VIPSolutionsGive step-by-step solution with explanation and final answer:Consider the block diagram shown in Fig. 2. Find the equivalent transfer function < Hs)
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Give step-by-step solution with explanation and final answer:
Consider the block diagram shown in Fig. 2. Find the equivalent transfer function <
Hs)
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Asked by: Test User 100199
Created at: 2025-05-17 04:35:29
Asked by: Test User 100199
Created at: 2025-05-17 04:35:29
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Let's solve the block diagram step by step to find the transfer function \(\frac{C(s)}{R(s)}\).
---
## **Step 1: Label the Signals**
Let the signals at each block be:
- After summing junction (input): \(E(s)\)
- After \(G_1(s)\): \(A(s)\)
- After \(G_2(s)\): \(B(s)\)
- Output: \(C(s)\)
---
## **Step 2: Write Equations for Each Signal**
### **At the First Summing Point:**
\[
E(s) = R(s) - H(s)C(s)
\]
### **After \(G_1(s)\):**
\[
A(s) = G_1(s)E(s)
\]
### **At the Second Summing Point (before \(G_2(s)\)):**
\[
\text{Input to } G_2(s) = A(s) + G_3(s)A(s) = A(s)[1 + G_3(s)]
\]
### **After \(G_2(s)\):**
\[
B(s) = G_2(s) \times \text{Input to } G_2(s) = G_2(s)A(s)[1 + G_3(s)]
\]
### **At the Third Summing Point (before output):**
\[
C(s) = B(s) + G_4(s)C(s)
\]
\[
C(s) - G_4(s)C(s) = B(s)
\]
\[
C(s)[1 - G_4(s)] = B(s)
\]
\[
C(s) = \frac{B(s)}{1 - G_4(s)}
\]
---
## **Step 3: Substitute Back**
Recall:
\[
B(s) = G_2(s)A(s)[1 + G_3(s)]
\]
\[
A(s) = G_1(s)E(s)
\]
\[
E(s) = R(s) - H(s)C(s)
\]
So,
\[
A(s) = G_1(s)[R(s) - H(s)C(s)]
\]
Thus,
\[
B(s) = G_2(s)[1 + G_3(s)]G_1(s)[R(s) - H(s)C(s)]
\]
And,
\[
C(s) = \frac{B(s)}{1 - G_4(s)}
= \frac{G_2(s)[1 + G_3(s)]G_1(s)[R(s) - H(s)C(s)]}{1 - G_4(s)}
\]
---
## **Step 4: Solve for \(\frac{C(s)}{R(s)}\)**
Bring all \(C(s)\) terms to one side:
\[
C(s) = \frac{G_2(s)[1 + G_3(s)]G_1(s)R(s)}{1 - G_4(s)} - \frac{G_2(s)[1 + G_3(s)]G_1(s)H(s)C(s)}{1 - G_4(s)}
\]
\[
C(s) + \frac{G_2(s)[1 + G_3(s)]G_1(s)H(s)C(s)}{1 - G_4(s)} = \frac{G_2(s)[1 + G_3(s)]G_1(s)R(s)}{1 - G_4(s)}
\]
\[
C(s)\left[1 + \frac{G_2(s)[1 + G_3(s)]G_1(s)H(s)}{1 - G_4(s)}\right] = \frac{G_2(s)[1 + G_3(s)]G_1(s)R(s)}{1 - G_4(s)}
\]
\[
C(s)\left[\frac{1 - G_4(s) + G_2(s)[1 + G_3(s)]G_1(s)H(s)}{1 - G_4(s)}\right] = \frac{G_2(s)[1 + G_3(s)]G_1(s)R(s)}{1 - G_4(s)}
\]
\[
C(s)\left[1 - G_4(s) + G_2(s)[1 + G_3(s)]G_1(s)H(s)\right] = G_2(s)[1 + G_3(s)]G_1(s)R(s)
\]
---
## **Step 5: Final Transfer Function**
\[
\frac{C(s)}{R(s)} = \frac{G_1(s)G_2(s)[1 + G_3(s)]}{1 - G_4(s) + G_1(s)G_2(s)[1 + G_3(s)]H(s)}
\]
---
### **Final Answer:**
\[
\boxed{
\frac{C(s)}{R(s)} = \frac{G_1(s)G_2(s)[1 + G_3(s)]}{1 - G_4(s) + G_1(s)G_2(s)[1 + G_3(s)]H(s)}
}
\]
---
If you need further simplification or a specific case, let me know!
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