当 \(r_2(t)=2\sin 2t\) 时,因
\[\Phi(\mathrm{j}2)=\left.\frac{1}{1-\omega^2+\mathrm{j}\omega}\right|_{\omega=2}=\frac{1}{\sqrt{(-3)^2+2^2}}\angle-146.3°\]
\[=0.277\angle-146.3°\]
相应稳态输出
\[c_{ss2}(t)=2\times0.277\sin(2t-146.3°)=0.554\sin(2t-146.3°)\]
由于是线性系统,故得
\[c_{ss}t=c_{ss1}(t)+c_{ss2}(t)=1+0.55\sin(2t-146.3°)\]
(2) 求 \(\gamma\)。
令 \(|G(\mathrm{j}\omega_c)|=1\),有
\[\frac{1}{\omega_c\sqrt{1+\omega_c^2}}=1\]
求出 \(\omega_c=0.786\)。故
\[\gamma=180°+\angle G(\mathrm{j}\omega_c)=180°-90°-\arctan\omega_c=51.8°\]
5-75 设单位负反馈系统的开环传递函数为 \(G(s)=\dfrac{\omega_n^2}{s(s+2\zeta\omega_n)}\),已知闭环频率特性的模 \(|\Phi(\mathrm{j}2)|=\dfrac{1}{2\zeta}\),相角裕度 \(\gamma=26.56°\),试求输入 \(r(t)=3\sin4t\) 时,系统的稳态输出 \(c_{ss}(t)\)。
解 系统闭环频率特性为
\[\Phi(\mathrm{j}\omega)=\left.\frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}\right|_{s=\mathrm{j}\omega}\]
\[|\Phi(\mathrm{j}\omega)|=\frac{\omega_n^2}{\sqrt{(\omega_n^2-\omega^2)^2+4\zeta^2\omega_n^2\omega^2}},\qquad \alpha(\omega)=-\arctan\frac{2\zeta\omega_n\omega}{\omega_n^2-\omega^2}\]
显然有
\[|\Phi(\mathrm{j}\omega_n)|=\frac{1}{2\zeta}\]
由题意,\(|\Phi(\mathrm{j}2)|=\dfrac{1}{2\zeta}\),故有 \(\omega_n=2\)。
因为
\[\gamma=180°-90°-\arctan\frac{\omega_c}{2\zeta\omega_n}=26.56°\]
求得
\[\omega_c=8\zeta\]
又因 \(|G(\mathrm{j}\omega_c)|=1\),即
\[\frac{\omega_n^2}{\omega_c\sqrt{\omega_c^2+4\zeta^2\omega_n^2}}=1\]
代入 \(\omega_n=2,\omega_c=8\zeta\),求得
\[\zeta=0.236\]
由题意,\(r(t)=3\sin4t\),必有
\[c_{ss}(t)=3|\Phi(\mathrm{j}4)|\sin(4t+\alpha(4))\]
由于 \(|\Phi(\mathrm{j}4)|=0.318,\qquad \alpha(\omega)|_{\omega=4}=-162.5°\)
故得
\[c_{ss}(t)=0.954\sin(4t-162.5°)\]
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