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Cleveland State University
MCE441: Intr. Linear Control Systems
Lecture 8: Final Value Theorem
Block Diagram Algebra
Prof. Richter
The Final Value Theorem
⊲The Final ValueTheorem
Block DiagramAlgebra
Block DiagramAlgebra
Block DiagramManipulations
Example
Solution
Solution
Example
Solution
Solution
2 / 11
� Provides information about the value of f(t) at steady state
(t = ∞)� The steady-state may not exist: the poles of sF (s) must lie
on the open left half of C (details later)� If limt→∞ f(t) exists, then
limt→∞ f(t) = lims→0 sF (s)
� Example: For F (s) = 1s(s+1) , we see that the only pole of
sF (s) lies on the left half plane. So
limt→∞
f(t) = lims→0
sF (s) = 1
� Check by inversion of F (s)
Block Diagram Algebra
The Final ValueTheorem
⊲Block DiagramAlgebra
Block DiagramAlgebra
Block DiagramManipulations
Example
Solution
Solution
Example
Solution
Solution
3 / 11
� The previous result allows us to multiply TF and input to getoutput in the Laplace domain
� Block diagrams are convenient representations of controlsystem structure
� The following basic operations are allowed in block diagramalgebra:1. Addition between signals. Result: another signal.2. Multiplication between a signal and a system. Result:another signal.
� Systems (TFs) are represented by blocks.� Signals are represented by arrows.
Block Diagram Algebra
The Final ValueTheoremBlock DiagramAlgebra
⊲Block DiagramAlgebra
Block DiagramManipulations
Example
Solution
Solution
Example
Solution
Solution
4 / 11
� A Summing junction or summing point is the symbol thatrepresents signal addition
� A branch point indicates that the output from a block isapplied to other blocks or summing junctions.
G(s)E(s) Y (s) = G(s)E(s)R(s)
+−
Summing point
Branch point
E(s) = R(s) − Y (s)
Y (s)
Block Diagram Manipulations
The Final ValueTheoremBlock DiagramAlgebra
Block DiagramAlgebra
⊲Block DiagramManipulations
Example
Solution
Solution
Example
Solution
Solution
5 / 11
Example
The Final ValueTheoremBlock DiagramAlgebra
Block DiagramAlgebra
Block DiagramManipulations
⊲ Example
Solution
Solution
Example
Solution
Solution
6 / 11
G1 G2
G4
G3
R(s) Y (s)
+−
++
+
Solution
The Final ValueTheoremBlock DiagramAlgebra
Block DiagramAlgebra
Block DiagramManipulations
Example
⊲ Solution
Solution
Example
Solution
Solution
7 / 11
Solution
The Final ValueTheoremBlock DiagramAlgebra
Block DiagramAlgebra
Block DiagramManipulations
Example
Solution
⊲ Solution
Example
Solution
Solution
8 / 11
Check that the required TF is(G1+1)G2
(G1+1)G2G4+1−G2G3
Example
The Final ValueTheoremBlock DiagramAlgebra
Block DiagramAlgebra
Block DiagramManipulations
Example
Solution
Solution
⊲ Example
Solution
Solution
9 / 11
G1 G3
G5 G4
R(s) Y (s)
+−
+−
G2
++
Solution
The Final ValueTheoremBlock DiagramAlgebra
Block DiagramAlgebra
Block DiagramManipulations
Example
Solution
Solution
Example
⊲ Solution
Solution
10 / 11
Solution
The Final ValueTheoremBlock DiagramAlgebra
Block DiagramAlgebra
Block DiagramManipulations
Example
Solution
Solution
Example
Solution
⊲ Solution
11 / 11
Check that the required TF isG1H1
1+G5(G1H1G4+G2), where H1 =
G31+G3
.