Chapter 5, Problem 5.7P

To determine

The expressions for $C\left(s\right),M\left(s\right)andE\left(s\right)$ in terms of $R\left(s\right)andD\left(s\right)$ for the given block diagram.

Answer to Problem 5.7P

The responses for $C\left(s\right),M\left(s\right)andE\left(s\right)$ are

$\begin{array}{l}E\left(s\right)=\frac{s}{\left(12s^3+11s^2+12s+6\right)}\left(\left(12s^2+11s+12\right)R\left(s\right)+\left(4s+1\right)D\left(s\right)\right),\\ M\left(s\right)=\frac{1}{\left(12s^3+11s^2+12s+6\right)}\left(6\cdot \left(3s+2\right)R\left(s\right)+\left(10s+6\right)D\left(s\right)\right)and\\ C\left(s\right)=\frac{1}{\left(12s^3+11s^2+12s+6\right)}\left(6R\left(s\right)-s\left(4s+1\right)D\left(s\right)\right)\end{array}$.

Explanation of Solution

Given:

The block diagram for the model with input $R\left(s\right)$ and the output as $C\left(s\right)$.

Concept Used:

The responses are observed at the sum points and the take-off points of the block diagram. Consequently, these expressions are employed for finding the expressions for $C\left(s\right),M\left(s\right)andE\left(s\right)$ in terms of $R\left(s\right)andD\left(s\right)$.

Calculation:

Following responses can be drawn from the block diagram:

$\begin{array}{l}E\left(s\right)=R\left(s\right)-C\left(s\right)\left(1\right)\\ M\left(s\right)=\frac{1}{\left(4s+1\right)}\left(\frac{6}{s}E\left(s\right)-10C\left(s\right)\right)\left(2\right)\\ C\left(s\right)=\frac{1}{\left(3s+2\right)}\left(M\left(s\right)-D\left(s\right)\right)\left(3\right)\end{array}$

Using equation (3), we get

$M\left(s\right)=\left(3s+2\right)C\left(s\right)+D\left(s\right)\left(4\right)$

From equations (2) and (4), we get

$\begin{array}{l}\left(3s+2\right)C\left(s\right)+D\left(s\right)=\frac{1}{\left(4s+1\right)}\left(\frac{6}{s}E\left(s\right)-10C\left(s\right)\right)\\ \Rightarrow \left(3s+2\right)\left(4s+1\right)C\left(s\right)+\left(4s+1\right)D\left(s\right)=\frac{6}{s}E\left(s\right)-10C\left(s\right)\\ \Rightarrow \left(12s^2+11s+12\right)C\left(s\right)=\frac{6}{s}E\left(s\right)-\left(4s+1\right)D\left(s\right)\\ \Rightarrow \left(12s^2+11s+12\right)C\left(s\right)=\frac{6}{s}\left(R\left(s\right)-C\left(s\right)\right)-\left(4s+1\right)D\left(s\right)\because equation\left(1\right)\\ \Rightarrow \left(12s^2+11s+12+\frac{6}{s}\right)C\left(s\right)=\frac{6}{s}R\left(s\right)-\left(4s+1\right)D\left(s\right)\\ \Rightarrow \left(12s^3+11s^2+12s+6\right)C\left(s\right)=6R\left(s\right)-s\left(4s+1\right)D\left(s\right)\\ \Rightarrow C\left(s\right)=\frac{1}{\left(12s^3+11s^2+12s+6\right)}\left(6R\left(s\right)-s\left(4s+1\right)D\left(s\right)\right)\left(5\right)\end{array}$

Now, from equation (1) on putting the value of $C\left(s\right)$, we have

$\begin{array}{l}E\left(s\right)=R\left(s\right)-C\left(s\right)\\ \Rightarrow E\left(s\right)=R\left(s\right)-\frac{1}{\left(12s^3+11s^2+12s+6\right)}\left(6R\left(s\right)-s\left(4s+1\right)D\left(s\right)\right)\\ \Rightarrow E\left(s\right)=R\left(s\right)-\frac{6}{\left(12s^3+11s^2+12s+6\right)}R\left(s\right)+\frac{s\left(4s+1\right)}{\left(12s^3+11s^2+12s+6\right)}D\left(s\right)\\ \Rightarrow E\left(s\right)=R\left(s\right)\left(\frac{12s^3+11s^2+12s+6-6}{\left(12s^3+11s^2+12s+6\right)}\right)+\frac{s\left(4s+1\right)}{\left(12s^3+11s^2+12s+6\right)}D\left(s\right)\\ \Rightarrow E\left(s\right)=\frac{s\left(12s^2+11s+12\right)}{\left(12s^3+11s^2+12s+6\right)}R\left(s\right)+\frac{s\left(4s+1\right)}{\left(12s^3+11s^2+12s+6\right)}D\left(s\right)\\ \Rightarrow E\left(s\right)=\frac{s}{\left(12s^3+11s^2+12s+6\right)}\left(\left(12s^2+11s+12\right)R\left(s\right)+\left(4s+1\right)D\left(s\right)\right)\left(6\right)\end{array}$

From equation (4), putting the value for $C\left(s\right)$, we get

$\begin{array}{l}M\left(s\right)=\left(3s+2\right)C\left(s\right)+D\left(s\right)\\ \Rightarrow M\left(s\right)=\left(3s+2\right)\frac{1}{\left(12s^3+11s^2+12s+6\right)}\left(6R\left(s\right)-s\left(4s+1\right)D\left(s\right)\right)+D\left(s\right)\\ \Rightarrow M\left(s\right)=\frac{6\left(3s+2\right)}{\left(12s^3+11s^2+12s+6\right)}R\left(s\right)+D\left(s\right)-\frac{s\left(4s+1\right)\left(3s+2\right)}{\left(12s^3+11s^2+12s+6\right)}D\left(s\right)\\ \Rightarrow M\left(s\right)=\frac{6\left(3s+2\right)}{\left(12s^3+11s^2+12s+6\right)}R\left(s\right)+D\left(s\right)\left(\frac{12s^3+11s^2+12s+6-12s^3-11s^2-2s}{\left(12s^3+11s^2+12s+6\right)}\right)\\ \Rightarrow M\left(s\right)=\frac{6\left(3s+2\right)}{\left(12s^3+11s^2+12s+6\right)}R\left(s\right)+D\left(s\right)\frac{\left(10s+6\right)}{\left(12s^3+11s^2+12s+6\right)}\\ \Rightarrow M\left(s\right)=\frac{1}{\left(12s^3+11s^2+12s+6\right)}\left(6\left(3s+2\right)R\left(s\right)+\left(10s+6\right)D\left(s\right)\right)\left(7\right)\end{array}$.

Conclusion:

The expressions for $C\left(s\right),M\left(s\right)andE\left(s\right)$ in terms of $R\left(s\right)andD\left(s\right)$ are as shown:

$\begin{array}{l}E\left(s\right)=\frac{s}{\left(12s^3+11s^2+12s+6\right)}\left(\left(12s^2+11s+12\right)R\left(s\right)+\left(4s+1\right)D\left(s\right)\right),\\ M\left(s\right)=\frac{1}{\left(12s^3+11s^2+12s+6\right)}\left(6\cdot \left(3s+2\right)R\left(s\right)+\left(10s+6\right)D\left(s\right)\right)and\\ C\left(s\right)=\frac{1}{\left(12s^3+11s^2+12s+6\right)}\left(6R\left(s\right)-s\left(4s+1\right)D\left(s\right)\right)\end{array}$.