User:Egm6321.f10.team2.vrasul/HW3

Given
Variable Definition:

$$\delta$$ = Aileron Angle (Deflection)

$$\phi$$ = Roll Angle

$$\omega$$ = Velocity of Roll Angle

$$Q$$ = Aileron Effectiveness

$$\tau$$ = Roll-Time Constant


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$$\dot{\phi}=\omega$$ (Eq 6.1)
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$$\dot{\omega}=-\tfrac{1}{\tau} \omega + \tfrac{Q}{\tau} \delta$$ (Eq 6.2)
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$$\dot{\delta}=u$$

(Eq 6.3)
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Find
Put above equations (Eq. 6.1 to 6.3) in the matrix form of:


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$$\dot{x}\left ( t \right)=\underline{A} \underline{x}\left(t \right) + \underline{B} \underline{u}\left(t \right) $$     (Eq 6.4)
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Solution
The above expressions (Eq. 6.1 to 6.3) must resemble Eq. 6.4; therefore, it is imperative that they are dependent on time ‘t.’

To bring Eq. 6.2 to this form $$\delta$$ will be expressed in terms of u(t).

Integrate Eq. 6.3 with respect to (wrt) time t yields:


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$$\delta=ut$$

(Eq 6.5)
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Eq. 6.2 can now be written in its intended form.


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$$\dot{\omega}=-\tfrac{1}{\tau} \omega + \tfrac{Q}{\tau} ut$$ (Eq 6.6)
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If Eq. 6.1 is true then the integral of $$ \dot{\omega}$$ will yield the solution to $$ \dot{\phi}$$
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Take the integral of Eq. 6.6:


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$${\omega}=-\tfrac{t}{\tau} \omega + \tfrac{Q}{2\tau} ut^2$$ (Eq 6.7)
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Therefore,


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$$\dot{\phi}=-\tfrac{t}{\tau} \omega + \tfrac{Q}{2\tau} ut^2$$ (Eq 6.8)
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Eq. 6.3 is simply a control signal and has no dependence on the roll time constant $$\tau$$. With this assumption it can be said that Eq 6.3 needs no further manipulation.

Eq. 6.1, 6.6 and 6.8 can now be written in matrix form of Eq. 6.4.


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$$

\begin{bmatrix} \dot{\phi}\\ \dot{\omega}\\ \dot{\delta} \end{bmatrix} = \underbrace{ \begin{bmatrix} \tfrac{1}{\tau} 0 0\\

0 \tfrac{1}{\tau} 0\\ 0 0 0

\end{bmatrix} }_{\underline{A}}

\underbrace{

\begin{bmatrix} \phi\\ \omega\\ \delta \end{bmatrix} }_{\underline{x}}

+ \underbrace{ \begin{bmatrix} \tfrac{Q t^2}{2\tau}\\ \tfrac{Q}{\tau}\\ 1 \end{bmatrix} }_{\underline{B}}

\underbrace{ \begin{bmatrix} u \end{bmatrix} }_{\underline{u}}

$$     (Eq 6.9)
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In conclusion, the form of Eq 6.1, 6.2 and 6.3 has been written in the form of 6.4. As stated previously the expression $$ \dot{\delta} $$ has no components associated to the roll time constant.

Given

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$$h_x + h_y y^\prime = 0 $$ (Eq 7.1)
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Find
Find the solution of Eq. 7.1.

Solution
For there to be an other solution to Eq. 7.1 the following must hold true:


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$$h_{xy} = h_{yx} $$ (Eq 7.2)
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That is, the derivative of,
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$$h_x = -h_y y^\prime$$ (Eq 7.3)
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with respect to y must yield $$h_{yx}$$.

Perform the above operation to Eq 7.3.


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$$h_{xy} = -h_y y^{\prime\prime}+ h_{yy} y^\prime$$ (Eq 7.4)
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Clearly Eq 7.4 does not equal Eq. 7.2. As state previously, for Eq 7.1 to hold true Eq 7.2 must be met. This can only be attained if h was indeed a constant thereby giving $$h_x$$ and $$h_y$$ a value of 0.