Nisse Solucionario Cap 5

F I V E

Reduction of Multiple Subsystems
SOLUTIONS TO CASE STUDIES CHALLENGES
Antenna Control: Designing a Closed-Loop Response
a. Drawing the block diagram of the system:

Pots

Pre amp

Power amp

Motor, load and gears

ui +

10 Π

K

150 s+150

-

0.16 s (s+1.32)

uo

Thus, T(s) =

76.39K s3+151.32s2+198s+76.39K

b. Drawing the signal flow-diagram for eachsubsystem and then interconnecting them yields:

ui

10 Π
pot

K
pre amp

150

1 s x3

0.8

1 s x2
-1.32 motor and load

1 s

x1

0.2

uo

gears -150 power amp -

10 Π
pot

Solutions to Case Studies Challenges 127

. x 1 = x2 . x 2 = - 1.32x2 + 0.8x3 . 10 x 3 = -150x3 +150K( (q i − 0.2x1 )) = -95.49Kx1 - 150x3 + 477.46Kθi

θo = 0.2x1 In vector-matrix notation,
.π

0 0 -95.49K

1 -1.32 0

0 0.8 -150
x +

0 0 477.46K
θ
i

x =

θo = 0.2

0

0

x

c. T1 =

76.39 ⎛ 10 ⎞ ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ ⎝ π ⎠ (K)(150)⎝ s ⎠ (0.8)⎝ s⎠ ⎝ s ⎠ (0.2) = s 3 −150 −1.32 ⎛1 ⎛1 ⎛1 ⎛ −10 ⎞ −76.39K ; GL 2 = ; GL 3 = (K)(150) ⎞ (0.8) ⎞ ⎞ (0.2) = ⎝ s⎠ ⎝ s ⎠ ⎝ s⎠ ⎝ π ⎠ s s s3

GL1 =

Nontouching loops: 198 GL1GL2 = 2 s ∆ = 1 - [GL1 + GL2 + GL3] + [GL1GL2] = 1 +∆1 = 1 T(s) = T1∆1 76.39K = 3 ∆ s +151.32s2 +198s+76.39K 10 150 1.32 76.39K 198 + 2 s + s + s3 s

d. The equivalent forward path transfer function is G(s) = s(s+1.32) . Therefore, 2.55 T(s) = 2 s +1.32s+2.55 The poles are located at -0.66 ± j1.454. ωn = 2.55 = 1.597 rad/s; 2ζωn = 1.32, therefore, ζ = 0.413.

π

0.16K

128 Chapter 5: Reduction of Multiple Subsystems

%OS = e −ζπ /

1−ζ2

x100 = 24% ; Ts = ζω

4

n

4 = 0.66 = 6.06 seconds; Tp =

π
ωn 1-ζ2

= 1.454 =

π

2.16 seconds; Using Figure 4.16, the normalized rise time is 1.486. Dividing by the natural frequency, 1.486 Tr = = 0.93 seconds. 2.55 e.

0.51K 0.51K f. Since G(s) = s(s+1.32) , T(s) = 2 . Also, ζ = s +1.32s+0.51K

%OS - ln ( 100 ) = 0.517 for 15% 2 + ln2 (%OS) π 100 1.32 1.32 overshoot;ωn = 0.51K ; and 2ζωn = 1.32. Therefore, ωn = = 2(0.5147) = 1.277 = 0.51K . 2ζ Solving for K, K=3.2.

UFSS Vehicle: Pitch-Angle Control Representation
a. Use the observer canonical form for the vehicle dynamics so that the output yaw rate is a state variable.
1 s 1 1 s x3 -1.483 -0.24897 1 1 s x2 1 s x1

u

1

-1

2

-0.125 x4

0.437

1

1 y

-2

-1

b. Using the signalflow graph to write the state equations:

Solutions to Case Studies Challenges 129

x1 = x2 Ý Ý x2 = −1.483x 2 + x 3 − 0.125x4 x3 = −0.24897x2 − (0.125* 0.437)x4 Ý Ý x4 = 2x1 + 2x 2 − 2x4 − 2u
In vector-matrix form:

1 0 0 ⎡0 ⎡0⎤ ⎤ ⎢ 0 −1.483 1 −0.125 ⎥ ⎢0⎥ Ý x= x+ u ⎢ 0 −0.24897 0 −0.054625⎥ ⎢0⎥ ⎢2 ⎢ −2 ⎥ ⎥ 2 0 −2 ⎣ ⎣ ⎦ ⎦ y = [1 0 0 0]x
c. Program:
numg1=-0.25*[1 0.437]; deng1=poly([-2-1.29 -0.193 0]); 'G(s)' G=tf(numg1,deng1) numh1=[-1 0]; denh1=[0 1]; 'H(s)' H=tf(numh1,denh1) 'Ge(s)' Ge=feedback(G,H) 'T(s)' T=feedback(-1*Ge,1) [numt,dent]=tfdata(T,'V'); [Acc,Bcc,Ccc,Dcc]=tf2ss(numt,dent)

Computer response:
ans = G(s) Transfer function: -0.25 s - 0.1093 -------------------------------------s^4 + 3.483 s^3 + 3.215 s^2 + 0.4979 s ans = H(s) Transfer function: -s ans = Ge(s)Transfer function: -0.25 s - 0.1093 -------------------------------------s^4 + 3.483 s^3 + 3.465 s^2 + 0.6072 s

130 Chapter 5: Reduction of Multiple Subsystems

ans = T(s) Transfer function: 0.25 s + 0.1093 ----------------------------------------------s^4 + 3.483 s^3 + 3.465 s^2 + 0.8572 s + 0.1093 Acc = -3.4830 1.0000 0 0 Bcc = 1 0 0 0 Ccc = 0 Dcc = 0 0 0.2500 0.1093 -3.4650 0 1.0000 0-0.8572 0 0 1.0000 -0.1093 0 0 0

ANSWERS TO REVIEW QUESTIONS
1. Signals, systems, summing junctions, pickoff points 2. Cascade, parallel, feedback 3. Product of individual transfer functions, sum of individual transfer functions, forward gain divided by one plus the product of the forward gain times the feedback gain 4. Equivalent forms for moving blocks across summing junctions and pickoff...