User:Egm6321.f10.team6.Kim.MK/hw3

= Problem 6 Roll control of rocket =

Given
System of coupled Linear 1st order ODE constant coefficient


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$$\begin{align} \underline{\dot{\textbf{x}}}=\underline{\textbf{A}}\cdot \underline{\textbf{x}}(t) + \underline{\textbf{B}}\cdot \underline{\textbf{u}}(t) \end{align}$$

(6.1)
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Consider a Rocket Prevent rolling by activating ailerons


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$$\begin{align} &{\delta} = {\textrm{aileron{~} angle(deflection)}} \\ &{\phi} = {\textrm{roll{~} angle}} \\ &{\omega} ={\textrm{roll{~} angular {~} velocity}} \\ &Q = {\textrm{aileron{~}effectiveness}} \\ &{\tau} = {\textrm{roll-time{~} constant}} \\ \end{align} $$


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$$\begin{align} &\dot{\phi} = {\omega}\\ \end{align}$$

(6.2)
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$$\begin{align} &\dot{\omega} = - {\frac{1}{\tau}}{\omega} + {\frac{Q}{\tau}}{\delta}\\ \end{align}$$

(6.3)
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$$\begin{align} &\dot{\delta} = u \left({\textrm{commmand{~} signal{~} to {~} aileon{~} actuators{~} ={~} control}}\right)\\ \end{align}$$

(6.4)
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Required
Put (6.2) - (6.4) in form of (6.1)

Solution
From(6.1), $$\begin{align} \dot{\phi},\dot{\omega},\dot{\delta}\end{align}$$ as regard as $$\begin{align} \dot{\textrm{x}}\end{align}$$, $$\begin{align} {\phi},{\omega},{\delta}\end{align}$$ as regard as $$\begin{align} \textrm{x(t)}\end{align}$$, put those three equations can be expressed as matrix form(6.5)


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$$\begin{align} \begin{bmatrix} \dot{\phi}\\ \dot{\omega}\\ \dot{\delta} \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0\\ 0 & -{\frac{1}{\tau}} & {\frac{Q}{\tau}}\\ 0 & 0 & 0 \end{bmatrix}

\begin{bmatrix} \phi\\ \omega\\ \delta \end{bmatrix} + \begin{bmatrix} 0\\ 0\\  1 \end{bmatrix} u \end{align}$$

(6.5)
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Given

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$$\begin{align} (6xy^2) y' + 2y^3 = h_{x} y' + h_{x} \end{align}$$

(9.1)
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Required
Find $$\begin{align} h(x,y) \end{align}$$ and $$\begin{align} {\phi}(x,y,y') \end{align}$$

Solution
2 choices: 1)1st choice, assume that $$\begin{align} h_{x} = 2 y^3 \end{align}$$, $$\begin{align} h_{y} = 6xy^2 \end{align}$$ then, $$\begin{align} h(x,y) = 2y^{3} x + k_{1} \end{align}$$, with integration constant $$\begin{align} k_{1} \end{align}$$. Hence,


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$$\begin{align} & \overline{\phi}=h(x,y)+\int{f(x,y,p)dp}=2x{{y}^{3}}+{{k}_{1}}+3{{p}^{5}}\cos ({{x}^{2}})={{k}_{2}} \\ & \phi =2x{{y}^{3}}+3{{p}^{5}}\cos ({{x}^{2}})=k \\ \end{align}$$

(9.2)
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, with integration constants $$\begin{align} k_{1}, k_{2}, k. \end{align}$$, $$\begin{align} \int{f(x,y,p)dp} \end{align}$$ from lecture note eq. (1) 17_3.

2) 2nd choice, assume that $$\begin{align} h_{y} y' = 2 y^3 \end{align}$$, $$\begin{align} h_{x} = (6xy^{2})y' \end{align}$$ At first, using $$\begin{align} h_{x} y' = 2 y^3 \end{align}$$


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$$\begin{align} &h_{y} = \frac{ 2 y^{3}}{y'}\\ &h = \int{\frac{2 y^{3}}{y'}dy}= \int{\frac{2 y^{3}}{\frac{\cancel{dy}}{dx}}\cancel{dy}}= \int{2y^{3}dx}\\ \end{align}$$

(9.3) then, $$\begin{align} h(x,y) = 2xy^3 + k_3 \end{align}$$, with integration constant $$\begin{align} k_{3} \end{align}$$ Thus, this will bring same result as (9.2).
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$$\begin{align} & \overline{\phi}=h(x,y)+\int{f(x,y,p)dp}=2x{{y}^{3}}+{{k}_{3}}+3{{p}^{5}}\cos ({{x}^{2}})={{k}_{4}} \\ & \phi =2x{{y}^{3}}+3{{p}^{5}}\cos ({{x}^{2}})=k \\ \end{align}$$

(9.4)
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, with integration constants $$\begin{align} k_{3}, k_{4}, k. \end{align}$$

Using $$\begin{align} h_{x} = (6xy^{2})y' \end{align}$$ also can reach same result as well.


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$$\begin{align} &h_{x} = (6xy^{2})y'\\ &h = \int{6xy^{2} {\frac{dy}{\cancel{dx}}}\cancel{dx}}= \int{6xy^{2}dy} = 6x \frac{y^{3}}{3} +k_{5} = 2 x y^{3} +k_{5}\\ \end{align}$$

(9.5) Since, this also $$\begin{align} h(x,y) = 2xy^3 + k_5 \end{align}$$, with integration constant $$\begin{align} k_{5} \end{align}$$
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We can say that 1st choice and 2nd choice are both can derive a same result of $$\begin{align} {\phi} \end{align}$$.