Question 6. An input of 25t’u(t) is applied to the input of a Type 3 unity feedback system, as shown in Figure P7.1, where G(s) = ? G _210(s + 4)(s + 6)(s + 11)(s + 13) S'(s + 7)(s + 14)(s + 19) Find the steady-state error in position. (Section: 7.3 R(S) + G(s) FIGURE P7.1

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Transcribed Image Text: 6. An input of 25t’u(t) is applied to the input of a Type 3 unity feedback system, as shown in Figure P7.1, where G(s) = ? G _210(s + 4)(s + 6)(s + 11)(s + 13) S'(s + 7)(s + 14)(s + 19) Find the steady-state error in position. (Section: 7.3 R(S) + G(s) FIGURE P7.1

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G(s)=\frac{210(s+4)(s+6)(s+111)(s+13)}{s^{3}(s+71)(s+14)(s+19)}\begin{array}{l}C(s)=E(s) \cdot G(s) \\ E(s)=R(s)-C(s) \\ E(s)=R(s)-E(s) G(s) \\ E(s)=\frac{R(s)}{1+G(s) .}\end{array}\begin{array}{l}R(t)=25 t^{3} u(t)=(25) \times 6 \frac{t^{3}}{(3 !)} u(t) \\ \downarrow \\ R(s)=\frac{150}{s 4} . \\ E(s)=\frac{150 / 54}{1+240 G(s)} \\ e_{s s}=\operatorname{ltSE}_{s \rightarrow 0} \operatorname{ls}^{\rightarrow} \\ =\operatorname{lt}_{s \rightarrow 0} \frac{s(150 / s 4)}{1+G(s)} \\ =\operatorname{lt}_{s \rightarrow 0} \frac{150}{s^{3}+\infty\left\{\frac{210(s+4)(s+6)(s+1))(s+13}{s \%(s+7)(s+14)(s+19)}\right.} \\ =\frac{150}{0+\frac{210(4)(6)(11)(13)}{7(14)(19)}} \\ f_{s y}=0.3875^{\circ} \\\end{array} ...