Colorado School of Mines CHEN403
Linear Open Loop Systems
Linear Open Loop Systems................................................................................................................................ 1
Transfer Function for a Simple Process ....................................................................................................... 1
Example Transfer Function — Mercury Thermometer ......................................................................... 2
Desirability of Deviation Variables............................................................................................................ 3
Transfer Function for Process with Multiple Inputs and/or Multiple Outputs ............................ 3
Example Transfer Function — Stirred Tank Heater ............................................................................... 5
Transfer Function of Process in Series ......................................................................................................... 8
Poles & Zeros of a Transfer Function ............................................................................................................ 9
Example — Poles & Zeros of a Transfer Function............................................................................ 12
Transfer Function for a Simple Process
f (t )
Dynamic
Process
Input
y (t )
f (s)
Output
G(s)
y(s)
Consider the simple process with one input & one output. The describing n-th order ODE
is:
an
dn y
d n−1 y
d2 y
dy
+
a
+
⋯
+
a
+ a1
+ a0 y = bf ( t )
n −1
2
n
n −1
2
dt
dt
dt
dt
Let us assume we are using deviation variables, so y′ ( 0) = 0 , and we are starting at
steady state, so:
d n−1 y
d2 y
dy
=
⋯
=
=
= 0.
n −1
2
dt t =0
dt t =0 dt t =0
Taking the Laplace transform of this gives:
an s n y ( s ) + an−1 s n−1 y ( s ) + ⋯ + a2 s 2 y ( s ) + a1 sy ( s ) + a0 y ( s ) = bf ( s )
y(s)
f (s)
=
an s + an−1 s
n
Transfer Functions
n −1
b
≡ G(s)
+ ⋯ + a2 s 2 + a1 s + a0
-1-
December 21, 2008
Colorado School of Mines CHEN403
where G ( s ) is defined as the transfer function and the simple diagram is called the block
diagram for the process.
Example Transfer Function — Mercury Thermometer
Make the following assumptions about the reading from a mercury thermometer:
•
•
•
•
All resistance to heat transfer is in a thin film around the bulb — i.e., neglect thermal
resistance of glass & mercury.
All thermal capacity is in the mercury.
Mercury always has uniform temperature.
The glass wall does not expand or contract.
The energy balance on thermometer will be:
dE d ( E + K + P ) dU dH
=
≈
≈
= hA (Ta − T )
dt
dt
dt dt
dT
mCˆ p
= hA (Ta − T ) for constant Cˆ p
dt
ˆ
mC p dT
= Ta − T
hA dt
dT
τ
− T = Ta
dt
where the time constant τ is:
τ≡
mCˆ p
hA
At steady state:
T * = Ta*
so in terms of deviation variables:
τ
dT ′
− T ′ = Ta′ where T ′ ( 0) = 0
dt
Taking the Laplace transform of this ODE gives:
Transfer Functions
-2-
December 21, 2008
Colorado School of Mines CHEN403
( τs + 1)T ′ = Ta′
so the transfer function is:
G(s) =
T′
1
=
Ta′ τs + 1
So, we would expect the heat transfer resistance around a thermometer to be a 1st order
system.
Desirability of Deviation Variables
If we didn’t use deviation variables the Laplace transform of the ODE would be:
τ
dT
+ T = Ta ⇒ τ ( sT − T ( 0 ) ) + T = Ta
dt
( τs + 1)T = Ta + τT (0)
1
τ
Ta +
T ( 0)
τs + 1
τs + 1
1
τ
=
Ta +
Ta*
τs + 1
τs + 1
T=
Now there are two inputs & two transfer functions: one for the driving function ( Ta ( t ) or
Ta ( s ) ) and one for the initial condition ( Ta* ).
Transfer Function for Process with Multiple Inputs and/or Multiple Outputs
What if there are multiple inputs and/or multiple outputs? We would associate a transfer
function with each pairing of an input & output. The block diagram for 2 inputs & 1 output
is:
f1 ( s )
G(s)
f1 ( s )
G1 ( s )
f2 ( s )
G2 ( s )
y(s)
f2 ( s )
+
+
y(s)
The overall relationship for y ( s ) would be:
Transfer Functions
-3-
December 21, 2008
Colorado School of Mines CHEN403
y ( s ) = G1 ( s ) ⋅ f1 ( s ) + G2 ( s ) ⋅ f2 ( s )
For n inputs and one output, then:
y ( s ) = G1 ( s ) ⋅ f1 ( s ) + G2 ( s ) ⋅ f2 ( s ) + G3 ( s ) ⋅ f3 ( s ) + ⋯ + Gn ( s ) ⋅ fn ( s )
n
y ( s ) = ∑ Gi ( s ) ⋅ fi ( s )
i =1
The block diagram for 2 inputs & 2 outputs is:
f1 ( s )
G1,1 ( s )
+
+
y1 ( s )
G2,1 ( s )
f2 ( s )
G2,1 ( s )
G2,2 ( s )
+
+
y2 ( s )
The overall relationship for the y ( s ) functions would be:
y1 ( s ) = G1,1 ( s ) ⋅ f1 ( s ) + G1,2 ( s ) ⋅ f2 ( s )
y2 ( s ) = G2,1 ( s ) ⋅ f1 ( s ) + G2,2 ( s ) ⋅ f2 ( s )
For n inputs and m outputs, then:
yi ( s ) = Gi ,1 ( s ) ⋅ f1 ( s ) + Gi ,2 ( s ) ⋅ f2 ( s ) + Gi ,3 ( s ) ⋅ f3 ( s ) + ⋯ + Gi ,n ( s ) ⋅ fn ( s )
n
yi ( s ) = ∑ Gi , j ( s ) ⋅ f j ( s ) for i = 1,2,3, … , m .
j =1
or in matrix notation as:
y ( s) = G(s ) f ( s )
Transfer Functions
-4-
December 21, 2008
Colorado School of Mines CHEN403
where y ( s ) is a column vector of length m , f ( s ) is a column vector of length n , and G ( s )
is a m × n rectangular matrix. G ( s ) is called the transfer function matrix.
Example Transfer Function — Stirred Tank Heater
F0, T0, ρ0
h, A, T, ρ
Fs, Ts
F1, T1, ρ1
The material balance on this system will be:
d ( hρ )
dm
= F0ρ0 − F1ρ1 ⇒ A
= F0ρ0 − F1ρ
dt
dt
assuming constant cross-sectional area, A . The energy balance is:
dE
= F0ρ0 Hˆ 0 − F1ρ1 Hˆ 1 + Q = F0ρ0 Hˆ 0 − F1ρHˆ + Q
dt
Remember, within the tank:
dE d (U + K + P ) dU dH
=
≈
≈
dt
dt
dt dt
So:
(
)
ˆ
dH d ρVH
=
= F0ρ0 Hˆ 0 − F1ρHˆ + Q
dt
dt
If we assume that the enthalpy can be expressed as:
Transfer Functions
-5-
December 21, 2008
Colorado School of Mines CHEN403
Hˆ = Cˆ p (T − Tref ) + Hˆ ref
then with Hˆ ref = 0 & Tref = 0 :
(
)
d
ρVCˆ pT = F0ρ0Cˆ pT0 − F1ρCˆ pT + Q
dt
d
Cˆ p ( ρVT ) = F0ρ0Cˆ pT0 − F1ρCˆ pT + Q
dt
d
Q
( ρVT ) = F0ρ0T0 − F1ρT + ˆ
dt
Cp
If we assume ρ ≈ constant , then:
A
dh
= F0 − F1
dt
and:
ρA
d
Q
( hT ) = F0ρT0 − F1ρT + ˆ
dt
Cp
F
F
d
Q
( hT ) = 0 T0 − 1 T + ˆ
dt
A
A
ρAC p
TA
dh
dT
Q
+ hA
= F0T0 − F1T +
dt
dt
ρCˆ p
Inserting the material balance:
T ( F0 − F1 ) + hA
hA
dT
Q
= F0T0 − F1T +
dt
ρCˆ p
dT
Q
= F0 (T0 − T ) +
where h = h( t ) .
dt
ρCˆ p
If we make the assumption that dh / dt = 0 then V = hA = constant & F0 = F1 , so:
V
dT
Q
= F0 (T0 − T ) +
dt
ρCˆ p
If we are using steam for the heating medium, then we could relate the rate of heat added,
Q , to the steam temperature, Ts , as:
Transfer Functions
-6-
December 21, 2008
Colorado School of Mines CHEN403
Q = UA (Ts − T ) .
So:
V
UA (Ts − T )
dT
= F0 (T0 − T ) +
dt
ρCˆ p
dT 
UA 
UA
+  F0 + ɶ T = F0T0 +
Ts
dt 
ρC p 
ρCˆ p
F
dT  F0
UA 
UA
Ts
+ +
T = 0 T0 +
dt  V ρVCˆ p 
V
ρVCˆ p

dT  1
1
+  + K T = T0 + KTs
τF
dt  τF

dT
1
+ aT = T0 + KTs
dt
τF
V
where:
1 F0
UA
1
≡ ,K≡
, and a ≡ + K .
τF V
τF
ρVCˆ p
At steady state:
aT * =
1 *
T0 + KTs*
τF
so:
dT ′
1
+ aT ′ = T0′ + KTs′
dt
τF
where the deviation variables are defined as:
T ′ ≡ T − T * , T0′ ≡ T0 − T0* , and Ts′ ≡ Ts − Ts* .
Note that this equation shows how the stirred tank fluid temperature is affected by changes
in the other temperatures.
Transfer Functions
-7-
December 21, 2008
Colorado School of Mines CHEN403
In this Chapter we will convert this equation into one involving transfer functions. Taking
the Laplace transform of the equation gives:
sT ′ − T ′ ( 0) + aT ′ =
1
T0′ + KTs′
τF
1
T0′ + KTs′
τF
1
( s + a )T ′ = T0′ + KTs′
τF
1 τF
K
T′ =
T0′ +
Ts′
s +a
s +a
sT ′ + aT ′ =
This shows that we have two transfer functions:
T ′ = G0 ( s )T0′ + Gs ( s )Ts′
where:
G0 ( s ) ≡
1/ τF
K
and Gs ( s ) ≡
s +a
s +a
A block diagram for the stirred tank heater can be drawn as follows.
1/ τF
s+a
T0′ ( s )
+
+
T ′( s )
K
s+a
Ts′ ( s )
Transfer Function of Process in Series
y1(s)
f(s)
G1(s)
y2(s)
G2(s)
Gn(s)
yn(s)
If there are a series of transfer functions, then:
Transfer Functions
-8-
December 21, 2008
Colorado School of Mines CHEN403
yn ( s ) = Gn ( s ) yn−1 ( s )
= Gn ( s ) Gn−1 ( s ) yn−2 ( s )
= Gn ( s ) Gn−1 ( s ) ⋯ G1 ( s ) f ( s )
yn ( s )
f (s)
n
= ∏ Gi ( s )
i =1
Poles & Zeros of a Transfer Function
According to definition of the transfer function:
y(s)
f (s)
= G(s)
where:
G(s) =
Q(s)
P (s)
and where Q ( s ) and P ( s ) are usually polynomials in s (time delays will introduce
exponential terms, however). In general, the order of Q ( s ) will be less than that of P ( s ) .
The roots of the numerator Q ( s ) are referred to as the zeros of the transfer function. At
the zeros, G ( s ) becomes zero. The roots of the denominator P ( s ) are referred to as the
poles of the transfer function. At the poles, G ( s ) becomes infinite.
We can get a qualitative sense of the response of a system by knowing the poles. Let:
f (s) =
r (s)
q(s)
Since:
G(s) =
Q(s)
P( s )
then:
Transfer Functions
-9-
December 21, 2008
Colorado School of Mines CHEN403
y(s) = G(s)⋅ f (s) =
Q(s) r (s)
P (s) q(s)
Let’s let the roots of P ( s ) be denoted as pi . Then, if P ( s ) is a polynomial of order n and
there are N non-repeating roots and M repeating roots (each one repeating mi times),
then:
N
M
i =1
i =1
P ( s ) = ∏ ( s − pi ) ⋅ ∏ ( s − pi )
mi
and:
y(s) =
Q(s)
N
r (s)
M
∏(s − p ) ⋅∏(s − p )
i
i =1
q(s)
mi
i
i =1
When split into partial fractions, each of the factors in the denominator will lead to a
separate term. Splitting up the factors of the transfer function (while leaving the
denominator from the input function aside for now) gives:
mi −1
N
M
C
y(s) = ∑ i + ∑
i =1 s − pi
i =1
∑ D (s − p )
j ,i
j
i
j =0
( s − pi )
mi
M  mi −1
D j ,i
Ci
y(s) = ∑
+ ∑ ∑
mi − j
i =1 s − pi
i =1  j =0 ( s − pi )

N
+
r* (s)
q( s)
.
 r* (s)
+
.
 q(s)

Note that for the repeated root, the numerator can be a polynomial of order up to one less
the order of denominator. Also, each repeated root can have a different order. The only
requirement on the number of roots is that they have to add up to n , i.e.:
M
n = N + ∑ mi .
i =1
When we invert the Laplace transforms, then:
N C  N
L−1  ∑ i  = ∑ C i exp ( pi t ) .
 i =1 s − pi  i =1
Transfer Functions
- 10 -
December 21, 2008
Colorado School of Mines CHEN403
 M  mi −1
D j ,i
L−1  ∑  ∑
mi − j

 i =1  j =0 ( s − pi )
  M  mi −1


D j ,i

  = ∑  ∑ L−1 

mi − j
  i =1  j =0

s
p
−


(
)
i




M
 mi −1
 D 
= ∑ exp ( pi t )  ∑ L−1  mij−,i j  
i =1
s

 j =0
 mi −1
= ∑ exp ( pi t )  ∑
 j =0 ( mi
i =1

M
 mi −1
= ∑ exp ( pi t )  ∑
 j =0 ( m
i =1
i

M
 ( m − j − 1)!  
L−1  i mi − j
 
− j − 1 )!
s


.
D j ,i

t mi − j −1 

− j − 1 )!

D j ,i
Note that the roots pi are important for the long-time characteristics of the solution. For
the real non-repeating roots:
•
•
•
If pi < 0 , then exp ( pi t ) → 0 as t → ∞ . This exponential decay leads to a zero
contribution from this pole.
If pi > 0 , then exp ( pi t ) → ∞ as t → ∞ . This exponential growth leads to a explosive
contribution from this pole.
If pi = 0 , then exp ( pi t ) = 1 for all t . This constant term should not lead to any
instability.
For the complex non-repeating roots (which will occur as a complex conjugate pair), then
pi can be expressed as α i ± βi i . These roots will give rise to terms of the form
exp ( α i t ) sin ( βi t + φi ) . Now, the important term with regards to stability is the real portion
of the root, α i :
•
•
•
If α i < 0 , then exp ( αi t ) sin ( βi t + φi ) → 0 as t → ∞ . This exponential decay leads to a
zero contribution from this pole.
If α i > 0 , then exp ( α i t ) sin ( βi t + φi ) → ∞ as t → ∞ . This exponential growth leads to
a explosive contribution from this pole.
If α i = 0 , then exp ( αi t ) sin ( βi t + φi ) = sin ( βi t + φi ) for all t . This term will lead to a
stable oscillation.
For the repeating roots, the situation is similar. The polynomial term will always grow
towards infinity as t → ∞ , so the behavior of the exponential term will dictate the overall
behavior.
•
If pi < 0 or α i < 0 , then the exponential term will go to zero as t → ∞ and the entire
term will also go to zero. This exponential decay leads to a zero contribution from
this pole.
Transfer Functions
- 11 -
December 21, 2008
Colorado School of Mines CHEN403
•
•
If pi > 0 or α i > 0 , then the exponential term will grow to infinity as t → ∞ and the
entire term will also grow to infinity. This exponential growth leads to an explosive
contribution from this pole.
If pi = 0 or α i = 0 , then the polynomial term will dictate the behavior for t → ∞ .
This polynomial term will lead to an explosive contribution from this pole..
So, in general:
•
•
•
If α i < 0 , stable contribution from this pole.
If α i > 0 , unstable contribution from this pole.
If α i = 0 , stable contribution only if non-repeated root — unstable contribution if
repeated root.
Example — Poles & Zeros of a Transfer Function
Given the transfer function:
G(s) =
Q(s)
P (s)
=
Q(s)
s + 3s + 5s 2 + 4s + 2
4
3
find the zeros & determine if stable.
The following chart shows the characteristics of P ( s ) vs. s . Note that there are no real
roots.
Transfer Functions
- 12 -
December 21, 2008
Colorado School of Mines CHEN403
10
9
8
7
6
P (s ) 5
4
3
2
1
0
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
s
Can factor P ( s ) to get:
P ( s ) = s 4 + 3s3 + 5s 2 + 4s + 2 = ( s 2 + s + 1 )( s2 + 2s + 2)
From this, we find that the roots are:
r=
−1 ± 1 − 4 ⋅ 1 ⋅ 1
1
3
i
=− ±
2
2 2
r=
−2 ± 22 − 4 ⋅ 1 ⋅ 2
= −1 ± i
2
Since the real portion of the roots are all negative, the system is stable.
Transfer Functions
- 13 -
December 21, 2008