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# Lecture 13 : RL Circuit | ||
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> 看到了老师手写的教案非常感动 | ||
> | ||
> 但是到场人数不到20人 | ||
> | ||
> 难绷 | ||
## Natural Response of RL Circuit(RL电路的自然响应) | ||
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### RL电路的UI关系 | ||
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当电容或电感具有初始能量时,即使没有外部信号输入,电路中的电压和电流也会发生变化。这种变化称为电路的自然响应。 | ||
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只有由电路上的初始能量决定的自然响应,称为电路的自然响应。其他的还有瞬态响应(Transient Response)、零输入响应(Zero Input Response)、无源响应(Source-Free Response)等。 | ||
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对于一个RL电路,我们可以推得: | ||
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![RL Circuit](Lecture13.assets/1733123563677.png) | ||
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$$ | ||
i(t)+\frac{L}{R}\frac{di(t)}{dt}=0 | ||
$$ | ||
$$ | ||
\Rightarrow \frac{di(t)}{dt} = -\frac{R}{L}i(t) | ||
$$ | ||
$$ | ||
\Rightarrow \frac{d}{dt}ln(i(t)) = -\frac{R}{L} | ||
$$ | ||
$$ | ||
\Rightarrow ln(i(t)) = -\int\frac{R}{L}dt = -\frac{R}{L}t + K | ||
$$ | ||
$$ | ||
\Rightarrow i(t) = e^{ln(i(t))} = e^{-\frac{R}{L}t + K} = e^{-\frac{R}{L}t}\cdot e^K = A\cdot e^{-\frac{R}{L}t} | ||
$$ | ||
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可以得出,当t在0时,有 $i(0) = A\cdot e^0 = A$ ,即为电感器上的电流。或者可以说 $A=i(0)=I_0$ 。 | ||
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通过这个来计算电压的大小,可以得到: | ||
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$$ | ||
v(t)=L\frac{di(t)}{dt} = L\cdot (-\frac{R}{L})\cdot I_0\cdot e^{-\frac{R}{L}t} = -R\cdot I_0\cdot e^{-\frac{R}{L}t} | ||
$$ | ||
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功率则是: | ||
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$$ | ||
p(t) = v(t)\cdot i(t) = -R\cdot I_0\cdot e^{-\frac{R}{L}t}\cdot I_0\cdot e^{-\frac{R}{L}t} = -R\cdot I_0^2\cdot e^{-\frac{2R}{L}t} | ||
$$ | ||
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### RL电路的时间常数 | ||
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同样的,我们可以定义一个时间常数 $\tau$ 来描述电路的自然响应。 $\tau$ 的大小越大,电路的自然响应越慢。对于一个RL电路,我们把它定义为: | ||
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$$ | ||
\tau = \frac{L}{R} | ||
$$ | ||
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在这种情况下,电流、电压、功率的大小可以分别写为: | ||
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$$ | ||
i(t) = I_0\cdot e^{-\frac{t}{\tau}} | ||
$$ | ||
$$ | ||
v(t) = -R\cdot I_0\cdot e^{-\frac{t}{\tau}} | ||
$$ | ||
$$ | ||
p(t) = -R\cdot I_0^2\cdot e^{-\frac{2t}{\tau}} | ||
$$ | ||
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和RC电路一样,当 $t=\tau$ 时,电感器上的电流下降到原来的 $1/e$ 。把这个时间常数代入电流的表达式中,可以得到: | ||
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$$ | ||
i(\tau) = I_0\cdot e^{-1} = \frac{I_0}{e} | ||
$$ | ||
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在 $t=5\tau$ 时,电流下降到原来的 $1/e^5$ 。此时电流的大小为原本的 $1/e^5$ 倍,小于1%的 $I_0$ 。**在大多数情况下,五倍时间常数的时间后的静态相应可以被忽略不计**。 | ||
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电流的衰减速率在 $t=0$ 时最快,随着时间的增加,电流的衰减速率逐渐减小。在 $t=\infty$ 时,电流的衰减速率为0。 | ||
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> 还是那个老生常谈的,在计算时间常数是,所需要的电阻值和电感值可能是整个电路等效电阻和等效电感的值。 | ||
> | ||
> 也就是说 $\tau = \frac{L_{eq}}{R_{eq}}$ 。 | ||
## Step Response of RL Circuit(RL电路的阶跃响应) | ||
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### UI关系 | ||
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电路对直流信号输入(或者说是阶跃信号)的响应称为电路的阶跃响应。对于一个RL电路,它的阶跃响应是指在电感器上的电压和电流的变化情况。当电路中有一个电感器,电路中的电流不会瞬间变化,而是会逐渐变化。 | ||
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![Step Responce : Circuit](Lecture13.assets/1733125467948.png) | ||
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我们可以根据电路图写出这样的节点方程: | ||
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$$ | ||
-I_s + i(t) + \frac{L}{R}\frac{di(t)}{dt} = 0 | ||
$$ | ||
$$ | ||
\Rightarrow \frac{d}{dt}\ln|i(t)-I_s| = -\frac{R}{L} | ||
$$ | ||
$$ | ||
|i(t)-I_s| = \pm e^{-\frac{R}{L}t + K} = \pm e^{-\frac{R}{L}t}\cdot e^K = \pm A e^{-\frac{R}{L}t} \ (A=e^K) | ||
$$ | ||
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于是,电流的响应可以写为: | ||
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$$ | ||
i(t) = I_s + A e^{-\frac{R}{L}t} | ||
$$ | ||
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此时 $A=i(0)-I_s$ ,即为电感器上的电流。 | ||
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同样把 $\tau=\frac{L}{R}$ 作为电路的时间常数,可以得到: | ||
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$$ | ||
i(t) = I_s + (I_0-I_s)e^{-\frac{t}{\tau}} | ||
$$ | ||
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同样的,当 $t\rightarrow\infty$ 时,电流的大小趋于 $I_s$ 。这也可以通过其微分方程来解释: | ||
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$$ | ||
\frac{di(t)}{dt} + \frac{R}{L}i(t) = \frac{R}{L}I_s | ||
$$ | ||
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当 $t\rightarrow\infty$ 时,电流的变化率为0,即 $\frac{di(t)}{dt}=0$ 。所以电流的大小趋于 $I_s$ 。 | ||
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### 时间常数 | ||
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同样的,我们可以定义一个时间常数 $\tau$ 来描述电路的阶跃响应。 $\tau$ 的大小越大,电路的阶跃响应越慢。对于一个RL电路,我们把它定义为: | ||
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$$ | ||
\tau = \frac{L}{R} | ||
$$d | ||
![SR for Diff tau](Lecture13.assets/1733126302350.png) | ||
> 超级车轱辘话反复说 | ||
> | ||
> 我都不好意思再在之后的RLC电路中再抄一遍这个话了 | ||
--- | ||
## Summary | ||
![Summary of RL Circuit](Lecture13.assets/1733125144544.png) | ||
> 老师在课上提到Moodle上给的PPT并不好理解,而他手写的PDF阅读资料更好理解一些,所以他建议我们看他手写的资料 | ||
> | ||
> 打开一看甚至是扫描全能王扫描的,是真用心了 | ||
> | ||
> 课后还问手写的PDF对我们有没有帮助,人确实够好 | ||
> | ||
> 唉 |
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