C2: Hydraulic Position Control
System
Yang Lin
yal596@mail.usask.ca
Rm. 1B15
Objectives
• Study the use of Bode plots to determine
the open-loop and closed-loop transfer
functions of an electro-hydraulic position
control system from experimental
sinusoidal frequency response data
Procedure Outline
• Review
– Frequency Response
– Common Bode Plots
– Study the electro-hydraulic position control
system
• Experiment 1: Use function generator to
apply sinusoidal input, use output to create
Bode Plots and Transfer Functions.
Procedure Outline
• Experiment 2: Use signal analyzer to
create Bode Plots, using plots create
Transfer Functions.
• Experiment 3: Use signal analyzer to
measure close loop response and
compare this to the theoretical response.
Review Outline
• Frequency Response
• Bode Plots
• Experimental System
Review Outline
• Frequency Response
• Bode Plots
• Experimental System
Frequency Response
Figure 1: General SISO System
But what if we want to study the system in the frequency domain?
What action must the input and output create to measure frequency?
ei t   A sint 
xt   B sin t   
Are these the same?
Frequency Response (Con’t)
Figure 2: Input-Output response relationship
Will the Amplitude of the input and output be the same?
Will the frequency be the same?
Frequency Response (Con’t)
Laplace Transform:
G( s) 
X ( s)
Ei ( s)
Figure 3: Plant diagram.
How do we change from continuous to frequency domain?
s → jω
Magnitude Frequency Response:
Phase Frequency Response:
G( j) 
G( j )  
B
A
Review Outline
• Frequency Response
• Bode Plots
• Experimental System
Bode Plots
To find the Magnitude (M), substitute in different values of ω and solve the
following:
M  20 logG j 
To find the Phase Shift (φ), substitute in different values of ω and solve the
following:
  tan1 G( j)
But what if the transfer function has imaginary and real terms, or perhaps a
numerator and denominator?
Bode Plot: Magnitude
Numerator and Denominator Present:
 Num( j ) 

M  20 logG ( j )   20 log
 Den( j ) 
M  20 logNum( j )   20 logDen( j ) 
Real and Imaginary Terms Present:
M  20 log G ( j )  20 log Im j  Re

M  20 log Im2  Re 2

Bode Plot: Phase
Numerator and Denominator Present:
 G( j)  NUM  DEN
Real and Imaginary Components Present:
Im
Example:
(2,1)
G(j2) = 2j + 1
φ
Re
2
  tan    63.43
1
1
Bode Plots: Case 1
Gs   K
20 log|K|
M
dB
M  20 log G j   20 log K
0
rad/s
0 degrees
0
rad/s
  tan1 0  0
Bode Plots: Case 2
1
G s  
s
20
-20 dB/decade
dB
0
0.1
rad/s
1
10
M 1  20 log
1
1
 20 log   0
1j
1
M 0.1  20 log
M 10  20 log
-20
Phase
angle
1
 1 
 20 log
  20
0.1 j
0
.
1


1
1
 20 log   20
10 j
 10 
  tan1 0  tan1   90
0
-90 degrees
Bode Plots: Case 3
G s  
K
s
-20 dB/decade
20 log|K|
0
1 rad/sec
1
M G( j )   M K   M  
s
 
   K    1 s  90
0
-90 degrees
Bode Plots: Case 4
G s  
1
1
M G (0)   20 log
 20 log  0
0j
1
1
1
0.1n j
n
M G ( jn )   20 log
1
n j
n
M G ( j10n )   20 log
1
n
0.1  1
2
1
1
10n j
1
 20 log
 20 log
1
12  12
 20 log
1
s
n
n
M G ( j 0.1n )   20 log
1
2
0
 3.01dB  0
1
102  12
1
n
 20dB
-20 dB/decade
Bode Plots: Case 4
 0

 1  0
 n 
 0   1   
G s  
 j

 1  0  90  90
 n

  j    1   
 j 0.1n
  j 0.1n    1   
 n
1
s
n
1

 1  5.71  0

 jn

 1  45
 n

  jn    1   
 j10n 
  j10n    1   
 1  84.29  90

n


0.1n
n
-45
-90
10n
Bode Plots: Case 5
n 2
Gs   2
2
s  2n s  n
Can we simplify?
What does ξ really mean?
Can we let ξ equal some value to help simplify the equation?
ξ=1
n 2
n 2
n2
1
Gs   2
 2


2
2
2
s  n   s  12
s  2n s  n
s  2n s  n
 

n


Bode Plots: Case 5
The equation from the last slide may be written as:



 1  1 
G s  



2
 s  1  s   1  s   1 
 
 
n
n


n


1
n
0.1n
n
-90
-180
-40 dB/decade
10n
All we need to do is double both of the plots from Case 4
Bode Plots: Case 6
G s  
100
ss  10
Let’s simplify the above equation into forms we have studied,


100
 1  1 
10
G s  
 
s s  10  s  s  1 
 10 
Review Outline
• Frequency Response
• Bode Plots
• Experimental System
Experimental System
Function
Generator
Amplifier
Valve
Recorder
Load
Transducer
Experimental System (Con’t)
Function
Amplitude
Generator
Valve
Load
Transducer
G(s)
Recorder
G( s) 
X mm 
X mm ( s)
Ei ( s)
Xv
Xv

K v 0.078V / m m
Xv
G( s) 
Ei  K v
Experiment 1:
•
•
•
•
•
Obtain transducer sensitivity, Kv.
Apply sinusoidal input and obtain output.
Create Bode plots from output data.
Obtain open-loop TF.
Predict closed-loop TF.
Experiment 1 (Con’t)
Open-Loop TF:
What form will our TF take? (max slope, max angle)
K
G (s) 
s
How can we find K?
@ ω = 1,
G ( j )  20 log K
G  j 
K  10
20
Calculate K for our experimental data
Experiment 1 (Con’t)
Closed-Loop TF:
Ei(s)
G(s)
Kv(s)
What form will our TF take?
Gs 
T ( s) 
1  Gs K v
What is our Closed-Loop TF?
Experiment 2:
•
•
•
•
•
Setup FFT signal analyzer.
Obtain data and import to Excel.
Obtain Bode Plots.
Obtain open-loop TF.
Predict closed-loop TF.
Experiment 2 (Con’t)
Open-Loop TF
What is the final phase shift?
What is the maximum magnitude slope?
How do these two properties help describe the order of the system?
Kn2
G( s)  2
s s  2 n s  n2


We must find K, ωn, ζ
Experiment 2 (Con’t)
Finding ωn:
-Again, if we break up the assumed TF into its smallest components we get,
n2
1
G( s )  K   2
s s  2 n s  n2


@ ωn, <φ3 = -90
@ all ω, <φ2 = -90
@ all ω, <φ1 = 0
<G(jωn) = <φ1 + <φ2 + <φ3
Experiment 2 (Con’t)
0
-90
@ ωn, -90
<G(jωn) = <φ1 + <φ2 + <φ3
Therefore @ ωn the phase angle should be -180
Experiment 2 (Con’t)
Finding K:
-If we break up the assumed TF into its smallest components we get,
n2
1
G( s )  K   2
s s  2 n s  n2


@ ω < ωn, |M3| = 0 dB
@ ω = 1, |M2| = 0 dB
@ any ω, |M1| = 20 log |K|
|G(jω)| = |M1| + |M2| + |M3|
Experiment 2 (Con’t)
0
0 @ω=1
|G(j*1rad/s)| = |M1| + |M2| + |M3|
Therefore,
|G(j*1rad/s)| = |M1| = 20 log |K|
Experiment 2 (Con’t)
Finding ζ:
-We have two Bode plots to choose from to help solve for ζ, magnitude and
phase.
-In order to use either plot we must make our transfer function comparable
by changing to the frequency domain, ie. replace s→jω,
Kn2
G j  
 j  j 2  2n  j   n2


Notice we are still left with two unknowns, ζ and ω
But only one equation!
Experiment 2 (Con’t)
-Since we are trying to find ζ, we need to replace ω with a suitable value.
Replacing ω→ωn,
Kn2
G jn  
 jn  jn 2  2n  jn   n2

Kn2
G jn  
 jn 1n2  2 jn2  n2

Kn2
Kn2
G jn  

2
 jn  2 jn 1 2n3

G jn   
K
2 n





Experiment 2 (Con’t)
-Which Bode plot should we compare to?
Using the Magnitude plot, G  jn   20 log
 
K
2 n
K
2n 10
G  jn 
20
Experiment 2 (Con’t)
Closed-Loop TF:
Ei(s)
G(s)
Gs 
T ( s) 
1  Gs K v
Kv(s)
Due to no zeros in G(s), the closed-loop TF takes one of two forms:
K cl
T1 * s  

s
 s
 s
  1  1  1
 T1  T2  T3 
Must be all real poles
T2 * s  

K cln2
s  2
2
  1 s  2 n s  n
T 

1 real pole, others can be real or imaginary
Experiment 2 (Con’t)
-To find which form to use we must factor out T(s).
We have:
-1 real pole
-2 imaginary poles
Therefore we must find Kcl, T, ζ and ωn.
T2 * s  

K cln2
s  2
2
  1 s  2 n s  n
T 

Experiment 2 (Con’t)
Finding ωn and ζ:
-We can equate the poles from the second order system
-Using a calculator/Matlab the poles may be found, with the imaginary poles
being,
s  a  bj
Therefore,
s  a  jbs  a  jb  s2  2n s  n2


s 2  2as  a2  b2  s 2  2n s  n2
With,
a 2  b2  n2
and
2a  2 n
Finding Kcl:
-Equate the numerator of T2*(s) to the predicted numerator of T(s).
Experiment 2 (Con’t)
-Now we must observe the effects of closing the loop:
What happens to the poles?
What happens to the damping ratio?
What happens to the natural frequency?
Experiment 3:
• Close the loop in the system.
• Get Bode Plot from the signal analyzer
• Draw asymptotes derived from the
theoretical model onto the Bode Plot of the
experimental results
• Compare results