Consider the transfer function7.19Y (s)s 6(7.283)G(s)s25s6U (s)(a) By rewriting Eq. (7.283) in the form1Y (s)S+6G(s)U (s)s 3S+2find a series realization of G(s) as a cascade of two first-order systems(b) Using a partial-fraction expansion of G(s), find a parallel realization ofG(s)(c) Realize G(s) in control canonical form

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Asked Jun 12, 2019
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Feedback Control of Dynamic Systems (7th edition) - Problem 7.19

Consider the transfer function
7.19
Y (s)
s 6
(7.283)
G(s)
s25s6
U (s)
(a) By rewriting Eq. (7.283) in the form
1
Y (s)
S+6
G(s)
U (s)
s 3
S+2
find a series realization of G(s) as a cascade of two first-order systems
(b) Using a partial-fraction expansion of G(s), find a parallel realization of
G(s)
(c) Realize G(s) in control canonical form
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Consider the transfer function 7.19 Y (s) s 6 (7.283) G(s) s25s6 U (s) (a) By rewriting Eq. (7.283) in the form 1 Y (s) S+6 G(s) U (s) s 3 S+2 find a series realization of G(s) as a cascade of two first-order systems (b) Using a partial-fraction expansion of G(s), find a parallel realization of G(s) (c) Realize G(s) in control canonical form

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Expert Answer

Step 1
Part a
Y(s)
G(s) U(S
1s+6
(1)
s+3s+2
Suppose
G(s) F(s) F(s)
(2)
Compare equation (1) and equation (2)
we get
1
F(s)
S+6
(s)
And
s+2
s+3
1
s+6
Y
U
s24
s+2
Z
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Part a Y(s) G(s) U(S 1s+6 (1) s+3s+2 Suppose G(s) F(s) F(s) (2) Compare equation (1) and equation (2) we get 1 F(s) S+6 (s) And s+2 s+3 1 s+6 Y U s24 s+2 Z

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Step 2
From above figure
х3-2x +и
6x
Put value of z in x, expression
x3x 6x
3-2x + бх,
Express above equation in canonical form
x+
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From above figure х3-2x +и 6x Put value of z in x, expression x3x 6x 3-2x + бх, Express above equation in canonical form x+

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Step 3
Since we get from figure
Write this in canonical form
y =[0 1x
So, canonical form can be written as,
-2 6
|x+
-2 0x0u Ahd y=[o 1]x
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Since we get from figure Write this in canonical form y =[0 1x So, canonical form can be written as, -2 6 |x+ -2 0x0u Ahd y=[o 1]x

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