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March 3rd, 2019, 03:38 AM  #1 
Newbie Joined: Mar 2019 From: Singapore Posts: 3 Thanks: 0  Solving a system of equations
I have the following equations: $\displaystyle c_0 = \frac{T^2b_1 + 2Q_RR_0T + 2Q_RRT + 4Q_RR_0RC + 2b_1RCT}{2Q_RT + 4Q_RRC}$ $\displaystyle c_1 = \frac{T^2b_1  4Q_RR_0RC}{Q_RT + 2Q_RRC}$ $\displaystyle c_2 = \frac{T^2b_1  2Q_RR_0T  2Q_RRT + 4Q_RR_0RC  2b_1RCT}{2Q_RT + 4Q_RRC}$ $\displaystyle a_1 = \frac{8Q_RRC}{2Q_RT + 4Q_RRC}$ $\displaystyle a_2 = \frac{2Q_RT + 4Q_RRC}{2Q_RT + 4Q_RRC}$ How do I go about solving for $\displaystyle R, C, R_0$ and $\displaystyle b_1$ in terms of $\displaystyle c_0, c_1, c_2, a_1$ and $\displaystyle a_2$, given that $\displaystyle Q_R$ and $\displaystyle T$ are constants? This problem originates from a transfer function, where $\displaystyle c_0, c_1, c_2, a_1$ and $\displaystyle a_2$ are coefficients and $\displaystyle R, C, R0$ and $\displaystyle b_1$ are system parameters, with $\displaystyle T$ being the Sample Time and $\displaystyle Q_R$ is a known system quantity. Last edited by babagreensheep; March 3rd, 2019 at 04:10 AM. 
March 3rd, 2019, 10:59 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,427 Thanks: 1314 
You've got 5 equations and 4 variables you want to solve for. The system is over determined. If we omit $a_2$ we can solve the system as $R=\dfrac{2 \left(a_1^2 c_0+2 a_1 c_0a_1 c_1+c_0c_1+c_2\right)}{a_1 \left(a_1+2\right){}^2}$ $C=\dfrac{a_1^3 (T)2 a_1^2 T}{4 \left(a_1^2 c_0+2 a_1 c_0a_1 c_1+c_0c_1+c_2\right)}$ $R_0= \dfrac{c_0+c_1c_2}{2 a_1}$ $b_1=\dfrac{\left(c_0+c_1+c_2\right) \text{QR}}{\left(a_1+2\right) T}$ If $a_2$ is consistent with this then this is the solution. If not there is no solution. 
March 3rd, 2019, 06:47 PM  #3  
Newbie Joined: Mar 2019 From: Singapore Posts: 3 Thanks: 0  Quote:
Also, how do I ensure $a_2$ is consistent with this? Last edited by babagreensheep; March 3rd, 2019 at 07:02 PM.  
March 3rd, 2019, 07:55 PM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 2,427 Thanks: 1314  Quote:
Evaluate $a_2$ using the values solved for. It either matches the value you are given for it or not.  
March 3rd, 2019, 08:11 PM  #5 
Senior Member Joined: Sep 2016 From: USA Posts: 609 Thanks: 378 Math Focus: Dynamical systems, analytic function theory, numerics 
These aren't linear equations so this isn't a problem you can solve with linear algebra. Use a numerical root finder like Newton's method for this.

March 4th, 2019, 04:48 AM  #6  
Newbie Joined: Mar 2019 From: Singapore Posts: 3 Thanks: 0  Quote:
$RC = (T)\dfrac{1}{2}\dfrac{a_1^3+2a_1^2}{a_1^3+4a_1^2+4a _1}$ Substituting that into the original equation for $a_2$, I can work it out to: $a_2 = 1  \dfrac{4Q_RT}{2Q_RT+4Q_R(T)\dfrac{1}{2}\dfrac{a_1^3+2a_1^2}{a_1^3+4a_1^2+4a _1}}$ which after some simplification works out to be: $a_2 = 1  a_1$ which is consistent with the original equations. Thus, I can say that this is the solution right?  
March 4th, 2019, 07:27 AM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,413 Thanks: 1024  
March 4th, 2019, 08:49 AM  #8  
Senior Member Joined: Sep 2015 From: USA Posts: 2,427 Thanks: 1314  Quote:
 

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