page 76

• Sometimes we don’t want to see the initial changes (transients), but we are more interested in the long term final value (steady state). We can use the final value theorem to find this,

H( t → ∞ ) = lim ( 1 – B) H( B)

B → 1

• Consider the previous function,

On	=		0.085B		0.005B	Sn
			-------------------------		-------------------
			1 – 0.915B		( 1 – B) 2
On →		∞	=

5.6 STEADY STATE ERROR

•When we examine a controller, we may look at the value of the error function into the controller.

•If the value of the error function does not become zero at infinity, the system is unstable.

•We can calculate the error value using the function,

page 77

er = rn – cn

• If we consider the example from before,

In

=

0.005B

-------------------Sn

( 1 – B) 2

O

= G( B) I

=

0.085B

0.005B

S

n

n

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

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

n

1 – 0.915B

( 1 – B) 2

Now we can write the error equation,

er

=

rn

– cn

=

0.005B

Sn

–

0.085B

0.005B

Sn

(-------------------1 – B) 2

1-------------------------– 0.915B

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

( 1 – B) 2

e

=

0.005B

S

–

0.085B

0.005B

S

=

1 –

0.085B

0.005B

r

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

n

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

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

n

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

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

S

n

(

1 – B) 2

1 – 0.915B

( 1 – B) 2

1 – 0.915B

(

1 – B) 2

e

=

1 – 0.915B – 0.085B

0.005B

S

=

( 1 – B) 0.005B

S

r

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

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

n

n

1 – 0.915B

( 1 – B) 2

( 1 – 0.915B) ( 1 – B) 2

e

=

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

0.005B

S

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

0.00546B

S

r

( 1 – 0.915B) ( 1 – B)

n

( 1.093 – B) ( 1 – B)

n

e

=

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

0.00546B

S

r

B2 – 2.093B + 1.093

n

Next, use long division to convert to an equation,

0.005B + 0.00957B2 + 0.0137B3 + 0.0175B4 + …

1.093 – 2.093B + B2

0.00546B

0.00546B – 0.01047B2 + 0.005B3

0.01047B2 – 0.005B3

0.01047B2 – 0.02B3 + 0.00957B4

0.015B3 – 0.00957B4

0.015B3 – 0.0287B4 + 0.0137B5

ETC..

The magnitude of the error seems to grow without bounds. If this continues to grow the system will become unstable.

page 78

At this point we should consider using the final value theorem to find if the error settles to zero,

e( t → ∞	)	=	lim ( 1 – B) e( B)		= lim	0.00546B
						( 1 – B) ---------------------------------------------
			B →	1	B → 1	( 1.093 – B) ( 1 – B)
e( t →	∞	)		0.00546( 1)
			=	-------------------------------	= 0.0587
				( 1.093 – ( 1) )

Here the final steady state error is finite, but defined, so the system is stable. Note that the output will never match the input.

5.7 PRACTICE PROBLEMS

1.Develop a discrete equation for the following transfer functions. Determine stability and realizability.

a)

5 + 3B + B2 G = ----------------------------

c 2 – 3B + B2

b)

1 – B G = --------------------

c – B + B2

c)

2 – B G = -------------------

c ( 3 – B) 2

2. Develop a differential equation for both the mechanical and the electrical system below. Find the transfer function for the module using the backshift operator ‘B’ for a time step of T = 1 sec.

page 79

OR

L = 1H

N

R = 1 ohm

K

s

= 1

---

m

F - input

Vin

-

M = 1 kg

+

Vout

y - output

C = 1F

K

= 2

Ns

d

------

m

4. a) Given the following transfer function, and the input function, find the resulting output for the first 5 time steps, if T=0.5 seconds.

G =	cn	0.5( 1 – B)	r( t) = 2t
	---- =	------------------------( B – 4)
	rn

b) What will the steady state output be for the system in part a)?

5. Develop a process model for one of the systems below. Assume the system starts at rest.

L=1

x

R=1

F

C=1

Vout

Kd=1

-

M=1

Vin

+

*note input is displacement ‘x’

a)Write the differential equations for one of the systems above.

b)Convert the equation to a transfer function using the backshift operator (use T=1sec).

page 80

c)Assume there is a step input of magnitude 1. Find the output function for the system in terms of the backshift operator. Do not convert the output function to numerical values in time.

d)Determine the steady state value for the system. Is the value consistent with what you would expect from the system? Explain.

	ANS.	Vin'' + Vin' =		–Vout				x'' + x' = –F
	a)
		Vin					0.368B + 0.264B2			x
	b)	Vout	= –1						=	--
						1----------------------------------------------------– 1.368B + 0.368B2
										F
	c)	F = Vout =					1 – 1.368B + 0.368B2
							0.368---------------------------------------------------------------------B – 0.632B2 + 0.264B3
	d)	F, Vout →		∞

e)Yes, it will grow without bounds, and the components are limited by finite bounds.

6.Given the following output function,

O( B) =	10B	Sn
	--------------------------------------------
	2
	0.5 – 0.9B + 0.5B

a)Find the steady state response as a function of time using the tables (assume T=0.2sec).

b)Find the first three values of the output in time using long division. Check that these values agree with the solution found in part a).

ANS.
a)	O( t) = 45.886 sin ( 2.255t) Sn
b)	20, 36, 44.8