Question 13.7: Consider the unity feedback system shown in Fig. 13.10. The ...

Consider the unity feedback system shown in Fig. 13.10. The system open-loop transfer function is

T_{OL}(s)=\frac{K}{s(s+2)}  ,     (13.37)

where K ≥ 0 is the open-loop gain. The closed-loop transfer function is

T_{CL}(s)=\frac{K}{(s^{2}+2s+K )}  .     (13.38)

 

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

Hence the closed-loop system characteristic equation is

s^{2}+2s+K=0  ,    (13.39)

and hence the roots are

s_{1} =-1-\sqrt{1-K}     ,    s_{1} =-1-\sqrt{1+K}         (13.40)

As the open-loop gain varies from zero to infinity, the roots change as follows:

_ For K = 0: s_{1} = −2   and s_{2} = 0 .

_ For K increasing from 0 to 1: Both roots remain real, with  s_{1}  moving from −2 to −1 and s_{2} moving from 0 to −1.

_ For K between 1 and infinity: The roots are complex conjugate, s_{1} = -1-j\sqrt{K-1}   and s_{2} = -1+j\sqrt{K-1} .

The migration of roots for 0 ≤ K < ∞ is shown in Fig. 13.11. Once the root locus is plotted, it is easy to determine the locations of the characteristic roots necessary for a desired system performance. For instance, if the desired damping ratio for the system considered in this example is, ξ= 0.7, the roots must be located at points A_{1} and  A_{2} , which correspond to

s_{1} = -1+j  ,   s_{2} = -1-j .

The characteristic equation for these roots takes the form

(s + 1 − j )(s + 1 + j ) = 0

Or

s^{2}  + 2s + 2 = 0. .   (13.41)

When Eq. (13.41) is compared with the characteristic equation in terms of K [Eq. (13.39)], the required value of the open-loop gain is found to be K = 2.

13.11

Related Answered Questions