Electrical Engineering

I like Artur.

Lec 1

Conductor - Electrical resistance - Opposition to the flow of the current

       -  𝞀(Resistivity)[Ω·m] / 𝞂(Conductivity)[Sm/m]    reciprocal

R = 𝞀l/A = l/(𝞂A)
Passive/Active Linear/non-linear Lumped(集总)/Distributed DC/AC
PASSIVE/ACTIVE 无源/有源 **Element that stores energy. Voltage and current through conductors connecting elements does not vary. Such circuit are called lumped.

Equivalent circuit

A simplified version of an electrical circuit that retains the electrical characteristics of the original circuit.

One port network is a two terminal electrical network in which, current enters through one terminal and leaves through another terminal. Resistors, inductors and capacitors are the examples of one port network because each one has two terminals.
Similarly, two port network is a pair of two terminal electrical network in which, current enters through one terminal and leaves through another terminal of each port. Two port network representation is shown in the following figure.

Energy source

*AC source uses lowercase letters

Inductor L=Ψ/I Ψ(Magnetic Linkage)=Nφ
*Magnetic Linkage 磁链

The inverse value of the resistance is a conductance in [Simens] R=U/I g=I/U
Transformation of EMF source into current source or backwards at P35

Circuit topology

Definition of node, junction, branch and loop at P53,54
Transformation between star and delta at P57,58

Wheatstone bridge

As the order |R1 R2 |, the bridge reaches balance when R1·R3=R2·R4
|R3 R4|

Lec 2

Circuit Analysis

Two approaches to circuit analysis: matrix-based topological

FUNDAMENTAL：

Kirchhoff’s current law(KCL)

The current flowing in a certain node equals to the current flowing out
-The algebraic sum of all branch currents flowing into any junction must be zero.

Kirchhoff’s voltage law(KVL)

The algebraic sum of the branch vantages around any closed path in a network must be zero.
-The voltage between two nodes is independent of the path along which it is accumulated.

I.Consider the topology of the circuit. Find junctions, branches with and without current sources
II.Choose the direction of currents flow through each branch. The direction must be the same as the direction of the voltage across passive elements, or as the direction of the current source.

|KCL Equation|=number of Junctions -1
|KVL Equation|=|branches|-|current source|-|KCL Equation|

A loop contains no other loops is known as mesh.
*It’s better to choose meshes.

Mesh-currents P34,40
When current source exists, choose direction arbitrarily.
Nodal-voltages P49,55 Reference node(ground) is important.

Superposition Theorem

The response (v/c) in any branch of a linear circuit having more than one source equals the algebraic sum of the responses caused by each source acting alone, while all the other sources are replaced by their internal resistances.

Thevenin’ s theorem

Find Thevenin Resistance by removing all voltage sources and load resistor
Find Thevenin Voltage by plugging in the voltages.
Use these to find the current flowing through the load.

Lec 3

AC current
*be careful with arctan/arctan2

Advantages of AC current

​ More economical power transmission (transformers)
​ Easily converted into DC (AC adapter)
​ low maintenance costs of high-speed motors
​ Sinusoidal form is not changed by basic components (R,L,C)

Disadvantage

​ modern consumer electronics operate internally on DC
​ Skin effect results in increasing the effective resistance of the conductor
​ Requires increased insulation
​ High voltage - dangerous to work with
F within range [50,60]
​ reduce skin effect to increase f
​ prevent incandescent lamp from flickering at lower f
​ precent the influence of centrifugal forces.、

Root mean square
Effective equivalent heat. divide by 1/2*sqrt(2)

Phasor Analysis - British scientist Oliver Heaviside

Using complex representation
X=X1+jX2 |X|=sqrt(X1^2+X2^2) tg(a)=Im(X)/Re(X)=X2/X1

Exponential form/ Analysis representation
X=|X|e^(ja)
|X|e^(ja)=|X| cos a+j |X| sin a , where X1=|X|cos a, X2=|X|sina
|X|=Xm, a=ωt+𝚿 Usually we use the phasors with the magnitude values, as it is presented in the lecture 3. But some times, to simplify the look, and I repeat, just the look of the power equation, we will use the effective values for phasors.
Phasor of current is ‘in-phase’ with phasor of voltage 𝚿_u=𝚿_i
p(t)=UrIr(1-cos(2ωt)) Active Power P=UrIr [W] the average power consumed by a resistance over on cycle

Inductor - X_l inductive reactance
U=ωLI 𝚿_u=𝚿_i+Pi/2
X_l=ωL
p(t)=UIsin(2ωt) Reactive Power Q=U_lI_l=Ul^2/X_l
*[VAr] is the portion of electricity that helps establish and sustain the magnetic field of an inductor ( volt-amperes reactive)

Capacitor -X_c capacitive reactance
I=ωCU 𝚿_i=𝚿_u+pi/2
X_c=1/(ωC)
p(t)=UIsin(2ωt) Reactive Power Q=UcIc=Uc^2/X_c
** the average reactive power of a inductor/capacitor over one cycle is zero

Power triangle
Complex power (volt-ampere [VA])
S=P+jQ=Se^(j*𝝋)
P=Σ{P_Rn} Q=Σ{Q_Ln-Q_Cn}
S=UI cos(𝝋)+J UI sin (𝝋)
Apparent power(VA) S=|S|=sqrt(P^2+Q^2)
Complex power (in phasor form) S=Se^j(𝚿_u-𝚿_i)=Ue^(j𝚿_u)Ie^(-j𝚿_i)

Power factor
cos(𝝋)=P/S=P/sqrt(P^2+Q^2)

*Reactive power compensation is one of the most effective ways to reduce consumed electric energy and improve power quality.
reducing / avoiding
costs, network loss
penalty charges from utilities for excessive consumption of reactive power
increasing /improving
system capacity and save costs on new installations, power availability
system power factor, voltage regulation in the network

Lec 4

Frequency response

Analysis of a circuit with varying frequency of a sinusoidal sources is called the frequency response of a circuit .
A circuit performed the frequency selection is called filters, because of their ability to filter out certain input signals on the basis of frequency.
Transfer function H(j*ω) is the magnitude and phase of the output voltage to the magnitude and phase of the input voltage of a circuit.

Re(H)={Re(A)Re(B)+Im(A)Im(B)} / {Re(B)^2+Im(B)^2} // jω
Im(H)={Im(A)Re(B)-Im(B)Re(A)} / {Re(B^2)+Im(B)^2} // jω
H(ω)=|H(jω)|=sqrt(Re(H)^2+Im(H)^2)
H(ω)=sqrt{ {Re(A)^2+Im(A)^2} / {Re(B)^2+Im(B)^2} }
𝝋(ω)=arctg(Im(H)/Re(H))
=arctg( {Im(A)Re(B)-Im(B)Re(A)} / {Im(A)Im(B)+Re(A)*Re(B)})

Group delay

𝛕(ω) [S] is the time delay of the amplitude envelopes of the various sinusoidal components of a signal through a device under test, and is a function of frequency for each component.
𝛕(ω) = -d𝝋(ω) / dω = - { d(Im(H)/Re(H)) / dω } / {1+{Im(H)/Re(H)}^2I
𝛕(ω) = \frac { d(Im(H))/ dω Re(H) - Im(H) d(Re(H))/ dω } { Re(H)^2 * (1+ (Im(H)/Re(H))^2 ) }

Case of RC

​ Uc
H_uc = \frac {1} {jωCR+1} Magnitude = \sqrt{\frac{1}{1+(ωCR)^2}}
𝝋(ω)= arctg(-ωCR)
𝛕(ω) = \frac{CR}{1+(ωCR)^2}
The transfer function magnitude is decreased by the factor 0.707 from its maximum value is called cut-off frequency f_c
A low-Pass filter passes signals at frequencies lower than the cutoff frequency from the input to the output.
​ Ur
H_ur = \frac{jωCR}{jωCR+1} Magnitude = \sqrt{\frac{(ωCR)^2}{1+(ωCR)^2}}
𝝋(ω) =arctg(\frac{1}{ωCR})
𝛕(ω) = \frac{CR}{(ωCR)^2+1}
A high-pass filler passes signals at frequencies higher than the cutoff frequency from the input to the output

Case of RL

​ Ul
H_ul = \frac {jωL} {jωL+R} Magnitude = \sqrt{ \frac {(ωL)^2} {R^2+(ωL)^2} }
𝝋(ω) = arctg( R / (ωL) )
𝛕(ω) = \frac {RL} {(ωL)^2+R^2}
​ Ur
H_ur = \frac {R} {jωL+R} Magnitude = \sqrt { \frac {R^2} {R^2+(ωL)^2} }
𝝋(ω) = arctg(-ωL/R)
𝛕(ω) = \frac {RL} {(ωL)^2+R^2}

Lec 5

When resonant, the impedance of inductor and capacitor cancels ( metaphor ) each other, their reactance are equal in value, opposite in sign.

In series circuit

At resonance, the total circuit impedance is minimum.
ωL-1/(ωC)=0 ω=1/ sqrt{LC}
It’s purely resistive and the phase difference between input current and voltage sinusoids is zero.

Voltage magnification

The voltages across capacitor and inductor can be greater than the supply voltage.

Q factor

It is the ratio of the energy stored in the oscillating resonator to the energy dissipated per cycle by damping process.
The magnification is defined by Q factor.
𝞀 - characteristic resistance of series RLC-circuit
𝞀 = ω_0L = 1/{ω_0C} = \sqrt{L/C}
Q_s = 𝞀/R = U_c/U_in = U_l/U_in = Lω_0/R
=\sqrt{L/C} /R
*The Q factor represents the effect of electrical resistance at the resonance condition.

Maximum power transfer

Internal impedance Z_s, which is connected to a load Z_load.
It occurs from source to load when the Z_load is equals to the complex conjugate of source impedance Z_s
jX_s=-jX_load
*共轭复数

In parallel circuit

At resonance, the total circuit impedance is maximum.
The imaginary part of complex impedance of a parallel RLC-circuit tends to infinity.
Z_lc=Im(Z_in)=jωL/{1-j(ω^2LC)} ->∞
Phase shift between input current and voltage is also 0.

Current magnification

The current flowing through the capacitor and inductor can be greater than the supply current.

Q factor

Q_p = R/𝞀 = I_l/I_in = I_c/I_in
= R/ sqrt{L/C}

Practical applications of resonant circuits

​ Band-pass or band-stop filter with high selectivity.
​ Maximum power transfer
​ Build oscillators to produce sinusoids of a given frequency
​ To model electric systems to avoid over voltage.
​ To model physical systems such as buildings and bridges to avoid the large amplitude responses.

Multiple resonance networks

​ jω=s -> Z_sL,Z_c=1/(sC)
Z_in(S)=\frac{ bns^n+bn-1s^(n-1)+…+b0 } {a(n-1)s^(n-1)+…..+a1S+1 }
The values of S_bn that causes the numerator to be zero are the zeroes, and the values of S_am that causes the denominator to be zero are the poles.

Resonant frequency

​ ω0= Im(S_b) of a series resonance
​ ω0= Im(S_a) of a parallel resonance

Lec 6

Three phase Circuit
I think basically like what slides said…

Lec 7

First order transient

The voltage across capacitor and current flow through an inductor can not change instantaneously, and these values( right before the switching )are called initial conditions.

Practical applications of transient analysis.

​ 1. To determine the effect of Turing energy sources ON and OFF to prevent the faults.
​ 2. To calculate the circuit response on changing excitation.
​ 3. To know the length in time of transients.
​ 4. To produce damped oscillations in a circuit containing two reactance elements, for example, for design of electronic signal generator.

Homogeneous and particular solutions

​ 1. find the initial conditions.
​ 2. Write the system of differential equations for the case when transient is started. Transform the system to the one differential equation with one unknown variable
​ 3. Find the particular solution of the differential equation.
​ 4. Find the homogeneous solution of differential equation.
​ 5. Write the total solution as the sum of the particular and homogeneous solutions; use the initial conditions to solve the remaining constants; plot the derived function.

y=y_Steady state+ y_Transient
The former satisfies the inhomogeneous differential equation for a particular excitation. Derived by the SS circuit analysis methods.
The later one satisfies the homogeneous one with zero excitation. It depends only on the internal energy storage properties of the circuit and not on the external inputs.

RC-circuit

Steady state, as it is named, when the transient over.
Transient solution: in KVL, set the sum to zero.
Forced and natural response
U_c=E(1-e^(-t/τ))+U_0e^(-t/τ) τ=RC [second]
, where E is the external input, thus this part a.k.a. forced response.
As RCd(y_tr)/dt+y_tr=0, we substitute that a1d(y_tr)/dt+a0y_tr=0,
we have a1SAe^(st)+a0Ae^(st)=0 —> Ae^(st)(a1S+a0)=0
here, a1S+a0 is the first order polynomial expression, aka. Characteristic equation.
s - the root of characteristic equation. We got root s=-1/RC <=> 𝜏=1/abs(s)=RC, which is time constant
Now we know Uc and we’re able to calculate I(t)=C(dUc)/dt=C(U0-E)se^(st)=(E-U0)/R e^(-t/𝜏)

RL-circuit

Generally, the UL(∞)=0.
Transient solution, set the original differential equation to zero. RI+Ldi/dt=0
Then using general integration we have I(t)=Ae^(-Rt/L), using initial conditions to solve it.
RAe^(st)+Ld(Ae^(st))/dt=0 -> RAe^(st)+LsAe^(st)=0
then we have Ae^(st)(R+Ls)=0
Root- s=-R/L, and 𝜏=L/R. Thus Ul(t)=E*e^(-Rt/L)

Heaviside’s Operational calculus

The residue theorem provides an effective technique for inverse transformation.
Res(F(s)e^(st))=lim{s->s_k}(F(s)e^(st)(S-Sk))

DC source:

​ e(t)=E <> E(s)=E/s e(t) | J(t) <> E(s) | J(s)

Sinusoidal:

​ e(t)=EmSin(ɷt)<>E(s)=(Em*ω)/(ω^2+s^2)

Differential

​ Dy/dt<>sY(s)-Y(0) d^2y/dt^2<>s^2Y(s)-sy(0)-dy(0)/dt d^ny/dt^n<>s^nY(S)-Σ(k=1->n)s^(n-k)y^(k-1)(0)
​ , thus for zero initial conditions: d^ny/dt^n<>s^n Y(S)

UL(t)=Ldi/dt<> LsI(s)-LI(0)=UL(s)， where Ls can be substitute as Resistor and LI(0) is power source. note that the equivalent power source’s direction is the same as voltage across inductor.

I(t)=Cdu/dt<>CsU(s)-Cu(0)=I(s)
Uc(s)=(I(s)+cu(0))/SC=I(s)/sc + u(0)/s, where 1/(CS) as resistor and U(0)/s is power source.
*the direction is the opposite to voltage across capacitor.
IF initial conditions are zero, capacitor and inductor can be described by impedance in s-domain
Z_L=sL Z_c=1/(SC)

Algorithm

​ 1. find the initial conditions;
​ 2. Transform the network: replace the time-domain current and voltage sources by their Laplace transforms; transform them all to s-domain; write the system of equations in s-domain;
​ 3. Find the currents and voltages as fractionally rational expressions in s-domain ;
​ 4. Use the residue theorem to derive the currents and voltages in time domain;

Lec 8

Second-order circuit Basically they’re same.

Usually we make the Uc the only variable in differential equation and U_l becomes second-order differential equation.
When deal with transient solution, again set the original differential equation to 0

Then the characteristic equation( Polynomial) :
LCs^2+Rcs+1=0
Obviously, there are two roots of this quadratic equation.
can be presented as: s1,2= -R/2L±√[(R/2L)^2-1/(LC)] *S1S2=1/(LC)
make a little bit change : s1,2= -𝝳±√(𝝳^2-ω0^2)
Here, 𝝳(delta) - damping factor
ω0 - resonance frequency
𝞀 - characteristic resistance | √(L/C)
s1,2= 𝝳 {-1 ± √[1-(2𝞀/R)^2]} | new form

Thus we derive three different cases of transient

​ R>2𝞀 - roots are real -> over damped transient solution
​ R=2𝞀 - real repeated roots -> critically damped ~
​ R<2𝞀 - complex conjugate roots -> underdamped ~
​ For the first case, we have Y_tr=A1e^(s1t)+A2e^(s2t)
​ For the second one, we have Y_tr=A1e^(st)+A2e^(st)t
​ For the last s1,2=-𝝳±jω, we have Y_tr=Ae^(-𝝳t)sin(ωt+ψ)<>A1e^[(-𝝳+jω)t]+A2e^[(-𝝳-jω)t]
​
For an underdamped transient solution, in plots
​ all parameters are in the form of decaying sinusoidal functions.
In residue theorem when ns=1 (multiplicity of the roots 重根) we can use the original methods to execute.
However when ns>1 the formula for the residue should be expressed as
​ Res[F(s)]_sk=1/ [(n-1)!] lim[s->sk] d(ns-1)/ d[s^(ns-1)] [F(s)e^(st) (s-sk)^ns]