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Problem 1
a) The transfer function of this process can be expressed as the product of three first order lag transfer functions. The AR and phase angles of a general 1st^ order lag are:
2 2
K
AR
τ ω +
and φ = tan −^1 ( −τω) (S1.1)
Thus, applying the principle of superposition we get:
2 2 2
AR
ω + ω + ω +
(S1.2)
φ = tan −^1 ( 8− ω + ) tan −^1 ( 2− ω + ) tan −^1 ( −ω) (S1.3)
b) Asymptotically as w goes to infinity, the AR is approximated by
3 1 1 AR 8 2
ω ω ω
(S1.4)
while for ω going to zero, AR goes to 3. Thus, the corner frequency will be obtained by solving the equations
3 1 1 3 0. 8 2
= → ω = ω ω ω
(S1.5)
Taking logarithms in the asymptotic expression for AR, the asymptote slope is –3.
c) The Bode plots are obtained computationally (i.e. give an array of values for ω and find the corresponding phase angle and amplitude ratios from the above formulae). They are shown in figure 1.
10 -
10 -
10 -
10 -
10 -
10 -
10 -
100
101
0.001 0.01 0.1 1 10 100
AR
w
0
0.001 0.01 0.1 1 10 100
φ
w Problem 1: Bode plots
Problem 2
a) The transfer function of the first process can be viewed as a product of 3 transfer functions: one 1st^ order lead and two 1st^ order lags. The transfer function of the second process can be viewed as a product of 3 functions: two 1 st^ order lags and one transfer function with a positive zero. The AR and phase angles of a general 1st^ order lag are:
2 2
K
AR
τ ω +
and φ = tan −^1 ( −τω) (S2.1)
The AR and phase angles of a general 1st^ order lead (G(s)=τs+1) are:
AR = τ ω +^2 2 1 and φ = tan −^1 ( τω) (S2.2)
The AR and phase angles of a general process with one positive zero (G(s)=-τs+1) are:
AR = τ ω +^2 2 1 and φ = tan −^1 ( −τω) (S2.3)
Applying the principle of superposition to the first process one obtains:
2 2 2 2 2 2
AR 10 1
= ω + ω + ω +
(S2.4)
φ = tan −^1 ( 0.3− ω +) tan −^1 ( 0.3− ω +) tan −^1 (10 ω) (S2.5)
Applying the principle of superposition to the second process we get:
2 2 2 2 2 2
AR 10 1
= ω + ω + ω +
(S2.6)
φ = tan −^1 ( 0.3− ω +) tan −^1 ( 0.3− ω +) tan −^1 ( 10− ω) (S2.7)
b) In both cases the AR is the same. As ω goes to zero, AR goes to 1. As ω goes to
infinity AR goes to
AR 10
= ω = ω ω ω
. From this we see that the asymptote
slope is –1 in a log-log plot, while the corner frequency is obtained by solving the equation:
1 1 1000 1000 AR 1 10 0.3 0.3 9 9
= = ω = → ω = ω ω ω
(S2.8)
c) The Bode plots are obtained computationally (i.e. give an array of values for w and find the corresponding phase angle and amplitude ratios from the above formulae). They are illustrated in figure 2.
Solving (S3.3) for ω provides the crossover frequency ωCO:
ωCO = 0.36733 rad / min (S3.5)
Substituting (S3.5) into (S3.3) provides the AR which at the crossover frequency:
AR( ωCO ) = 0.2629 (S3.6)
Thus, the ultimate period and ultimate gain are respectively
u CO
P 17.
π = = ω
min/cycle (S3.7)
and:
u CO
K 3.
AR( )
ω
(S3.8)
Therefore, according to the Ziegler-Nichols controller tuning technique, the settings for the PI controller are:
Kc=Ku /2.2=1.729 and τI=Pu /1.2=14.254 (S3.9)
while for the PID controller they are:
Kc=Ku /1.7=2.2375 and τI=Pu /2=8.5525 and τD=Pu /8=2.138 (S3.10)
Implementing these two controllers using Simulink (see the block diagram in figure 3) for a unit step change in the set-point yields the following ISE, IAE and ITAE indices:
Figure 3: Simulink block diagram
ISE IAE ITAE PI 7.315 10.74 113. PID 6.373 9.542 84.
