Skip to content

new test

你好,world!

Liam

07/22/2018

\begin{equation}
x^n+y^n=z^n
\end{equation}

$$\frac{{\partial \frac{{\partial L}}{{\partial \left( {\frac{{\partial {x_i}}}{{\partial {a_k}}}} \right)}}\partial {x_i}}}{{\partial {a_k}\partial L}} = \frac{{\partial \frac{{\partial L}}{{\partial {a_k}}}\partial {x_i}}}{{\partial \left( {\frac{{\partial {x_i}}}{{\partial {a_k}}}} \right)\partial L}} = \frac{{\partial \frac{{\partial L}}{{\partial L}}\partial {a_k}}}{{\partial \left( {\frac{{\partial {x_i}}}{{\partial {x_i}}}} \right)\partial {a_k}}} = 1$$

1. 一个简单弹簧振子,质点质量为m,弹簧劲度系数为K,力阻为R。在外力f(t)作用下振动。 (1)写出振子的(位移)传递函数、格林函数,力学品质因素和力阻抗,以及受迫振动时的一般解。
\vbox{}位移的频谱等于力的频谱乘以传递函数:\[x(t) = \mathscr{F}^{-1}H(\omega)\mathscr{F}f(t)\]
传递函数:\[ H(\omega) = \frac{1}{-m\omega^2-i\omega R +K} \]
位移是外力和格林函数 (冲击响应) 的卷积:\[x(t) = \frac{g(t)}{m}\otimes f(t) = \frac{1}{m}\int_{ – \infty }^t f(t’)g(t-t’)dt’\]
格林函数:\[g(t-t’) = \frac{\exp(-\gamma(t-t’))\sin(\omega_0(t-t’))}{\omega_0}\]
对比时域和频域的两种方法, 得到算子间的关系:
\[\begin{aligned}\frac{g(t)}{m}\otimes&=\mathscr{F}^{-1}H(\omega)\mathscr{F}\\\mathscr{F}g\otimes &= mH\mathscr{F}\\\mathscr{F}g\mathscr{F} &= mH\mathscr{F}\\\mathscr{F}g &= mH\end{aligned}\]
我们称 $G = mH$ 为稳态格林函数, 表示外力 $ m \exp(-i\omega t)$ 产生的稳态响应.
\vbox{}运动方程:\[m \ddot x + R \dot x + Kx=f(t) \]
力学品质因数:\[Q_m= \frac{\omega_0 m}{R}=\frac{\omega_0 m}{2m\gamma}=\frac{\pi}{T_0\gamma}\]
力阻抗:\[Z_m = R – i(\omega m-\frac{K}{\omega})\]
\vbox{}(2)如外力为高斯包络调制信号: $f(t)=A \exp(-a^2 t^2) \cos⁡(\omega’t)$ , 求振子位移响应 $x(t)$ .
\vbox{}力的频谱:\[\begin{aligned}F(\omega) &= \int_{ – \infty }^{+ \infty} f(t)e^{i\omega t}dt\\&=A \int_{ – \infty }^{+ \infty} e^{-a^2t^2} cos(\omega’t) e^{i\omega t}dt\\&=A \frac{1}{2} \int_{ – \infty }^{+ \infty} e^{-a^2t^2} (e^{i\omega’t}+e^{-i\omega’t}) e^{i\omega t}dt\\&=A \frac{1}{2}(e^{-\frac{(\omega+\omega’)^2}{4a^2}}+e^{-\frac{(\omega-\omega’)^2}{4a^2}}) \int_{ – \infty }^{+ \infty} e^{-a^2t^2} dt\\&=A \frac{\sqrt{\pi}}{2a}(e^{-\frac{(\omega+\omega’)^2}{4a^2}}+e^{-\frac{(\omega-\omega’)^2}{4a^2}}) \end{aligned}\]
位移:\[\begin{aligned}x(t) &= \mathscr{F}^{-1}H(\omega)F(\omega)\\&= \int_{ – \infty }^{+ \infty} H(\omega)F(\omega)e^{-i\omega t} d\omega\\&=A \int_{ – \infty }^{+ \infty} \frac{\sqrt{\pi}(e^{-\frac{(\omega+\omega’)^2}{4a^2}}+e^{-\frac{(\omega-\omega’)^2}{4a^2}})}{2a(-m\omega^2-i\omega R +K)}e^{-i\omega t} d\omega\end{aligned}\]
\vbox{}(3)如要使系统振动响应为高斯包络调制信号形式, 即 $x(t)=x_0 \exp(-a^2 t^2)\cos⁡ (\omega’t)$ , 求外力 $f(t)$ .
\vbox{}带入时域的运动方程:\[m \ddot x + R \dot x + Kx=f(t) \]
得到力:\[\begin{aligned}f( t ) =& (Kx_0-m(2x_0a^2+\omega’^2x_0- 4x_0a^4t^2)-R2x_0a^2)\exp(-a^2t^2)\cos ( \omega’t )-\\&(m4x_0a^2\omega’t+Rx_0\omega’)\exp ( – a^2t^2 )\sin ( \omega’t )\end{aligned}\]2. 习题2.1与2.7。(课本84-85页) 对于一维质点系的振动系统, 我们可以将运动方程写为:
\[ m \ddot x +R \dot x + Kx = f \]
其中 \[\begin{aligned}m &=\text{diag}(m_1,\cdots,m_n)\\R &=\text{diag}(R_1,\cdots,R_n)\\x &= (x_1,\cdots,x_n)^T\\f &= (f_1,\cdots,f_n)^T\\K &= \left( {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{{K_0} + {K_1}}&{ – {K_1}}&0\\{ – {K_1}}&{{K_1} + {K_2}}&{ – {K_2}}\\0&{ – {K_2}}&{{K_2} + {K_3}}\end{array}}& \cdots \\ \vdots & \ddots \end{array}} \right)\end{aligned}\]
解这个方程, 我们需要先求出振动模式和简正坐标. 在简正坐标下, 各个振动模式正交, 所以我们可以运用第一题中的时域和频域两种方法求解.
为了求出振动模式, 令 $ f = 0$ . 质点的运动可以写为 $ x = x_0 \exp(-\gamma t) \exp(-i\omega t)$ .
令 $ \omega’ = \omega-i\gamma$ , 运动方程变为:
\[\begin{aligned}m \ddot x +R \dot x + Kx &= 0\\-\omega’^2mx -i\omega’Rx + Kx &= 0,\ \text{令}m’ = m – \frac{R}{i\omega’}\\-\omega’^2m’x + Kx &= 0,\ \text{令}\omega_0^2 = Km’^{-1}\\\omega_0^2 &= \omega’^2x\end{aligned}\]
代入题目中的条件, 得到:\[\left( {\begin{array}{*{20}{c}}{\omega {‘^2} – 2{\omega _0}^2}&{{\omega _0}^2}&0\\{{\omega _0}^2}&{\omega {‘^2} – 2{\omega _0}^2}&{{\omega _0}^2}\\0&{{\omega _0}^2}&{\omega {‘^2} – 2{\omega _0}^2}\end{array}} \right)\left( \begin{array}{l}{x_1}\\{x_2}\\{x_3}\end{array} \right) = {\bf{0}}\]
解得:
\[\begin{aligned}{\omega {‘} ^{\left( 1 \right)}} &= {\omega _0}\\{\omega{‘} ^{\left( 2 \right)}} &= \sqrt {1 + \frac{{\sqrt 2 }}{2}} {\omega _0}\\{\omega{‘} ^{\left( 2 \right)}} &= \sqrt {1 – \frac{{\sqrt 2 }}{2}} {\omega _0}\end{aligned}\]

3. 习题2.5。 等效导纳图:

No comments yet.

Leave a Reply

Your email address will not be published. Required fields are marked *

SidebarComments (0)