Motivation: to solve the normal Sinusoidal Signal sources more efficiently
为了更加简便得求解一般形式的正弦波
求解正弦输入的电路 (AC) Steady response
$$
L\frac{di}{dt}+Ri = V_m cos(\omega t+\phi)
\
i = \underbrace{\frac{-V_m}{\sqrt{R^2+\omega^2L^2}}cos(\phi-\theta)e^{(-R/L)t}}{\text{Transient response}}+\underbrace{\frac{-V_m}{\sqrt{R^2+\omega^2L^2}}cos(\omega t+\phi-\theta)}{\text{Steady
state response}}
$$
- Steady state solution is sinusoidal稳态是正弦形式的
- Response frequency = source frequency响应频率和输入频率相同
- Magnitude & phase of response differs from that of source响应的幅度和相位与源的响应不同（相位不同）
LTI: Linear Time Invariant (LTI) circuit线性时不变
WHY?
natural phenomenon are sinusoidal.
very easy signal to generate and transmit.
very easy to handle mathematically.
Sinusoidal signals
$v(t) = V_mcos(\omega t+\theta)$$V_m$peak value$\omega$angular frequency$\theta$phrase angle$f = \frac{1}{T} = \frac{\omega}{2\pi}$frequency
Root-Mean-Square(RMS)有效值
Def: RMS value of a current is equal to the value of the direct current that would produce the same average power dissipation in a resistance load
$$
V_t= V_mcos(\omega t + \phi) \
V_{rms} = \sqrt{\frac{1}{T}\int_0^Tv^2(t)dt} = \sqrt{\frac{1}{T}\int_0^Tcos ^2(\omega t +\phi)dt} = \frac{V_m}{\sqrt{2}}
$$
Complex Number
Phaser
Complex representation of the magnitude and phase of a sinusoid.
$$
V(t) = V_m cos(\omega t + \phi) = Re(V_m e^{j\phi}\cdot e^{j\omega t})= Re(\widetilde{V}e^{j\omega t})
\
\widetilde{V} = V_m e^{j\phi}
\
\Rightarrow v(t) = Re(\widetilde{V}e^{j\omega t})=V_m cos(\omega t + \phi)
\
\widetilde{V}_{rms} = \frac{V_m}{\sqrt{2}} e^{j\phi}
$$
Sinusoid-Phasor Transformation
$$
v(t)\leftrightarrow \widetilde{V} \Rightarrow \frac{dv(t)}{dt}\leftrightarrow j\omega \widetilde{V} \
v(t)\leftrightarrow \widetilde{V} \Rightarrow \int v(t) dt\leftrightarrow \frac{ \widetilde{V}}{j\omega}
$$
EXAMPLE
- 正弦
- 单频
- 稳态
注意条件
