Question

The transfer function of the following signal flow graph is \( \frac{x_3(s)}{x_1(s)} \):

• A. \(\frac{(a - b)(c + d)}{1 - ab}\)
• B. \(\frac{ac + bc}{1 - cd}\)
• C. \(\frac{ac + ad}{1 - cd}\)
• D. \(\frac{(a - b)(c + d)}{1 - cd}\)

Hint:

Apply Mason’s Gain Formula: Find forward paths and loop gains carefully.

Answer 1

The Correct Option is C

Using Mason's Gain Formula: \[ T = \frac{\sum \text{(Gain of each forward path)} \times \Delta_{\text{path}}}{\Delta} \] There is one forward path: \( x_1 \xrightarrow{a} x_2 \xrightarrow{c} x_3 \) and \( x_1 \xrightarrow{b} x_2 \xrightarrow{d} x_3 \). So total forward gain: \( ac + ad \) Since loops are not touching the path, overall: \[ T = \frac{ac + ad}{1 - cd} \]