root locus - 11.14

11.5 PRACTICE PROBLEM SOLUTIONS

1.

+

+

1

3.0

--------------------

( D + 1) 2

-

-

KdD

+

3.0

-------------------------------------

( D + 1) 2 + K

d

D

-

3.0

---------------------------------------------------

D2 + D( K

d

+ 2)

+ 4.0

2

D( Kd

+ 2) + 4.0

– K

d

– 2 ±

( K

d

+ 2) 2

– 4( 4.0)

D +

= 0

D =

--------------------------------------------------------------------------

2

Kd

roots

D =

– Kd

– 2 ±

Kd2 + 4Kd

– 12

----------------------------------------------------------------

0

-1 +/- 1.732j

2

1

-1.5 +/- 1.323j

Critical points: (this is simple for a quadratic)

2

-2.000, -2.000

The roots becomes positive when

5

-0.628, -6.372

0 >

– Kd – 2±

Kd2 + 4Kd – 12

10

-0.343, -11.657

100

-0.039, -102.0

2 + K

> ±

K2 + 4K

– 12

1000

-0.004, -1000

d

d

d

16 > 0

0 > – Kd – 2

Kd > –2

The roots becomes complex when

0 > K2		+ 4K		d	– 12
	d
Kd =	– 4 ±		16 – 4			( –12) Kd = –6, 2
	----------------------------------------------				2

Gains larger than -2 will result in a stable system. Any gains between -4 and -2 will result in oscillations.

root locus - 11.15

2.

a)	200K
	---------------------------------------
	D2 + 2D + 200K
b)	roots =			– 2 ± 4 – 4( 200K)			= – 1 ±		1 – 200K
			-----------------------------------------------
			2										Im
		K		roots
	0			0,-2		K=0.005
	0.001			-0.1,-1.9										Re
	0.005			-1,-1
	0.1			etc.			-2		-1
	1
	5
	10

c) D2 + 2D + 200K = D2 + 2ζω nD + ω n2	ω n = 1	K = 0.005
From the root locus graph this value is critically stable.

root locus - 11.16

3.

a)D2 + D + 20K = 0

D =		–--------------------------------------------1 ± 1 – 4( 20K)
		2
	K		roots
0			0.000, -1.000
1/80			-0.500, -0.500
1			-0.5 +/- 4.444j
10			-0.5 +/- 14.13j
1000			-0.5 +/- 141.4j

For complex roots 1 – 80K < 0

For negative real roots (stable)

– 1 ± 1 – 80K <

------------------------------------ 0 2

± 1 – 80K < 1

b)

Matching the second order forms,

2ω nξ = 1

ω n2 = 20K

The gain can only be used for the natural frequency

20	20	= 2.22
K = -----	= -----
ω n2	32

K >	1
	-----
	80

K > 0

root locus - 11.17

c)

θ o

20( 10)

---- =

----------------------------------------

θ d

D2 + D + 20( 10)

θ··o + θ·d + θ d200 = 200θ

d

Homogeneous:

A2 + A + 200 = 0

A =

– 1 ± 1 – 4( 200)

A =

– 0.5 ±

14.1j

-------------------------------------------

2

θ o( t)

=

C1e–0.5t sin ( 14.1t + C2)

Particular:

θ = A

0 + 0 + A200 = 200( 1rad)

A = 1rad

θ o( t)

= 1rad

Initial Conditions (assume at rest):

θ o( t)

= C1e–0.5t sin ( 14.1t + C2)

+ 1rad

θ o( 0)

= C1( 1) sin ( 14.1( 0) + C2) + 1rad = 0

C1 sin ( C2)

=

–1rad

(1)

θ '

o

( t)

=

–0.5C

1

e–0.5t sin ( 14.1t + C )

– 14.1C

1

e–0.5t cos ( 14.1t + C )

2

2

0

= –0.5C1 sin ( C2) – 14.1C1 cos ( C2)

14.1 cos ( C2) = –0.5 sin ( C2)

14.1

----------

C2 = –1.54

–0.5 = tan ( C2)

C1 =

–1rad

–1rad

1.000rad

------------------ =

------------------------- =

sin ( C2)

sin ( –1.54)

θ o( t)

= ( e–0.5t sin ( 14.1t – 1.54) + 1) ( rad)

root locus - 11.18

d)	θ	o	20( 0.01)
	---- =		--------------------------------------------
	θ	d	D2 + D + 20( 0.01)
		θ··o + θ·d + θ d0.2 = 0.2θ d
		Homogeneous:
			A2 + A + 0.2 = 0
			A =	– 1 ± 1 – 4( 0.2)			A = –0.7236068, –0.2763932
				------------------------------------------
				2
			θ o( t)	= C1e–0.724t + C2e–0.276t
		Particular:
			θ = A
			0 + 0 + A0.2 = 0.2( 1rad)				A = 1rad
			θ o( t)	= 1rad
		Initial Conditions (assume at rest):
			θ o( t)	= C1e–0.724t	+ C2e–0.276t		+ 1rad
			θ o( 0)	= C1e–0.724t	+ C2e–0.276t		+ 1rad = 0
						C1 + C2 = –1rad		(1)
			θ 'o( t)	= – 0.724( C1e–0.724t)		– 0.276( C2e–0.276t)
			C1 = –0.381C2
			– 0.381C2 + C2 = –1rad				C2 = –1.616rad
			C1 = –0.381( –1.616rad)			= 0.616rad
		θ o( t)	= ( 0.616) e–0.724t + ( –1.616) e–0.276t + 1rad

4.

(ans. X +

-

X +

-

X

X

root locus - 11.19

KpD + Ki + KdD

2

Y

3

------------------------------------------

D

-------------

D + 9

4

Y

3KpD + 3Ki

+ 3KdD2

----------------------------------------------------

D( D + 9)

4

Y

3KpD + 3Ki + 3KdD2

-------------------------------------------------------------------------------------------

D( D + 9) + 12K

D + 12K

+ 12K

D2

p

i

d

Y

3KpD + 3Ki

+ 3KdD2

----------------------------------------------------------------------------------------------

D2( 12K

d

+ 1) + D( 9 + 12K ) + ( 12K )

p

i

root locus - 11.20

b)D2( 12Kd + 1) + D( 9 + 12Kp) + ( 12Ki) = 0

	– 9 – 12K		p	±	( 9 + 12K ) 2	– 4( 12K			d	+ 1) 12K		i
D =					p
	--------------	-----------	---	---------	------2---(---12------K----d----+-----1---)------------------				----	----------	-----------
Kd		roots
-100		-0.092, 0.109
-10		-0.241, 0.418
-1		-0.46, 2.369
-0.1		-0.57, 105.6
0		-0.588, -20.41
1		-0.808 +/- 0.52j
10		-0.087 +/- 0.303j
100		-0.0087 +/- 0.1j

Stable for, – 9 – 12Kp ± ( 9 + 12Kp) 2 – 4( 12Kd + 1) 12Ki < 0

± ( 9 + 12Kp) 2 – 4( 12Kd + 1) 12Ki < 9 + 12Kp

( 9 + 12Kp) 2 – 4( 12Kd + 1) 12Ki < ( 9 + 12Kp) 2

–4( 12Kd + 1) 12Ki < 0

Kd >	–1
	-----
	12

Becomes complex at,

0 > ( 9 + 12Kp) 2 – 4( 12Kd + 1) 12Ki 576KdKi > ( 9 + 12Kp) 2 – 48Ki

	>	( 9 + 12K ) 2	– 48K	i
Kd		p			> 0.682
		----------------------------------------------576KdKi		Kd

root locus - 11.21

c)

Kp = 1

Ki = 1

Kd = 1

Y

=

3KpD + 3Ki

+ 3KdD2

--

----------------------------------------------------------------------------------------------

X

D

2

( 12Kd + 1) + D( 9 + 12Kp) + ( 12Ki)

Y

=

3D2 + 3D + 3

=

3

2

D2 + D + 1

--

2

-----------------------------------------

-----

--------------------------------------------------

X

D 13 + D21 + 12

13

D + D1.615 + 0.923

final gain

=

20log

3

=

–12.7

-----

13

initial gain

=

20log

3

=

–12.0

-----

12

for the numerator,

ω

n

=

1

= 1

1

=

0.5

ξ = ---------

2ω n

ω

d

= ω

n

1 – ξ 2

=

1 – 0.52

= 0.866

for the denominator,
ω	n	=	0.923 =	0.961	ξ	=	1.615		= 0.840
							------------
							2ω	n
ω	d	= ω	n 1 – ξ 2	= 0.961	1 – 0.8402			= 0.521
-12dB

root locus - 11.22

	Y	=		3KpD + 3Ki + 3KdD2
	---		-------	--------------------------	-------------------------------------------------------------
	X		2	( 12Kd + 1) + D( 9 + 12Kp) + ( 12Ki)
			D
	ω n	=		12Ki	= 2
				12-------K----d----+-----1-
	2ξω		n =	9 + 12Kp	= 20.5( 2)
				---------------------
				12Kd + 1

12Ki = 48Kd + 4

24Kd = 7 + 12Kp

At this point there are two equations and two unknowns, one value must be selected to continue, therefore,

Kp = 10
24Kd	= 7 + 12Kp = 7 + 12( 10) = 127			Kd	= 5.292
12Ki	= 48Kd + 4 = 48( 5.292) + 4 = 258.0			Ki	= 21.5
Now to check for stability
D2( 12( 5.292) + 1) + D( 9 + 12( 10) )			+ ( 12( 21.5) ) = 0
64.504D2 + 129D + 258 = 0
D =	– 129 ± 1292 – 4( 64.5) 258	=	– 1 ± 1.73j
	--------------------------------------------------------------------
	2( 64.5)

root locus - 11.23

e) Cannot be found without an assumed input and initial conditions

f)

Y

=

3( 0) D + 3( 0) + 3( 1) D2

---

X

D2( 12( 1) + 1) + D( 9 + 12( 0) ) + ( 12( 0) )

Y

=

3D2

---

X

13D

2

+ 9D

Y( 13D2 + 9D) = X( 3D2)

··

·

··

X = t

·

··

Y13 + Y9 = X3

X = 1

X = 0

··

·

9

= 0

Y + Y-----

13

It is a first order system,

Y( t) = C

–-

-9---t

1

e

13 + C

2

Y( 0) = 0

Y'( 0)

= 0

starts at rest/undeflected

0 = C11 + C2

C1 = –C2

9

Y'( t) = –-----

9

C e–13-----t

13

1

0 = –

-----9

C 1

C1

= 0

13

1

no response

C2

= 0

root locus - 11.24

5.

X	+			1			+				0.1		Y
			5 +						--------------------------------------
		K		D---	+ D				D	2	+ 10D + 100
		-					-
										10
										-----
		0.01									D
X	+	5D + 1			2				0.1				Y
					+ D		----------------------------------------------------------------
		K -----------------------------		D			D	2	+ 10D + 100 + 0.1			10
												-----
		-										D
		0.01

X

+

K( 0.5D + 0.1 + 0.1D2)

Y

-------------------------------------------------------

-

D3 + 10D2 + 100D + 1

X

0.01

Y

K( 0.5D + 0.1 + 0.1D2)

-------------------------------------------------------------------------------------------------------------------------------------

D3 + 10D2 + 100D + 1 + 0.01( K( 0.5D + 0.1 + 0.1D2) )

X

Y

K( 0.5D + 0.1 + 0.1D2)

----------------------------------------------------------------------------------------------------------------------------------------------

D3 + D2( 10 + 0.001K) + D( 100 + 0.005K) + ( 1 + 0.001K)