Mixing
LTI System Time Constant Representation - Problem with quadratic formula
$begingroup$
I've got a problem with factorising the denominator of an LTI system. The system is a simple boost converter and my current transfer function is
$$ U_O = frac{lambda' cdot U_I - I_O cdot L cdot s}{C cdot L cdot s^2 + lambda'^2}. $$
I want to solve for the systems poles next and and bring them into time constant representation. I thus use the common approach
$$ C cdot L cdot s^2 + lambda'^2 overset{!}{=} 0 $$
which can be solved with the quadratic formula
$$ s = pm frac{sqrt{-4 cdot C cdot L cdot lambda'^2}}{2 cdot C cdot L} = pm frac{lambda'}{underbrace{sqrt{C cdot L}}_{T_{CL}}} j = frac{lambda'}{T_{CL}} j $$
and leads to the factorised form
$$ (s + frac{lambda'}{T_{CL}} j) cdot (s - frac{lambda'}{T_{CL}} j). $$
This result is obviously wrong since expanding the term doesn't give the same result and I'm not able to wrap my head around why this is the case.
Factorising the denominator with the third binomial formula gives the correct factors instead
$$ (T_{CL} cdot s + lambda' j) cdot (T_{CL} cdot s - lambda' j) $$
and these are already in time constant representation, which leads to the fully factorised transfer function in time constant representation
$$ U_O = frac{lambda' cdot U_I - I_O cdot L cdot s}{lambda'^2 cdot underbrace{(frac{T_{CL}}{lambda'} cdot s + j)}_p cdot underbrace{(frac{T_{CL}}{lambda'} cdot s - j)}_{p^*}}. $$
Where did I mess up the quadratic formula?
My only approach is that the coefficients aren't
$$
begin{split}
a &= C cdot L \
b &= 0 \
c &= lambda'^2
end{split}
$$
and I'm not solving for $s$, but that the coefficients are
$$
begin{split}
a &= 1 \
b &= 0 \
c &= lambda'^2
end{split}
$$
and I'm solving for $T_{CL} cdot s$ with respect to the quadratic equation
$$ {underbrace{(T_{CL} cdot s)}_x}^2 + lambda'^2. $$
Please give some advice on how to deal with such situations, I'm seriously confused and couldn't find an answer browsing through the material about quadratic equations. Different problems are far to over represented.
dynamical-systems quadratics
asked Jan 29 at 0:01
SimonSimon
31
$endgroup$
add a comment |
$begingroup$
I've got a problem with factorising the denominator of an LTI system. The system is a simple boost converter and my current transfer function is
$$ U_O = frac{lambda' cdot U_I - I_O cdot L cdot s}{C cdot L cdot s^2 + lambda'^2}. $$
I want to solve for the systems poles next and and bring them into time constant representation. I thus use the common approach
$$ C cdot L cdot s^2 + lambda'^2 overset{!}{=} 0 $$
which can be solved with the quadratic formula
$$ s = pm frac{sqrt{-4 cdot C cdot L cdot lambda'^2}}{2 cdot C cdot L} = pm frac{lambda'}{underbrace{sqrt{C cdot L}}_{T_{CL}}} j = frac{lambda'}{T_{CL}} j $$
and leads to the factorised form
$$ (s + frac{lambda'}{T_{CL}} j) cdot (s - frac{lambda'}{T_{CL}} j). $$
This result is obviously wrong since expanding the term doesn't give the same result and I'm not able to wrap my head around why this is the case.
Factorising the denominator with the third binomial formula gives the correct factors instead
$$ (T_{CL} cdot s + lambda' j) cdot (T_{CL} cdot s - lambda' j) $$
and these are already in time constant representation, which leads to the fully factorised transfer function in time constant representation
$$ U_O = frac{lambda' cdot U_I - I_O cdot L cdot s}{lambda'^2 cdot underbrace{(frac{T_{CL}}{lambda'} cdot s + j)}_p cdot underbrace{(frac{T_{CL}}{lambda'} cdot s - j)}_{p^*}}. $$
Where did I mess up the quadratic formula?
My only approach is that the coefficients aren't
$$
begin{split}
a &= C cdot L \
b &= 0 \
c &= lambda'^2
end{split}
$$
and I'm not solving for $s$, but that the coefficients are
$$
begin{split}
a &= 1 \
b &= 0 \
c &= lambda'^2
end{split}
$$
and I'm solving for $T_{CL} cdot s$ with respect to the quadratic equation
$$ {underbrace{(T_{CL} cdot s)}_x}^2 + lambda'^2. $$
Please give some advice on how to deal with such situations, I'm seriously confused and couldn't find an answer browsing through the material about quadratic equations. Different problems are far to over represented.
dynamical-systems quadratics
asked Jan 29 at 0:01
SimonSimon
31
$endgroup$
add a comment |
$begingroup$
I've got a problem with factorising the denominator of an LTI system. The system is a simple boost converter and my current transfer function is
$$ U_O = frac{lambda' cdot U_I - I_O cdot L cdot s}{C cdot L cdot s^2 + lambda'^2}. $$
I want to solve for the systems poles next and and bring them into time constant representation. I thus use the common approach
$$ C cdot L cdot s^2 + lambda'^2 overset{!}{=} 0 $$
which can be solved with the quadratic formula
$$ s = pm frac{sqrt{-4 cdot C cdot L cdot lambda'^2}}{2 cdot C cdot L} = pm frac{lambda'}{underbrace{sqrt{C cdot L}}_{T_{CL}}} j = frac{lambda'}{T_{CL}} j $$
and leads to the factorised form
$$ (s + frac{lambda'}{T_{CL}} j) cdot (s - frac{lambda'}{T_{CL}} j). $$
This result is obviously wrong since expanding the term doesn't give the same result and I'm not able to wrap my head around why this is the case.
Factorising the denominator with the third binomial formula gives the correct factors instead
$$ (T_{CL} cdot s + lambda' j) cdot (T_{CL} cdot s - lambda' j) $$
and these are already in time constant representation, which leads to the fully factorised transfer function in time constant representation
$$ U_O = frac{lambda' cdot U_I - I_O cdot L cdot s}{lambda'^2 cdot underbrace{(frac{T_{CL}}{lambda'} cdot s + j)}_p cdot underbrace{(frac{T_{CL}}{lambda'} cdot s - j)}_{p^*}}. $$
Where did I mess up the quadratic formula?
My only approach is that the coefficients aren't
$$
begin{split}
a &= C cdot L \
b &= 0 \
c &= lambda'^2
end{split}
$$
and I'm not solving for $s$, but that the coefficients are
$$
begin{split}
a &= 1 \
b &= 0 \
c &= lambda'^2
end{split}
$$
and I'm solving for $T_{CL} cdot s$ with respect to the quadratic equation
$$ {underbrace{(T_{CL} cdot s)}_x}^2 + lambda'^2. $$
Please give some advice on how to deal with such situations, I'm seriously confused and couldn't find an answer browsing through the material about quadratic equations. Different problems are far to over represented.
dynamical-systems quadratics
asked Jan 29 at 0:01
SimonSimon
31
$endgroup$
I've got a problem with factorising the denominator of an LTI system. The system is a simple boost converter and my current transfer function is
$$ U_O = frac{lambda' cdot U_I - I_O cdot L cdot s}{C cdot L cdot s^2 + lambda'^2}. $$
I want to solve for the systems poles next and and bring them into time constant representation. I thus use the common approach
$$ C cdot L cdot s^2 + lambda'^2 overset{!}{=} 0 $$
which can be solved with the quadratic formula
$$ s = pm frac{sqrt{-4 cdot C cdot L cdot lambda'^2}}{2 cdot C cdot L} = pm frac{lambda'}{underbrace{sqrt{C cdot L}}_{T_{CL}}} j = frac{lambda'}{T_{CL}} j $$
and leads to the factorised form
$$ (s + frac{lambda'}{T_{CL}} j) cdot (s - frac{lambda'}{T_{CL}} j). $$
This result is obviously wrong since expanding the term doesn't give the same result and I'm not able to wrap my head around why this is the case.
Factorising the denominator with the third binomial formula gives the correct factors instead
$$ (T_{CL} cdot s + lambda' j) cdot (T_{CL} cdot s - lambda' j) $$
and these are already in time constant representation, which leads to the fully factorised transfer function in time constant representation
$$ U_O = frac{lambda' cdot U_I - I_O cdot L cdot s}{lambda'^2 cdot underbrace{(frac{T_{CL}}{lambda'} cdot s + j)}_p cdot underbrace{(frac{T_{CL}}{lambda'} cdot s - j)}_{p^*}}. $$
Where did I mess up the quadratic formula?
My only approach is that the coefficients aren't
$$
begin{split}
a &= C cdot L \
b &= 0 \
c &= lambda'^2
end{split}
$$
and I'm not solving for $s$, but that the coefficients are
$$
begin{split}
a &= 1 \
b &= 0 \
c &= lambda'^2
end{split}
$$
and I'm solving for $T_{CL} cdot s$ with respect to the quadratic equation
$$ {underbrace{(T_{CL} cdot s)}_x}^2 + lambda'^2. $$
Please give some advice on how to deal with such situations, I'm seriously confused and couldn't find an answer browsing through the material about quadratic equations. Different problems are far to over represented.
dynamical-systems quadratics
dynamical-systems quadratics
asked Jan 29 at 0:01
SimonSimon
31
asked Jan 29 at 0:01
SimonSimon
31
asked Jan 29 at 0:01
SimonSimon
31
asked Jan 29 at 0:01
SimonSimon
31
asked Jan 29 at 0:01
SimonSimon
31
31
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This result is obviously wrong since expanding the term doesn't give the same result
I think you mean that you don't get back the original denominator. But this is ok. Remember that, when you have a quadratic equation, the quadratic formula gives you the roots of the equation. In your case, the quadratic equation is
$$ CLs^2 + lambda^2 = 0. $$
Note that we can divide and multiply both sides of the equation by any number that we want. Let's go ahead and divide both sides of the equation by $CL$:
$$ s^2 + frac{lambda^2}{CL} = 0. $$
And this is exactly what you get when you multiply out the following:
$$ (s + frac{lambda'}{T_{CL}} j)(s - frac{lambda'}{T_{CL}} j) = 0. $$
This shows you that you can multiply quadratic equation by a scalar and you will get the same roots.
answered Jan 29 at 0:25
NicNic8NicNic8
4,64831123
$endgroup$
$begingroup$
But how can I ensure that I get the "correct" form, the one that expands back to the original denominator. That is really important for problems like this. Doing a backwards calculation and adding a correction factor is always possible but not very elegant I guess.
$endgroup$
– Simon
Jan 29 at 0:29
$begingroup$
Ok, it seems to me that you are trying to factor the polynomial. I think the way to do this is to find the roots using the quadratic formula and then scale appropriately so that you get the original form back after multiplying out. That's the best way that I can think to do it.
$endgroup$
– NicNic8
Jan 29 at 17:47
$begingroup$
Thanks for your help man, I'm honestly embarrassed that I didn't realise this myself. Well... don't we all have moments of weakness?
$endgroup$
– Simon
Feb 9 at 21:05
$begingroup$
@Simon Of course!
$endgroup$
– NicNic8
Feb 11 at 18:32
add a comment |
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1 Answer
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$begingroup$
This result is obviously wrong since expanding the term doesn't give the same result
I think you mean that you don't get back the original denominator. But this is ok. Remember that, when you have a quadratic equation, the quadratic formula gives you the roots of the equation. In your case, the quadratic equation is
$$ CLs^2 + lambda^2 = 0. $$
Note that we can divide and multiply both sides of the equation by any number that we want. Let's go ahead and divide both sides of the equation by $CL$:
$$ s^2 + frac{lambda^2}{CL} = 0. $$
And this is exactly what you get when you multiply out the following:
$$ (s + frac{lambda'}{T_{CL}} j)(s - frac{lambda'}{T_{CL}} j) = 0. $$
This shows you that you can multiply quadratic equation by a scalar and you will get the same roots.
answered Jan 29 at 0:25
NicNic8NicNic8
4,64831123
$endgroup$
$begingroup$
But how can I ensure that I get the "correct" form, the one that expands back to the original denominator. That is really important for problems like this. Doing a backwards calculation and adding a correction factor is always possible but not very elegant I guess.
$endgroup$
– Simon
Jan 29 at 0:29
$begingroup$
Ok, it seems to me that you are trying to factor the polynomial. I think the way to do this is to find the roots using the quadratic formula and then scale appropriately so that you get the original form back after multiplying out. That's the best way that I can think to do it.
$endgroup$
– NicNic8
Jan 29 at 17:47
$begingroup$
Thanks for your help man, I'm honestly embarrassed that I didn't realise this myself. Well... don't we all have moments of weakness?
$endgroup$
– Simon
Feb 9 at 21:05
$begingroup$
@Simon Of course!
$endgroup$
– NicNic8
Feb 11 at 18:32
add a comment |
$begingroup$
This result is obviously wrong since expanding the term doesn't give the same result
I think you mean that you don't get back the original denominator. But this is ok. Remember that, when you have a quadratic equation, the quadratic formula gives you the roots of the equation. In your case, the quadratic equation is
$$ CLs^2 + lambda^2 = 0. $$
Note that we can divide and multiply both sides of the equation by any number that we want. Let's go ahead and divide both sides of the equation by $CL$:
$$ s^2 + frac{lambda^2}{CL} = 0. $$
And this is exactly what you get when you multiply out the following:
$$ (s + frac{lambda'}{T_{CL}} j)(s - frac{lambda'}{T_{CL}} j) = 0. $$
This shows you that you can multiply quadratic equation by a scalar and you will get the same roots.
answered Jan 29 at 0:25
NicNic8NicNic8
4,64831123
$endgroup$
$begingroup$
But how can I ensure that I get the "correct" form, the one that expands back to the original denominator. That is really important for problems like this. Doing a backwards calculation and adding a correction factor is always possible but not very elegant I guess.
$endgroup$
– Simon
Jan 29 at 0:29
$begingroup$
Ok, it seems to me that you are trying to factor the polynomial. I think the way to do this is to find the roots using the quadratic formula and then scale appropriately so that you get the original form back after multiplying out. That's the best way that I can think to do it.
$endgroup$
– NicNic8
Jan 29 at 17:47
$begingroup$
Thanks for your help man, I'm honestly embarrassed that I didn't realise this myself. Well... don't we all have moments of weakness?
$endgroup$
– Simon
Feb 9 at 21:05
$begingroup$
@Simon Of course!
$endgroup$
– NicNic8
Feb 11 at 18:32
add a comment |
$begingroup$
This result is obviously wrong since expanding the term doesn't give the same result
I think you mean that you don't get back the original denominator. But this is ok. Remember that, when you have a quadratic equation, the quadratic formula gives you the roots of the equation. In your case, the quadratic equation is
$$ CLs^2 + lambda^2 = 0. $$
Note that we can divide and multiply both sides of the equation by any number that we want. Let's go ahead and divide both sides of the equation by $CL$:
$$ s^2 + frac{lambda^2}{CL} = 0. $$
And this is exactly what you get when you multiply out the following:
$$ (s + frac{lambda'}{T_{CL}} j)(s - frac{lambda'}{T_{CL}} j) = 0. $$
This shows you that you can multiply quadratic equation by a scalar and you will get the same roots.
answered Jan 29 at 0:25
NicNic8NicNic8
4,64831123
$endgroup$
This result is obviously wrong since expanding the term doesn't give the same result
I think you mean that you don't get back the original denominator. But this is ok. Remember that, when you have a quadratic equation, the quadratic formula gives you the roots of the equation. In your case, the quadratic equation is
$$ CLs^2 + lambda^2 = 0. $$
Note that we can divide and multiply both sides of the equation by any number that we want. Let's go ahead and divide both sides of the equation by $CL$:
$$ s^2 + frac{lambda^2}{CL} = 0. $$
And this is exactly what you get when you multiply out the following:
$$ (s + frac{lambda'}{T_{CL}} j)(s - frac{lambda'}{T_{CL}} j) = 0. $$
This shows you that you can multiply quadratic equation by a scalar and you will get the same roots.
answered Jan 29 at 0:25
NicNic8NicNic8
4,64831123
answered Jan 29 at 0:25
NicNic8NicNic8
4,64831123
answered Jan 29 at 0:25
NicNic8NicNic8
4,64831123
answered Jan 29 at 0:25
NicNic8NicNic8
4,64831123
4,64831123
$begingroup$
But how can I ensure that I get the "correct" form, the one that expands back to the original denominator. That is really important for problems like this. Doing a backwards calculation and adding a correction factor is always possible but not very elegant I guess.
$endgroup$
– Simon
Jan 29 at 0:29
$begingroup$
Ok, it seems to me that you are trying to factor the polynomial. I think the way to do this is to find the roots using the quadratic formula and then scale appropriately so that you get the original form back after multiplying out. That's the best way that I can think to do it.
$endgroup$
– NicNic8
Jan 29 at 17:47
$begingroup$
Thanks for your help man, I'm honestly embarrassed that I didn't realise this myself. Well... don't we all have moments of weakness?
$endgroup$
– Simon
Feb 9 at 21:05
$begingroup$
@Simon Of course!
$endgroup$
– NicNic8
Feb 11 at 18:32
add a comment |
$begingroup$
But how can I ensure that I get the "correct" form, the one that expands back to the original denominator. That is really important for problems like this. Doing a backwards calculation and adding a correction factor is always possible but not very elegant I guess.
$endgroup$
– Simon
Jan 29 at 0:29
$begingroup$
Ok, it seems to me that you are trying to factor the polynomial. I think the way to do this is to find the roots using the quadratic formula and then scale appropriately so that you get the original form back after multiplying out. That's the best way that I can think to do it.
$endgroup$
– NicNic8
Jan 29 at 17:47
$begingroup$
Thanks for your help man, I'm honestly embarrassed that I didn't realise this myself. Well... don't we all have moments of weakness?
$endgroup$
– Simon
Feb 9 at 21:05
$begingroup$
@Simon Of course!
$endgroup$
– NicNic8
Feb 11 at 18:32
$begingroup$
But how can I ensure that I get the "correct" form, the one that expands back to the original denominator. That is really important for problems like this. Doing a backwards calculation and adding a correction factor is always possible but not very elegant I guess.
$endgroup$
– Simon
Jan 29 at 0:29
$begingroup$
But how can I ensure that I get the "correct" form, the one that expands back to the original denominator. That is really important for problems like this. Doing a backwards calculation and adding a correction factor is always possible but not very elegant I guess.
$endgroup$
– Simon
Jan 29 at 0:29
$begingroup$
Ok, it seems to me that you are trying to factor the polynomial. I think the way to do this is to find the roots using the quadratic formula and then scale appropriately so that you get the original form back after multiplying out. That's the best way that I can think to do it.
$endgroup$
– NicNic8
Jan 29 at 17:47
$begingroup$
Ok, it seems to me that you are trying to factor the polynomial. I think the way to do this is to find the roots using the quadratic formula and then scale appropriately so that you get the original form back after multiplying out. That's the best way that I can think to do it.
$endgroup$
– NicNic8
Jan 29 at 17:47
$begingroup$
Thanks for your help man, I'm honestly embarrassed that I didn't realise this myself. Well... don't we all have moments of weakness?
$endgroup$
– Simon
Feb 9 at 21:05
$begingroup$
Thanks for your help man, I'm honestly embarrassed that I didn't realise this myself. Well... don't we all have moments of weakness?
$endgroup$
– Simon
Feb 9 at 21:05
$begingroup$
@Simon Of course!
$endgroup$
– NicNic8
Feb 11 at 18:32
$begingroup$
@Simon Of course!
$endgroup$
– NicNic8
Feb 11 at 18:32
add a comment |
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