motivation
which should be easy to solve according to the lecture 2.
Linear property 线性定理
What if
- $I_0 = 12A$?
- $V_0=90V$?
Will $I=aI_0+bV_0$？
Excitation(source)->Response (voltage/current)
$$
x \to \fbox{system}\to f(x)
$$
A linear circuit consists of only linear elements (resistors， capacitors and inductors), linear dependent sources, and independent sources.含有且仅有线性元件与电源
Homogeneity property 齐次性
$$
\alpha x \to \fbox{system}\to \alpha f(x)
$$
Example
Find f,g,where $I=f(I_0,V_0),P = g(I_0,V_0)$
solution:
Mesh analysis
$$
-(I-I_0)R_1-IR_2-V_0=0 \\
-I(R_1+R_2) = V_0-I_0R_1 \\
I=\frac{R_1}{R_1+R_2}I_0+\frac{-1}{R_1+R_2}V_0
$$
$$
P = UI = V_{R_1}I = (I_0-I)R_1I\\
P = (\frac{R_1}{R_1+R_2}I_0+\frac{1}{R_1+R_2}V_0) R_1 (\frac{R_1}{R_1+R_2}I_0+\frac{-1}{R_1+R_2}V_0) \\= a'I_0^2+b'V_0I_+c'V_0^2
$$
Super position Theorem 叠加定理
What if
- $I_0 = 0A, I'=?$
- $V_0=0V,I''=?$
Will $I=I'+I''$？
The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone./用独立源计算（就是将其他电源完全关闭状态下测算
Additivity property
$$
x_1+x_2\to\fbox{system}\to f(x)+f(x2)
$$
Proof
$$ I' = \frac{R_1}{R_1+R_2}I_0 \\ I'' = \frac{-1}{R_1+R_2}V_0 \\ I = I'+I'' = \frac{R_1}{R_1+R_2}I_0+\frac{-1}{R_1+R_2}V_0 = aI_0+bV_0 $$
Note:
- Linear Circuit
- one independent source at a time
- dependent source are left alone inside the circuit
Advantages:
- simplify the analyzing process
- Evaluate the sensitivity of response to a specific source
Super position Theorem is a inference of linear property
Example
Find $i$ using Super position Theorem
- $i$'s contribution is from the independent voltage source:
$$
10v-2Ωi'-1Ωi'-2i'=0 \\
i' = 2A
$$
- $i$'s contribution is from the independent current source:
$$
-2Ωi''-V=0\\
V-(i''+5A)1Ω-2i''=0
i' = -1A
$$
$$
i= i‘+i'' =1A
$$
Thevenin’s theorem/戴维南定理 • Source transformation电源变换定理 • Norton’s theorem/诺顿定理
What if $R_2 = 1Ω,I=?$
What if $R_2 = 5Ω,I=?$
Substitution Theorem
If we know the voltage across/ the current trough any branch of a network, the branch can be replaced by a voltage or current source that will make the same voltage and current through that branch.
proof
core: KCL,KVL 等物理关系在原电路中保持不变
Thevenin’s theorem/戴维南定理
motivation
complex linear system (CLS) which contains resistors independent sources dependent sources only.
Equivalent circuit for CLS Handle variable loads.
proof:
$$
U' = U_{oc} = U_{Th} \\
U'' = -R_{eq}i \\
U= U'+U'' =U_{Th}+R_{eq}i \\
U = U_{Th}-R_{Th}i
$$
proof2:
In CLS assuming M independent voltage sources
and N independent currunt sources
$$
U= Ai+\sum^{M}{j=1}B_jV{S_j}+\sum^N_{k=1}C_kI_{S_k}\\
U= Ai+CONST \\
$$
Find $CONST$ if $i=0$, $U=U_{oc} =U_{Th}$
Find $A$ if $V_{S_j},I_{S_k}=0,A=U/i=-R_{Th}$
$$U = U_{Th}-R_{Th}i$$
Note
- $V_{Th}= U_{oc}$
- to get $R_{Th}$
- equivalent resistance with all indeoendent sources turned off (does not work for circuits with dependent sources)只适用于无非独立源
- calculate shout circuit current (isc)-> $R_{Th} = \frac{V_{oc}}{i_{sc}}$
- Deactivate all independent sources, add external source $V_{ex}$, slove current $i_{ex}$ $R_{Th} = \frac{V_{ex}}{i_{ex}}$
- 任意取(u,i)算系数
Example1
Example2
source transformation and norton Theorem
$$
I_{sc} = I_{N}\\
R_N = R_{Th} \\
I_N = \frac{V_{Th}}{R_{Th}} \\
$$
Maximum Power Transfer Theorem
$$
P_L = I^2R_L=(\frac{V}{R_s+R_L})^2R_L=\frac{V^2}{R_s^2/R_L+2R_s+R_L}\\
when \space R_L = \pm R_S, P_L \space gets \space its \space maximum \space power
$$
