Chapter 5, Problem 5.1P

To determine

The transfer function $\frac{X(s)}{F(s)}$ for block diagram shown in figure (i).

Answer to Problem 5.1P

The response is as follows:

$\frac{X(s)}{F(s)}=\frac{6}{(s+2)}$.

Explanation of Solution

Given:

The given block diagram is as shown in figure (i).

Concept Used:

The given block diagram is reduced in the simpler form by solving the loops.

Calculation:

Given block diagram is as shown below in figure (i)

The block is modified by naming the response after the constant gain block of 6 as shown in figure (ii).

For this figure (ii), we have

$\begin{array}{l}\left(Y(s)+4X(s)\right)\cdot \frac{1}{s}=X(s)\\ \Rightarrow X(s)=\frac{1}{\left(s-4\right)}\cdot Y(s)\end{array}$

Therefore, the block diagram in figure (ii) can be reduced to that shown in figure (iii).

Also, the two blocks shown in figure (iii) are in series so their gains can be multiplied as shown in figure (iv).

Thus, from the figure (iv), we get

$\begin{array}{l}\left(F(s)-X(s)\right)\cdot \frac{6}{\left(s-4\right)}=X(s)\\ \Rightarrow \left(F(s)-X(s)\right)=\frac{\left(s-4\right)}{6}X(s)\\ \Rightarrow F(s)=\frac{\left(s+2\right)}{6}X(s)\\ \Rightarrow \frac{X(s)}{F(s)}=\frac{6}{\left(s+2\right)}\end{array}$

So, the transfer function for the block diagram shown in figure (i) is

$\frac{X(s)}{F(s)}=\frac{6}{\left(s+2\right)}$.

Conclusion:

The obtained transfer function for the given block diagram is as

$\frac{X(s)}{F(s)}=\frac{6}{\left(s+2\right)}$.