Operating Systems Chapter 3 Homework Find the range of K for which all the roots of the following

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(d) The Routh array is,

(e)

a(s) = s4+ 6s2+ 25 = 0

Two coefficients (those of s3and s) are missing so there are roots

=)Two sign changes in the first column of the array: 2 roots not in

the LHP.

check:

55. Find the range of Kfor which all the roots of the following polynomial

are in the LHP:

s5+ 5s4+ 10s3+ 10s2+ 5s+K= 0:

Use Matlab to verify your answer by plotting the roots of the polynomial

in the s-plane for various values of K.

Solution:

The Routh array is,

s5: 1 10 5

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For stability: all elements in first the first column of the Routh array must

be positive. That results in the following set of constraints:

Root Locus

Real Axis

-3 -2.5 -2 -1.5 -1 –0.5 00.5 1

-1.5

-0.5

0.5

1.5

2

K=0

K= 3. 88

56. The transfer function of a typical tape-drive system is given by

KG(s) = K(s+ 4)

s[(s+ 0:5)(s+ 1)(s2+ 0:4s+ 4)];

where time is measured in milliseconds. Using Routh’s stability crite-

rion, determine the range of Kfor which this system is stable when the

characteristic equation is 1 + KG(s) = 0.

Solution:

The Routh array is,

s5: 1:0 5:1 2 + K

where

For stability we must have all the elements in the first column of the routh

array to be positive, and that results in the following set of constraints:

57. Consider the closed-loop magnetic levitation system shown in Figure 3.66.

(a) Compute the transfer function from the input (R) to the output

(Y).

(b) Assume Ko= 1. Determine the conditions on the system parame-

ters (a; K; z; p), to guarantee closed-loop system stability.

Solution:

(a) The transfer function is

(b) With K= 1 we have Denominator (s) = s3+ps2+ (Ka2)s+

Kz pa2;constructing the Routh array we obtain

58. Consider the system shown in Fig. 3.67.

(a) Compute the closed-loop characteristic equation.

(b) For what values of (T; A)is the system stable? Hint : An approximate

answer may be found using

eT s 

=1T s

or

eT s 

=1T

2s

1 + T

2s

for the pure delay. As an alternative, you could use the computer

Matlab (Simulink) to simulate the system or to find the roots of the

system’s characteristic equation for various values of Tand A.

Solution:

3116 CHAPTER 3. DYNAMIC RESPONSE

(a) The characteristic equation is,

The Routh’s array is,

For stability we must have all the coefficients in the first column be

positive. The following Simulink diagram simulates the closed-loop

system.

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59. Modify the Routh criterion so that it applies to the case in which all the

poles are to be to the left of when  > 0. Apply the modified test to

the polynomial

finding those values of Kfor which all poles have a real part less than 1.

Solution:

The Routh’s array is,

60. Suppose the characteristic polynomial of a given closed-loop system is

computed to be

3118 CHAPTER 3. DYNAMIC RESPONSE

For stability, the elements in the first column must all be positive. This

Problem 3.60: s-plane region for stability.

61. Overhead electric power lines sometimes experience a low-frequency, high-

amplitude vertical oscillation, or gallop, during winter storms when the

line conductors become covered with ice. In the presence of wind, this ice

can assume aerodynamic lift and drag forces that result in a gallop up to

several meters in amplitude. Large-amplitude gallop can cause clashing

conductors and structural damage to the line support structures caused by

the large dynamic loads. These effects in turn can lead to power outages.

Assume that the line conductor is a rigid rod, constrained to vertical

motion only, and suspended by springs and dampers as shown in Fig. 3.68.

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Figure 3.68: Electric power-line conductor

A simple model of this conductor galloping is

m•y+D() _yL()v

( _y2+v2)1=2+Tn

`y= 0;

where

m=mass of conductor;

y= conductor0s vertical displacement;

D= aerodynamic drag force;

L= aerodynamic lift force;

v= wind velocity;

= aerodynamic angle of attack = tan1( _y=v);

T= conductor tension;

n= number of harmonic frequencies;

`= length of conductor:

Assume that L(0) = 0 and D(0) = D0(a constant), and linearize the

equation around the value y= _y= 0. Use Routh’s stability criterion to

show that galloping can occur whenever

@L

@ +D0<0:

Solution:

m•y+“D() _yL()v

p_y2+v2#+Tn

l2y= 0;

3120 CHAPTER 3. DYNAMIC RESPONSE

Let x1=yand x2= _y= _x1

@f2

Now,

so,