Final Value Theorem Notes PDF

Preview

Summary

Final Value Theorem...

Description

School of Computing, Engineering & Built Environment

Maths for Control Engineering 4

Final Value Theorem

Dr Calum A. Macdonald

The Final Value Theorem (FVT) If a function f (t ) and its derivative f (t ) are both Laplace transformable and

lim t

f ( t)

exists then lim t

f (t)



lim s 0

s F( s )

where F ( s ) is the Laplace transform (LT) of f (t ) .

In terms of F ( s ) the FVT applies when:

(i).

F(s) has no poles in the right-half plane (RHP).

(ii).

F(s) has no poles on the imaginary axis except for at most one pole at the origin.

The FVT provides a method to find the steady state value of a system response without inverting the LT. For example, suppose we solve an ODE using LTs then, provided it is applicable, the FVT can be used to determine the long-term behaviour ( t   ) of the solution from its LT, without needing to invert the LT. Example For the following transfer functions determine whether the FVT applicable? Where the FVT does apply calculate the final value of the transfer function.

(i).

F( s ) 

4 s

(ii).

G( s ) 

4 s2

(iii).

H (s ) 

2 s ( s  3)

(iv).

K (s ) 

2 s ( s  3)

(v).

M( s) 

(vi).

N( s ) 

3 s  9

(vii).

P( s ) 

s (s

2

73  4 s  20)

2s  3 s  5s  6 2

(viii). R( s ) 

2

(s

2

2s  3 .  5s  6)(s  3) 2

(i).

F( s ) 

lim s 0

4 has a single pole at the origin, i.e. at s  0 , and so the FVT does apply. s

s F( s) 

lim s0

s

4 s



4.

lim 4 Also, f ( t )  L  1    4 and 4  t  s 

(ii).

G( s ) 

4 confirming the FVT result above.

4 has two poles at the origin, i.e. at s  0 , and so the FVT does not apply. s2

lim  4  4 t does not exist so the final value is not defined. Also, L 1  2   4 t and t s 

(iii).

H (s ) 

2 has one pole at the origin and one pole in the left-half plane at s ( s  3)

s   3 and so the FVT does apply lim s 0

s H (s ) 

lim

lim 2s 2 2    s  0 s (s  3 ) 0  3 s  0 (s  3)

2 . 3

Alternatively,

   2 2 1  3 2  h( t )  L 1  (1  e 3 t ) and     L  s s s s ( 3 ) 3 ( 3 ) 3      

lim 2 (1  e 3 t )  t 3

2 (1  0 ) 3



2 confirming the FVT result above. 3

3

(iv).

K( s ) 

2 has a pole in the right-half plane, i.e. at s  3 and so the FVT s( s  3)

does not apply.

  3  2 2 1  Alternatively, k( t )  L 1     L   3  s ( s  3)   s (s  3 )   

2 2 3t (e  1 ) . ( 1  e 3t )  3 3

lim

2 3t ( e  1) does not exist so the final value is not defined. t 3

(v).

M( s) 

s (s

2

73 has one pole at the origin and two poles in the LHP, at  4 s  20)

s   2  4 j and s   2  4 j and so the FVT applies.

Note: A quadratic with positive coefficients will always have roots with negative real parts.

lim s 0

s M (s ) 

lim s  0 s (s

2

73 s  4s  20) 

lim 73 73 73   2 s  0 (s  4s  20) 0  0  20 20

We omit finding the inverse LT as the calculations are rather lengthy.

(vi).

N (s ) 

3 has two poles on the imaginary axis i.e. at s   3 j and s  3 j so s 9 2

the FVT does not apply. lim   3  sin (3 t ) does not exist, Alternatively, n( t )  L 1  2   sin (3 t ) and t s  9 

as sin (3 t ) oscillates between  1 and 1, so the final value is not defined. 4

(vii).

P( s ) 

2s  3 2s  3 .  s  5s  6 ( s  3 )( s  2 ) 2

P(s) has two poles in the left-half plane i.e. at s  3 and s   2 so the FVT applies.

lim s 0

s P( s ) 

0 (0  3) 0 s(2 s  3)    0. 0  0  6 6 s  0 s  5s  6 lim

2

We omit finding the inverse LT as the calculations are rather lengthy.

(viii).

R( s ) 

2s  3 2s  3  ( s  5 s  6)( s  3) ( s  6)( s  1)( s  3) 2

R(s) has one pole in the right-half plane at s  1 and so the FVT does not apply

Alternatively note that the quadratic s 2  5 s  6 has one negative coefficient and so one of its roots must be positive and lie in the right half plane. Hence, the FVT does not apply. Incorrect use of the FVT Define F ( s ) 

lim s 0

15 . ( s  2)

s F( s) 

lim

3s 0  0.  s  0 ( s  2) 0  2

lim 3   2t However, f ( t )  L  1  3 e 2 t does not exist.   3e and t   ( s 2 )   

We therefore have a contradiction as

lim s 0

s F ( s )  0 but

lim t

f ( t ) does not exist.

The problem is that F ( s ) has a pole in the RHP at s  2 and so the FVT does not apply.

5...