ⎡1⎤ ⎡1⎤ ⎢ 2 ⎥ = ⎢1⎥ , y (1) = [1 2] ⎢1⎥ = 3 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎡1 0 ⎤ x(2) = ⎢ ⎥ ⎣0 0.5⎦
⎡1⎤ ⎡ 1 ⎤ ⎡1 ⎤ ⎢1⎥ = ⎢0.5⎥ , y(2) = [1 2 ] ⎢0.5⎥ = 2 ⎣⎦ ⎣ ⎦ ⎣ ⎦
⎡1 0 ⎤ ⎡ 1 ⎤ ⎡ 1 ⎤ ⎡ 1 ⎤ x(3) = ⎢ =⎢ , y (3) = [1 2] ⎢ ⎥ ⎢ ⎥ ⎥ ⎥ = 1.5 ⎣ 0 0.5⎦ ⎣ 0.5⎦ ⎣ 0.25⎦ ⎣ 0.25⎦
(f) A = [1 0;0 .5]; B = [2; 1]; C = [1 2]; x=[1; 2]; u = 0; for k = 0:3 x1 = A*x + B*u; y = C*x; [k,y] x = x1; end
2.123. The system described by the equations ⎡ 1.1 1 ⎤ ⎡ ⎤ ⎥ x (k ) + ⎢ 1 ⎥ u(k ) x ( k + 1) = ⎢⎢ ⎥ ⎢ 1 ⎥ ⎢⎣ ⎥⎦ ⎢⎣ −0.3 0 ⎥⎦
y ( k ) = ⎡⎢ 1 −1 ⎤⎥ x ( k ) ⎣ ⎦ T is excited by the initial conditions x (0) = ⎡⎢⎣ −1 2 ⎤⎥⎦ with u ( k ) = 0 for all k.
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(a) Use (289) to solve for x ( k ), k ≥ 0. (b) Find the output y ( k ) . (c) Show that Φ( k ) in part (a) satisfies the property Φ(0) = I . (d) Show that the solution in part (a) satisfies the given initial conditions. (e) Use an iterative solution of the state equations to show that the values y ( k ) , for k = 0, 1, 2, and 3, in part (b) are correct.
Solution: (a)
⎡ z − 1.1 −1⎤ zI − A = ⎢ ; zI − A = Δ = z 2 − 1.1z + 0.3 = ( z − 0.5)( z − 0.6) ⎥ z⎦ ⎣ 0.3 ( zI − A)−1 =
1 ⎤ 1⎡ z ⎢ Δ ⎣ −0.3 z − 1.1⎥⎦
⎛ ⎡ z ⎜ ⎢ ( z − 0.5)( z − 0.6) Φ(k ) = / −1[ z ( zI − A) −1 ] = / −1 ⎜ z ⎢ ⎜ ⎢ −0.3 ⎜ ⎢ ( z − 0.5)( z − 0.6) ⎝ ⎣ ⎛ ⎡ −5 6 + ⎜ ⎢ z .5 z .6 − − = / ⎜z⎢ 3 3 − ⎜ ⎢ + ⎜ ⎢ ⎝ ⎣ z − .5 z − .6 −1
⎡ −5(0.5)k + 6(0.6)k =⎢ k k ⎣⎢ 3(0.5) − 3(0.6)
−10 + z − .5 6 + z − .5
1 ⎤⎞ ⎟ ( z − 0.5)( z − 0.6) ⎥ ⎟ ⎥ z − 1.1 ⎥⎟ ⎟ ( z − 0.5)( z − 0.6) ⎥⎦ ⎠
10 ⎤ ⎞ ⎟ z − .6 ⎥ ⎟ ⎥ −5 ⎥ ⎟ ⎟ z − .6 ⎥⎦ ⎠
−10(0.5)k + 10(0.6)k ⎤ ⎥ 6(0.5)k − 5(0.6)k ⎦⎥
⎡ −5(0.5)k + 6(0.6)k ∴ x(k ) = Φ(k )x(0) = ⎢ k k ⎣⎢ 3(0.5) − 3(0.6)
−10(0.5)k + 10(0.6)k ⎤ ⎡ −1⎤ ⎡ −15(0.5)k + 14(0.6)k ⎤ ⎥⎢ ⎥ = ⎢ ⎥ 6(0.5)k − 5(0.6)k ⎦⎥ ⎣ 2 ⎦ ⎣⎢ 9(0.5)k − 7(0.6)k ⎦⎥
⎡ −15(0.5)k + 14(0.6)k ⎤ k k (b) y(k ) = Cx(k ) = [1 −1] ⎢ ⎥ = −24(0.5) + 21(0.6) k k 9(0.5) 7(0.6) − ⎣⎢ ⎦⎥ ⎡ −5 + 6 −10 + 10 ⎤ ⎡1 0 ⎤ = =I 6 − 5 ⎦⎥ ⎣⎢ 0 1 ⎦⎥ ⎣ 3−3
(c) Φ(0) = ⎢
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⎡ −15 + 14 ⎤
⎡ −1⎤
(d) x(k ) k =0 = ⎢ ⎥=⎢ ⎥ ⎣ 9−7 ⎦ ⎣ 2 ⎦ (e) From (b),
y (0) = −3
y (2) = 1.56
y (1) = 0.6
y (3) = 1.536
⎡ −1⎤ y (0) = Cx(0) = [1 −1] ⎢ ⎥ = −3 ⎣2⎦ ⎡ 1.1 1 ⎤ ⎡ −1⎤ ⎡ 0.9 ⎤ ⎡ 0.9 ⎤ x(1) = ⎢ = ⎢ ⎥ ; y (1) = [1 −1] ⎢ ⎥ = 0.6 ⎥ ⎢ ⎥ ⎣ −0.3 0 ⎦ ⎣ 2 ⎦ ⎣ 0.3⎦ ⎣ 0.3⎦ ⎡ 1.1 1 ⎤ ⎡0.9 ⎤ ⎡ 1.29 ⎤ ⎡ 1.29 ⎤ x(2) = ⎢ ⎥ ⎢ 0.3⎥ = ⎢ −0.27 ⎥ ; y(2) = [1 −1] ⎢ −0.27 ⎥ = 1.56 − 0.3 0 ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡ 1.1 1 ⎤ ⎡ 1.29 ⎤ ⎡ 1.149 ⎤ ⎡ 1.149 ⎤ x(3) = ⎢ =⎢ ; y (3) = [1 −1] ⎢ ⎥ ⎢ ⎥ ⎥ ⎥ = 1.536 ⎣ −0.3 0 ⎦ ⎣ −0.27 ⎦ ⎣ −0.387 ⎦ ⎣ −0.389 ⎦
MATLAB: A = [1.1 1;–0.3 0]; B = [1; 1]; C = [1 –1]; x=[–1; 2]; u = 0; for k = 0:3 x1 = A*x + B*u; y = C*x; [k,y] x = x1; end
2.124. Let Φ( k ) be the state transition matrix for the equations x ( k + 1) = Ax ( k )
Show that Φ( k ) satisfies the difference equation Φ( k + 1) = AΦ( k )
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Solution: x(k + 1) = Ax(k ); x(k ) = Φ(k )x(0) ∴ Φ(k + 1)x(0) = AΦ(k )x(0)
Since this is true for any x(0), ∴ Φ(k + 1) = AΦ(k )
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link full download: https://bit.ly/2F5sVgB Language: English ISBN10: 0132938316 ISBN13: 9780132938310 ISBN13: 9780132938310 digital con...
Published on Mar 17, 2019
link full download: https://bit.ly/2F5sVgB Language: English ISBN10: 0132938316 ISBN13: 9780132938310 ISBN13: 9780132938310 digital con...