Question

Determine the transfer function \(\dfrac{C(s)}{R(s)}\) of the system shown below.

• A. \[ \frac{G_{1}G_{2}G_{3}G_{4}} {(1+G_{1}G_{2}H_{1})(1+G_{3}G_{4}H_{2})+G_{2}G_{3}H_{3}} \]
• B. \[ \frac{G_{1}G_{4}} {(1+G_{1}G_{2}H_{1}H_{2})+(1+G_{2}G_{3}G_{4}H_{3})} \]
• C. \[ \frac{G_{1}G_{3}G_{4}} {(1+G_{1}G_{2}H_{1})(1+G_{3}G_{4})+(1+G_{2}G_{3}H_{2})} \]
• D. \[ \frac{G_{1}G_{2}G_{3}} {(1+G_{1}G_{2}H_{1})(1+G_{3}G_{4})+(G_{2}G_{3}H_{3}+G_{4})} \]

Hint:

For block diagrams containing multiple feedback loops: \[ T=\frac{\sum P_k\Delta_k}{\Delta} \] using Mason's Gain Formula is usually the fastest method. Always identify:
• Forward paths
• Individual loop gains
• Non-touching loops before calculating \(\Delta\).

Answer 1

The Correct Option is A

Concept: The transfer function of a complicated feedback network can be obtained using Mason's Gain Formula. \[ T=\frac{\sum P_k\Delta_k}{\Delta} \] where
• \(P_k\) = Forward path gain
• \(\Delta = 1-\text{(sum of loop gains)}+\text{(sum of products of non-touching loops)}\)
• \(\Delta_k\) = Cofactor associated with the \(k^{th}\) forward path

Step 1: Determine the forward path gain.
There is only one forward path from input \(R(s)\) to output \(C(s)\): \[ R(s)\rightarrow G_1\rightarrow G_2\rightarrow G_3\rightarrow G_4 \] Hence \[ P_1=G_1G_2G_3G_4. \]

Step 2: Identify individual feedback loops.
Loop through \(H_1\): \[ L_1=-G_1G_2H_1 \] Loop through \(H_2\): \[ L_2=-G_3G_4H_2 \] Loop through \(H_3\): \[ L_3=-G_2G_3H_3 \]

Step 3: Find non-touching loops.
The loops involving \(H_1\) and \(H_2\) do not touch each other. Therefore \[ L_1L_2 = (-G_1G_2H_1)(-G_3G_4H_2) = G_1G_2G_3G_4H_1H_2. \] No other pair of loops is non-touching.

Step 4: Compute \(\Delta\).
Using Mason's formula, \[ \Delta = 1-(L_1+L_2+L_3)+L_1L_2. \] Substituting, \[ \Delta = 1+G_1G_2H_1+G_3G_4H_2+G_2G_3H_3 +G_1G_2G_3G_4H_1H_2. \] Rearranging, \[ \Delta = (1+G_1G_2H_1)(1+G_3G_4H_2) + G_2G_3H_3. \]

Step 5: Determine \(\Delta_1\).
Since all loops touch the forward path, \[ \Delta_1=1. \]

Step 6: Apply Mason's Gain Formula.
\[ \frac{C(s)}{R(s)} = \frac{P_1\Delta_1}{\Delta} = \frac{G_1G_2G_3G_4} {(1+G_1G_2H_1)(1+G_3G_4H_2)+G_2G_3H_3}. \]

Step 7: Final Answer.
\[ \boxed{ \frac{C(s)}{R(s)} = \frac{G_1G_2G_3G_4} {(1+G_1G_2H_1)(1+G_3G_4H_2)+G_2G_3H_3} } \] Hence, \[ \boxed{\text{Correct Option (A)}} \]