# Common Constants in Electrical Power Systems Engineering

Speed of electric current (in 12-gauge copper wire) ≈ v ≈ 300 m/μs

Vacuum permittivity (electric constant) = ε0 =8.85418782 × 10-12 m-3 kg-1 s4 A2

# WIL: Making Quick References

Learning 1:
Quick references are sometimes also referred to as reference cards, reference sheets, crib sheets, and cheat sheets.

Learning 2:
Quick references used more in the early stages. If information is used often enough, the reference eventually gets to the mind.

Learning 3:
Having physical quick reference sheets helps with memorization using the method of loci.

Learning 4:
Yes search engines exist, committing some information to memory still helps though because of 1) current low bandwidth between the information found from search engines and information we need as well as 2) ongoing limited development, adoption, and accessibility of “ultra-high bandwidth brain-machine interfaces to connect humans and computers”(NEURALINK)

# WIL: Making Math Models

Learning 1:
A good general structure to present thought process of creating a math model of a system (especially physical) is:
1. Nomenclature
2. Diagram(s)
3. System equation(s)

# DC Motor Math Model

1 Nomenclature

2 Diagram

3 System equations

3.1 Electrical

Back emf is directly proportional to rotational speed of shaft:

$e \propto \dot{\theta} \rightarrow e = K_{e} \dot{\theta}$

Where:

e = back emf

$\dot{\theta}$ = rotational speed of stator

$K_{e}$ = constant

From Kirchhoff’s voltage law:

The directed sum of the potential differences (voltages) around any closed loopis zero.

$\sum_{k=1}^{n} V_{k} = 0$

Therefore:

$L \frac{di}{dt} + Ri = V - K_{e} \dot{\theta}$

3.2 Mechanical

Torque is directly proportional to current and electric field strength:

Assuming armature controlled motor therefore electric field strength is constant

$T \propto i \rightarrow T = K_{t}i$

Where:

T = torque

i = current

$K_{t}$ = constant

From Newton’s Second Law for Rotation:

If more than one torque acts on a rigid body about a fixed axis, then the sum of the torques equals the moment of inertia times the angular acceleration

$\sum_{k=1}^{n} T_{k} = I \Ddot{\theta}$

Therefore:

$J \Ddot{\theta} + b\dot{\theta} = K_{t}i$

4 Transfer function

In SI units, the motor torque and back emf constants are equal

Laplace transform equation:

$F(s) = \mathcal{L} \left[ f(t) \right] = \int_{0}^{\infty} f(t)e^{-st}dt$

General Laplace transforms:

$\mathcal{L} [\dot y(t)] = sY(s) - y(0)$

$\mathcal{L} {\Ddot y(t)} = s^2Y(s) - sy(0) - y(0)$

4.1 Electrical

$\mathcal{L} \left[L \frac{di}{dt} + Ri\right] = \mathcal{L} \left[V - K_{e}\right]$

Assuming i(0) = 0

$sI(s)L + RI(s) = V(s) - K_{e}s \theta(s)$

$I(s)(Ls + R) = V(s) - K_{e}s \theta(s)$

4.2 Mechanical

$\mathcal{L} \left[J \Ddot{\theta} + b\dot{\theta}\right] = \mathcal{L} \left[K_{t}i\right]$

Assuming

$\theta(0) = 0$

$s(Js+b) \ominus(s) = K_{t}I(s)$

Open loop transfer function given input is voltage and output is rotational speed:

$G(s) = \frac{\dot\ominus(s)}{V(s)}$

Voltage

From Equation $I(s)(Ls + R) = V(s) - K_{e}s \theta(s)$

$V(s) = I(s)(Ls+R)+K_{e}s\ominus(s)$

Rotational speed:

$\dot\ominus(s) = \ominus(s)s$

From Equation $s(Js+b) \ominus(s) = K_{t}I(s)$ & Equation $\dot\ominus(s)=\ominus(s)s$

$\dot\ominus(s) = \frac{K_{e}I(s)}{Js+b}$

Open loop transfer function for speed control:

From Equation $V(s)=I(s)(Ls+R)+K_{e}s\ominus(s)$ & Equation $\dot\ominus(s)=\frac{K_{e}I(s)}{Js+b}$:

$G(s) = \frac {I(s)K_{e}} {I(s)((Js+b)[(Ls+R)+K_{t} \dot \ominus(s)])}$

From Equation $s(Js+b) \ominus(s) = K_{t}I(s)$

$K_{t} = \frac{\dot \ominus(s)(Js+b)} {I(s)}$

From Equation $K_{t} = \frac{\dot \ominus(s)(Js+b)} {I(s)}$  & Equation $\dot\ominus(s)=\ominus(s)s$

$G(s) = \frac {K_{e}} {(Js+b)(Ls+R)+K_{t}^2}$

Open loop transfer function for position control:

From Equation $\dot\ominus(s)=\ominus(s)s$  & Equation $G(s)=\frac{K_{e}}{(Js+b)(Ls+R)+K_{t}^2}$

$G(s) = \frac {sK_{e}} {(Js+b)(Ls+R)+K_{t}^2}$

# Common Laplace Transforms

$F(s) = L \left\{ f(t) \right\} = \int_{0}^{\infty} f(t)e^{-st} dt$

Common laplace transforms

$\text{a is a constant}$

$L \left\{ a \right\} =$ $\frac{a}{s}$

$L \left\{ t\right\} =$ $\frac{1}{s^2}$

$L \left\{ t^n \right\} =$ $\frac{n!}{s^{n+1}}$

$L \left\{ e^{-at}\right\} =$ $\frac{1}{s+a}$

$L \left\{ te^{-at}\right\} =$ $\frac{1}{(s+a)^2}$

$L \left\{ 1 - e^{-at}\right\} =$ $\frac{a}{s(s+a)}$

$L \left\{ sin(\omega t)\right\} =$ $\frac{\omega}{s^2 + \omega^2}$

$L \left\{ cos(\omega t)\right\} =$ $\frac{s}{s^2 + \omega^2}$

$L \left\{ 1 - cos(\omega t)\right\} =$ $\frac{\omega^2}{s(s^2 + \omega^2)}$

$L \left\{ \omega t sin(\omega t) \right\} =$ $\frac{2 \omega^2 s}{(s^2 + \omega^2)^2}$

$L \left\{ sin(\omega t) - \omega t cos(\omega t) \right\} =$ $\frac{2 \omega^3}{(s^2 + \omega^2)^2}$

$L \left\{ sin(\omega t + \phi) \right\} =$ $\frac{sin(\phi) + \omega cos(\phi) }{s^2 + \omega^2}$

$L \left\{ e^{-at} sin(\omega t) \right\} =$ $\frac{\omega}{(s + a)^2 + \omega^2}$

$L \left\{ e^{-at} cos(\omega t) \right\} =$ $\frac{s+a}{(s + a)^2 + \omega^2}$

$L \left\{ e^{-at} (cos(\omega t) - \frac{a}{\omega}sin(\omega t )) \right\} =$ $\frac{s}{(s + a)^2 + \omega^2}$

$L \left\{ e^{-at} + \frac{a}{\omega}sin(\omega t ) - cos(\omega t) \right\} =$ $\frac{a^2 + \omega^2}{(s + a)(s^2 + \omega^2)}$

$L \left\{ sinh(\beta t \right\} =$ $\frac{\beta}{s^2- \beta^2}$

$L \left\{e^{-at} sinh(\beta t \right\} =$ $\frac{\beta}{(s+a)^2- \beta^2}$

$L \left\{cosh(\beta t) \right\} =$ $\frac{s}{s^2 - \beta^2}$

$L \left\{e^{-at} sinh(\beta t \right\} =$ $\frac{s+a}{(s+a)^2- \beta^2}$

$L \left\{e^{-at} f(t) \right\} =$ $F(s+a)$

$L \left\{tf(t) \right\} =$ $-F(s)$

$L \left\{ x \right\} =$ $\bar{x}$

$L \left\{ \dot{x} \right\} =$ $s \bar{x} - x(0)$

$L \left\{ \ddot{x} \right\} =$ $s^2 \bar{x} - sx(0) - \dot{x}(0)$

$\text{Unit Step Function:}$

$L \left\{ u(t-a) \right\} = s^{-1}e^{-as}$

$L \left\{ f(t-a)u(t-a) \right\} = F(s)e^{-sa}$

$\text{Dirac Delta Function:}$

$L \left\{ \delta (t) \right\} =$ $1$

$L \left\{ \delta (t-a) \right\} =$ $e^{-as}$

$\text{Convolution Integral:}$

$\int_{0}^{t} f(z) g(t-z) dz = F(s)G(s)$

$\int_{0}^{t} f(t - z)g(z) dz = F(s)G(s)$