User:Fozolo

Lab 5a Tips

Pre-Lab

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 $\omega _{r}$

$v_{a}(t)=V_{a}u(t)$

$\omega _{r}(t)={\mathcal {L}}^{-1}\left\{{\frac {\Omega (s)}{V_{a}(s)}}V_{a}(s)\right\}$

---

Lab 5a Solutions

Pre-Lab

(1) $T_{e}-T_{L}=J{\frac {dw_{r}}{dt}}+B_{m}w_{r}$

(2) $v_{a}=r_{a}i_{a}+L_{aa}{\frac {di_{a}}{dt}}+k_{v}w_{r}$

(3) $T_{e}=k_{T}i_{a}$

State Space Form (2 points)

Solve equations (1) and (2) for ${\frac {dw_{r}}{dt}}$ and ${\frac {di_{a}}{dt}}$.

${\frac {dw_{r}}{dt}}={\frac {1}{J}}\left[-B_{m}w_{r}+k_{t}i_{a}-T_{L}\right]$

${\frac {di_{a}}{dt}}={\frac {1}{L_{aa}}}\left[-k_{v}w_{r}-r_{a}i_{a}+V_{a}\right]$

$\left[{\begin{array}{c}{\frac {d\omega _{r}}{dt}}\\{\frac {di_{a}}{dt}}\end{array}}\right]=\left[{\begin{array}{cc}-{\frac {B_{m}}{J}}&{\frac {k_{T}}{J}}\\-{\frac {k_{v}}{L_{aa}}}&-{\frac {r_{a}}{L_{aa}}}\end{array}}\right]\left[{\begin{array}{c}\omega _{r}\\i_{a}\end{array}}\right]+\left[{\begin{array}{cc}-{\frac {1}{J}}&0\\0&{\frac {1}{L_{aa}}}\end{array}}\right]\left[{\begin{array}{c}T_{L}\\V_{a}\end{array}}\right]$

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)

$T_{e}-T_{L}=J{\frac {dw_{r}}{dt}}+B_{m}w_{r}$

$v_{a}=r_{a}i_{a}+k_{v}w_{r}$

$T_{e}=k_{T}i_{a}$

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 $v_{a}(t)=V_{a}u(t)$. 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

$\tau _{m}={\frac {Jr_{a}}{B_{m}r_{a}+k_{T}k_{v}}}$

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

$v_{a}={\frac {Jr_{a}}{k_{T}}}{\frac {d\omega _{r}}{dt}}+{\frac {B_{m}r_{a}+k_{T}k_{v}}{k_{T}}}\omega _{r}$

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

$K_{1}+K_{2}\cos(\omega _{e}t)=-{\frac {Jr_{a}}{k_{T}}}A\omega _{e}\sin(\omega _{e}t+\phi )+{\frac {B_{m}r_{a}+k_{T}k_{v}}{k_{T}}}\left(\omega _{ro}+A\cos(\omega _{e}t+\phi )\right)$

Applying some trigonometry the above can be rewritten as

$K_{1}+K_{2}\cos(\omega _{e}t)={\frac {B_{m}r_{a}+k_{T}k_{v}}{k_{T}}}\omega _{ro}+{\sqrt {\left({\frac {Jr_{a}\omega _{e}}{k_{T}}}A\right)^{2}+\left({\frac {B_{m}r_{a}+k_{T}k_{v}}{k_{T}}}A\right)^{2}}}\cos(\omega _{e}t)$

$\omega _{ro}$ (4 points)

$\omega _{r0}={\frac {k_{T}K_{1}}{B_{m}r_{a}+k_{T}k_{v}}}$

$A$ (3 points)

$A={\sqrt {\frac {(k_{T}K_{2})^{2}}{(Jr_{a}\omega _{e})^{2}+(B_{m}r_{a}+k_{T}k_{v})^{2}}}}$

$J$ (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

$a_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\cos(nx)\,dx,\quad n\geq 0$

and

$b_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\sin(nx)\,dx,\quad n\geq 1$

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

$T_{e}=-3L_{B}\{i_{as}^{2}\sin \left(6\theta _{rm}\right)+i_{bs}^{2}\sin \left[6\left(\theta _{rm}-20\,^{\circ }\right)\right]+i_{cs}^{2}\sin \left[6\left(\theta _{rm}+20\,^{\circ }\right)\right]\}$

$\left[{\begin{array}{ccc}\;\,&\;\,&\;\,\\&&\\&&\end{array}}\right]\left[{\begin{array}{c}e_{as}\\e_{bs}\\e_{cs}\end{array}}\right]=\left[{\begin{array}{c}e_{ab}\\e_{cb}\\0\end{array}}\right]$