Problem 1
ySP (s) +
k = 1.6
−
5
(s + 1)(2s + 1)
y(s)
The closed loop transfer function can be written as follows:
8
y(s)
8
89
(s + 1)(2s + 1)
G CL (s) =
=
= 2
=
2
8
ySP (s) 1 +
2s + 3s + 9 2 9s + 1 3s + 1
(s + 1)(2s + 1)
(S1.1)
Thus, comparing (S1.1) to the standard form of 2nd order systems, one obtains:
τ2 = 2 9
⇒ τ = 2 3 = 0.471
(S1.2)
2ξτ = 1 2 ⇒ ξ = 2 4 = 0.354
(S1.3)
k = 8 9 = 0.889
(S1.4)
Since ξ is smaller than 1, the system is underdamped.
a) From textbook, we know that:
⎛ −πξ ⎞
max dev final value y max − y(∞)
⎟ = 0.305 =
OS = exp ⎜
=
⎜ 1 − ξ2 ⎟
final
value
y(∞)
⎝
⎠
(S1.5)
The output of the system for a step change of magnitude 0.1 in the set point is:
y(s) =
0.1
89
2
s 2 9s + 1 3s + 1
(S1.6)
Applying the final value theorem, yields:
y(∞) = lim sy(s) = lim
s →0
s →0
0.8 9
0.8
=
= 0.0889
2 9s + 1 3s + 1 9
2
(S1.7)
Hence, the maximum value of the response is:
y max = y(∞)(1 + OS) = 0.116
(S1.8)
b) For a servo problem, the offset is given by:
offset = new set point − y(∞) = 0.1 − 0.0899 = 0.0111
(S1.9)
c) From the textbook, the period of the oscillation is:
Τ = 2π ω where ω = 1 − ξ 2 τ
(S1.10)
Therefore, the value of the period of the oscillation Τ is:
Τ = 3.17 min
(S1.11)
Problem 2
h (ft)
The resistance of a liquid to a hydrostatic pressure can
be defined as the rate of changing of the liquid level due
to the change of the output low rate. Thus:
dh
R=
dq
R
(S2.1)
q
(ft3/min)
Assuming linear resistances, R1 and R2 can be directly evaluated by plotting h (ft) versus
q (ft3/min) and estimating the slope of the straight line. Hence, one obtains:
R1 = R 2 = 0.5
(S2.2)
Since the cross sections are equal to 2, the time constant of the two tanks are both equal
to τ1 = A1R1 = 1 (min) and τ2 = A 2 R 2 = 1 (min) , while the static gains of two tanks are
equal to k1 = R1 = 0.5 (min ft 2 ) and k 2 = R 2 R1 = 1 (min ft 2 ) . Thus, the transfer
functions of the two tanks are the following:
G1 (s) =
0.5
s +1
⇒ G 2 (s) =
1
s +1
(S2.3)
Employing a proportional controller, the corresponding transfer function is:
G s (s) = k c
(S2.4)
Plotting the change in pressure to the valve versus the change in flow provides the
transfer function for the final control element:
G f (s) =
dP 0.1
=
= 0.1
dq
1
(S2.5)
With no lag in the measuring device dynamics, the corresponding transfer function is:
G m (s) = 1
(S2.6)
Therefore, the block diagram is the following:
a)
controller
ySP (s)
valve
kc
q(s)
0.1
tank 1
tank 2
0.5
s +1
1
s +1
y(s)
Measuring
device
1
b) The close loop transfer function is:
0.05 k c
0.05 k c
2
1 + 0.05 k c
y(s)
(s + 1)
G CL (s) =
=
=
1
2
ySP (s) 1 + 0.05 k c
s2 +
s +1
2
1 + 0.05 k c
1 + 0.05 k c
(s + 1)
(S2.7)
Comparing the closed loop transfer function to the standard form of a 2nd order system
transfer function one obtains:
τ = 1 1 + 0.05 k c
(S2.8)
ξ = 1 1 + 0.05 k c
(S2.9)
k = 0.05 k c (1 + 0.05 k c )
(S2.10)
For a critically damped 2nd order system, ξ=1 and therefore:
k c(CD) = 0
(S2.11)
Therefore, the critical damping cannot occur.
c) For interacting tanks, the transfer function between the output of the second tank and
the input of the first tank is given by:
G(s) =
y(s)
R2
0.5
=
= 2
2
q(s) τ1τ2s + (τ1 + τ2 + A1R 2 )s + 1 s + 3s + 1
Thus, the close loop transfer function becomes:
(S2.12)
0.05 k c
0.05 k c
2
1 + 0.05 k c
y(s)
G CL (s) =
= s + 3s + 1 =
0.05
k
1
3
ySP (s) 1 +
c
s2 +
s +1
2
1 + 0.05 k c
s + 3s + 1 1 + 0.05 k c
(S2.13)
Comparing the closed loop transfer function to the standard form of a 2nd order system
transfer function one obtains:
τ = 1 1 + 0.05 k c
(S2.14)
3
2 1 + 0.05 k c
(S2.15)
ξ=
k = 0.05 k c (1 + 0.05 k c )
(S2.16)
For a critically damped 2nd order system, ξ=1 and therefore:
k c(CD) = 25 psi/ft
(S2.17)
Hence, for interacting capacities the critical damping does occur.
d) Assuming kc=1.5kc(CD), one obtains:
k c = 1.5k c(CD) = 37.5 psi/ft
(S2.18)
Thus, the natural period of the oscillations, damping factor and gain becomes:
τ = 0.590 (min)
(S2.19)
ξ = 0.885
(S2.20)
k = 0.652 (ft)
(S2.21)
For a step change of 1/12 ft in the se point, ysp(s) is equal to 1/(s12). Therefore, one
obtains:
y(s) =
0.652
0.054
ySP (s) =
2
0.348s + 1.043s + 1
(0.348s + 1.043s + 1)s
(S2.22)
2
Since ξ is smaller than 1, from textbook we have:
⎧⎪
⎡
⎛ 1 − ξ2
1
−ξt τ1
−1
2 t
y(t) = 0.054 ⎨1 −
e
sin ⎢ 1 − ξ
+ tg ⎜
⎜
τ1
⎢⎣
1 − ξ2
⎪⎩
⎝ ξ
= 0.054 − 0.116e −1.5t sin(0.8t + 28o )
⎞ ⎤ ⎫⎪
⎟⎥ ⎬ =
⎟
⎠ ⎥⎦ ⎪⎭
(S2.23)
Problem 3:
a) The block diagram is the following:
mA
mA
mA
Controller
mA
m3/min
psgi
Transducer
Delay
mA
Valve
Kg/m3
Process
Composition
Transmitter Data
Kg/m3
Line
Kg/m3
b) Controller (pure PID)
Transfer function is: Gc ( s) = K c (1 +
1
+ τ D s)
τIs
Transducer
~
p ( s) p v ( s) − 3 15 − 3 12 3
=
=
=
=
Transfer Function is: GT ( s) = ~v
p ( s)
p ( s) − 4 20 − 4 16 4
Control Valve
Linearizing qA around an operative point:
q A = q A0 + 0.03 ln(20)20 ( pv 0 −3) / 12 ( p v − pv 0 )
Choosing qA0=0.17 and pV0=3, gives: q A = q A0 + 0.0025ln(20)(p v − p v0 )
q (s) q (s) − 0.17
Transfer Function is: G v (s) = A = A
= 0.0025ln(20)
p v (s)
p v (s) − 3
Also, considering delay: G v (s) = 0.0025ln(20)e − s
Process
dc
V
= q A c A + q F c F − (q A + q F )c . Note q A + q F ≈ q F .
dt
dc
Hence V
+ qF c = q Ac A + qF cF .
dt
At steady state: q Fcs = q As c A + q Fc Fs
Defining c~ = c − c s , q~A = q A − q As , c F = c F − c Fs yields:
dc
dc
or
V + q Fc = c A q A + q Fc F
τp + c = k p q A + k d c F where τP = V / q F ,
dt
dt
K P = c A / q F and K d = 1 .
Taking Laplace transform gives:
kd
kP
c(s) =
q A (s) +
c F (s) = G P (s)q A (s) + G D (s)c F (s)
τPs + 1
τPs + 1
Therefore, process can be expanded as
illustrated in figure.
Kg/m3
GD(s)
Trasmission Line
Transfer function is: G L (s) = e − t Ls ,
m3/min
Kg/m3
GP(s)
where t L = ⎡⎣ 20π(0.5) 2 ⎤⎦ / 4q F
Composition Transmitter Data
c~ ( s ) c m ( s ) − 4 20 − 4 16
Transfer Function is: GCTA ( s ) = ~m
=
=
=
c ( s)
c( s)
200
200