Le Zhang lfz062@mail.usask.ca
Rm. 2B60

• Study the open-loop performance of an electro-hydraulic position control system.
• Study the use of Bode plots to create the open and closed loop transfer functions for the electro-hydraulic position control system using experimental data.

• Review
– Study the electro-hydraulic position control system
– Frequency Response
– Common Bode Plots
• Experiment 1: Use function generator to apply sinusoidal input, use output to create
Bode Plots and Transfer Functions.

• 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.

• Frequency Response
• Bode Plots
• Experimental System

• Frequency Response
• Bode Plots
• Experimental System

Figure 1: General SISO System e i

A sin x
 

B sin
  t
  
Are these the same?

Frequency Response (Con’t)
Figure 2: Input-Output response relationship

But what if we want to study the system in the frequency domain?

Frequency Response (Con’t)
Laplace Transform:
G ( s )

X ( s )
E i
( s )
Figure 3: Plant diagram.
How do we change from continuous to frequency domain?
Magnitude Frequency Response:
Phase Frequency Response:
 s → jω
G ( j

)

B
A
(

)
 
(

)
  i
(

)
 

• If the ω changes, the magnitude of output signal (B) and the phase ( φ)of output signal will change.
• As a consequence, the value of |G(j ω )| and ∠ G(j ω) will change.

• Frequency Response
• Bode Plots
• Experimental System

M dB
0
20 log|K |
Bode Plots: Case 1
G
 

K
M

20 log G
 

20 log K rad/s
0
0 degrees rad/s
  tan

1
 

0


dB
20
0
-20
Bode Plots: Case 2
G

1 s
-20 dB/decade M

20 log
1
1 j

20 log
1
1

0
0.1
rad/s
1 10
M
M

20 log
1
0 .
1 j

20 log
1
0 .
1

20

20 log
1
10 j

20 log
1
10
 
20
Phase angle
0
  tan

1  tan

1  
90
-90 degrees

20 log|K|
0
1 rad/sec
Bode Plots: Case 3
G
 

K s
-20 dB/decade
M

G ( j

)


M
 

M
1 s
0
-90 degrees
    
  s
 
90

Bode Plots: Case 4
M

G ( 0 )


20 log
1
0
 n j

1

20 log
1
1

0
M

G ( j 0 .
1
 n
)


20 log
0 .
1

1 n
 n j

1

20 log
1
0 .
1 2

1 2

0
M

G ( j
 n
)


20 log

 n n j
1

1

20 log
1
1 2

1 2
 
3 .
01 dB

0
M

G ( j 10
 n
)


20 log
10

 n n
1 j

1

20 log
1
10 2

1 2
 
20 dB
G
 s
 n
1

1
 n
-20 dB/decade

Bode Plots: Case 4



 j 0 .
1
 n
 


0
 n


1


0
 

 j

 n

1



0

90
 
90

 

 j 0 .
1
 n
 n

1


 
5 .
71
 
0

  n
 

 j
 n
 n

1


 
45


 j 10
 n
  
 

 j 10
 n
 n

1


 
84 .
29
  
90

-45
-90
G
 s
 n
1

1
0 .
1
 n
 n
10
 n

Bode Plots: Case 5
G
 s
2 

2
 n n
2 s
  n
2
Can we simplify?
What does ξ really mean?
Can we let ξ equal some value to help simplify the equation?
ξ = 1
G
 s
2 
 n
2
 n
2 s
  n
2
 s
2 
 n
2
2
 n s
  n
2

 n
2
 s
  n

2

1 s
 n

1 
2

Bode Plots: Case 5
The equation from the last slide may be written as:
G

1 s
 n

1
 2



 s
 n
1

1






  s
 n
1

1



 n
-90
-180
0 .
1
 n
 n
10
 n
-40 dB/decade
All we need to do is double both of the plots from Case 4

• Frequency Response
• Bode Plots
• Experimental System

Function
Generator
Amplifier Valve Actuator Transducer
Recorder

Function
Generator
Experimental System (Con’t)
Load Amplitude Valve
G(s)
Recorder
Transducer
G ( s )

X mm
( s )
E i
( s )
X mm

X v
K v

X v
0 .
078 V / mm
G ( s )

E i
X v

K v

• Obtain transducer sensitivity, K v
.
• Apply sinusoidal input and obtain output.
• Create Bode plots from output data.
• Obtain open-loop TF.
• Predict closed-loop TF.

• Setup spectrum analyzer.
• Obtain data and import to Excel.
• Obtain Bode Plots.
• Obtain open-loop TF.
• Predict closed-loop TF.

• 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