EE 4314 Homework2 Solutions

8/4/2019 EE 4314 Homework2 Solutions

1/18

EE 4314 Homework2 Solutions

3.32

a). y + 2n

y +n

2y = 0 , y 0

( )= y

0, y 0

( )= 0 (1)

Take the laplace transform of equation (1), we get:

s2Y s( ) s y0 + 2n sY s( )+n

2Y s( ) = H s( )

then

G s( ) =sy

0

s2+ 2n s+n

2

=

k1

s s1

+

k1

*

s s1

*

where

s1= n + jn 1

2

s2= n jn 1

2

k1=

nej cos

1( )

2n 12e

j / 2

then

y t( ) =ent

2 12e

j n 12

t+

2cos1

+ e j n 1

2t+

2cos1

y t( ) = y0e

t

12sin dt cos

1( )

b).dy t( )

dt= 0 t=

n

d


2/18

tMax

=

2

d

n

y t( )tMax

yn = y0e nd

1

2sin cos

1( )

yn =y

01

2

12

end (2)

For = lny

0

yn=d

From (2): y1= y

0e

d lny0

yn

= d

For lny

1

y1

lnyi

yi

We have yn= y

n1 y

n,

yn = y0end y

0e n1( ) d

= y0end 1 e

d( ) ,

yn

yn=

y0end

y0end 1 e

d

( )

yn

yn=

yi

yifor all i, n.

3.43

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

The Routh array is,

s4 : 1 32 100

s3 : 8 80

s2 : 22 100

s1 : 80-800/22=43.6

s0 : 100


3/18

No roots not in the LHP.

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

The Routh array is,

s5 : 1 30 344

s4 : 10 80 480

s3 : 22 296

s2 : -545 480

s1 : 490

s0 : 480

2 roots not in the LHP.

(c) s4 + 2s3 + 7s2 2s+ 8 = 0

The Routh array is,

s4 : 1 7 8

s3 : 2 -2

s2 : 8 8

s1 : -4

s0 : 8


(d) s3 + s2 + 20s+ 78 = 0

The Routh array is,

s3 : 1 20

s2 : 1 78

s1

: -58s0 : 78


(e) s4 + 6s2 + 25 = 0


4/18

The Routh array is,

s4 : 1 6 25

s3 : 4 12

s2 : 3 25

s1 : 12-100/3=-21.3

s0 : 25


3.44

s5+ 5s

4+10s

3+10s

2+ 5s+K= 0

The Routh array is,

s5 : 1 10 5

s4 : 5 10 K

s3 : a

1 a

2

s2 : b

1 K

s1 : c

1

s0 : K

where

a1=

5 10( ) 1 10( )5

= 8

a2=

5 5( )1 K( )5

=

25K

5

b1=

a1( ) 10( ) 5( ) a2( )

a1

=

55 +K

8

c1=

b1( ) a2( ) a1( )K

b1

=

K2+ 350K1375( )5 55+K( )

For stability: all elements in the first column must be positive. The following

conditions must be satisfied.


5/18

1. b1=

55+K

8> 0 K> 55

2. c1=

K2+ 350K1375( )5 55+K

( )

> 055 < K< 3.88

3. K> 0

By intersecting all 3 conditions, we have

0


6/18

3.46

The transfer function for the above system is

LetK

1

=

1

K

Y s( )R s( )

=

K1

s+ z( )s+ p( )

K0

s2 a

2( )

1+K

1s+ z( )

s+ p( )K

0

s2 a

2( )

=

K1K

0s+ z( )

s3+ ps

2+ K

1K

0 a

2( )s+ K1K0z pa2


7/18

Assume K0=1, we can construct Routh array as follow

s3

: 1K

1 a

2

s2 : p K

1z pa

2

s1 :

K1z + pa

2+K

1p pa

2

p=

K1z +K

1p

p

s0 : K

1z pa

2



p > 0,

K1pK

1z > 0 if K

1> 0 p > z

K1z pa

2> 0 ifK

1> 0 z >

pa2

K1

=Kpa2

3.47

(a)Y s( )R s( )

=

AesT

s s+1( )

1+Ae

sT

s s+1( )

=

AesT

s2+ s+ Ae

sT

The characteristic equation is: s2 + s+ AesT

(b) Using eTs 1Ts, the characteristic equation is

s2+ 1TA( )s+ A = 0

The Routh array is,

s2 : 1 A

s1 : 1-TA 0s0 : A


8/18



A > 0,

TA


9/18

scatter(real(r), imag(r), 'x')

end

hold offgrid on

3.22

Using block-diagram algebra.

Define dummy variables r1, r

2and r

inas shown in the figure below.

We have

r1= G

2rin (r

1+ r

2)H

2( )G4 (1)

r2= G

3rin r

1+ r

2( )H2( )G5 (2)

Letr

0

= r1

+ r2

(1)+ (2);r0=G

2G

4rinG

4H

2r0+G

3G

5rinG

5H

2r0

1+G4H

2+G

5H

2( )r0 = G2G4 +G3G5( )rin


10/18

r0

rin

=

G2G

4+G

3G

5

1+G4H

2+G

5H

2( )

Define G7 =(G2G4 +G3G5)G1G6

1+G4H2 +G5H2( ) (3)

Define dummy variable e3

as shown in the figure below.

We now have,

e3= R s( ) G7H4e3 G7H3e3

e3=

R s( )1+ G

7H

4+ G

7H

3( )

Since

Y s( ) = e3G7

We now have,

Y s( )R s( )

=

G7

1+ G7H

4+ G

7H

3( )(4)

Substitute (3) into (4), we get


11/18

Y s( )R s( )

=

G1G

6G

2G

4+ G

3G

5( )1+ G

2H

4+ G

2H

5( )

1+G

1G

6H

4G

2G

4+ G

3G

5( )1+ G

2

H4

+ G2

H5( )

+

G1G

6H

3G

2G

4+ G

3G

5( )1+ G

2

H4

+ G2

H5( )

=

G1G

6G

2G

4+G

3G

5( )1+G

2H

4+G

2H

5( )+G1G6H4 G2G4 +G3G5( )+G1G6H3 G2G4 +G3G5( )

Using Masons rule,

G =Your

Yin

=

Gk

k

k=1

N

Gk=1

=G1G

6G

2G

4

Gk=2

=G1G

6G

3G

5

k=1

=1

k= 2

=1

=1 Li + LiLj LiLjLk ++ 1( )m+

Li=1

=G4H

2

Li= 2

=G5H

2

Li=1Lj=1 =G1G2G4G6H4


Li= 3Lj= 3 =G1G2G4G6H3


therefore

=1G4H

2G

5H

2+G

1G

2G

4G

6H

4+G

1G

3G

5G

6H

4+G

1G

2G

4G

6H

3+G

1G

3G

5G

6H

3


12/18

Y s( )R s( )

=

G1G

2G

4G

6

+

G1G

3G

5G

6

=

G1G

2G

4G

6+G

1G

3G

5G

6

1G4H

2G

5H

2+G

1G

2G

4G

6H

4+G

1G

3G

5G

6H

4+G

1G

2G

4G

6H

3+G

1G

3G

5G

6H

3

4.2

(a) ForK=10 and y = 10r , we have:

Case a:

Y

R

= 1K

3

1= 0.01

Case b:

Y

R=

K

1+ 2K

3

2= 0.364

Case c:

Y

R=

K3

1+ 3K

3 3 = 0.099

(b) Sensivity SK

G , G =Y

R

Case a:

dG

dK= 3

1K

2

SK

G=

K

G

dG

dK=

K

1K

33

1K

2( ) = 3

Case b:

SK

G= 0.646


13/18

Case c:

SK

G= 0.03

Case c is the least sensitive.

(c)

Case b:

SK

G= 2.354

Case c:

SK

G= 0.99

The closed-loop system is much more sensitive to errors in the

feedback path than in the forward path.

4.4 G s( ) =A

s s+ a( )

(a) T s( ) =G s( )

1+ G s( )=

A

s s+ a( )

1+A

s s+ a( )

=

A

s2+ as+ A

dT

dA=

s2+ as+ A( ) A

s2+ as+ A( )

2

SAT =

A

T

dT

dA=

s s+ a

( )s s+ a( )+ A

(b)

dT

da=

sA

s2+ as+ A( )

2


14/18

a

T

dT

da=

a s2+ as+ A( )

A

sA

s2+ as+ A( )

2

Sa

T=

as

s s+ a( )+ A

(c)

T s( ) =G s( )

1+ G s( )

dT

d=

G s( )2

1+ G s( )( )2

T

dT

d=

1+ G( )G

G2

1+ G( )2=

G

1+ G

ST=

A

s s+ A( )

1+A

s s+ a( )

=

A

s s+ a( )+ A

- If a = A =1, the transfer function is most sensitive to variations in a andA near=1 rad/sec

- The steady-state response is not affected by variations in A and a.- The steady-state response is heavily dependent on since ST 0( ) =1.0

4.9

(a)For a unity feedback system to be Type 1 the open loop transferfunction must have a pole at s = 0. Thus in this case, since G has nosuch pole, it is necessary for D to have a pole at s = 0.


15/18

(b)Y s( ) =

1

s2+ D s( )+ K

W s( )

lims0

s1

s2+ D s( )+ K

1

s= 0

if and only if

lims0

s1

D s( ) =

if and only if =1 since D(s) has a pole at the origin. Therefore

the system will reject step disturbances with zero error.

4.19

(a) System (a):E= R Y =

R s( )1+ G s( )

E s( ) =s 4s+1( )

4s2+ s+ K

0K

1

R s( )

Using final value theorem:

ess,ramp =1

K1

=

1

Kv K

1=Kv =1

System (b):

E= R Y=1+G K

2G

1+GR

E s( ) =4s+1+ K

3K

01K

2( )4s+1+ K

3K

0

R s( )


16/18

Using final value theorem:

ess,ramp =1+K3K0 1K2( )

1+K3

K0

= 0

forK0=11+K

31K

2( ) = 0

ess,ramp =4

1+K3

=

1

Kv

forKv=1 K

3= 3

K2=

4

3 ,K

3=

3

(b)

Let K0=K

0+ K

0, system (a):

ess,ramp = lims0

ss 4s+1( )

4s2+ s+K

0+ K

0( )1

s= 0

System (b):

ess,ramp =1+K3 K0 + K0( ) 1K2( )

1+ K3 K0 + K0( ) K0=1=

K0

1+ 3 1+ K0( ) 0

System type of system (b) is not robust. The system (a) is preferred over

system (b) because it is more robust to parameter changes.


17/18

4.20

(a)

T s( ) =Y s( )R s( )

=

10 kp

s+ kI

( )s s+1( ) s+10( )

1+ 10 kp s+ kI( )

s s+1( ) s+10( )

E s( ) = 1T s( )( )R s( )

=

s s+1( ) s+10( )+10 kp s+ kI( )10 kp s+ kI( )s s+1( ) s+10( ) +10 kp s+ kI( )

1

s2

For=1,

ess = lims0

ss s+1( ) s+10( )

s s+1( ) s+10( ) +10 kp s+ kI( )

1

s2

=

10

10kI

=

1

kI

for characteristic equation: s3 +11s2 +10s+10 kps+ kI( ) = 0

The Rouths array is,

s3 : 1 10 1+ kp( )

s2 : 11 10k

I

s1 :

110 1+ kp( )10kI11

s0

:10k

I

For stability: all elements in the first column must be positive. The

following conditions must be satisfied;

kI> 0 and 111+ kp( ) kI > 0 .


18/18

(b) = 0.9

E s( ) =s s+1( ) s+10( )+ 9 kp s+ kI( )10 kp s+ kI( )

s s+1( ) s+10( )+ 9 kp s+ kI( )

R s( )

=

s s+1( ) s+10( ) kp kIs s+1( ) s+10( )+ 9 kp s+ kI( )

R s( )

Using final value theorem, for R s( ) =1

s2

ess

(c) for R s( ) =1

s

lims0

sE s( ) = lims0

s s+1( ) s+10( ) kp s kIs s+1( ) s+10( ) + 9 kp s+ kI( )

= k

I

9kI

= 1

9

The system is type 0. Kp is defined such that ess =1

1+Kp . Kp = 8.

Without the magnitude an equivalent result is that Kp = 10 .

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EE 4314 Homework2 Solutions