考研851 自动控制原理
题海 · solution · p.408

T=1;t=0:1:20;

sys1=tf([0.9817,0],[4,0.9817-4.0733,0.0733],T);

step(sys1,t); axis([0,20,0,1.2]); grid;

figure(2)

sys2=tf([9.817,0],[4,9.817-4.0733,0.0733],T);

step(sys2,t); grid;

(2)\(T\)\(K\)值范围的影响。如果\(T\)待定,则有

\[G(z)=\frac{K(1-\mathrm{e}^{-4T})z}{4(z-1)(z-\mathrm{e}^{-4T})}\]
\[\Phi(z)=\frac{K(1-\mathrm{e}^{-4T})z}{4z^2+[K(1-\mathrm{e}^{-4T})-4(1+\mathrm{e}^{-4T})]z+4\mathrm{e}^{-4T}}\]

\(z=\dfrac{w+1}{w-1}\),得

\[K(1-\mathrm{e}^{-4T})w^2+8(1-\mathrm{e}^{-4T})w+8(1+\mathrm{e}^{-4T})-K(1-\mathrm{e}^{-4T})=0\]

列出劳斯表

\(w^2\) \(K(1-\mathrm{e}^{-4T})\) \(8(1+\mathrm{e}^{-4T})-K(1-\mathrm{e}^{-4T})\)
\(w^1\) \(8(1-\mathrm{e}^{-4T})\) \(0\)
\(w^0\) \(8(1+\mathrm{e}^{-4T})-K(1-\mathrm{e}^{-4T})\)

可见,当\(0<K<\dfrac{8(1+\mathrm{e}^{-4T})}{1-\mathrm{e}^{-4T}}\)时,系统稳定。此时若\(T\)减小,则\(K\)值范围将增大。

图:自控原理题海_p408_fig1

图7-39 \(T=0.1,K=10\)时系统的 单位阶跃响应(MATLAB)

\(K=10,T=1\)时系统不稳定,阶跃响应曲线如图7-38(b)所示;而当\(T=0.1\)时,使系统稳定的\(K\)值范围为\(0<K<40.75\),取\(K=10\),系统单位阶跃响应曲线如图7-39所示。

MATLAB文本:exe730b.m

T=0.1;t=0:0.1:2;

sys=tf([10(1-exp(-0.4)),0],[4,6-14exp(-0.4),4*exp(-0.4)],T);

step(sys,t); grid;

7-31 设采样系统的结构图如图7-40所示。(1)试求系统在参考输入\(r(t)\)和扰动输入\(n(t)\)作用下的总输出响应\(C(z)\);(2)若要求系统输出能充分反映输入,并尽可能少受扰动的影响,在理论上对\(D_1(z)\)\(D_2(z)\)有何要求?(3)求\(\mathscr{Z}\left[\dfrac{1-\mathrm{e}^{-Ts}}{s}\cdot G_1(s)G_2(s)\right]\),其中\(G_1(s)=\dfrac{\mathrm{e}^{-0.2s}}{s+1}\),\(G_2(s)=\dfrac{1}{s}\),采样周期\(T=1\)

(1)总输出响应\(C(z)\)。由于

\[C(s)=[N(s)+B_1(s)]G_2(s)\]
\[B_1(s)=B_2^*(s)G_h(s)G_1(s),\quad B_2(z)=R(z)D_2(z)+E(z)D_1(z)\]
\[E(z)=R(z)-C(z)\]

于是

\[C(s)=N(s)G_2(s)+B_2^*(s)G_h(s)G_1(s)G_2(s)\]

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