1.2 Algebra with Block Diagrams
Table 10-1 on page 495 gives several rules for simplifying block diagrams. The goal is to
find the open-loop transfer function C(s)/R(s) that is equivalent to a possibly complex block
diagram containing several transfer functions and loops. To see the process, consider the
following example:
R(s)
G1
C(s)
G2
H1
Example: The block diagram above gives a schematic layout for a controller. There is a feedforward component (the loop around G1) and two feedback loops. Find the overall transfer
function C(s)/R(s) for the system.
Solution: First, consider the two summation blocks at the center of the diagram. There appears to be no “rule” governing these in Table 10-1, but we might suppose that the order of
summation is unimportant. To see whether this is true, let us write expressions for the contents of each wire.
C
R(s)
E
G1
G1E
G1E - C
G1E - C + E
G2
C(s)
E
H1
1
Recall that the output of a transfer function block is the input multiplied by the transfer
function, or algebraically
𝑉𝑜𝑢𝑡 = 𝐺 ∙ 𝑉𝑖𝑛
Now reverse the order of summation
C
R(s)
E
G1
G1E
G1E + E
G1E - C + E
G2
C(s)
E
H1
As you can see, the result is the same. There is now a familiar-looking feedback loop around
G2 on the right, that we can reduce using Rule Number 5.
R(s)
E
G1
G1E
G1E + E
G2
1  G2
C(s)
E
H1
There is a feedforward loop around G1 on the left, for which no rule seems to serve. Again,
writing the contents above each wire, we see that the output of the summation block is
𝑉𝑜𝑢𝑡 = 𝐺1 𝐸 + 𝐸
= 𝐸(𝐺1 + 1)
Thus, we can replace the feedforward loop with an equivalent transfer function
2
R(s)
C(s)
G2
1  G2
1 + G1
H1
Since the two transfer functions on top are in series, the equivalent can be found through
multiplication
R(s)
C(s)
1  G1
G2
1  G2
H1
And finally, we can use Rule 5 to simplify the feedback loop
R(s)
1  G1
G2
1  G2
1  G1
1
G2 H1
1  G2
C(s)
Or equivalently,
R(s)
1  G1 G2
1  G2  1  G1 G2 H1
C(s)
3