User:Fozolo

: $$

\left[

\begin{array}{c}

\frac{d\omega_r}{dt} \\

\frac{di_a }{dt}

\end{array}

\right]

=

\left[

\begin{array}{cc}

A_{11} & A_{12} \\

A_{21} & A_{22}

\end{array}

\right]

\left[

\begin{array}{c}

\omega_r \\

i_a

\end{array}

\right]

+

\left[

\begin{array}{cc}

B_{11} & B_{12} \\

B_{21} & B_{22}

\end{array}

\right]

\left[

\begin{array}{c}

u_1 \\

u_2

\end{array}

\right]

: $\backslash mathcal\{L\}\; \backslash \{\; af(t)\; +\; bg(t)\; \backslash \}\; =\; aF(s)\; +\; bG(s)$

: $\backslash mathcal\{L\}\; \backslash \{\; f\text{'}(t)\; \backslash \}\; =\; sF(s)\; -\; f(0)$

: $\backslash Omega(s)\; =\; \backslash frac\{()V\_a(s)+()T\_L(s)\}\{s+()\}$

: $\backslash left.\; \backslash frac\{\backslash Omega(s)\}\{V\_a(s)\}\; \backslash right|\_\{T\_L(s)=0\}\; =?$

: $\backslash left.\; \backslash frac\{\backslash Omega(s)\}\{T\_L(s)\}\; \backslash right|\_\{V\_a(s)=0\}\; =?$

: $v\_a(t)\; =\; V\_a\; u(t)$

: $\backslash omega\_r(t)\; =\; \backslash mathcal\{L\}^\{-1\}\; \backslash left\backslash \{\; \backslash frac\{\backslash Omega(s)\}\{V\_a(s)\}V\_a(s)\; \backslash right\backslash \}$

----

= Lab 5a Solutions =

:(1) $T\_e\; -\; T\_L\; =\; J\backslash frac\{d\; w\_r\}\{dt\}\; +\; B\_m\; w\_r$

:(2) $v\_a\; =\; r\_a\; i\_a\; +\; L\_\{aa\}\backslash frac\{d\; i\_a\}\{dt\}\; +\; k\_v\; w\_r$

:(3) $T\_e\; =\; k\_T\; i\_a$

Solve equations (1) and (2) for $\backslash frac\{d\; w\_r\}\{dt\}$ and $\backslash frac\{d\; i\_a\}\{dt\}$.

: $\backslash frac\{d\; w\_r\}\{dt\}\; =\; \backslash frac\{1\}\{J\}\backslash left[-B\_m\; w\_r\; +\; k\_t\; i\_a\; -\; T\_L\; \backslash right]$

: $\backslash frac\{d\; i\_a\}\{dt\}\; =\; \backslash frac\{1\}\{L\_\{aa\}\}\backslash left[\; -k\_v\; w\_r\; -\; r\_a\; i\_a\; +\; V\_a\; \backslash right]$

: $$

\left[

\begin{array}{c}

\frac{d\omega_r}{dt} \\

\frac{di_a }{dt}

\end{array}

\right]

=

\left[

\begin{array}{cc}

-\frac{B_m}{J} & \frac{k_T}{J} \\

-\frac{k_v}{L_{aa}} & -\frac{r_a}{L_{aa}}

\end{array}

\right]

\left[

\begin{array}{c}

\omega_r \\

i_a

\end{array}

\right]

+

\left[

\begin{array}{cc}

-\frac{1}{J} & 0 \\

0 & \frac{1}{L_{aa}}

\end{array}

\right]

\left[

\begin{array}{c}

T_L \\

V_a

\end{array}

\right]

: $T\_e\; -\; T\_L\; =\; B\_m\; \backslash Omega\_r$

: $V\_a\; =\; r\_a\; I\_a\; +\; k\_v\; \backslash Omega\_r$

: $T\_e\; =\; k\_T\; I\_a$

: $T\_e\; -\; T\_L\; =\; J\backslash frac\{d\; w\_r\}\{dt\}\; +\; B\_m\; w\_r$

: $v\_a\; =\; r\_a\; i\_a\; +\; k\_v\; w\_r$

: $T\_e\; =\; k\_T\; i\_a$

Combine equations (1)-(3) to eliminate $i\_a$. First solve (1) and (3) for $i\_a$

: $i\_a\; =\; \backslash frac\{1\}\{k\_T\}\; \backslash left(\; J\; \backslash frac\{d\backslash omega\_r\}\{dt\}\; +\; B\_m\; \backslash omega\_r\; +\; T\_L\; \backslash right)$

Then substitute the result into (2).

: $v\_a\; =\; \backslash frac\{r\_a\}\{k\_T\}\; \backslash left(\; J\; \backslash frac\{d\backslash omega\_r\}\{dt\}\; +\; B\_m\; \backslash omega\_r\; +\; T\_L\; \backslash right)\; +\; k\_v\; \backslash omega\_r$

Convert the resulting equation to the frequency domain through application of Laplace transforms. Note that we choose the capital form of $\backslash omega$ ($\backslash Omega$), when in the frequency domain. Also, it is safe to assume $\backslash omega\_r(t)\; =\; 0$.

: $V\_a(s)\; =\; \backslash frac\{r\_a\}\{k\_T\}\; \backslash left(\; Js\backslash Omega\_r(s)\; +\; B\_m\backslash Omega\_r(s)\; +\; T\_L(s)\; \backslash right)\; +\; k\_v\; \backslash Omega\_r(s)$

Solving the resulting equation for $\backslash Omega\_r(s)$ yeilds

: $\backslash Omega\_r(s)\; =\; \backslash frac\{\backslash frac\{k\_T\}\{Jr\_a\}V\_a(s)\; -\; \backslash frac\{1\}\{J\}T\_L(s)\; \}\{s+\; \backslash frac\{B\_m\; r\_a\; +\; k\_T\; k\_v\}\{Jr\_a\}\}$

Finally, solve the above equation for the transfer functions

: $\backslash left.\; \backslash frac\{\backslash Omega\_r(s)\}\{V\_a(s)\}\; \backslash right|\_\{T\_L(s)\; =\; 0\}\; =\; \backslash frac\{\backslash frac\{k\_T\}\{Jr\_a\}\}\{s+\; \backslash frac\{B\_m\; r\_a\; +\; k\_T\; k\_v\}\{Jr\_a\}\}$

and

: $\backslash left.\; \backslash frac\{\backslash Omega\_r(s)\}\{T\_L(s)\}\; \backslash right|\_\{V\_a(s)\; =\; 0\}\; =\; \backslash frac\{-\backslash frac\{1\}\{J\}\}\{s+\; \backslash frac\{B\_m\; r\_a\; +\; k\_T\; k\_v\}\{Jr\_a\}\}$

Using the first transfer function above, solve for $\backslash omega\_r(t)$ given $v\_a(t)\; =\; V\_au(t)$. In other words, solve the following

: $\backslash omega\_r(t)\; =\; \backslash mathcal\{L\}^\{-1\}\; \backslash left\backslash \{\; \backslash frac\{\backslash Omega(s)\}\{V\_a(s)\}\; \backslash cdot\; V\_a(s)\; \backslash right\backslash \}$.

: $\backslash omega\_r(t)\; =\; \backslash mathcal\{L\}^\{-1\}\; \backslash left\backslash \{\; \backslash frac\{\backslash frac\{k\_T\}\{Jr\_a\}\}\{s+\; \backslash frac\{B\_m\; r\_a\; +\; k\_T\; k\_v\}\{Jr\_a\}\}\; \backslash cdot$

\frac{V_a}{s} \right\}

We use a Laplace transform table to look up the transform for an exponential approach

: $\backslash mathcal\{L\}^\{-1\}\; \backslash left\backslash \{\; \backslash frac\{\backslash alpha\}\{s(s+\backslash alpha)\}\; \backslash right\backslash \}\; =\; \backslash left(\; 1\; -\; e^\{-\backslash alpha\; t\}\backslash right)\; \backslash cdot\; u(t)$

then if we let

: $\backslash alpha\; =\; \backslash frac\{B\_m\; r\_a\; +\; k\_T\; k\_v\}\{Jr\_a\}$

we can express $\backslash omega\_r(t)$ as

: $\backslash omega\_r(t)\; =\; \backslash mathcal\{L\}^\{-1\}\; \backslash left\backslash \{\; \backslash frac\{k\_T\; V\_a\}\{B\_m\; r\_a\; +\; k\_t\; k\_v\}\; \backslash cdot\; \backslash frac\{\backslash alpha\}\{s(s+\backslash alpha)\}\; \backslash right\backslash \}\; =\; \backslash frac\{k\_T\; V\_a\}\{B\_m\; r\_a\; +\; k\_t\; k\_v\}\; \backslash left(\; 1\; -\; e^\{-\backslash alpha\; t\}\backslash right)\; \backslash cdot\; u(t)$.

Given

: $\backslash omega\_r(t)\; =\; \backslash omega\_\{r,ss\}\backslash left(\; 1-e^\{-t/\backslash tau\_m\}\; \backslash right)$

we have

: $\backslash omega\_\{r,ss\}\; =\; \backslash frac\{k\_T\; V\_a\}\{B\_m\; r\_a\; +\; k\_t\; k\_v\}$

and

: $\backslash tau\_m\; =\; \backslash frac\{J\; r\_a\}\{B\_m\; r\_a\; +\; k\_T\; k\_v\}$

= Lab 5b Solutions =

:(1) $k\_T\; i\_a\; -\; T\_L\; =\; J\; \backslash frac\{d\backslash omega\_r\}\{dt\}\; +\; B\_m\; \backslash omega\_r$

:(2) $v\_a\; =\; r\_a\; i\_a\; +\; k\_v\; \backslash omega\_r$

:(3) $v\_a\; =\; K\_1\; +\; K\_2\; \backslash cos(\backslash omega\_e\; t)$

:(4) $\backslash omega\_r\; =\; \backslash omega\_\{ro\}\; +\; A\; \backslash cos(\backslash omega\_e\; t\; +\; \backslash phi)$

Solving equation (1) for $i\_a$ and substituting into equation (2) results in

: $v\_a\; =\; \backslash frac\{r\_a\}\{k\_T\}\; \backslash left(\; J\; \backslash frac\{d\backslash omega\_r\}\{dt\}\; +\; B\_m\backslash omega\_r\; \backslash right)\; +\; k\_v\; \backslash omega\_r$

Simplifying

: $v\_a\; =\; \backslash frac\{J\; r\_a\}\{k\_T\}\; \backslash frac\{d\backslash omega\_r\}\{dt\}\; +\; \backslash frac\{B\_m\; r\_a\; +\; k\_T\; k\_v\}\{k\_T\}\; \backslash omega\_r$

Now equations (3) and (4) can be substituted into the above equation to produce

: $K\_1\; +\; K\_2\; \backslash cos(\backslash omega\_e\; t)\; =\; -\backslash frac\{J\; r\_a\}\{k\_T\}\; A\; \backslash omega\_e\; \backslash sin(\backslash omega\_e\; t\; +\; \backslash phi)\; +\; \backslash frac\{B\_m\; r\_a\; +\; k\_T\; k\_v\}\{k\_T\}\; \backslash left(\; \backslash omega\_\{ro\}\; +\; A\; \backslash cos(\backslash omega\_e\; t\; +\; \backslash phi)\; \backslash right)$

Applying some trigonometry the above can be rewritten as

: $K\_1\; +\; K\_2\; \backslash cos(\backslash omega\_e\; t)\; =\; \backslash frac\{B\_m\; r\_a\; +\; k\_T\; k\_v\}\{k\_T\}\; \backslash omega\_\{ro\}\; +\; \backslash sqrt\{\backslash left(\; \backslash frac\{J\; r\_a\; \backslash omega\_e\}\{k\_T\}\; A\; \backslash right)^2\; +\; \backslash left(\; \backslash frac\{B\_m\; r\_a\; +\; k\_T\; k\_v\}\{k\_T\}\; A\; \backslash right)^2\}\; \backslash cos(\backslash omega\_e\; t)$

: $\backslash omega\_\{r0\}\; =\; \backslash frac\{k\_T\; K\_1\}\{B\_m\; r\_a\; +\; k\_T\; k\_v\}$

: $A\; =\; \backslash sqrt\{\; \backslash frac\{(k\_T\; K\_2)^2\}\{(J\; r\_a\; \backslash omega\_e)^2\; +\; (B\_m\; r\_a\; +\; k\_T\; k\_v)^2\}\; \}$

: $J\; =\; \backslash frac\{1\}\{r\_a\; \backslash omega\_e\}\; \backslash sqrt\{\backslash left(\; \backslash frac\{k\_T\; K\_2\}\{A\}\backslash right)^2\; -\; (B\_m\; r\_a\; +\; k\_T\; k\_v)^2\}$

= Post-Lab 6b =

: $\backslash left|\; e\_\{ab\}\; \backslash right|\; =\; \backslash sqrt\{3\}\; \backslash ,\; \backslash omega\_r\; \backslash lambda\_m^\text{'}$

= Lab 7a =

The Fourier series of a 2π-periodic function ƒ(x) that is integrable on [−π, π], is given by

:$\backslash frac\{a\_0\}\{2\}\; +\; \backslash sum\_\{n=1\}^\backslash infty\; \backslash ,\; [a\_n\; \backslash cos(nx)\; +\; b\_n\; \backslash sin(nx)]$

where

:$a\_n\; =\; \backslash frac\{1\}\{\backslash pi\}\backslash int\_\{-\backslash pi\}^\backslash pi\; f(x)\; \backslash cos(nx)\backslash ,\; dx,\; \backslash quad\; n\; \backslash ge\; 0$

and

:$b\_n\; =\; \backslash frac\{1\}\{\backslash pi\}\backslash int\_\{-\backslash pi\}^\backslash pi\; f(x)\; \backslash sin(nx)\backslash ,\; dx,\; \backslash quad\; n\; \backslash ge\; 1$

In question 2, you are being asked to find the fundamental component of the fourier series of the functions vas, vbs, and vcs. The fundamental component is the component with the lowest freqency, specifically:

: $v\_\{as\}(\backslash theta\_r)\; \backslash approx\; \backslash frac\{a\_0\}\{2\}\; +\; a\_1\; \backslash cos(\backslash theta\_r)\; +\; b\_1\; \backslash sin(\backslash theta\_r)$

To find the coefficients an and bn from the equations above, the integral must be broken down into the sum of integrals over continuous regions.

:$$

\begin{array}{l}

\int\limits_{ -\pi}^{ \pi} f\, dx =

\int\limits_{ -\pi}^{-\frac{2\pi}{3}} f\, dx\, +

\int\limits_{-\frac{2\pi}{3}}^{-\frac{ \pi}{3}} f\, dx\, +

\int\limits_{-\frac{ \pi}{3}}^{ 0} f\, dx\, \\

\qquad \qquad \qquad +

\int\limits_{ 0}^{ \frac{ \pi}{3}} f\, dx\, +

\int\limits_{ \frac{ \pi}{3}}^{ \frac{2\pi}{3}} f\, dx\, +

\int\limits_{ \frac{2\pi}{3}}^{ \pi} f\, dx

\end{array}

= Other =

: $$

T_e = -3 L_{B}

\{

i_{as}^2\sin\left(6\theta_{rm}\right) +

i_{bs}^2\sin\left[6\left(\theta_{rm} - 20\,^{\circ}\right)\right] +

i_{cs}^2\sin\left[6\left(\theta_{rm} + 20\,^{\circ}\right)\right]

\}

: $$

\left[

\begin{array}{ccc}

\;\, & \;\, & \;\, \\

& & \\

& &

\end{array}

\right]

\left[

\begin{array}{c}

e_{as} \\

e_{bs} \\

e_{cs}

\end{array}

\right]

=

\left[

\begin{array}{c}

e_{ab} \\

e_{cb} \\

0

\end{array}

\right]