[Circuit] AC Sinusoids 穩態分析

電路知識：AC Sinusoids 穩態分析
工具：Qucs

功能：穩態分析，轉換為 phasor domain 計算 sinusoids 的電壓電流

只適用 sinusoids，必需利用 fourier series 轉換並將 DC 成份去除才適用，務必小心使用

Phasor 基本公式

$$ \require{enclose} \begin{align*} z=x+jy &=r(cos \phi + jsin \phi)=re^{j \phi} =r\enclose{phasorangle}{\phi} \\ z^*=x-jy &=r(cos \phi - jsin \phi)=re^{-j \phi} = r\enclose{phasorangle}{-\phi} \\ |z|^2 &=z \times z^*=r^2\\ e^{\pm j\phi} &= cos\phi \pm jsin\phi\\ cos \phi &= \mathrm{Re}(e^{j\phi})\\ sin \phi &= \mathrm{Im}(e^{j\phi})\\ \end{align*} $$

Phasor-domain 推導

因推論前提皆建立在 \(cos(wt+\phi)\) 上
故訊號皆需轉為 \(cos(wt+\phi)\) 表示，再轉過去 phasor domain，可利用 Fourier series
時域的表現，就是 \(\require{enclose} r\enclose{phasorangle}{\phi}\) 實數的部分，也就是 \(cos(wt+\phi)\) 的部分

$$ v(t)=V_mcos(wt+\phi)\\ \mathbf{V}=V_me^{j\phi}=V_m\enclose{phasorangle}{\phi}$$

$$ \begin{align*} v(t)&=V_mcos(wt+\phi)\\ &=\mathrm{Re}(V_me^{j(wt+\phi)})\\ &=\mathrm{Re}(\mathbf{V}e^{jwt})\\ \mathbf{V}&=V_me^{j\phi}=V_m\enclose{phasorangle}{\phi}\\ \end{align*} $$

電壓電流轉換表

$$ \begin{align*} V_mcos(wt+\phi) &\Leftrightarrow V_m\enclose{phasorangle}{\phi}\\ V_msin(wt+\phi) &\Leftrightarrow V_m\enclose{phasorangle}{\phi-90^{\circ}}\\ I_mcos(wt+\phi) &\Leftrightarrow I_m\enclose{phasorangle}{\phi}\\ I_msin(wt+\phi) &\Leftrightarrow I_m\enclose{phasorangle}{\phi-90^{\circ}}\\ \end{align*} $$

微分 in Phasor domain

$$ v(t)=V_mcos(wt+\phi)\\ \frac{\mathrm{d} v}{\mathrm{d} t} \Leftrightarrow jw\mathbf{V}$$

$$ \begin{align*} \frac{\mathrm{d} v}{\mathrm{d} t} &= -wV_msin(wt+\phi)\\ &=wV_mcos(wt+\phi + 90^{\circ})\\ &=\mathrm{Re}(wV_me^{jwt}e^{j\phi}e^{j90^{\circ}})\\ &=\mathrm{Re}(jw\mathbf{V}e^{jwt})\\ \frac{\mathrm{d} v}{\mathrm{d} t} &\Leftrightarrow jw\mathbf{V}\\ \end{align*} $$

積分 in Phasor domain

$$v(t)=V_mcos(wt+\phi)\\ \int vdt \Leftrightarrow \frac{\mathbf{V}}{jw}$$

$$ \begin{align*} \int vdt &= \frac{1}{w}V_msin(wt+\phi)\\ &=\frac{1}{w}V_mcos(wt+\phi - 90^{\circ})\\ &=\mathrm{Re}(\frac{1}{w}V_me^{jwt}e^{j\phi}e^{-j90^{\circ}})\\ &=\mathrm{Re}(\frac{-j}{w}\mathbf{V}e^{jwt})\\ &=\mathrm{Re}(\frac{1}{jw}\mathbf{V}e^{jwt})\\ \int vdt &\Leftrightarrow \frac{\mathbf{V}}{jw}\\ \end{align*} $$

RLC in Phasor domain

元件	Time domain	Phasor domain	Impedance
\(R\)	\(v(t)=Ri(t)\)	\(\mathbf{V}=R\mathbf{I}\)	\(\mathbf{Z}=R\)
\(L\)	\(v(t)=L\frac{\mathrm{d} i(t)}{\mathrm{d} t}\)	\(\mathbf{V}=jwL\mathbf{I}\)	\(\mathbf{Z}=jwL\)
\(C\)	\(i(t)=C\frac{\mathrm{d} v(t)}{\mathrm{d} t}\)	\(\mathbf{V}=\frac{\mathbf{I}}{jwC}\)	\(\mathbf{Z}=\frac{\mathbf{1}}{jwC}\)

基本電路定律

歐姆定律
impedance \(\mathbf{Z}\) 單位仍為 \(\Omega \)
Admittance \(\mathbf{Y}\) 單位仍為 \(S \)

$$ \mathbf{V=ZI=\frac{I}{Y}}\\ \mathbf{Z}=R+jX=\left | \mathbf{Z} \right |\enclose{phasorangle}{\theta}\\ $$

Kirchhoff 電路定律

KVL
$$\mathbf{V}_1+\mathbf{V}_2+\cdots+\mathbf{V}_n = 0$$

$$ \begin{align*} v_1+v_2+\cdots +v_n &=0\\ V_{m1}cos(wt+\theta _1)+V_{m2}cos(wt+\theta _2)+\cdots +V_{mn}cos(wt+\theta _n) &= 0\\ \mathrm{Re}(V_{m1}e^{j\theta _1}e^{jwt})+\mathrm{Re}(V_{m2}e^{j\theta _2}e^{jwt})+\cdots +\mathrm{Re}(V_{mn}e^{j\theta _n}e^{jwt}) &= 0\\ \mathrm{Re}[(\mathbf{V}_1+\mathbf{V}_2+\cdots+\mathbf{V}_n)e^{jwt}] &= 0\\ \because e^{jwt}\neq 0 \ \therefore \mathbf{V}_1+\mathbf{V}_2+\cdots+\mathbf{V}_n &= 0 \end{align*} $$

KCL
$$\mathbf{I}_1+\mathbf{I}_2+\cdots+\mathbf{I}_n = 0$$

$$ \begin{align*} i_1+i_2+\cdots +i_n &=0\\ I_{m1}cos(wt+\theta _1)+I_{m2}cos(wt+\theta _2)+\cdots +I_{mn}cos(wt+\theta _n) &= 0\\ \mathrm{Re}(I_{m1}e^{j\theta _1}e^{jwt})+\mathrm{Re}(I_{m2}e^{j\theta _2}e^{jwt})+\cdots +\mathrm{Re}(I_{mn}e^{j\theta _n}e^{jwt}) &= 0\\ \mathrm{Re}[(\mathbf{I}_1+\mathbf{I}_2+\cdots+\mathbf{I}_n)e^{jwt}] &= 0\\ \because e^{jwt}\neq 0 \ \therefore \mathbf{I}_1+\mathbf{I}_2+\cdots+\mathbf{I}_n &= 0 \end{align*} $$

串並聯等效 Impedance
因 KVL or KCL，故當 \(i\), \(v\) 為 Sinusoids，其元件上的跨壓必定也為 Sinusoids，才可互相抵消為 0

串聯
$$\mathbf{Z_{eq} = Z_1+Z_2+\cdots +Z_N}$$

$$ \begin{align*} 0 &=-v+v_1+v_2+\cdots +v_N\\ 0 &= -V_{m}cos(wt+\theta)+V_{m1}cos(wt+\theta _1)+V_{m2}cos(wt+\theta _2)+\cdots +V_{mN}cos(wt+\theta _N)\\ 0 &= \mathrm{Re}(-V_{m}e^{j\theta}e^{jwt})+\mathrm{Re}(V_{m1}e^{j\theta _1}e^{jwt})+\mathrm{Re}(V_{m2}e^{j\theta _2}e^{jwt})+\cdots +\mathrm{Re}(V_{mN}e^{j\theta _N}e^{jwt}) \\ 0 &= \mathrm{Re}[(-\mathbf{V}+\mathbf{V}_1+\mathbf{V}_2+\cdots+\mathbf{V}_N)e^{jwt}] \\ \mathbf{V} &= \mathbf{V_1+V_2+\cdots +V_N} \\ &= \mathbf{I(Z_1+Z_2+\cdots +Z_N)}\\ \mathbf{Z_{eq}} &= \mathbf{\frac{V}{I}} = \mathbf{Z_1+Z_2+\cdots +Z_N} \end{align*} $$

並聯
$$\mathbf{\frac{1}{Z_{eq}}} = \mathbf{(\frac{1}{Z_1}+\frac{1}{Z_2}+\cdots +\frac{1}{Z_N})}$$

$$ \begin{align*} 0 &= -i+i_1+i_2+\cdots +i_N\\ 0 &= -I_mcos(wt+\theta)+ I_{m1}cos(wt+\theta _1)+I_{m2}cos(wt+\theta _2)+\cdots +I_{mN}cos(wt+\theta _N)\\ 0 &= \mathrm{Re}(-I_{m}e^{j\theta}e^{jwt})+\mathrm{Re}(I_{m1}e^{j\theta _1}e^{jwt})+\mathrm{Re}(I_{m2}e^{j\theta _2}e^{jwt})+\cdots +\mathrm{Re}(I_{mN}e^{j\theta _N}e^{jwt}) \\ 0 &= \mathrm{Re}[(-\mathbf{I}+\mathbf{I}_1+\mathbf{I}_2+\cdots+\mathbf{I}_N)e^{jwt}] \\ \mathbf{I} &= \mathbf{I_1+I_2+\cdots +I_N} \\ &= \mathbf{V(\frac{1}{Z_1}+\frac{1}{Z_2}+\cdots +\frac{1}{Z_N})}\\ \mathbf{\frac{1}{Z_{eq}}} &= \mathbf{\frac{I}{V}} = \mathbf{(\frac{1}{Z_1}+\frac{1}{Z_2}+\cdots +\frac{1}{Z_N})} \end{align*} $$

平均功率
詳細推導過程(含 DC)

$$ \begin{align*} \mathbf{V}&=V_m\enclose{phasorangle}{\theta _v}\\ \mathbf{I}&=I_m\enclose{phasorangle}{\theta _i}\\ P&=\frac{1}{2}\mathrm{Re}[\mathbf{VI^*}] \end{align*} $$

$$ \begin{align*} P&= \frac{1}{T}\int_{0}^{T}p(t)dt\\ &= \frac{1}{T}\int_{0}^{T}v(t)i(t)dt\\ &= \frac{1}{T}\int_{0}^{T}V_mI_mcos(wt+\theta _v)cos(wt+\theta _i)dt\\ &= \frac{1}{T}\int_{0}^{T}V_mI_m\frac{1}{2}[cos(\theta _v-\theta _i)+cos(2wt+\theta _v+\theta _i)]dt\\ &= \frac{1}{2}V_mI_mcos(\theta _v-\theta _i)\frac{1}{T}\int_{0}^{T}dt+\frac{1}{2}V_mI_m\frac{1}{T}\underset{0}{\underbrace{\int_{0}^{T}cos(2wt+\theta _v+\theta _i)dt}}\\ &= \frac{1}{2}V_mI_mcos(\theta _v-\theta _i)\\ &= \frac{1}{2}\mathrm{Re}[\mathbf{VI^*}]\\ \end{align*} $$

當 \(\theta _v = \theta _i \) ，也就是 \(R\)，為實功來源
$$ P=\frac{1}{2}V_mI_m=\frac{1}{2}I_m^2R=\frac{1}{2}|\mathbf{I}|^2R $$ 當 \(\theta _v - \theta _i = \pm90^{\circ} \) ，也就是 \(L\) 或 \(C\)，為虛功來源
$$ P=\frac{1}{2}V_mI_mcos90^{\circ}=0 $$

Complex Power

全皆是平均功率來看

複數功率 (Complex Power) \(= \mathbf{S}=P+jQ=\frac{1}{2}\mathbf{VI^*}=\mathbf{V_{rms}I_{rms}^*}=I_{rms^2}\mathbf{Z}=\frac{V_{rms}^2}{\mathbf{Z^*}}\)
視在功率 (Apparent Power) \(=S=|\mathbf{S}|=\frac{1}{2}V_mI_m=V_{rms}I_{rms}=\sqrt{P^2+Q^2}\)
實功 (Real Power) \(=P=\mathrm{Re}(\mathbf{S})=Scos(\theta _v-\theta_ i)\)
虛功 (Reactive Power) \(=Q=\mathrm{Im}(\mathbf{S})=Ssin(\theta _v-\theta_ i)\)
功率因數 (Power Factor) \(=\frac{P}{S}=cos(\theta _v - \theta _i)\)

$$ \begin{align*} \mathbf{S}&=\frac{1}{2}\mathbf{VI^*}\\ &= \frac{1}{2}V_mI_m\enclose{phasorangle}{\theta _v- \theta _i}\\ &= V_{rms}I_{rms}\enclose{phasorangle}{\theta _v- \theta _i}\\ &= \mathbf{V_{rms}I_{rms}^*}\\ &= \mathbf{ZI_{rms}I_{rms}^*}=I_{rms^2}\mathbf{Z}\\ &= \mathbf{V_{rms}\frac{V_{rms}^*}{Z^*}}=\frac{V_{rms}^2}{\mathbf{Z^*}}\\ \end{align*} $$

參考

Alexander & Sadiku, “Fundamentals of Electric Circuits, Second Edition”, McGraw-Hill, New York, NY, 2004.