Question

In the signal flow graph shown below, the transfer function is _______.

 

• A. 3.75
• B. -3
• C. 3
• D. -3.75

Hint:

In signal flow graphs, always calculate forward gain and loop gain separately. Then apply Mason’s rule properly.

Answer 1

The Correct Option is C

Step 1: Identify forward path gain.
The forward path from $R$ to $C$ is: \[ R \xrightarrow{3} \xrightarrow{2} \xrightarrow{1} C \Rightarrow \text{Total forward path gain} = 3 \times 2 \times 1 = 6 \] Step 2: Identify feedback loop.
There is one feedback loop: from output node $C$ back to the input with gain $-3$.
Step 3: Use Mason’s Gain Formula:
\[ T = \frac{P}{1 - L} \] Where:
$P = $ forward path gain = 6
$L = $ loop gain = $-1$ (because loop: $3 \times -1 = -3$ and then divided by 3 → net $-1$) \[ T = \frac{6}{1 - (-1)} = \frac{6}{2} = 3 \] Conclusion: $\boxed{3}$ is the transfer function.