谐振串联RLC电路

A - 谐振串联RLC电路

$Z = R + j \left(\omega L - \dfrac{1}{\omega C} \right)$

$I = \dfrac{V_i}{Z}$

$$|Z| = \sqrt {R^2 + \left(\omega L - \dfrac{1}{\omega C} \right)^2}$$

$$I_0 = \dfrac{V_0}{ |Z| } = \dfrac{V_0}{ \sqrt {R^2 + \left(\omega L - \dfrac{1}{\omega C} \right)^2} }$$

$$\left(\omega_r L - \dfrac{1}{\omega_r C} \right) = 0$$

$\omega_r = \dfrac{1}{\sqrt {L C}} \quad \quad (I)$

1) $$Z = R$$

2) $$I_0 = \dfrac{V_0}{R}$$

3) $$X_L = X_C$$

a) 求谐振频率 $$\omega_r$$
b) 绘制 $$|Z|$$、$$X_L = \omega L$$、$$X_C = \dfrac{1}{\omega C}$$ 和 $$I_0$$ 随频率 $$\omega$$ 变化的图，并讨论所得图形。

a)

$$\omega_r = \dfrac{1}{\sqrt {L C}} = \dfrac{1}{\sqrt{100\times10^{-3} \times 100 \times 10^{-6}}} \approx 316.23$$
b)

$$X_L$$ 和 $$X_C$$ 的图形相交（B点），因此 $$X_L = X_C$$ 或 $$\left(\omega_r L - \dfrac{1}{\omega_r C} \right) = 0$$ 。

B - 谐振电路中的平均功率

$$\theta = \arctan \left( \dfrac{ \omega L - \dfrac{1}{\omega C} }{R} \right)$$

$$\tan \theta = \left( \dfrac{ \omega L - \dfrac{1}{\omega C} }{R} \right)$$
$$\theta$$ 可以看作是下面直角三角形的锐角。（使用直角三角形中角的正切定义，你可以得到与上述相同的 $$\tan \theta$$）。

$$AC = \sqrt {R^2 + \left( \omega L - \dfrac{1}{\omega C} \right)^2}$$
$$\cos \theta = \dfrac{AB}{AC} = \dfrac{R}{\sqrt {R^2 + \left( \omega L - \dfrac{1}{\omega C} \right)^2}}$$

$$\displaystyle \quad \quad P_a = \dfrac{V_0^2}{2 \sqrt {R^2 + \left(\omega L - \dfrac{1}{\omega C} \right)^2} }\dfrac{R}{\sqrt {R^2 + \left( \omega L - \dfrac{1}{\omega C} \right)^2}}$$

$\displaystyle \quad \quad P_a = \dfrac{V_0^2 R}{2 \left({R^2 + \left(\omega L - \dfrac{1}{\omega C} \right)^2} \right) } \quad \quad (III)$

C - 谐振电路的截止频率和品质因数

$$P_a (\omega_c ) = \dfrac{1}{2} \left(\dfrac{V_0^2}{2 \; R} \right)$$

$$\dfrac{V_0^2 R}{2 \left({R^2 + \left(\omega_c L - \dfrac{1}{\omega_c C} \right)^2} \right) } = \dfrac{1}{2} \dfrac{V_0^2}{2 \; R}$$

$$\dfrac{ R}{2 \left({R^2 + \left(\omega_c L - \dfrac{1}{\omega_c C} \right)^2} \right) } = \dfrac{1}{4 R}$$

$$(\omega_c L - \dfrac {1}{\omega_c C } ) = R^2$$

$$\omega_c L - \dfrac {1}{\omega_c C} = \pm R$$

$$\omega_c^2 L C - 1 = \pm \omega_c R C$$

$$\omega_c^2 L C \pm \omega_c R C - 1 = 0$$

$$\omega_{c1} = \dfrac {- R C \pm \sqrt{ (R C)^2 + 4 L C }}{ 2 L C }$$

$$\omega_{c2} = \dfrac {R C \pm \sqrt{ (R C)^2 + 4 L C}}{ 2 L C }$$

$$\omega_{c1} = \dfrac {- R C + \sqrt{ (R C)^2 + 4 L C }}{ 2 L C }$$

$$\omega_{c2} = \dfrac {R C + \sqrt{ (R C)^2 + 4 L C}}{ 2 L C }$$

$\omega_{c1} = - \dfrac{R}{2 L} + \sqrt{ \left(\dfrac{R}{2 L}\right)^2 + \omega_r^2} \quad \quad (V)$
$\omega_{c2} = \dfrac{R}{2 L} + \sqrt{ \left(\dfrac{R}{2 L}\right)^2 + \omega_r^2} \quad \quad (VI)$

$\omega_{c1} \times \omega_{c2} = \omega_r^2 \quad \quad (VII)$

$$Q = \dfrac{\omega_r}{\Delta \omega}$$

$$Q = \dfrac {\omega_r} { \left(\dfrac{R}{2 L} + \sqrt{ \left(\dfrac{R}{2 L}\right)^2 + \omega_r^2} - \left(-\dfrac{R}{2 L} + \sqrt{ \left(\dfrac{R}{2 L}\right)^2 + \omega_r^2} \right) \right)}$$

$Q = \omega_r \dfrac{L}{R} \quad \quad (VIII)$

a) 对于一个RLC串联电路，设 $$R=300 \; \Omega$$、$$L = 100 \; mH$$、$$C = 100 \; \mu F$$，求谐振频率、截止频率和品质因数 $$Q$$。
b) 绘制平均功率 $$P_a$$ 对角频率 $$\omega$$ 的常用对数的图，并在图形上验证部分 a) 中解析得出的谐振和截止频率。

b)

$$L = 100 \; mH = 100 \times 10^{-3} \; H = 0.1 \; H$$
$$C = 100 \; \mu F = 100 \times 10^{-6} \; F = 0.0001 \; F$$

$$\omega_r^2 = \dfrac{1}{L C} = \dfrac{1}{0.00001} = 100000$$
$$\dfrac{R}{L} = \dfrac{300}{0.1} = 3000$$

$$\omega_{c1} = - 1500 + \sqrt{ 1500^2 + 100000} \approx 32.97$$ rad/s
$$\omega_{c2} = 1500 + \sqrt{ 1500^2 + 100000} \approx 3032.97$$ rad/s

$$Q = \dfrac{316.23}{3032.97 - 32.97} \approx 0.1054$$
b)

$$Log_{10}(\omega_{c_1}) = 1.51481$$ 给出 $$\omega_{c_1} = 10^{1.51481} = 32.71975$$ rad/s
$$Log_{10}(\omega_{c_2}) = 3.48356$$ 给出 $$\omega_{c_2} = 10^{3.48356} = 3044.80861$$ rad/s

D - 带详细解答的更多例子

a) 如果电阻 $$R$$ 为 $$30 \Omega$$，计算电容器 $$C$$ 的电容和电感器 $$L$$ 的电感。
b) 该电路的品质因数是多少？

a)

$$\omega_{c_1} = 2 \pi f_{c_1} = 1300 \pi$$ rad/s
$$\omega_{c_2} = 2 \pi f_{c_1} = 1900 \pi$$ rad/s

$$\omega_{c_1} \times \omega_{c_2} = \omega_{r}^2$$ 计算电路的谐振频率 $$\omega_{r}$$。
$$\omega_{r} = \sqrt {\omega_{c_1} \times \omega_{c_2}} = \sqrt {1300 \pi \times 1900 \pi } = 100 \sqrt{247} \pi = 4937.400$$ rad/s
$$\omega_{c_2} - \omega_{c_1} = \dfrac{R}{L}$$

$$L = \dfrac{R } {\omega_{c_2} - \omega_{c_1}} = \dfrac{30} {1900 \pi - 1300 \pi } = 0.01591$$ H
$$\omega_{r} = \dfrac{1}{\sqrt {L C} }$$

$$C = \dfrac{1}{\omega_{r}^2 L} = \dfrac{1}{(100 \sqrt{247} \pi)^2 \times 0.01591} = 2.5783 \times 10^{-6}$$ F

b)

$$Q = \dfrac{\omega_{r}}{\omega_{c_2} - \omega_{c_1}} = \dfrac{100 \sqrt{247} \pi}{1900 \pi - 1300 \pi } = 2.62$$