Question 13.6: Determine the open-loop gain necessary for the closed-loop s...

The open-loop system transfer function is

T_{OL}(s)  =\frac{K}{6s^{3}+11s^{2}+6s+1  }  .     (13.25)

When s=j \omega , the sinusoidal transfer function is found to be

T_{OL}(j \omega )  =\frac{K}{(1-11\omega ^2 ) +j6 \omega (1-\omega ^2 )   }    ,     (13.26)

or, equally,

T_{OL}(j \omega)  =\frac{K[(1-11 \omega^2)-j6 \omega (1- \omega^2 ) ] }{[(1-11 \omega^2 )^2 +36 \omega^2 (1- \omega^2 )^2] }.  (13.27)

The frequenc \omega _{p} for which the phase angle is −180° is found by the solution of

Im[ T_{OL}(j \omega)] =0 .     (13.28)

Substituting into Eq. (13.28) the expression for the imaginary part of the open-loop transfer function from Eq. (13.27) yields

\omega_{P} = 1rad/s.     (13.29)

The real part of the open-loop transfer function for this frequency is

Re[ T_{OL}(j \omega_p) ] =- \frac{K}{10} .     (13.30)

Substituting Eq. (13.30) into the condition for the gain margin, K_{g} \geq 1.2 , yields

\frac{10}{K} \geq 1.2  ,  K\leq 8.3333 .

Now, to satisfy the phase margin requirement, the phase angle of the open-loop transfer function for the frequency at which the magnitude is 1 should be

\angle T_{OL}(j\omega _{g} ) =-180^{◦} +45^{◦} =-135^{◦}  .    (13.31)

From the expression for the open-loop transfer function [Eq. (13.26)], the phase angle of  T_{OL}(j\omega ) for \omega _{g} is

\angle T_{OL}(j\omega _{g} ) = \tan ^{-1} (0)-\tan ^{-1} \left[\frac{6\omega _{g}(1-\omega ^{2}_{g} ) }{1-11\omega ^{2}_{g} } \right] .     (13.32)

Comparing the right-hand sides of Eqs. (13.31) and (13.32) yields

-\tan ^{-1} \left[\frac{6\omega _{g}(1-\omega ^{2}_{g} ) }{1-11\omega ^{2}_{g} } \right] =-135^{◦}  .     (13.33)

Taking the tangents of both sides of Eq. (13.33) yields a cubic equation for \omega _{g}  :

6\omega ^{3}_{g} +11\omega ^{2}_{g} -6\omega _{g}-1=0 ,     (13.34)

which has the solution

\omega _{g}=0.54776  rad/s.

Now use Eq. (13.24) to obtain

\left|T_{OL}(j\omega _{g} ) \right| =1 .    (13.35)

The magnitude of T_{OL}(j\omega) for this system is

\left|T_{OL}(j\omega _{g} ) \right| =\frac{K}{\sqrt{(1-11\omega ^{2}_{g})^{2}+36\omega ^{2}_{g}(1-\omega ^{2}_{g} )^{2}   } } .    (13.36)

Combine Eqs. (13.35) and (13.36) to obtain the solution for K that satisfies the phase condition: K = 3.27.

This value is smaller than the value of K obtained from the gain condition: Therefore the open-loop gain must be equal to 3.27 (or less) to meet both the gain and the phase angle conditions.