Question

For the block diagram shown in the figure, the correct expression for the transfer function \( G_d = \frac{y_2(s){d(s)} \) is:} \includegraphics[width=0.7\linewidth]{q48 CE.PNG}

• A. \( \frac{-G_{p1}G_{c2}}{(1 + G_{c1}G_{c2}G_{p1})(1 + G_{c2}G_{p2})} \)
• B. \( \frac{-G_{p1}G_{c2}}{1 + G_{c2}G_{p2} + G_{c1}G_{c2}G_{p1}G_{p2}} \)
• C. \( \frac{-G_{p1}G_{c2}}{1 + G_{c2}G_{p2} + G_{c1}G_{c2}G_{p1}} \)
• D. \( \frac{1}{1 + G_{c2}G_{p2} + G_{c1}G_{c2}G_{p1}G_{p2}} \)

Hint:

When analyzing block diagrams, simplify each feedback loop step-by-step, reducing inner loops first, and apply superposition for disturbance transfer functions.

Answer 1

The Correct Option is C

Step 1: Identify the transfer function relationship. The transfer function \( G_d = \frac{y_2(s)}{d(s)} \) involves analyzing the impact of the disturbance \( d(s) \) on the output \( y_2(s) \). Step 2: Determine the effect of \( d(s) \) through the block diagram. 1. The disturbance \( d(s) \) directly affects the first process block \( G_{p1} \). The output of this block is: \[ y_1(s) = G_{p1} d(s). \] 2. This output \( y_1(s) \) enters the second control loop with \( G_{c1} \) and \( G_{c2} \). Step 3: Include feedback paths. The feedback loop affects the transfer function. Using block diagram reduction techniques: 1. The first feedback loop has a feedback gain of \( G_{c1}G_{c2}G_{p1} \). The closed-loop transfer function for this segment is: \[ \frac{1}{1 + G_{c1}G_{c2}G_{p1}}. \] 2. The second feedback loop has a gain of \( G_{c2}G_{p2} \). The total output \( y_2(s) \) after combining the loops is: \[ y_2(s) = \frac{-G_{p1}G_{c2}}{1 + G_{c2}G_{p2} + G_{c1}G_{c2}G_{p1}} d(s). \] Step 4: Conclusion. The transfer function \( G_d \) is: \[ G_d = \frac{-G_{p1}G_{c2}}{1 + G_{c2}G_{p2} + G_{c1}G_{c2}G_{p1}}. \]