My Math Forum Solving a system of equations

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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: 2,575 Thanks: 1423 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 $R=\dfrac{2 \left(a_1^2 c_0+2 a_1 c_0-a_1 c_1+c_0-c_1+c_2\right)}{a_1 \left(a_1+2\right){}^2}$ $C=\dfrac{a_1^3 (-T)-2 a_1^2 T}{4 \left(a_1^2 c_0+2 a_1 c_0-a_1 c_1+c_0-c_1+c_2\right)}$ $R_0= \dfrac{-c_0+c_1-c_2}{2 a_1}$ $b_1=\dfrac{\left(c_0+c_1+c_2\right) \text{QR}}{\left(a_1+2\right) T}$ If $a_2$ is consistent with this then this is the solution. If not there is no solution. Thanks from Denis, topsquark, SenatorArmstrong and 1 others
March 3rd, 2019, 06:47 PM   #3
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 Originally Posted by romsek 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 $R=\dfrac{2 \left(a_1^2 c_0+2 a_1 c_0-a_1 c_1+c_0-c_1+c_2\right)}{a_1 \left(a_1+2\right){}^2}$ $C=\dfrac{a_1^3 (-T)-2 a_1^2 T}{4 \left(a_1^2 c_0+2 a_1 c_0-a_1 c_1+c_0-c_1+c_2\right)}$ $R_0= \dfrac{-c_0+c_1-c_2}{2 a_1}$ $b_1=\dfrac{\left(c_0+c_1+c_2\right) \text{QR}}{\left(a_1+2\right) T}$ If $a_2$ is consistent with this then this is the solution. If not there is no solution.
Thanks for the reply! May I know how you arrived at this solution in a bit more detail?

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
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 Originally Posted by babagreensheep Thanks for the reply! May I know how you arrived at this solution in a bit more detail? Also, how do I ensure $a_2$ is consistent with this?
I dumped it into mathematica.

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: 667 Thanks: 439 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. Thanks from topsquark and SenatorArmstrong
March 4th, 2019, 04:48 AM   #6
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Quote:
 Originally Posted by romsek 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 $R=\dfrac{2 \left(a_1^2 c_0+2 a_1 c_0-a_1 c_1+c_0-c_1+c_2\right)}{a_1 \left(a_1+2\right){}^2}$ $C=\dfrac{a_1^3 (-T)-2 a_1^2 T}{4 \left(a_1^2 c_0+2 a_1 c_0-a_1 c_1+c_0-c_1+c_2\right)}$ $R_0= \dfrac{-c_0+c_1-c_2}{2 a_1}$ $b_1=\dfrac{\left(c_0+c_1+c_2\right) \text{QR}}{\left(a_1+2\right) T}$ If $a_2$ is consistent with this then this is the solution. If not there is no solution.
I have worked out the following:

$RC = (-T)\dfrac{1}{2}\dfrac{a_1^3+2a_1^2}{a_1^3+4a_1^2+4a _1}$

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
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Quote:
 Originally Posted by babagreensheep I have the following equations: $\displaystyle c_0 = \frac{T^2b_1 + 2Q_RR_0T + 2Q_RRT + 4Q_RR_0RC + 2b_1RCT}{2Q_RT + 4Q_RRC}$
Have you got that equation memorized?

March 4th, 2019, 08:49 AM   #8
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 Originally Posted by babagreensheep I have worked out the following: $RC = (-T)\dfrac{1}{2}\dfrac{a_1^3+2a_1^2}{a_1^3+4a_1^2+4a _1}$ 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?
yep

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