User:Egm6321.f10.team03.sigillo/Hwk3

= Problem 5 - Generalization of a SC-L1-ODE-CC to SC-L1-ODE-VC =

Given
Given is the solution for $$\underline{x}(t)$$ of a first order linear ODE for a coupled system with constant coefficients (SC-L1-ODE-CC) out of the lecture notes in meeting 14 on page 14-1 and 14-2, Eq. (4) :
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$$\underline{x}(t)={{e}^{\underline{A}\cdot \left( t-{{t}_{0}} \right)}}\cdot \underline {x}({{t}_{0}})+\int\limits_^{t}{{{e}^{\underline{A}\cdot \left( t-\tau \right)}}\cdot \underline{B}\cdot \underline{u}\left( \tau  \right)d\tau }$$
 * $$\,(Eq. 5.1) $$
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Where $$\underline{x}$$ is a n x 1 vector, $$\underline{u}$$ a m x 1 vector, $$\underline{A}$$ is a n x n matrix and $$\underline{B}$$ is a n x m matrix.

Find
The general form for SC-L1-ODEs with varying coefficient (SC-L1-ODE-VC) has to be found based on the given SC-L1-ODE-CC

Solution
Following the approach of the integrating factor method and comparing the general form for L1-ODE-VC given in Eq. (3) on page 14-1 in meeting 14, it can be seen (and calculated through the intergration factor method - refer to Problem 3-) that the generalized form for (Eq. 5.1) can be achieved by integrating the argument of the exp(.)-function in the limits $$\,{{t}_{0}}$$ and $$\,t$$ for the first term, $$\,\tau $$ and $$\,t$$ for the second term:
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$$\underline{x}(t)={{e}^{\int\limits_^{t}{\underline{A}(\tau )d\tau }}}\cdot \underline{x}({{t}_{0}})+\int\limits_^{t}{{{e}^{\int\limits_{\tau }^{t}{\underline{A}(s)ds}}}\cdot \underline{B}\cdot \underline{u}\left( \tau \right)d\tau }$$ Note: The integral of a matrix is calculated by integrating each element that composes the matrix.
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 * $$\,(Eq. 5.2) $$
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= Problem 10 - Generate exact N2-ODE's - Reverse Engineering=

Given
Given is the following function from meeting 17
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$$\phi \left( x,y,p \right)=3{{p}^{5}}\cos \left( {{x}^{2}} \right)+2x{{y}^{3}}=k$$ and Eq. (2) from meeting 17
 * $$\,(Eq. 10.1) $$
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$$\left( 6x{{y}^{2}} \right)\cdot {y}'+2{{y}^{3}}=\left( {{h}_{y}}+0 \right){y}'+{{h}_{x}}$$ on page 17-1 and 17-3 respectively in the lecture notes.
 * $$\,(Eq. 10.2) $$
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Find
Using (Eq. 10.2) the function $$\,h\left( x,y \right)$$ has to be found in order to get again the given function $$\,\phi $$ (reverse engineering) assuming $${{h}_{y}}\cdot {y}'=2{{y}^{3}}$$.

Solution
Applying the 'separation of variables'-method to the given assumption, $$\,h\left( x,y \right)$$ can be calculated:
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$${{h}_{y}}\cdot dy=2{{y}^{3}}\cdot dx$$
 * $$\,(Eq. 10.3) $$
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$$\int{{{h}_{y}}\cdot dy=\int{2{{y}^{3}}\cdot dx}}$$
 * $$\,(Eq. 10.4) $$
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$$h\left( x,y \right)=2{{y}^{3}}x+{{k}_{1}}(y)$$ Deriving $$\,h\left( x,y \right)$$ with respect to $$\,x$$ gives:
 * $$\,(Eq. 10.5) $$
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$$\,{{h}_{x}}=2{{y}^{3}}$$ At the same time it can be seen from (Eq. 10.2) that $$\,{{h}_{x}}$$ is equal to $$\left( 6x{{y}^{2}} \right)\cdot {y}'$$, so that the following is valid:
 * $$\,(Eq. 10.6) $$
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$${{h}_{x}}=2{{y}^{3}}=6x{{y}^{2}}\cdot {y}'$$ What implies that:
 * $$\,(Eq. 10.7) $$
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$${y}'=\frac{y}{3x}$$ and
 * $$\,(Eq. 10.8) $$
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$$\,{{k}_{1}}(y)=0$$ However, inserting the found experssion in (Eq. 10.5) with $$\,{{k}_{1}}(y)=0$$ into Eq. (1) of page 17-3 in meeting 17 gives:
 * $$\,(Eq. 10.9) $$
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$$\phi \left( x,y,p \right)=3{{p}^{5}}\cos \left( {{x}^{2}} \right)+2x{{y}^{3}}=k$$ Which is the same as the given function $$\,\phi \left( x,y,p \right)$$ = Problem 6 - General SC-L1-ODE-CC with state transition matrix $$\underline{\Phi} $$ =
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 * $$\,(Eq. 10.10) $$
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Given
The state equation for a coupled system like it's usual for a openloop control system is given as:
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$$\underline{\dot{x}}(t)=\underline{A}\underline{x}(t)+\underline{B}\underline{u}(t)$$ The state transition matrix $$\underline{\Phi} $$ has the following properties: and where I is the identity matrix.
 * $$\,(Eq. 6.1) $$
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Find
With (Eq. 6.1) through (Eq. 6.3), Eq. (4) on pp. 14-1 of the lecture notes, has to be found:
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$$\underline{x}(t)={{e}^{\underline{A}(t-{{t}_{0}})}}\cdot \underline{x}({{t}_{0}})+\int\limits_^{t}{{{e}^{\underline{A}(t-\tau )}}\underline{B}\underline{u}(\tau )\,d\tau }$$
 * $$\,(Eq. 6.4) $$
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Solution
In (Eq. 6.2) the homogeneous differential equation for $$\underline{\Phi} (t,{{t}_{0}})$$ is given. Thus replacing $$\underline{x}(t)$$ with $$\underline{\Phi} (t,{{t}_{0}})$$ in (Eq. 6.1) gives the inhomogeneous equation for $$\underline{\Phi} (t,{{t}_{0}})$$:
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$$\dot{\underline{\phi} }(t,{{t}_{0}})=\underline{A}\underline{\phi} (t,{{t}_{0}})+\underline{B}\underline{u}(t)$$
 * $$\,(Eq. 6.5) $$
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Regrouping the equation above results in:
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$$\underbrace{-\left( \underline{A}\underline{\phi} (t,{{t}_{0}})+\underline{B}\underline{u}(t) \right)}_{\underline{M}(\underline{\phi} ,\underline{u})}+\underbrace{1}_{\underline{N}(\underline{\phi} ,\underline{u})}\cdot \dot{\underline{\phi} }(t,{{t}_{0}})=0$$ (Eq. 6.6) meets the first condition of exactness, but the second condition $${{\underline{M}}_{\underline{\phi} }}={{\underline{N}}_{\underline{u}}}$$ needs to be checked:
 * $$\,(Eq. 6.6) $$
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$${{\underline{M}}_{\underline{\phi} }}=-\underline{A}\ne 0={{\underline{N}}_{\underline{u}}}$$ The second condition is not met, thus a function $$h(t)$$ to make (Eq. 6.6) exact needs to be found. Therefor (Eq. 6.6) will be rewritten as:
 * <p style="text-align:right;">$$\,(Eq. 6.7) $$
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$$-\underline{A}\underline{\phi} (t,{{t}_{0}})+1\cdot \dot{\underline{\phi} }(t,{{t}_{0}})=\underline{B}\underline{u}(t)$$ The integratin factor $$h(t)$$ is then given as:
 * <p style="text-align:right;">$$\,(Eq. 6.8) $$
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$$h(t)={{e}^{\int\limits_^{t}{-\underline{A}\,ds}}}={{e}^{-\underline{A}(t-{{t}_{0}})}}$$ Multiplying $$h(t)$$ throughout (Eq. 6.8) gives:
 * <p style="text-align:right;">$$\,(Eq. 6.9) $$
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$$-\underline{A}\underline{\phi} (t,{{t}_{0}})\cdot {{e}^{-\underline{A}(t-{{t}_{0}})}}+{{e}^{-\underline{A}(t-{{t}_{0}})}}\cdot \dot{\underline{\phi} }(t,{{t}_{0}})=\underline{B}\underline{u}(t)\cdot {{e}^{-\underline{A}(t-{{t}_{0}})}}$$ which is equal to
 * <p style="text-align:right;">$$\,(Eq. 6.10) $$
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$$\frac{d\left( {{e}^{-\underline{A}(t-{{t}_{0}})}}\cdot \underline{\phi} (t,{{t}_{0}}) \right)}{dx}=\underline{B}\underline{u}(t)\cdot {{e}^{-\underline{A}(t-{{t}_{0}})}}$$ Integrating (Eq. 6.11) on both sides between $${{t}_{0}}$$ and $$t$$ gives:
 * <p style="text-align:right;">$$\,(Eq. 6.11) $$
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$${{e}^{-\underline{A}(t-{{t}_{0}})}}\cdot \underline{\phi} (t,{{t}_{0}})-{{e}^{-\underline{A}({{t}_{0}}-{{t}_{0}})}}\cdot \underline{\phi} ({{t}_{0}},{{t}_{0}})={{e}^{\underline{A}{{t}_{0}}}}\cdot \int\limits_^{t}\cdot \underline{B}\underline{u}(t)\,d\tau $$ that is equal to
 * <p style="text-align:right;">$$\,(Eq. 6.12) $$
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$$\underline{\phi} (t,{{t}_{0}})={{e}^{\underline{A}(t-{{t}_{0}})}}\cdot \underline{\phi} ({{t}_{0}},{{t}_{0}})+\int\limits_^{t}{{{e}^{\underline{A}(t-\tau )}}\cdot }\underline{B}\underline{u}(t)\,d\tau $$ Replacing $$\underline{\Phi} (t,{{t}_{0}})$$ with $$\underline{x}(t)$$ gives:
 * <p style="text-align:right;">$$\,(Eq. 6.13) $$
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$$\underline{x}(t)={{e}^{\underline{A}(t-{{t}_{0}})}}\cdot \underline{x}({{t}_{0}})+\int\limits_^{t}{{{e}^{\underline{A}(t-\tau )}}\underline{B}\underline{u}(\tau )\,d\tau }$$ which is the same as the given solution in (Eq. 6.4)!
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 * <p style="text-align:right;">$$\,(Eq. 6.13) $$
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Contributing Members

 * Solved by Michele Sigilló
 * Posted by Michele Sigilló