Dynamic Behavior of First-order and Second-order Systems
1. Analysis of first-order systems
Consider the system shown in Figure 1 which consists of a tank of uniform cross
sectional area A to which is attached a flow resistance R. Assume that qo is related to the
head by linear relationship
h
qo =
(4. 1)
R
qi
V
h
q
A
Figure 1 Liquid level system
The change in liquid height inside the tank can be described by the following equation;
dh
h
=q−
dt
R
At steady state, this equation can be written as:
h
0 = qs − s
R
Subtracting Equation (4.3) from Equation (4.2) gives
d (h − hs )
(h − hs )
A
= ( q − qs ) −
dt
R
Defining the deviation variables as
Q = q − qs
A
H = h − hs
Equation (4.4) can be written as
dH H
A
+ =Q
dt
R
Chapter 4: Dynamic Behavior of First-order & Second-Order Systems
(4. 2)
(4. 3)
(4. 4)
(4. 5)
(4. 6)
1
Taking the L.T. of this equation gives
H ( s)
Q( s) =
+ AsH ( s )
(4. 7)
R
Equation (4.7) can be rearranged into the standard form of the first-order system to give
H (s)
R
=
Q( s ) τs + 1
(4. 8)
where τ = AR
In a general form, Equation (4.8) can be written in a general form:
Y ( s)
K
= G ( s) =
U ( s)
τs + 1
Where
Y ( s ) = l[ y (t )]
(4. 9)
(4. 10)
U ( s ) = l[u (t )]
G is called transfer function. It is a convenient way to represent linear dynamic models. It
relates one input and one output:
u(t)
U(s)
System
y(t)
Y(s)
Figure 2 General system representation
The term τ (time constant) and K (steady state gain) characterize the first-order system.
Note that the both parameters depend on operating conditions of the process and that the
transfer function does not contain the initial conditions explicitly.
Properties of transfer functions
Time constant of a process is a measure of the time necessary for the process to adjust to
a change in the input.
Steady state gain is the steady state change in output divided by the sustained change in
the input. It characterizes the sensitivity of the output to the change in input. The steadystate of a transfer function can be used to calculate the steady-state change in an output
due to a steady-state change in the input. For example, suppose we know two steady
states for an input, u, and an output, y. Then we can calculate the steady-state gain, K,
from:
Chapter 4: Dynamic Behavior of First-order & Second-Order Systems
2
K=
y2, ss − y1, ss
(4. 11)
u2, ss − u1, ss
Another important property of the transfer function is that the order of the denominator
polynomial (in s) is the same as the order of the equivalent differential equation. A
general nth-order differential equation has the form
dny
d n −1 y
dy
d mu
d m −1u
du
an n + an −1 n −1 + L + a1 + ao y = bm m + bm −1 m −1 + L + b1
+ bou
(4. 12)
dt
dt
dt
dt
dt
dt
Where u and y are input and output deviation variables respectively. The transfer function
obtained by L.T. is
m
Y ( s)
G(s) =
=
U ( s)
∑b s
i
∑a s
i
i
i =0
n
i
i =0
=
bm s m + bm −1s m −1 + L + bo
an s n + an −1s n −1 + L + ao
(4. 13)
The steady state gain of G(s) in Equation (4.13) is bo/ao, obtained by setting s = 0 in G(s).
Definition: The order of the transfer function is defined to be the order of the
denominator polynomial
Multiplicative rule
U
G1
Y
G2
Y = G1 ⋅ G 2 ⋅ U
(4. 14)
Additive rule
U1
G1
+
U2
+
G2
Y = G1U1 + G2U 2
Chapter 4: Dynamic Behavior of First-order & Second-Order Systems
(4. 15)
3
Dynamic response
The dynamic response for a step change in u(t) of magnitude M: U(s) = M/s
−
t
y (t ) = KM (1 − e τ )
(4. 16)
The transient response for a step change is shown in Figure 3.
M
u(t)
y(t)
KM
0
t
Figure 3 Response of a first-order system to a step change in the input.
Dynamic Characteristics
•
A first-order process is self-regulating. The process reaches a new steady state.
•
The ultimate value of the output is K for a unit step change in the input, or KM
for a step of size M. This can be seen from Equation (4.16), which
yields y → KM as t → ∞ . This characteristic explains the name steady state or
static gain given for the parameter K, since for any step change in the input the
resulting change in the output steady state is given by
Δ(output ) = KΔ(input )
(4. 17)
This Equation tells us by how much we should change the value of the input in
order to achieve a desired change in the output, for a process with given K. Thus,
to effect the same change in the output, we need:
A small change in the input if K is large (very sensitive systems)
A large change in the input if K is small. See Figure 5.
•
The value of y(t) reaches 63.2 % of its ultimate value when the time elapsed is
equal to one time constant τ. When the time elapsed is 2τ, 3τ, 4τ, the percent
Chapter 4: Dynamic Behavior of First-order & Second-Order Systems
4
response is 86.5, 95 and 98 respectively. From these facts, one can consider the
response essentially completed in three to four time constants.
•
The slope of the response at t = 0 is equal to 1.
d [ y (t ) / KM ]
= (e −t τ ) t = 0 = 1
d (t τ )
t =0
(4. 18)
This implies that if the initial rate of change of y(t) were to be maintained, the
response would reach its final value in one time constant (see the dashed line in
Figure 4). The corollary conclusions are:
The smaller the value of time constant, the steeper the initial response of the
system (Figure 5).
1.0
y
KM
0.632
0
1
t
10
0.0
τ
Figure 4 Response of a first-order system to a step change in the input.
y(t)
y(t)
K1M
K1
K2M
K2
KM
t1
K2 < K1
0
t2
Time
0
t1 < t2
Time
Figure 5 Effect of static gain, time constant on the response of first-order lag system.
Chapter 4: Dynamic Behavior of First-order & Second-Order Systems
5
2. Pure capacitive system (Integrator)
Consider the first order system described by Equation (4.19)
dy
a1 + ao y = bu (t )
dt
If ao = 0, then from Equation (4.19) we take
dy b
= u (t ) = Ku (t )
dt a1
Which gives a transfer function
Y ( s) K
G ( s) =
=
U ( s) s
(4. 19)
(4. 20)
(4. 21)
In this case, the process is called purely capacitive or pure integrator.
Dynamic response
For a step change in u of magnitude M: U(s) = M/s. Equation (4.21) yields
KM
G ( s) = 2
s
After inversion we find that
y (t ) = Kt
(4. 22)
(4. 23)
Notice that the output grows linearly with time in unbounded fashion (Figure 6). Thus,
y→∞
as
t→∞
Note that a pure capacitive process causes serious control problem because it can not
balance itself. For small change in the input, the output grows continuously. This attribute
is known as non-self-regulating process.
y(t)
K
t
Figure 6 Unbounded response of pure capacitive process
Chapter 4: Dynamic Behavior of First-order & Second-Order Systems
6
3. Analysis of second-order systems
A second-order system is one whose output, y(t), is described by a second-order
differential equation. For example, the following equation describes a second-order linear
system:
d2y
dy
+ a1 + ao y = bu (t )
2
dt
dt
If ao ≠ 0, then Equation (4.24) yields
a2
(4. 24)
d2y
dy
+ 2ξτ
+ y = Ku (t )
(4. 25)
2
dt
dt
Equation (4.25) is in the standard form of a second-order system, where
τ = natural period of oscillation of the system
ζ = damping factor
K = steady state gain
The very large majority of the second- or higher-order systems encountered in a
chemical plant come from multicapacity processes, i.e. processes that consist of two or
more first-order systems in series, or the effect of process control systems.
Laplace transformation of Equation (4.25) yields
K
(4. 26)
G ( s) = 2 2
τ s + 2ξτs + 1
τ2
Dynamic response
For a step change of magnitude M, U(s) = M/s, Equation (4.26) yields
KM
Y (s) =
2 2
s (τ s + 2ξτs + 1)
(4. 27)
The two poles of the second-order transfer function are given by the roots of the
characteristic polynomial,
τ 2 s 2 + 2ξτs + 1 = 0
(4. 28)
and they are
p1 = −
ξ 2 −1
ξ
+
and
τ
τ
p2 = −
ξ 2 −1
ξ
−
τ
τ
(4. 29)
Therefore, Equation (4.27) becomes
K τ2
Y (s) =
s ( s − p1 )( s − p2 )
(4. 30)
The form of the response of y(t) will depend on the location of the two poles in the
complex plane. Thus, we can distinguish three cases:
Chapter 4: Dynamic Behavior of First-order & Second-Order Systems
7
Case A: (over-damped response), when ζ > 1, we have two distinct and real poles.
In this case the inversion of Equation (4.30) by partial fraction expansion yields
⎡
⎛
ξ
t
t
y (t ) = K ⎢1 − e −ξt τ ⎜ cosh ξ 2 − 1 +
sinh ξ 2 − 1
2
⎜
τ
τ
⎢⎣
ξ −1
⎝
⎞⎤
⎟⎥
⎟⎥
⎠⎦
(4. 31)
Where cosh(.) and sinh(.) are the hyperbolic trigonometric functions defined by
e α − e −α
e α + e −α
and cosh α =
(4. 32)
2
2
Case B: (critically damped response), when ζ = 1, we have two equal poles (multiple
pole).
In this case, the inversion of Equation (4.30) gives the result
sinh α =
⎡ ⎛ t⎞
⎤
y (t ) = K ⎢1 − ⎜1 + ⎟e −t τ ⎥
(4. 33)
⎣ ⎝ τ⎠
⎦
Case C: (Under-damped response), when ζ < 1, we have two complex conjugate poles.
The inversion of Equation (4.30) in this case yields
⎡
⎤
1
y (t ) = K ⎢1 −
e −ξt τ sin(ωt + φ )⎥
1− ξ 2
⎢⎣
⎥⎦
where
ω=
1− ξ 2
τ
(4. 34)
and
⎡ 1− ξ 2 ⎤
φ = tan ⎢
⎥
⎢⎣ ξ ⎥⎦
Several remarks can be made concerning the responses shown in Figure 7:
1. Responses exhibiting oscillation and overshoot (y/KM > 1) are obtained only for
values of ζ less than one (Under-damped response). This response is initially
faster than the other two responses; however, it does not reach steady-state earlier
due to oscillations. Note that almost all under-damped responses are caused by the
interactions of the controllers with the process units they control.
2. Large values of ζ yield a sluggish (slow) response (Over-damped response). The
larger the value of ζ the more sluggish the system. Note that system response
resembles a little the response of a first-order system to a unit step input. But
when compared to a first-order response we notice that the system initially delays
to respond and then its response is rather sluggish (Figure 8). Over-damped are
the responses of multicapacity systems, which result from the combination of
first-order systems in series.
−1
Chapter 4: Dynamic Behavior of First-order & Second-Order Systems
8
3. The fastest response without overshoot is obtained for the critically-damped case
(ζ =1).
y(t)/KM
t/tau
Figure 7 Dimensionless response of second-order system to input step change.
Chapter 4: Dynamic Behavior of First-order & Second-Order Systems
9
1
First-order system
0.9
0.8
Second-Order system
0.7
y/KM
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
time
6
7
8
9
10
Figure 8 Comparison between first-order and second-order responses.
Characteristics of under-damped systems
Figure 9 Characteristics of an under-damped response.
- Overshoot: Is the ratio of a/b, where b is the ultimate value of the response and a is the
maximum amount by which the response exceeds its steady state value. It can be shown
that it is given by the following expression:
Chapter 4: Dynamic Behavior of First-order & Second-Order Systems
10
⎛
πξ
OS = exp⎜ −
⎜
1− ξ 2
⎝
⎞
⎟
⎟
⎠
(4. 35)
- Decay ratio: Is the ratio of the amount above the stead state value of two successive
peaks, c/a. it can be shown that it can be calculated by the following equation:
⎛
2πξ ⎞⎟
DR = exp⎜ −
⎜ 1− ξ 2 ⎟
⎝
⎠
-
(4. 36)
Rise time: tr is the the process output takes to first reach the new steady state value.
Time to first peak: tp is the time required for the output to reach its first maximum
value.
Settling time: ts is defined as the time required for the process output to reach and
remain inside a band whose width is equal to ± 5 % of the total change in the output.
Period: Equation (4.34) defines the radian frequency, to find the period of oscillation
P (i.e. the time elapsed between two successive peaks), use the well-known
relationship ω = 2π/P; thus:
2πτ
P=
(4. 37)
1 − ξ2
4. Dynamic systems with dead time
Time delay is inherent property of chemical processes. The transfer function of a time
delay of θ units is given by
G ( s ) = e − θs
(4. 38)
Consider a first-order system with a dead time θ between the input and the output. This
system can be represented by a series of two systems as shown in Figure 10. For the firstorder system, we have the following transfer function:
l[ y( t )]
K
(4. 39)
=
l[u( t )] τs + 1
while for the dead time we have:
l[ y( t − θ)]
= e − θs
l[ y( t )]
(4. 40)
Therefore, the transfer function between the input u(t) and the delayed output y(t-θ) is
given by:
l[ y( t − θ)] Ke − θs
=
l[ u ( t )]
τs + 1
(4. 41)
Same argument can be used with second-order systems.
Chapter 4: Dynamic Behavior of First-order & Second-Order Systems
11
l[ u( t )]
K
τs + 1
l[ y( t )]
e − θs
l[ y( t − θ)]
Figure 10 Block diagram for first-order process with dead time
Dynamic characteristics
The fitting of the first order plus time-delay model to the step response, using the
tangent method, requires the following steps, as shown in Figure 11:
Figure 11 Graphical analysis of the process reaction curve to obtain parameters of a
first-order plus time-delay model.
1. The process gain for the model is found by calculating the ratio of the change in the
steady-state value of y to the size of step change M in x.
2. A tangent is drawn at the inflection point of the step response; the intersection of the
tangent line and the time axis (where y = 0) is the time delay.
3. If the tangent is extended to intersect the steady-state response line (where y = KM),
the point of intersection corresponds to time t = θ + τ.
This method suffers from using only a single point to estimate the time constant. Use
of multiple points may provide a better estimate.
Chapter 4: Dynamic Behavior of First-order & Second-Order Systems
12
Polynomial approximations to e-θs
Processes with dead time are difficult to control because the output does not contain
information about current events.
The introduction of exponent form results in a non-rational transfer function, which
can not be put in the form of a ratio of two polynomials in s. Quite often the exponential
term is approximated by the first- (Equation 4.42) or second-order Padé approximations
(Equation 4.43)
θ
1− s
2
e − θs ≈
(4. 42)
θ
1+ s
2
e − θs
θ
θ2
1 − s + s2
2 12
≈
θ
θ2
1 + s + s2
2 12
(4. 43)
Approximation of higher-order systems
Systems with order that is higher than one can be represented by
n
G (s ) = ∏ G i (s ) =
i =1
K
n
∏ ( τis + 1)
(4. 44)
i =1
It can be approximated by low-order transfer function with dead time as follows
G (s ) =
Ke − θs
τ1s + 1
(4. 45)
where τ1 is the dominant time constant, i.e. dominated implies that the overall process
transient process is determined primarily by a few large time constants; that is response
modes that are much slower than the others. The dead time is defined as
n
θ = ∑ τi
(4. 46)
i =2
Dynamic systems with inverse response
The dynamic behavior of certain processes shows an initial response which opposes in
direction to what it ends up, Figure 12. Such behavior is called inverse response. Such
behavior is the net result of two opposing effects. Systems show an inverse response if
one of the roots of the numerator (i.e. one of the zeros of the transfer function) has
positive real part.
Chapter 4: Dynamic Behavior of First-order & Second-Order Systems
13
Figure 12 Inverse response
5. Examples
Non-interacting system
Two tanks are connected in series as shown in Figure 13. The time constants are τ2 = 1
and τ1 = 0.5; R1 = R2 = 1. Sketch the response of the level in tank 2 if a unit-step change
is made in the inlet flow rate to tank 1.
Solution
q(t)
h1
R1, q1(t)
h2
R2 q2(t)
Figure 13 Two-tank liquid level system: Noninteracting
Chapter 4: Dynamic Behavior of First-order & Second-Order Systems
14
A balance on tank 1 gives
dh1
dt
A balance on tank2 gives
dh
q1 − q2 = A2 2
dt
The flow head relationships are given by the following expressions
q − q1 = A1
q1 =
h1
R1
h
q2 = 2
R2
Equation (4.47) can be rearranged to give
H1 ( s)
R1
=
,
τ = AR
Q ( s ) τ 1s + 1
H1
R
Q1 ( s )
1
=
Q( s ) τ 1s + 1
(4. 47)
(4. 48)
(4. 49)
(4. 50)
Q1 =
In the same manner, the following equations van be obtained for tank2
H 2 (s)
R2
=
Q1 ( s ) τ 2 s + 1
(4. 51)
(4. 52)
Having the transfer function for each tank, we can obtain the overall transfer function
H2(s)/Q(s) by multiplying Equations (4.51) and (4.52) to eliminate Q1
H 2 (s)
1
R2
(4. 53)
=
⋅
Q1 ( s ) τ 1s + 1 τ 2 s + 1
Note that all roots here are real; thus, this is an over-damped system. By comparing this
equation with the general form of the over-damped system, it follows that
τ 1 = τ (ξ + ξ 2 − 1)
τ 2 = τ (ξ − ξ 2 − 1)
(4. 54)
It can be proved that H2(s) is
H 2 (t ) = 1 − (2e − t − e −2t )
(4. 55)
Generalization for several noninteracting systems in series
Consider n noninteracting first-order systems as represented by the block diagram of
Figure 14. The block diagram is equivalent to the relationships
Chapter 4: Dynamic Behavior of First-order & Second-Order Systems
15
X 1 (s)
k1
=
X o ( s ) τ 1s + 1
X 2 (s)
k2
=
X 1 (s) τ 2 s + 1
(4. 56)
LLLLLL
X n (s)
kn
=
X n −1 ( s ) τ n s + 1
To obtain the overall transfer function, we simply multiply together the individual
transfer functions; thus
n
X n ( s)
k
=∏ i
X o ( s ) i =1 τ i s + 1
Xo
(4. 57)
k1
τ 1s + 1
X1
k2
τ 2s +1
Xn-1
X2
kn
τ ns +1
Xn
Figure 14 Noninteracting first-order systems
Interacting system
q(t)
h1
R1, q1(t)
h2
R2 q2(t)
Figure 15 Two-tank liquid level system: Interacting
Solution
The balances on tanks 1 and 2 are the same as before and are given by Equations (4.47)
and (4.48). However, the flow-head relationship for tank I is now
(h − h )
q1 = 1 2
(4. 58)
R1
Chapter 4: Dynamic Behavior of First-order & Second-Order Systems
16
The flow-head relationship for R2 is the same as before. Expressing Equation (4.58) in
terms of deviation variables gives
(H − H 2 )
Q1 = 1
(4. 59)
R1
Transforming model equations gives
Q( s ) − Q1 ( s ) = A1sH1 ( s )
Q1 ( s ) − Q2 ( s ) = A2 sH 2 ( s )
R1Q1 ( s ) = H1 ( s ) − H 2 ( s )
(4. 60)
R2Q2 ( s ) = H 2 ( s )
The combination of these equations gives
H 2 (s)
R2
=
2
Q( s ) τ 1τ 2 s + (τ 1 + τ 2 + A1 R2 ) s + 1
(4. 61)
Notice that the difference between the transfer function for noninteracting system and the
interacting system is the presence of the term A1R2 in the coefficient of s.
To understand the effect of interaction on the transient of a system, consider a two-tank
system for which the time constants are equal, if the tanks are noninteracting, the transfer
function relating inlet flow to outlet flow is
2
Q2 ( s ) ⎛ 1 ⎞
=⎜
⎟
Q( s ) ⎝ τs + 1 ⎠
The unit-step response for this transfer function is
t
Q2 (t ) = 1 − e −t τ − e −t τ
τ
(4. 62)
(4. 63)
If the tanks are interacting, the overall transfer function, according to Equation (4.61), is
(assuming A1 = A2)
Q2 ( s )
1
= 2
(4. 64)
Q( s ) τ + 3τs + 1
Notice that R2A2 = τ2 = τ. By application of the quadratic formula, the denominator of
this transfer function can be written as
Q2 ( s )
1
=
(4. 65)
Q( s ) (0.38τs + 1)(2.62τs + 1)
For this example, we see that the effect of interaction has been to change the effective
time of the interacting system. One time constant has become considerably larger and the
other smaller than the time constant of either tank in the noninteracting system. The
response of Q2(t) to a unit step change in Q(t) for interacting case [Equation (4.65)] is
Q2 (t ) = 1 + 0.17e −t 0.38τ − 1.17e −t 2.62τ
Chapter 4: Dynamic Behavior of First-order & Second-Order Systems
(4. 66)
17
1
0.9
0.8
Noninteracting
0.7
0.6
Q2
Interacting
0.5
0.4
0.3
0.2
0.1
0
0
0.5
1
1.5
2
2.5
t
3
3.5
4
4.5
5
Figure 16 Effect of interaction on step response of two-tank system
From Figure 16, it can be seen that interaction slows up the response. This is
understandable, when inlet to the first tank is stepped, the flow q1 will be reduced by the
build-up of level in tank 2.
In general, the effect of interaction on a system containing two first-order lags is to
change the ratio of effective time constants in the interacting system. In terms of transient
response, this means that the interacting system is more sluggish than the noninteracting
system.
Fitting of a first order system
Figure 17 gives response of the temperature T in a continuous stirred-tank reactor to a
step change in feed flow rate w from 120 to 125 kg/min. Find an approximate first-order
model for the process for these operating conditions.
Solution
First note that Δw = M = 125-120 = 5 kg/min. Since ΔT = T(∞) – T(0) = 160 – 140 =
20 ºC, the process gain is
o
20o C
C
ΔT
K=
=
=4
kg min
Δw 5 kg min
Chapter 4: Dynamic Behavior of First-order & Second-Order Systems
18
The time constant obtained from the graphical construction shown in Figure 17 is 5 min.
note that this result agrees with the time constant when 63.2% of the response is
complete, that is
T = 140+0.632*20 = 152.6 ºC
Consequently, the desired process model is
T (s)
4
=
W ( s ) 5s + 1
u(t)
125
w (kg/min)
120
y(t)
160
152.6
140
0
5
t
Figure 17 Temperature response of a stirred-tank reactor for a step change in feed
flow rate.
Chapter 4: Dynamic Behavior of First-order & Second-Order Systems
19