User:Fozolo

= Lab 5a Tips =

State-Space Form


\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] $$

Transfer Functions

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


 * $$ \mathcal{L} \{ f'(t) \} = sF(s) - f(0) $$


 * $$ \Omega(s) = \frac{V_a(s)+T_L(s)}{s+} $$


 * $$ \left. \frac{\Omega(s)}{V_a(s)} \right|_{T_L(s)=0} =?$$


 * $$ \left. \frac{\Omega(s)}{T_L(s)} \right|_{V_a(s)=0} =?$$

Step response of $$\omega_r$$

 * $$ v_a(t) = V_a u(t)$$


 * $$ \omega_r(t) = \mathcal{L}^{-1} \left\{ \frac{\Omega(s)}{V_a(s)}V_a(s) \right\} $$

Post-Lab
= Lab 5a Solutions =

Pre-Lab

 * (1) $$T_e - T_L = J\frac{d w_r}{dt} + B_m w_r$$
 * (2) $$v_a = r_a i_a + L_{aa}\frac{d i_a}{dt} + k_v w_r$$
 * (3) $$T_e = k_T i_a$$

State Space Form (2 points)
Solve equations (1) and (2) for $$\frac{d w_r}{dt}$$ and $$\frac{d i_a}{dt}$$.
 * $$\frac{d w_r}{dt} = \frac{1}{J}\left[-B_m w_r + k_t i_a - T_L \right]$$


 * $$\frac{d i_a}{dt} = \frac{1}{L_{aa}}\left[ -k_v w_r - r_a i_a + V_a \right]$$



\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] $$

Steady State Form (2 points)

 * $$T_e - T_L = B_m \Omega_r$$
 * $$V_a = r_a I_a + k_v \Omega_r$$
 * $$T_e = k_T I_a$$

Simplify (2 points)

 * $$T_e - T_L = J\frac{d w_r}{dt} + B_m w_r$$
 * $$v_a = r_a i_a + k_v w_r$$
 * $$T_e = k_T i_a$$

Transfer Functions(2 points)
Combine equations (1)-(3) to eliminate $$i_a$$. First solve (1) and (3) for $$i_a$$
 * $$i_a = \frac{1}{k_T} \left( J \frac{d\omega_r}{dt} + B_m \omega_r + T_L \right)$$

Then substitute the result into (2).
 * $$v_a = \frac{r_a}{k_T} \left( J \frac{d\omega_r}{dt} + B_m \omega_r + T_L \right) + k_v \omega_r $$

Convert the resulting equation to the frequency domain through application of Laplace transforms. Note that we choose the capital form of $$\omega$$ ($$\Omega$$), when in the frequency domain. Also, it is safe to assume $$\omega_r(t) = 0$$.
 * $$V_a(s) = \frac{r_a}{k_T} \left( Js\Omega_r(s) + B_m\Omega_r(s) + T_L(s) \right) + k_v \Omega_r(s) $$

Solving the resulting equation for $$\Omega_r(s)$$ yeilds
 * $$\Omega_r(s) = \frac{\frac{k_T}{Jr_a}V_a(s) - \frac{1}{J}T_L(s) }{s+ \frac{B_m r_a + k_T k_v}{Jr_a}} $$

Finally, solve the above equation for the transfer functions
 * $$ \left. \frac{\Omega_r(s)}{V_a(s)} \right|_{T_L(s) = 0} = \frac{\frac{k_T}{Jr_a}}{s+ \frac{B_m r_a + k_T k_v}{Jr_a}}

$$ and
 * $$ \left. \frac{\Omega_r(s)}{T_L(s)} \right|_{V_a(s) = 0} = \frac{-\frac{1}{J}}{s+ \frac{B_m r_a + k_T k_v}{Jr_a}}

$$

$$\omega_{r,ss}$$ and $$T_m$$ (2 points)
Using the first transfer function above, solve for $$\omega_r(t)$$ given $$v_a(t) = V_au(t)$$. In other words, solve the following
 * $$ \omega_r(t) = \mathcal{L}^{-1} \left\{ \frac{\Omega(s)}{V_a(s)} \cdot V_a(s) \right\} $$.


 * $$ \omega_r(t) = \mathcal{L}^{-1} \left\{ \frac{\frac{k_T}{Jr_a}}{s+ \frac{B_m r_a + k_T k_v}{Jr_a}} \cdot

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

We use a Laplace transform table to look up the transform for an exponential approach
 * $$ \mathcal{L}^{-1} \left\{ \frac{\alpha}{s(s+\alpha)} \right\} = \left( 1 - e^{-\alpha t}\right) \cdot u(t)$$

then if we let
 * $$ \alpha = \frac{B_m r_a + k_T k_v}{Jr_a} $$

we can express $$\omega_r(t)$$ as
 * $$ \omega_r(t) = \mathcal{L}^{-1} \left\{ \frac{k_T V_a}{B_m r_a + k_t k_v} \cdot \frac{\alpha}{s(s+\alpha)} \right\} = \frac{k_T V_a}{B_m r_a + k_t k_v} \left( 1 - e^{-\alpha t}\right) \cdot u(t)$$.

Given
 * $$ \omega_r(t) = \omega_{r,ss}\left( 1-e^{-t/\tau_m} \right) $$

we have
 * $$\omega_{r,ss} = \frac{k_T V_a}{B_m r_a + k_t k_v}$$

and
 * $$\tau_m = \frac{J r_a}{B_m r_a + k_T k_v}$$

= Lab 5b Solutions =

Pre-Lab

 * (1) $$k_T i_a - T_L = J \frac{d\omega_r}{dt} + B_m \omega_r$$
 * (2) $$v_a = r_a i_a + k_v \omega_r$$
 * (3) $$v_a = K_1 + K_2 \cos(\omega_e t)$$
 * (4) $$\omega_r = \omega_{ro} + A \cos(\omega_e t + \phi)$$

Solving equation (1) for $$ i_a$$ and substituting into equation (2) results in
 * $$ v_a = \frac{r_a}{k_T} \left( J \frac{d\omega_r}{dt} + B_m\omega_r \right) + k_v \omega_r$$

Simplifying
 * $$ v_a = \frac{J r_a}{k_T} \frac{d\omega_r}{dt} + \frac{B_m r_a + k_T k_v}{k_T} \omega_r $$

Now equations (3) and (4) can be substituted into the above equation to produce
 * $$K_1 + K_2 \cos(\omega_e t) = -\frac{J r_a}{k_T} A \omega_e \sin(\omega_e t + \phi) + \frac{B_m r_a + k_T k_v}{k_T} \left( \omega_{ro} + A \cos(\omega_e t + \phi) \right) $$

Applying some trigonometry the above can be rewritten as
 * $$K_1 + K_2 \cos(\omega_e t) = \frac{B_m r_a + k_T k_v}{k_T} \omega_{ro} + \sqrt{\left( \frac{J r_a \omega_e}{k_T} A \right)^2 + \left( \frac{B_m r_a + k_T k_v}{k_T} A \right)^2} \cos(\omega_e t) $$

$$ \omega_{ro} $$ (4 points)

 * $$\omega_{r0} = \frac{k_T K_1}{B_m r_a + k_T k_v}$$

$$ A $$ (3 points)

 * $$A = \sqrt{ \frac{(k_T K_2)^2}{(J r_a \omega_e)^2 + (B_m r_a + k_T k_v)^2} } $$

$$ J $$ (3 points)

 * $$J = \frac{1}{r_a \omega_e} \sqrt{\left( \frac{k_T K_2}{A}\right)^2 - (B_m r_a + k_T k_v)^2}$$

= Post-Lab 6b =
 * $$\left| e_{ab} \right| = \sqrt{3} \, \omega_r \lambda_m^'$$

= Lab 7a =

Post-Lab
The Fourier series of a 2π-periodic function ƒ(x) that is integrable on [−π, π], is given by
 * $$\frac{a_0}{2} + \sum_{n=1}^\infty \, [a_n \cos(nx) + b_n \sin(nx)]$$

where
 * $$a_n = \frac{1}{\pi}\int_{-\pi}^\pi f(x) \cos(nx)\, dx, \quad n \ge 0$$

and
 * $$b_n = \frac{1}{\pi}\int_{-\pi}^\pi f(x) \sin(nx)\, dx, \quad n \ge 1$$

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:
 * $$ v_{as}(\theta_r) \approx \frac{a_0}{2} + a_1 \cos(\theta_r) + b_1 \sin(\theta_r) $$

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] $$