Fundamentals of Control Systems

We have already tried our best to solve this problem. But we can’t calculate the equivalent transfer function of that system by using block diagram algebra. So we would like to ask your permission to solve this problem by using the traditional method. 

 After some transformations, we obtain the following block diagram

 

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Homework report 01
Course: Fundamentals of Control Systems
Class: DD12KST1
Date: 08/09/2014
Group 12:
	Student	Student’s code
Phan Trùng Dương	4120 0648
Nguyễn Trọng Tuấn	4120 4295
Nguyễn Duy Vĩnh	4120 4579
Hồ Đắc Thuận	4120 3697	
Problem 1.
4
2
1
3
Moving the pickoff point ƒ behind the block G2(s):
GA
 GA(s) = feedback loop (G2(s), G3(s)):
GB
GB(s) = cascade (G1(s), GA(s)):
G(s) = feedback loop (GB(s), ())
The equivalent transfer function of the system:

†
„
ƒ
‚
Moving the pickoff point ‚ ahead the block G1(s).
Moving the summing point „ ahead the block G2(s) then interchanging the summing points „ and ƒ.
ƒ
„

†
GA
GB
GA(s) = feedback loop (G2(s),G4(s)): 
The equivalent transfer function of the system:
We have already tried our best to solve this problem. But we can’t calculate the equivalent transfer function of that system by using block diagram algebra. So we would like to ask your permission to solve this problem by using the traditional method. J
	After some transformations, we obtain the following block diagram:
The equivalent transfer function of the system:
Problem 2.
¬ :
 GR(s) = feedback loop (GA(s), G3(s)): 
The equivalent transfer function:
¬ 
Interchanging the summing points ƒ and „:
GA(s) = feedback loop (G2(s), G3(s).G5(s)): 
Moving the summing point „ ahead the block G1(s) then interchanging it with the summing point ‚:
GC(s) = cascade (G1(s), GA(s)): 
	GD(s) = feedback loop (GC(s), G3(s)): 
	GN(s) = cascade (-GB(s), GD(s)): 
	The equivalent transfer function of the system:
Problem 3.
- G4
Figure 1.
R(s)
1
G1
1
G2
1
Y(s)
- G3
- G5
G5
Figure 2.
- G4
1
1
1
1
1
G3
G2
G1
Y(s)
R(s)
- G4
Figure 3. 
G5
G2
G1
1
1
1
1
Y(s)
R(s)
- G3
- G6
Figure 4.
G5
¬ :
1
1
1
G2
G1
Y(s)
R(s)
- G3
- G3
¬ 
G2
1
G1
1
1
R(s)
Y(s)
- G4
1
- G3 G5

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