EL PROCEDIMIENTO DE RECONOCIMIENTO DE GRADOS Y TÍTULOS

% Plotting the pitch attitude subplot(3,2,i):

Plot(t,y);

grid on;

title(titleText);

text( 0.5, 0.3, overshootText);

text( 0.5, 0.1, risetimeText);

end

Step 6 of 7

The following are the plots of pitch attitude verses time for various values of X - For a: = 3.5:

Figure 2: Step response of autopilot system for K — Z.S

150

A measure of the degree of instability in an unstable aircraft response is the amount of time it takes for the amplitude of the time response to double (see Fig), given some nonzero initial condition.

(a) For a first-order system, show that the tim e to double is bi2

where p is the pole location in the RHP.

(b) For a second-order system (with two complex poles in the RHP), show that lo2

*2 = —;:— ■

*2 = —;:— ■fai2

Figure Time to double Time Amplitude 2A Anqditude

______

I te w

L *2 »

Step-by-step solution

step 1 of 4

For the 1* order system.

Let the Transfer Function is 0 ( e ) = ———

s — p By taking Inverse L ^ la c e Transform we can get

g {t)= k e ’ ‘

Now consider the initial time as /g

And consider at t = tgthe an^litude o f the response is as ‘A’

Then,

g ( g = t e < ' = A ... (1)

Step 2 of 4

Now consider that at time t s ^ the an^litude is a s ' 2 A’

Then,

g W = f e * = 2 A ... (2)

By taking the ration o f equation (2) to equation (1), we can get f e "

2 A _ t e * A “ t e ” _ 2

Given that the increased time as Therefore

ln 2 Where = ti- to

Step 3 of 4 ^

(b)

For the second order system,

_ a Let the Transfer Function as O fs) = -*--- -=--- r

Then, the respons e is given by taking the Inverse Ls^lace Transform o f 0{s) a g (<) = go ^ ^ sin [ a ^ +cos"‘ i )

A tt let the amplitude of the response be 'A'.

Then, theresponeis g (^ ) = gn ^ s in f a t ^ + cos“* ^ )= A

Where cos"* ^ = cos"* |^ |+ ^

Therefore g ( < b ) = - g o - 4 — - s in ( ® A + cos"“|{|) = A ... (3) i H i f

Step 4 of 4

At t s let the amplitude of the response be 2A'.

Then, tile respone is g (O = So ^ ^ ^ (<Vi

Where cos'* ^ = cos”* |^|+ ?r

Therefore g(t^) = - g , ^ ^ ^ sin + cos"* |^|) = 2 A ... (4)

From the equation (3) |g(A))| = “ So = A .. •(5)

By taking ratio o f equation (6) to equation (5), we can get 2 A wfc-41

Here a - = \{ \a ,= - { a ^ Therefore

Problem 3.53PP

Suppose that unity feedback is to be applied around the listed open-loop systems. Use Routh’s stability criterion to determine whether the resulting closed-loop systems will be stable.

(b) = ^

W «” <»> =

Step-by-step solution

(a)

The transfer function of open-loop system is.

' ' + 2 5 ^ + 3 5 + 4 ) First, find the characteristic equation.

I + ^ G ( s ) = 0

5 ( i ’ + 2 j* + 3 » + 4 ) j* + 2j* + 3j’ + 8s+ 8 = 0

To determine the Routh array, first arrange the coefficients of the characteristic polynomial in two rows, beginning with first and second coefficients and followed by the even numbered and odd- numbered coefficients.

Write the Routh array for the polynomial.

1 3 8

Evaluate the variables a,b,cnoAd - 2 x 3 - 8 x l

According to Routh array a system is stable If and only If all the elements in the first column of the Routh array are positive.

From the obtained Routh array, it is clear that the sign changes twice in the first column, so there are two poles not in the left hand side of the plane, hence the given closed-loop system Is unstable.

Step 5 of 11

(b)

The transfer function of open-loop system is.

XG(s) S* ( j + l)

First, find the characteristic equation.

I + ^ G ( j ) = 0

5 *(j+ I ) + 2 ( 5 + 4 ) = 0 5’ + 5 * + 2j+ 8 » 0

Write the Routh array for the polynomial.

1 2 1 8 - 6 8

>7 of 11

According to Routh array a system is stable If and only If all the elements in the first column of the Routh array are positive.

(c)

The transfer function of open-loop system is.

First, find the characteristic equation.

I-flC G (s ) = 0

Write the Routh array for the polynomial.

1 3 4

According to Routh array a system is stable If and only If all the elements in the first column of the Routh array are positive.

Use Routh’s stability criterion to determine how many roots with positive real parts the following equations have;

(a) s4 + 8s3 + 32s2 + 80s + 100 = 0

(b) s5 + 10s4 + 30s3 + 80s2 + 344s + 480 = 0

(d) S3 + s2 + 20s + 78 = 0

(e) s4 + 6s2 + 25 = 0

Step-by-step solution

step 1 of 5

(a)

The characteristic equation is,

* * + &t’ + 3 2 i’ + 8 0 s + 1 0 0 = 0

Write the Routh array for the polynomial.

s*: 1 3 2 100 s ’ : 8 8 0

22 100

s' : 43.6 s": 100

Since there are no sign changes in the first column of Routh array, the number of roots with the positive real parts is zero.

Step 2 of 5 ^

(b)

The characteristic equation is,

s* + 1 Oj^ + 3 0 s’ + 80s^ + 3 4 4 s + 4 8 0 = 0

Write the Routh array for the polynomial.

344 4 8 0

1 30

10 80

s’ : 22 296

s’ : - 5 4 .5 480 s': 4 9 0 s’ : 4 8 0

Since there are two sign changes in the first column of Routh array, the number of roots with the positive real parts is two.

Step 3 of 5

(c)

The characteristic equation is,

s * + 2 s ’ + 7 s ’ - 2 s + 8 = 0

Write the Routh array for the polynomial.

1 7 8 2 - 2 j ’ ; 8 8

»>: - 4 j * : 8

Since there are two sign changes in the first column of Routh array, the number of roots with the positive real parts is two.

Step 4 of 5

(d)

The characteristic equation is,

*’ + i ’ + 2 0 * + 78 = 0

Write the Routh array for the polynomial.

1* : 1 20 1 78 - 5 8 78

Since there are two si'^n Ghan*^es in the first c-oiui positive real parts a two.

1 nf Rniith arrai/ tha niimhpr nf mnt« u/ith thp

Step 5 of 5 ^

(e)

The characteristic equation is,

sV 6j“ + 2 5 - 0 5 * + Os ’ + + O r+ 2 5 = 0

Write the Routh array for the polynomial.

t ’ : 1 6 25

i * : 4 12

3 100

25 i * : 1 2 - -2 1 3 i * : 25

From the characteristic equation, it is clear that there two coefficients missing so there are roots outside of the LHP. Since there are two sign changes in the first column of Routh array, the number of roots with the positive real parts is two.

Problem 3.55PP

Find the range of K for which all the roots of the following polynomial are in the LHP:

s5 + 5s4 + 10s3 + 10s2 + 5s + /< = 0.

Use Matlab to verify your answer by plotting the roots of the polynomial in the s-plane for various values of K.

Step-by-step solution

step 1 of 2

s’ +Ss’ +lOs’ +lOs’ SrHtN)

oiep I ui ^

8^+58^+10s^+10s^5s^^c=0

1 10 5

5 10 k

8 -5

j+ (k -2 5 )+ 8 0 +8

■1 -8 L , .( fc - 2 5 ) ^ + 8 0 ( k - 2 5 ) l k-25+80

* k

Step 2 of 2

— - i( k - 2 5 ) > 0 k+55 5^ ^ 64k+j(k-25)(lc4-55) <0 3201rtk’ +30k-(25x55) <0

=>k^+350k-1375 <0

■ |-353.85>k>3.85l

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

__________________________

j[(j + OJ)(j+ l)(j2 + 0 ^ + 4 ) ] ’

where time is measured in miiiiseconds. Using Routh’s stability criterion, determine the range of K for which this system is stabie when the characteristic equation is 1 + KG(s) = 0.

Step-by-step solution

step 1 of 6

Step 1 of 6

Consider the transfer function of a typical tape drive system.

, , __________g ( j + 4 )__________

s + 0.5)( j + 1)( + 0.4j+ 4)J

Step 2 of 6

Consider characteristic equation.

l + ^ T G (j) = 0 ...(2)

Step 3 of 6

Substitute Equation (1) in Equation (2).

,

K (s+ 4)

1 + f ^ ~ Q

4|^(»+ 0 .5 ) ( 4 + l ) ( s + 0 .4 s + 4 ) ] j [ ( j + 0 . 5 ) ( j + l ) ( » ’ + 0 . 4 s + 4 ) ] + ^ r ( j + 4 ) = 0 s ’ + l.9 s ‘ + 5 . l s ’ + 6 .6j’ + 2 s +jA :+ 4A: = 0 s ’ + 1.9s‘ + 5 .1 i’ + 6 .2 s ’ + ( 2 + A : ) i + 4 X = 0

Step 4 of 6

Thus, the characteristic equation is + 1.9 s* + 5 .L $ ^+ 6 .2 » ^ + ( 2 + Jl)4+4AT = 0 Apply Routh array for this polynomial

Step 5 of 6

5 .1 2 + K

1.9

6.2

F ig u re 1

Step 6 of 6

The system is stable if the equation satisfies the following conditions;

• All the terms in the first column of the Routh’s array should have a positive sign.

• The first column of Routh’s array should not posses any sign change.

From the above statement, the stability conditions are, / : + 3 . 6 3 > 0

^ > - 3 . 6 3 And

- 8 .4 3 < ^ : < 0.78

Thus, the stability condition is |0< J^T <

0.781-the conditions on 0.781-the system parameters (a, ^ f f ^ ) to guarantee closed-loop system stability.

Figure Magnetic levitation system

Re ^a

( f + p )

Step-by-step solution

Step 1 of 6

Consider transfer function of the system.

Step-by-step solution

step 1 of 6

Consider transfer function of the system.

21 -y+ p

K , s + p j ' - o “

{ s+ p )(s‘ - a ‘ )+ K K ,{s+ z) __________ K K , { s + z ) __________

**j*+ps’ + (A3T**

, - o’ +JCKjZ - po*

Thus, the transfer function of the system is K K ,{ s* z) 5^ + ps^ + (kK ^ - < ^ ]s+ K KqZ - p<^

Step 2 of 6

Step-by-step solution

step 1 of 6

Consider transfer function of the system.

21 5 + p s ^ - a *

j + p j ’ - o ’

( » + p ) ( j’ - f l ’ ) + A ^ r , ( i+ z ) __________K K , { s + z )__________

j * + ps’ + (A3Tj - o’ ) z + JCKjZ - pc’

Thus, the transfer function of the system is ___________K K ^(s* z)___________

+ ps^ + ( K K ^ - i^ ) s + K K f ^ z - p ( ^

Step 2 of 6

KK(gz-pd^

K K a z - p d i

Figure I

Step 6 of 6

The system is stable if the equation satisfies the following condition.

• All the terms in the first column of the Routh’s array should have same sign.

• The first column of Routh’s array should not posses any sign change.

• All the terms in the first column of the Routh’s array should be greater than zero.

Therefore, for stability all the elements in the first column to be positive and we obtain the following constraints.

From the above statement, write the stability conditions.

p > 0

K K ^ p - K K ^ > 0 i f K > 0 ^ p > z K K ^ - p a ‘ > 0 i / K > 0 = z > - ^

KK^

Therefore, for stability all the elements in the first column to be positive system condition on the

system parameters is \ p >zla n d , V 2 -K -K ,

Consider the system shown in Fig.

Figure Control system A J ( f + I )

(a) Compute the closed-loop characteristic equation.

(b) For what values of (7, is the system stable? H in t: An approximate answer may be found using

i + h '

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 7 and A.

Step-by-step solution

step 1 of 2

(a) Y (s) s(s+ l) 1+A e ^ s(s+ l)

s(s+ l)+ A e'^

CH Equation:

s (s+ l)4 A (l-T s) = 0

(By taking the ^proxim ation of exponential term) s^+s-ATs+A=0

|s^+(l-A T)s+A =0|

Step 2 of 2

0>)

1 A

1-AT 0 A 1-A T>0 A > 0 T < —

A A > 0 & T <

-Problem 3.59PP

Modify the Routh criterion so that it applies to the case in which ali the poies are to be to the left of - a when a > 0. Apply the modified test to the polynomial

S3 + (6 + K)s2 + ( 5 + 6K)s + 5K = 0.

finding those values of K for which all poles have a real part less than -1.

Step-by-step solution

step 1 of 1

Shift the root to origin and then Routh’s Criteria.

- s te p 1 o n

Shift the root to origin and then Routh’s Criteria.

Replacing s by s-a.

For the given Equation

( s - l f +(64*;)(s-l)“ +(5+6k)(s-l)+3k=0

=>(s=-l+3s-3s’ )+ (sH l-2 s)(6 4 k )+ (5 + 6 fc)(!-l)+ 5 k = 0

=> s’+ s ' (-343c«) +s (3-12-2k+5+€k) -l+€4k-5.61d-5k=0 s*+(lri-3)!^+ s(4M )= 0

s’ 1 s’ Icf3 0 s’ 4k-4 s" 0

Find constraints on the two gains K^ and K2 that guarantee a stable closedloop system, and plot the allowable region(s) in the (/Cl, K2) plane.You may wish to use the computer to help solve this problem.

Suppose the characteristic polynomial of a given closed-loop system is computed to be s4 + (11 +K2)s3 + (121 +K■\)s2 + {K^ + K1K2 + 110K2 + 210)s + 11/C1 +100 = 0.

Step-by-step solution

step 1 of 3

Step 1 of 3

The characteristic polynomial of the closed loop system is.

j ^ + ( i i + A : 2 ) j ’ + ( i 2 i + i c , y + ( x , + A : , A : 2 + i i o ^ : 2 + 2 i o ) i + i i K | + 1 0 0 = 0 Write the Routh array for the polynomial.

s-.*

I I2 I + £ , HAT,+100

i i+a:2 ac, +a:|A:2+iiojc2+21o o

s^: a

IIA:,+100

s : b

iia:, + ioo Where.

^ ( ii+ A r 2 ) ( i2 i+ A :i) - ( A : i+ A :|A : 2 + iiO A : 2 + 2 io ) ( i) l I + ATj

133i+i i a:,+ 1 2ia: 2 +a:,a: 2 -a:i-AT,AC;- 1 1 0 ^ 2 - 2 1 0 11+AC2

10AC,+MAC;+ 1121 II + AC2 And,

^ a(AC| + AC|AC2 + 110AC2+2l0)-(llAC, + 100)(ll+A:2) a

+AC,AC2+iioa:2+ 210)-(1 lAC,+100)(11+AC2) V. l l + AT;

)

___________________________________________

io / : i + n A : a + i i 2 i 1 1 + ^2

(ioa: , -1-1 1^2 + ^1^2-*-iioa:2-(-21o) - (i i+a:2)(iia:,+ioo)(i i-i-a:2)

io a: , +i i/ : 2 +i i21

step 2 of 3

For a stable system, first column elements of Routh array must be greater than zero.

Therefore, i u a: 2 >o

And,

ii/:,-i-ioo>o

\ \ K t > - m -1 0 0 K t >

-11

Therefore, the two gain of the system are AC, and AC2> -11

' 11 ^

Step 3 of 3

Draw the allowable region in the (ac„a:2 ) plane.

i 1C2

•lO O /ld

-150 -120 -90 -AO -30 ^

10 0

A llo w a b le re g io n

0 30 60 90 120

...

1 -50

i 1

-100

1 1 -150

1 1

•200

1 -250

Hence, the plot is drawn.

Problem 3.61 PP

Overhead electric power lines sometimes experience a low-frequency, highamplitude 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. A simple model of this conductor galloping is

(» 2 + ,2 )l/2 Where

m = mass of conductor, ir“fnr’c \/ortinal riicnla^om ont

m = mass of conductor,

y = conductor’s vertical displacement, D = aerodynamic drag force, L = aerodynamic lift force,

V =

wind velocity,

a = aerodynamic angle of attack = — 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 =

^ =0. Use Routh’s stability criterion to show that galloping can occur whenever

^ + D b < 0 .

Figure Electric power-line conductor

Step-by-step solution

step 1 of 2

dx dx^

D ( a ) ! ! ^ - H a ) V

1 \ dD d a _ , .

D ( a ) j : , - L ( a ) r

Step 2 of 2

Now

So,

-1 1 -ao

d a m

d a

D { a ) x j - L { a ) V

If S is the sensitivity of the unity feedback system to changes in the plant transfer function and T is the transfer function from reference to output, show that S + T =

Step-by-step solution

step 1 of 1 ^

Given that S is the sensitivity o f the unity fe edback system to changes in the plant transfer function and T is the transfer function &om reference to output.

Given that S is the sensitivity o f the unity fe edback sjrstem to changes in the plant transfer function and T is the transfer function from reference to output.

We know that

^ ^

S + T = — 5

s + r =

1 + 01D,i 1 + ODd 1 + GfDd

T =

1 + GZ)„

G D .

| g + r = i|

Problem 4.02PP

We define the sensitivity of a transfer function G to one of its parameters k as the ratio of percent change in G to percent change in k.

^ d G jG d \n G k dG

* “ d t / T “ d t o t “ G ^ '

The purpose of this problem is to examine the effect of feedback on sensitivity. In particular, we would like to compare the topologies shown in Fig. for connecting three amplifier stages with a gain of - K into a single amplifier with a gain of -10.

(a) For each topology in Fig., compute j8/ so that if /C = 10, V = -10R.

(b) For each topology, compute when G = (Use the respective j8/values found ii part(a).) Which case is the least sensitive?

(b) For each topology, compute when G = (Use the respective j8/values found in part(a).) Which case is the least sensitive?

(c) Compute the sensitivities of the systems in Fig.(b, c) to j82 and /S3. Using your results, comment on the relative need for precision in sensors and actuators.

Figure Three-amplifier topologies

Step-by-step solution

step 1 of 14

(a)

Refer to Figure 4.24 (a) in the text book.

The transfer function is.

r = - f i , K ‘ R

Compare above equation with y = -1 0 Jt- f i, K ’ = 10

Substitute 10 for K in the above equation.

^ 10’

= 10- ’

= 0.01

Therefore, the value of is |o,q i

|-Step 2 of 14

Refer to Figure 4.24 (b) in the text book.

Determine the transfer function of the system.

Compare above equation with y = -1 0

Substitute 10 for K in the above equation.

Refer to Figure 4.24 (c) in the text book.

Determine the transfer function of the system.

Y (-iC )(-A :)(-iC )

-Substitute 10 for K in the above equation.

A = - 10*

Determine the sensitivity, . ,0 dG/G

dK/K

Step 5 of 14

Refer to Figure 4.24 (a) in the text book.

The transfer function is.

Refer to Figure 4.24 (b) in the text book.

The transfer function is.

Substitute 10 for K and 0.3642 for 0^ in the above equation.

5 ?

Refer to Figure 4.24 (c) in the text book.

The transfer function is.

Substitute 10 for Kand 0.099 for in the above equation.

1+(10)’ (0.099) 3 1+99

= 0.03

Therefore, the sensitivity 5 ^ is [q^03| . The topology in (c) is least sensitive.

Step 10 of 14

Refer to Figure 4.24 (b) in the text book.

The transfer function is,

R (i+a t aA ' + ^a J U +j^aJ AT’

(1 + ATA)’

Write the expression for the sensitivity of the system in Figure 4.24 (b) to ■

^ J P A d G

Determine the sensitivity, . , J p A d G

Substitute 10 for Kand 0.3642 for in the above equation.

. 3 ( 1 0 ) ( 0 .3 6 4 2 )

Refer to Figure 4.24 (c) in the text book.

The transfer function is.

Y ( - K ) ( - K ) ( - K ) R l- ( - K ) { - K ) ( - K ) P , G = — K '

\ + K %

Write the expression for the sensitivity of the system in Figure 4.24 (c) to .

* l o J d A

Determine the sensitivity, 5 ^ .

, j p ^ ^

Substitute 10 for Kand 0.099 for 0^ in the above equation.

j , . ( 1 0 ) ’ ( 0 .0 9 9 )

The results indicate that the closed loop system is much sensitive to errors in the feedback path than that in the fonward path.

Hence, sensors should have more precision than actuators.

Compare the two structures shown in Fig. with respect to sensitivity to changes in the overall gain due to changes in the amplifier gain. Use the relation

as the measure. Select H1 and H2 so that the nominal system outputs satisfy F1 = F2, and assume /CH1 > 0.

Figure Block diagrams

I h p ' - — { ! ] — Q —

^ T— --- 1

Step-by-step solution

step 1 of 10

In document Pontificia Universidad Católica del Perú Facultad de Derecho. Programa de Segunda Especialidad en Derecho Administrativo (página 15-22)

EL PROCEDIMIENTO DE RECONOCIMIENTO DE GRADOS Y TÍTULOS

150

oiep I ui ^

,

6.2

j*+ps’ + (A3T

, - o’ +JCKjZ - po*

s*-.

s^: a

)

ii/:,-i-ioo>o

V =

-1 1 -ao

^ T— --- 1

**j*+ps’ + (A3T**

s-.*