1234567898
8.1
a)
For step response, M s τD s +1 τ s +1 Ya′ ( s ) = K c D U ′( s ) = K c M s (ατ D s + 1) ατ D s + 1
input is u ′(t ) = M
Ya′ (s ) =
U ′( s ) =
,
K c Mτ D KcM + ατ D s + 1 s (ατ D s + 1)
Taking inverse Laplace transform y ′a (t ) =
K c M −t /( ατ D ) e + K c M (1 − e −t /( ατ D ) ) α
As α →0 ∞
e − t /( ατ D ) dt + K c M α t =0
y a′ (t ) = K c Mδ(t ) ∫
y a′ (t ) = K c Mδ(t )τ D
+ KcM
Ideal response, KM τ s + 1 Yi′( s ) = Gi ( s )U ′( s ) = K c M D = KcMτD + c s s
yi′ (t ) = K c Mτ D δ(t ) + K c M Hence y a′ (t ) → y i′ (t ) as α → 0 For ramp response, input is u ′(t ) = Mt
,
U ′( s ) =
M s2
Solution Manual for Process Dynamics and Control, 2nd edition, Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp.
8-1
τDs +1 τ s +1 Ya′ ( s ) = K c D U ′( s ) = K c M 2 ατ D s + 1 s (ατ D s + 1) Ya′ (s ) =
K c Mτ D K M + 2 c s (ατ D s + 1) s (ατ D s + 1) − ατ D 1 1 ατ D (ατ D ) 2 K M = K c Mτ D − + + + c s 2 ατ D s + 1 s ατ D s + 1 s
Taking inverse Laplace transform
[
]
[
y a′ (t ) = K c Mτ D 1 − e − t /( ατ D ) + K c M t + ατ D (e − t /( ατ D ) − 1)
]
As α →0 y a′ (t ) = K c Mτ D + K c Mt Ideal response, K c Mτ D K c M τ s + 1 Yi′(s ) = K c M D 2 = + 2 s s s
yi′ (t ) = K c Mτ D
+ K c Mt
Hence y a′ (t ) → y i′ (t ) as α → 0 b)
It may be difficult to obtain an accurate estimate of the derivative for use in the ideal transfer function.
c)
Yes. The ideal transfer function amplifies the noise in the measurement by taking its derivative. The approximate transfer function reduces this amplification by filtering the measurement.
8.2
a)
K1 K + K 2 τ1 s + K 2 P ′( s ) = + K2 = 1 E ( s ) τ1 s + 1 τ1 s + 1
8-2
K 2 τ1 K + K s + 1 2 = ( K1 + K 2 ) 1 τ1 s + 1
b)
Kc = K1 + K2
K2 = Kc − K1
→
τ1 = ατ D τD =
K 2 τ1 K ατ = 2 D K1 + K 2 K1 + K 2
or
1=
K 2α K1 + K 2
K1 + K 2 = K 2 α K 1 = K 2 α − K 2 = K 2 (α − 1) Substituting, K 1 = ( K c − K 1 )(α − 1) = (α − 1) K c − (α − 1) K 1 Then, α −1 K1 = K c α
c)
If Kc = 3
, τD = 2
K1 =
,
α = 0.1
− 0 .9 × 3 = −27 0 .1
K 2 = 3 − (−27) = 30 τ1 = 0.1 × 2 = 0.2 Hence K1 + K2 = -27 + 30 = 3
K 2 τ1 30 × 0.2 = =2 K1 + K 2 3 2s + 1 Gc ( s ) = 3 0 .2 s + 1
8-3
then,
8.3
a)
From Eq. 8-14, the parallel form of the PID controller is : 1 Gi ( s ) = K c′ 1 + + τ′D s τ′I s From Eq. 8-15, for α →0, the series form of the PID controller is:
1 Ga ( s ) = K c 1 + [τ D s + 1] τI s τ 1 = K c 1 + D + + τ D s τI τI s τ D τDs 1 = K c 1 + 1+ + τI τ τ 1 + D τ I s 1 + D τI τI
Comparing Ga(s) with Gi(s)
τ K c′ = K c 1 + D τI τ τ′I = τ I 1 + D τI τ′D =
b)
τD τ 1+ D τI
τ Since 1 + D τI K c ≤ K c′ ,
≥ 1 for all τD, τI, therefore τ I ≤ τ′I and τ D ≥ τ′D
c)
For Kc = 4, τI=10 min , τD =2 min
d)
K c′ = 4.8 , τ′I = 12 min , τ′D = 1.67 min Considering only first-order effects, a non-zero α will dampen all responses, making them slower.
8-4
8.4
Note that parts a), d), and e) require material from Chapter 9 to work. a)
System I (air-to-open valve) : Kv is positive. System II (air-to-close valve) : Kv is negative.
b)
System I : Flowrate too high → need to close valve → decrease controller output → reverse acting System II: Flow rate too high → need to close valve → increase controller output → direct acting.
c)
System I : Kc is positive System II : Kc is negative
d) System I : System II :
Kc
Kv
+ −
+ −
Kp Km + +
+ +
Kc and Kv must have same signs e)
Any negative gain must have a counterpart that "cancels" its effect. Thus, the rule: # of negative gains to have negative feedback = 0 , 2 or 4. # of negative gains to have positive feedback = 1 or 3.
8.5
a)
From Eqs. 8-1 and 8-2,
[
p(t ) = p + K c y sp (t ) − y m (t )
]
(1)
The liquid-level transmitter characteristic is ym(t) = KT h(t)
(2)
where h is the liquid level KT > 0 is the gain of the direct acting transmitter.
8-5
The control-valve characteristic is q(t) = Kvp(t)
(3)
where q is the manipulated flow rate Kv is the gain of the control valve. From Eqs. 1, 2, and 3
[
]
[
q(t ) − q = K v p (t ) − p = K V K c y sp (t ) − K T h(t ) KV K c =
]
q (t ) − q y sp − K T h(t )
For inflow manipulation configuration, q(t)> q when ysp(t)>KTh(t). Hence KvKc > 0 then for "air-to-open" valve (Kv>0), Kc>0 : and for "air-to-close" valve (Kv<0), Kc<0 :
reverse acting controller direct acting controller
For outflow manipulation configuration, KvKc <0 then for "air-to-open" valve, Kc<0 : and for "air-to-close" valve, Kc>0 : b)
direct acting controller reverse acting controller
See part(a) above
8.6
For PI control t 1 p (t ) = p + K c e(t ) + ∫ e(t*)dt * τI 0 t 1 p ′(t ) = K c e(t ) + ∫ e(t*)dt * τI 0
Since e(t) = ysp – ym
and
ym= 2
8-6
Then e(t)= -2 t 1 2 p ′(t ) = K c − 2 + ∫ (−2)dt * = K c − 2 − τI 0 τI
t
Initial response = − 2 Kc Slope of early response = − − 2 Kc = 6
−
2K c = 1.2 min-1 τI
2K c τI
→ Kc = -3 → τI = 5 min
8.7
a)
To include a process noise filter within a PI controller, it would be placed in the feedback path
b)
1 K c 1 + τI s
1 1 f s 23
c)
The TF between controller output P ′(s ) and feedback signal Ym(s) would be
8-7
P ′( s ) − K c (τ I s + 1) = Ym ( s ) τ I s (τ f s + 1) M s
For Ym ( s ) =
P ′( s ) =
The
Negative sign comes from comparator
− KcM τI
τI s +1 − KcM 2 = τI s (τ f s + 1)
A B C 2 + + s τ f s + 1 s
C term gives rise to an exponential. τ f s +1
To see the details of the response, we need to obtain B (= τI - τf) and A(=1) by partial fraction expansion. The response, shown for a negative change in Ym, would be Slope = KcM/τI "Ideal" PI
y Filtered PI -KcM
-KcM(1-τf/τI)
time
d)
τf
→ 0 , the two responses become the same. τI If the measured level signal is quite noisy, then these changes might still be large enough to cause the controller output to jump around even after filtering. Note that as
One way to make the digital filter more effective is to filter the process output at a higher sampling rate (e.g., 0.1 sec) while implementing the controller algorithm at the slower rate (e.g., 1 sec). A well-designed digital computer system will do this, thus eliminating the need for analog (continuous) filtering. 8-8
8.8 a)
From inspection of Eq. 8-26, the derivative kick = K c
b)
Proportional kick = K c ∆r
c)
e1 = e2 = e3 = …. = ek-2 = ek-1 = 0
τD ∆r ∆t
ek = ek+1 = ek+2 = …= ∆r p k −1 = p
τ ∆t p k = p + K c ∆r + ∆r + D ∆r τI ∆t ∆t p k +i = p + K c ∆r + (1 + i ) ∆r , τI
Kc
pk
τD ∆r ∆t
K c ∆r
Kc
p
k-1
c)
i = 1, 2, …
k
k+1
∆t ∆r τI
k+2
k+3
To eliminate derivative kick, replace (ek – ek-1) in Eq. 8-26 by (yk-yk-1).
8-9
8.9
a)
The digital velocity P algorithm is obtained by setting 1/τI = τD = 0 in Eq. 8-28 as
∆pk = Kc(ek – ek-1)
[
]
= K c ( y sp − y k ) − ( y sp − y k −1 ) = K c [ y k −1 − y k ]
The digital velocity PD algorithm is obtained by setting 1/τI = 0 in Eq. 828 as τ ∆pk = Kc [(ek – ek-1) + D (ek – 2ek-1 + ek-2)] ∆t τ = Kc [ (-yk + yk-1) + D (-yk – 2yk-1 + yk-2) ] ∆t In both cases, ∆pk does not depend on y sp . b)
c)
For both these algorithms ∆pk = 0 if yk-2 = yk-1 = yk. Hence steady state is reached with a value of y that is independent of the value of y sp . Use of these algorithms is inadvisable if offset is a concern. ∆t ( y sp − y k ) . τI Thus, at steady state when ∆pk = 0 and yk-2 = yk-1 = yk , yk = y sp and the offset problem is eliminated.
If the integral mode is present, then ∆pk contains the term Kc
8.10
a)
τDs P ′( s ) 1 = K c 1 + + E (s) τ I s ατ D s + 1 = Kc
( τ I s(ατ D s + 1) + ατD s + 1 + τD sτ I s ) τ I s (ατ D s + 1)
1 + (τ I + ατ D ) s + (1 + α)τ I τ D s 2 = Kc τ I s (ατ D s + 1)
8-10
Cross- multiplying
(
)
(ατ I τ D s 2 + τ I s ) P ′( s ) = K c 1 + (τ I + ατ D ) s + (1 + α )τ I τ D s 2 E ( s )
ατ I τ D
b)
d 2 e(t ) d 2 p ′(t ) dp ′(t ) de(t ) ( 1 ) + τ = K e ( t ) + ( τ + ατ ) + + α τ τ I c I D I D dt dt 2 dt 2 dt
τ s + 1 τ D s P ′( s ) = K c I E (s) τ I s ατ D s + 1 Cross-multiplying τ I s 2 (ατ D s + 1) P ′( s ) = K c ((τ I s + 1)(τ D s + 1) ) E ( s )
ατ I τ D c)
d 2 e(t ) d 2 p ′(t ) dp ′(t ) de(t ) + τ = K e ( t ) + ( τ + τ ) + τ τ I c I D I D dt dt dt 2 dt 2
We need to choose parameters in order to simulate: e.g.,
Kc = 2 ,
τI = 3
, τ D = 0.5
,
α = 0 .1
,
M=1
By using Simulink-MATLAB Step Response
22 Parallel PID with a derivative filter Series PID with a derivative filter
20 18 16 14
p'(t)
12 10 8 6 4 2 0
2
4
Time
6
8
10
Figure S8.10. Step responses for both parallel and series PID controllers with derivative filter.
8-11
8.11
a)
τ s +1 P ′( s ) (τ D s + 1) = K c I E (s) τI s τ I s P ′( s ) = K c ((τ I s + 1)(τ D s + 1) ) E ( s )
d 2 e(t ) dp′(t ) K c de(t ) + τI τD = e(t ) + (τ I + τ D ) dt dt τI dt 2 b)
With the derivative mode active, an impulse response will occur at t = 0. Afterwards, for a unit step change in e(t), the response will be a ramp with slope = K c (τ I + τ D ) / τ I and intercept = K c / τ I for t > 0 .
Impulse at t=0
p' slope =
Kc τI
t
8-12
K c (τ I + τ D ) τI
