Question

A control system is shown in the Figure.

Which option represents the correct transfer function of the system?

• A. \(\dfrac{1}{(s+4)^2}\)
• B. \(\dfrac{1}{(s+4)}\)
• C. \(\dfrac{2}{(s+4)}\)
• D. \(\dfrac{1}{(s^2+8s+17)}\)

Hint:

Check exactly where the second block's input line is tapped from. It shares its input with the first block, so its output equals \(C(s)\), which makes the inner feedback cancel to zero.

Answer 1

The Correct Option is B

Step 1: Draw the signal flow graph nodes.
Call the reference $R$, the error node $E$, the forward output $C$, the second block's output $D$, and the inner summing node output $F$, which feeds back to the error node.

Step 2: Write the branch gains.
$E$ to $C$: gain $\dfrac{1}{s+4}$. $E$ to $D$: gain $\dfrac{1}{s+4}$, since the second block taps the same node $E$, not $C$. $C$ and $D$ combine at the inner summer as $F=C-D$. Finally $E=R-F$.

Step 3: Substitute to collapse the inner loop.
Since both $C$ and $D$ equal $E\cdot\dfrac{1}{s+4}$, we get $F=E\cdot\dfrac{1}{s+4}-E\cdot\dfrac{1}{s+4}=0$ identically, for any $E$.

Step 4: Reduce the graph.
With $F\equiv0$, the feedback branch carries no signal at all, so effectively there is no loop; the loop gain of this inner loop is zero and Mason's gain formula reduces to just the forward path from $R$ to $E$ to $C$.

Step 5: Write the forward path gain.
$E=R$, since $F=0$, and $C=E\cdot\dfrac{1}{s+4}=\dfrac{R}{s+4}$.

Step 6: State the transfer function.
\[ \frac{C(s)}{R(s)}=\frac{1}{s+4} \] \[ \boxed{\dfrac{1}{s+4}} \]