User:ReCover/lab2

=2.4=

$$\angle e^{-jwT} = -wT$$

$$|e^{-jwT}| = 1 \,$$

$$1+K \left[1+\frac{1}{T_is} \right]\frac{0.4}{280s + 9.81}e^{-sT} = 0$$

$$\frac{0.4KT_is+0.4K}{280T_is^2+9.81T_is}e^{-sT}$$

$$$$

$$$$

= 2 =

$$G_R(s) = K \left[1+\frac{1}{T_is} \right]$$

$$\frac{G_R(s)G(s)}{1+G_R(s)G(s)} = \frac{K \left[1+\frac{1}{T_is} \right]\frac{0.4}{280s + 9.81}}{1+K \left[1+\frac{1}{T_is} \right]\frac{0.4}{280s + 9.81}} = $$

$$= \frac{0.4KT_is+0.4K}{280T_is^2+(9.81+0.4K)T_is+0.4K}$$

Karakteristisk ekv: $$280T_is^2+(9.81+0.4K)T_is+0.4K \,$$

Krav: $$K > 0 \,$$

$$D_1 = \begin{vmatrix} (9.81+0.4K)T_i \end{vmatrix} = (9.81+0.4K)T_i > 0 \Rightarrow T_i > 0$$

$$E(s) = R(s) - Y(s) \,$$

$$Y(s) = G(s)\cdot G_R(s)\cdot E(s)$$

$$E(s) = R(s) - G(s) \cdot G_R(s) \cdot E(s)$$

$$E(s) = \frac{1}{1+G_R(s) \cdot G(s)}R(s) = \frac{1}{1+K \left[1+\frac{1}{T_is} \right] \cdot \frac{0.4}{280s + 9.81}}R(s) =$$

$$= \frac{(280s+9.81)T_is}{280T_is^2+(9.81+0.4K)T_is+0.4K} R(s)$$

Stegstörning: $$L[ steg ] = \frac{1}{s}$$

$$E(s) = \frac{(280s+9.81)T_is}{280T_is^2+(9.81+0.4K)T_is+0.4K}\frac{1}{s}$$

$$\lim_{t \to \infty} E(t) = \lim_{s \to 0} sE(s) = \lim_{s \to 0} \frac{(280s+9.81)T_is}{280T_is^2+(9.81+0.4K)T_is+0.4K} = $$

$$ = \frac{0}{0.4K} = 0$$

Impuls: $$L[ impuls ] = 1 \,$$

$$E(s) = \frac{(280s+9.81)T_is}{280T_is^2+(9.81+0.4K)T_is+0.4K} \cdot 1$$

$$\lim_{t \to \infty} E(t) = \lim_{s \to 0} sE(s) = 0 \cdot \frac{0}{0.4K} = 0$$

=2.2=

$$\frac{G_R(s)G(s)}{1+G_R(s)G(s)} = \frac{B(s)}{s^2+2 \xi w_ns+w_n^2}$$

$$\frac{0.4KT_is+0.4K}{280T_is^2+(9.81+0.4K)T_is+0.4K} = \frac{\frac{0.4KT_is+0.4K}{280T_i}}{s^2+\frac{(9.81+0.4K)}{280}s+\frac{0.4K}{280T_i}}$$

$$\Rightarrow w_n^2 = \frac{0.4K}{280T_i} \Rightarrow w_n = \sqrt{\frac{0.4K}{280T_i}}$$

$$\Rightarrow 2 \xi w_ns = \frac{(9.81+0.4K)}{280}s \Rightarrow \xi = \frac{(9.81+0.4K)}{280} \cdot \frac{1}{\sqrt{\frac{0.4K}{280T_i}}} \cdot \frac{1}{2}$$

$$\xi = \frac{0.0472456 \cdot \sqrt{T_i} \cdot (0.4K+9.81)}{\sqrt{K}}$$

=1=

$$\frac{dV}{dt} = q_{in}(t)-q_{ut}(t) = \alpha u - \beta \sqrt{h}$$

$$q_{in}(t) = 4u \cdot 10^{-6}\ m^3/s$$ (från lab 1)

$$q_{ut}(t) = a\sqrt{2gh}$$

$$a = 0.14 \cdot 10^{-4}\ m^2$$ arbetspunkt (5 V, 0.1 m)

$$\alpha = 4\cdot 10^{-6}\ m^3/s$$

$$\beta = 0.14 \cdot 10^{-4} \sqrt{2g}$$

$$A \frac{d \Delta h}{dt} = \alpha \Delta u - \gamma \Delta h$$

$$A = 2.8 \cdot 10^{-3}\ m^2$$

$$\gamma = a \sqrt{\frac{g}{2h_0}} = 9.81 \cdot 10^{-5}$$

$$L \left[ A \frac{d \Delta h}{dt} = \alpha \Delta u - \gamma \Delta h \right] \Rightarrow AsH(s) = \alpha U(s) - \gamma H(s)$$

$$H(s) = \frac{\alpha}{As+\gamma} U(s)$$

$$G(s) = \frac{\alpha}{As+\gamma} = \frac{4\cdot 10^{-6}}{2.8 \cdot 10^{-3}s + 9.81 \cdot 10^{-5}} = \frac{0.4}{280s + 9.81}$$

$$E(s) = R(s) - Y(s) \,$$

$$Y(s) = G(s)\cdot K\cdot E(s)$$

$$E(s) = R(s) - G(s) \cdot K \cdot E(s)$$

$$E(s) = \frac{1}{1+K \cdot G(s)}R(s) = \frac{1}{1+K \cdot \frac{0.4}{280s + 9.81}}R(s) =$$

$$= \frac{280s + 9.81}{280s + 9.81 + 0.4K}R(s)$$

Karakteristisk ekv: $$280s + 9.81 + 0.4K \,$$

Hurwitz: $$D_1 = \begin{vmatrix} 9.81 + 0.4K \end{vmatrix} = 9.81 + 0.4K > 0 \Rightarrow K > -\frac{981}{40}$$

$$E(s) = G_2(s)Q(s) \,$$

$$E(s) = \frac{280s + 9.81}{280s + 9.81 + 0.4K}Q(s)$$

Stegstörning: $$L[ steg ] = \frac{1}{s}$$

$$E(s) = \frac{280s + 9.81}{280s + 9.81 + 0.4K}\frac{1}{s}$$

$$\lim_{t \to \infty} E(t) = \lim_{s \to 0} sE(s) = \lim_{s \to 0} \frac{280s + 9.81}{280s + 9.81 + 0.4K} = \frac{9.81}{9.81+0.4K}$$

Impuls: $$L[ impuls ] = 1 \,$$

$$E(s) = \frac{280s + 9.81}{280s + 9.81 + 0.4K} \cdot 1$$

$$\lim_{t \to \infty} E(t) = \lim_{s \to 0} sE(s) = 0 \cdot \frac{9.81}{9.81+0.4K} = 0$$