How are these three related?
F(s)
Block
Diagram
+
1
M
-
-
1
s
1
s
Block Diagrams
X(s)
• A transfer function converts an input to
an output
Y(s)
U(s)
H(s) =
B
K
Transfer Function
State Variable Equations
1
X(s)
M
=
B
K
F(s)
2
s + s+
M
M
sX(s) = V(s)
B
K
1
sV(s) = − V(s) − X(s) + F(s)
M
M
M
Definitions - Block & Signal
U(s)
Y(s)
H(s)
A solid line (usually
with an arrow)
represents a signal or
variable
Y(s) = H(s)U(s)
• On a block diagram this is shown as
U(s)
Y(s)
H(s)
Definitions - Gain Block
X(s)
Y(s)
X(s)
K
Blocks contain transfer
functions (which transform
inputs to outputs)
Y(s)
6
When the transfer function
is a constant, the block is
often referred to as a gain
A gain can be either a constant parameter
(K) or a numerical value (6)
Definitions - Summer
X1(s)
X2(s)
X3(s)
+
-
Y(s) = +X1(s) -X2(s)+X3(s)
Z1(s)
Z1(s)
+
Signals (variables) can be added
(or subtracted) at a summer, which
is represented by the circle
Definitions - Pick-off
Note that there is
only one output
from a summer!
Z1(s)
Dot not always drawn!
Signals (variables) can be used or
accessed at any point along their “line”
Note that there is no loss or reduction in
the signal caused by the “pick off”
Addition is represented by a + symbol,
subtraction by the - symbol
1
Block Diagram Reduction
Series Blocks - Multiply Gains
Three primary situations for reducing
block diagrams to simpler forms:
–series blocks
–parallel blocks
–“feedback” blocks
Z(s)
1
M
Z(s)
Parallel Blocks - Add Gains
G1 (s)
+
G2 (s)
U(s)
- G1 (s) + G2 (s)
Z(s)
+
-
C(s)H(s)
Z(s) = −C(s )H(s) + R(s)
C(s) = G(s)Z(s)
G(s)
1
Ms
V(s)
+
G(s)
-
C(s)
-
H(s)
Y(s)
R(s)
Y(s)
Feedback Loop
R(s)
s
Feedback Loop
R(s)
U(s)
sV(s) 1 V(s)
C(s)
G(s)
1 + G(s)H(s)
Block Diagram Reduction
C(s)
H(s)
find G(s) =
F(s)
X(s)
F(s)
+
-
-
sV(s)
1
M
V(s)
1
s
1
s
X(s)
B
C(s)
G(s)
=
R(s) 1 + G(s)H(s)
K
2
Step #1
Step #2a
Re-draw the block diagram slightly,
Combine the two blocks in series,
Define Z = F(s) -BV(s) - KX(s)
Z
+
F(s)
-
1
Ms
-
V(s)
1
s
X(s)
F(s) +
Z
+
-
-
X(s)
Step #2c
Close the feedback loop,
Simplify the feedback loop,
1
Ms
1
1+
B
Ms
1
s
X(s)
F(s) +
1
Ms + B
-
K
K
Step #3a
Step #3b
Combine two blocks in series,
-
1
s
Feedback loop
Step #2b
F(s) +
X(s)
K
K
-
1
s
B
B
F(s) +
1
Ms
Simplifying the block,
 1  1 

 
 Ms + B   s 
X(s)
F(s) +
-
K
X(s)
1
Ms2 + Bs
K
Feedback loop
3
Step #4a
Step #4a
Close the feedback loop,
F(s)
Simplify the feedback loop,
1


 2

 Ms + Bs 
1


1+ 
 (K )
2
 Ms + Bs 
F(s)
X(s)
1
Ms + Bs + K
X(s)
2
From the definition of a block,
X(s)
1
=
F(s) Ms2 + Bs + K
Example #2
find
Example #2 - Step #1
Z(s)
U(s)
Combine G1 and G2 (parallel blocks)
G 1 (s)
U(s)
Z(s)
+
G3 (s)
U(s)
Z(s)
+
G1 (s) + G2 (s)
+
G3 (s)
G2 (s)
Combine in series
Example #2 - Step #2
Example #3
Combine the two parallel blocks
find
1
U(s)
(G1 (s) + G2 (s))G3 (s)
Z(s)
= 1 + (G1 (s) + G 2 (s))G 3 (s)
U(s)
+
Z(s)
Y(s)
U(s)
b1
U(s) +
-
-
1
a 2s
1
s
b0
+
+
Y(s)
a1
a0
4
Example #3 - Step #1
Re-draw slightly, then close
this feedback loop
+
U(s) -
1
s
b0
+
+
Y(s)
-
+
U(s) -
1
s
1
a 2s + a 1
a0
a0
Example #3 - Step #2
Note the definitions for Z(s),
need to move 1 pick-off
U(s) -
1
a 2s + a 1
b1
Y(s)
Cannot have a
“pick-off” in
in middle of
parallel path!
Z(s)
b1
1
Z(s)
1 s
b0
s
a0
+
Example #3 - Step #3
Z(s)
Z(s)
+
b0
Feedback loop
a1
+
Parallel path
Cannot have a “pick-off”
in middle of feedback loop!
b1
1
a 2s
+
Example #3 - Step #2a
Combine blocks in series
+
+
Y(s)
+
U(s) -
1
Z(s)
s
1
1
a 2s + a 1 Z(s) s
a0
Example #3 - Step #4
s
1
Z(s)
s
b0
b1
+
+
Y(s)
1
Z(s)
s
Example #3 - Step #5
Z(s)
Close feedback loop
+
U(s) -
1
a 2s2 + a1s
Combine blocks in series,
b1s
b0
+
+
1
a 2s2 + a1s + a 0
Y(s)
b1s + b 0
Y(s)
Combine parallel paths
a0
U(s)
Y(s)
b1s + b 0
=
U(s) a 2 s2 + a 1s + a 0
5
Example #3 - Step #2
Z(s)
Note the definitions for Z(s),
need to move 1 pick-off
1
a 2s + a 1
+
U(s) -
Example #3 - Step #3b
b1
1
Z(s)
1 s
b0
s
Z(s)
Remove the pick-off that
causes problems
+
+
Y(s)
1
a 2s + a 1
+
U(s) -
Z(s)
Z(s)
b1
1
Z(s)
1 s
b0
s
+
+
Y(s)
Move this one!
1
Z(s)
s
a0
a0
1
Z(s)
s
How can we still get this?
Example #3 - Step #3c
Z(s)
What do you do next?
1
a 2s + a 1
+
U(s) a0
Z(s)
s
a0
1
s
Z(s)
b1
1
Z(s)
1 s
b0
s
Z(s)
+
+
Y(s)
Add an integrator block!
6