linear feedback controller analysis - 18.5

18.2.2 Analysis of PID Controlled Systems With Laplace Transforms

1. We can rewrite the control equation as a ratio of output to input.

dverror

θ

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

gas = Kcverror + Ki∫verrordt + Kd

dt

θ

gas

Kc + Ki∫dt + Kd

d

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

----

verror

dt

Then do a Laplace transform

d

→

dx

----

s

-----

sx

dt

dt →

∫dt

=

1

∫xdt

=

x

--

-

s

s

The transfer function

L

θ gas

= Kc +

Ki

+ Kds

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

----

verror

s

linear feedback controller analysis - 18.6

2. We can also develop a transfer function for the car.

	F = Aθ gas					= 10θ			gas
		F			= 10		Transfer function for engine and transmission. (Laplace
	---------
	θ	gas					transform would be the same as initial value.)
							d2x		dv
	F =				Ma =	M
							-------		= M-----
							dt	2	dt
	F	=			d
	--				M----
	v				dt
			v		1
	L	---			------			Transfer function for acceleration of car mass
			F		= Ms

3. We want to draw the system model for the car.

θ gas		10	F		1	vactual
					------
					Ms

•The ‘system model’ is shown above.

•If θ gas is specified directly, this is called ‘open loop control’. This is not desirable, but much simpler.

•The two blocks above can be replaced with a single one.

θ gas		10	vactual
		------
		Ms

linear feedback controller analysis - 18.7

4.If we have an objective speed, and an actual speed, the difference is the ‘system error’

verror = vdesired – vactual

‘set-point’ - desired system operating point

5. Finally, knowing the error is verror, and we can control θ gas (the control variable), we can select a control system.

verror		Controller	θ gas

L		θ gas		= Kc	Ki	+ Kds
	------------				+ ----
		verror			s

*The coefficients can be calculated using classical techniques, but they are more commonly approximated by trial and error.

6. For all the components we can now draw a ‘block diagram’

vdesired	verror		K		+ K s	θ gas	10	F	1	vactual
		K + ----							------
	+	c		i	d				Ms
			s
	-

• A ‘negative feedback loop

that is the fundamental part of this ‘closed loop control system’

linear feedback controller analysis - 18.8

18.2.3Finding The System Response To An Input

•Even though the transfer function uses the Laplace ‘s’, it is still a ratio of input to

output.

•Find an input in terms of the Laplace ‘s’

vdesired

	100	step
		vdesired(t) = 100 for t >= 0 sec
		ramp
		vdesired(t) = 50t for t >= 0 sec
		t(sec)
		0

				Input type		Time function		Laplace function
				STEP		f( t)	= Au( t)	f( s)		A
										= --
										s
										A
				RAMP		f( t)	= Atu( t)	f( s)
										= ----
										s2	2
										Aω
				SINUSOID		f( t)	= A sin ( ω t) u( t)	f( s) =
										-----------------
										s2 + ω 2

PULSE

f( t)

= A( u( t) – u( t – t1) )

f( s)

=

etc......

linear feedback controller analysis - 18.9

Therefore to continue the car example, lets assume the input below,

vdesired( t)	= 100	t ≥ 0 sec
vdesired( s)	= L[ vdesired( t) ] =	100
		--------
		s

Next, lets use the input, and transfer function to find the output of the system.

				vactual
	vactual = v-----------------						vdesired
					desired
				s2( K )		+ s( K ) + K
								i			100
	=			d			c
vactual		-------------------------------------------------------------									--------
			2	M							s
		s					+ s( K ) + K
				----- + K
				10		d	c
									i

To go further, some numbers will be selected for the values.

	Kd = 10000
	Kc = 10000
	Ki = 1000
	M = 1000
	s2( 10000) + s( 10000)	+ 1000		100
	-----------------------------------------------------------------------			--------
vactual	=
	s2( 10100) + s( 10000)	+ 1000		s

linear feedback controller analysis - 18.10

At this point we have the output function, but not in terms of time yet. To do this we break up the function into partial fractions, and then find inverse Laplace transforms for each.

2

s2

+ s + 0.1

vactual

=

10

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

s( s2( 1.01)

+ s + 0.1)

Aside: We must find the roots of the equation, before we

can continue with the partial fraction expansion.

recall the quadratic formula,

ax2 + bx + c = 0

x =

--------------------------------------– b ± b2 – 4ac

2a

x =

– 1 ±

12

– 4( 1.01) ( 0.1)

= –0.113, –0.877

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

2

( 1.01)

v

=

102

s2

+ s + 0.1

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

actual

1.01

s( s + 0.113) s + 0.877

A

B

C

vactual

=

--s

+ s + 0.114 + s + 0.795--------------------- ---------------------

linear feedback controller analysis - 18.11

A = lim

s

102

s2 + s + 0.1

102

0.1

---------

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

= ---------

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

s → 0

1.01

s( s + 0.113) ( s + 0.877)

1.01

( 0.113) ( 0.877)

A = 99.9

102

s2

+ s + 0.1

B =

lim

( s + 0.113)

---------

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

s →

–0.113

1.01

s( s + 0.113) ( s + 0.877)

B =

102

( –0.113) 2

+ ( –0.113) + 0.1

=

0.264

---------

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

1.01

( –0.113) (

– 0.113 + 0.877)

102

s2

+ s + 0.1

C =

lim

( s + 0.877)

---------

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

s →

–0.877

1.01

s( s + 0.113) ( s + 0.877)

C =

102

( –0.877) 2

+ ( –0.877) + 0.1

=

–1.16

---------

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

1.01

( –0.877) (

– 0.877 + 0.113)

v

=

99.9

0.264

1.16

actual

--------- + ---------------------

– ---------------------

s

s + 0.113

s + 0.877

linear feedback controller analysis - 18.12

Next we use a list of forward/inverse transforms to replace the terms in the partial fraction expansion.

	f( t)					f( s)
	A					A
						--
						s
	At					A
						----
						s2
	Ae	–α t				A
						-----------
						s + α
	A sin ( ω		t)			Aω
					-----------------
					s2 + ω 2
–ξω	nt		2	ω	n	1 – ξ 2	for( ξ < 1)
e	sin ( ω nt		1 – ξ )	---------------------------------------
				s2 + 2ξω ns + ω n2

etc.

To finish the problem, we simply convert each term of the partial fraction back to the time domain.

vactual =	99.9---------	+	---------------------0.264	–	---------------------1.16
	s		s + 0.113		s + 0.877

vactual = 99.9 + 0.264e–0.113t – 1.16e–0.877t