Methods of determining if all roots of a polynomial have a negative real part
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So my question is basically summarized in the title.
The root of the question lies actually in application of the rules. Namely the stability of linear time invariant feedback systems is determined by concluding that all the the poles of rational polynomial function have negative real parts.
Now in control theory this is very cubersomly solved by use of Hurwitz determinants and constructing wierd tables with loads of special cases. All of this if founded on Routh-Hurwitz theorem.
My question is if there are any more elegant methods to determine any of following :
- Number of roots with negative real parts
- Number of roots with positive real parts
- Ascertain existence of roots with positive real parts ( as to say system is unstable ) ?
- Exclude the possibility that roots with negative/positive real parts exist?
polynomials roots control-theory
asked Nov 18 at 19:41
TheCoolDrop
16010
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show 2 more comments
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1
down vote
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So my question is basically summarized in the title.
The root of the question lies actually in application of the rules. Namely the stability of linear time invariant feedback systems is determined by concluding that all the the poles of rational polynomial function have negative real parts.
Now in control theory this is very cubersomly solved by use of Hurwitz determinants and constructing wierd tables with loads of special cases. All of this if founded on Routh-Hurwitz theorem.
My question is if there are any more elegant methods to determine any of following :
- Number of roots with negative real parts
- Number of roots with positive real parts
- Ascertain existence of roots with positive real parts ( as to say system is unstable ) ?
- Exclude the possibility that roots with negative/positive real parts exist?
polynomials roots control-theory
asked Nov 18 at 19:41
TheCoolDrop
16010
en.wikipedia.org/wiki/Sturm%27s_theorem Ive used this method many times to locate real roots in intervals. It will work as long as the degree of the polynomial is no so large as to hit a computer precision problem.
– herb steinberg
Nov 18 at 19:47
@herbsteinberg It's been a while, but doesn't Sturm only address the real roots in an interval? Here the OP wants the roots with real part in the specified range.
– lulu
Nov 18 at 19:49
@lulu The interval is specified by the user. It is not applicable for finding complex roots.
– herb steinberg
Nov 18 at 19:53
@herbsteinberg As I say, though, the OP is not trying to count the negative real roots. For that, I agree Sturm would get the job done. The OP is looking for complex roots with real parts in a specified interval. Unless I have misunderstood, of course. Anyway, I'm not aware of a good way to do the desired count (which, of course, doesn't mean there isn't one)
– lulu
Nov 18 at 19:59
@lulu actually also counting the number of roots with negative real parts would also be useful since total number of roots is capped by degree of the polynomial.Then I could ascertain that there are no roots with positive real parts. But strictly real roots would not be useful in general
– TheCoolDrop
Nov 18 at 20:17
|
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
So my question is basically summarized in the title.
The root of the question lies actually in application of the rules. Namely the stability of linear time invariant feedback systems is determined by concluding that all the the poles of rational polynomial function have negative real parts.
Now in control theory this is very cubersomly solved by use of Hurwitz determinants and constructing wierd tables with loads of special cases. All of this if founded on Routh-Hurwitz theorem.
My question is if there are any more elegant methods to determine any of following :
- Number of roots with negative real parts
- Number of roots with positive real parts
- Ascertain existence of roots with positive real parts ( as to say system is unstable ) ?
- Exclude the possibility that roots with negative/positive real parts exist?
polynomials roots control-theory
asked Nov 18 at 19:41
TheCoolDrop
16010
So my question is basically summarized in the title.
The root of the question lies actually in application of the rules. Namely the stability of linear time invariant feedback systems is determined by concluding that all the the poles of rational polynomial function have negative real parts.
Now in control theory this is very cubersomly solved by use of Hurwitz determinants and constructing wierd tables with loads of special cases. All of this if founded on Routh-Hurwitz theorem.
My question is if there are any more elegant methods to determine any of following :
- Number of roots with negative real parts
- Number of roots with positive real parts
- Ascertain existence of roots with positive real parts ( as to say system is unstable ) ?
- Exclude the possibility that roots with negative/positive real parts exist?
polynomials roots control-theory
polynomials roots control-theory
asked Nov 18 at 19:41
TheCoolDrop
16010
asked Nov 18 at 19:41
TheCoolDrop
16010
edited Nov 18 at 20:21
asked Nov 18 at 19:41
TheCoolDrop
16010
asked Nov 18 at 19:41
TheCoolDrop
16010
asked Nov 18 at 19:41
TheCoolDrop
16010
16010
en.wikipedia.org/wiki/Sturm%27s_theorem Ive used this method many times to locate real roots in intervals. It will work as long as the degree of the polynomial is no so large as to hit a computer precision problem.
– herb steinberg
Nov 18 at 19:47
@herbsteinberg It's been a while, but doesn't Sturm only address the real roots in an interval? Here the OP wants the roots with real part in the specified range.
– lulu
Nov 18 at 19:49
@lulu The interval is specified by the user. It is not applicable for finding complex roots.
– herb steinberg
Nov 18 at 19:53
@herbsteinberg As I say, though, the OP is not trying to count the negative real roots. For that, I agree Sturm would get the job done. The OP is looking for complex roots with real parts in a specified interval. Unless I have misunderstood, of course. Anyway, I'm not aware of a good way to do the desired count (which, of course, doesn't mean there isn't one)
– lulu
Nov 18 at 19:59
@lulu actually also counting the number of roots with negative real parts would also be useful since total number of roots is capped by degree of the polynomial.Then I could ascertain that there are no roots with positive real parts. But strictly real roots would not be useful in general
– TheCoolDrop
Nov 18 at 20:17
|
show 2 more comments
en.wikipedia.org/wiki/Sturm%27s_theorem Ive used this method many times to locate real roots in intervals. It will work as long as the degree of the polynomial is no so large as to hit a computer precision problem.
– herb steinberg
Nov 18 at 19:47
@herbsteinberg It's been a while, but doesn't Sturm only address the real roots in an interval? Here the OP wants the roots with real part in the specified range.
– lulu
Nov 18 at 19:49
@lulu The interval is specified by the user. It is not applicable for finding complex roots.
– herb steinberg
Nov 18 at 19:53
@herbsteinberg As I say, though, the OP is not trying to count the negative real roots. For that, I agree Sturm would get the job done. The OP is looking for complex roots with real parts in a specified interval. Unless I have misunderstood, of course. Anyway, I'm not aware of a good way to do the desired count (which, of course, doesn't mean there isn't one)
– lulu
Nov 18 at 19:59
@lulu actually also counting the number of roots with negative real parts would also be useful since total number of roots is capped by degree of the polynomial.Then I could ascertain that there are no roots with positive real parts. But strictly real roots would not be useful in general
– TheCoolDrop
Nov 18 at 20:17
en.wikipedia.org/wiki/Sturm%27s_theorem Ive used this method many times to locate real roots in intervals. It will work as long as the degree of the polynomial is no so large as to hit a computer precision problem.
– herb steinberg
Nov 18 at 19:47
en.wikipedia.org/wiki/Sturm%27s_theorem Ive used this method many times to locate real roots in intervals. It will work as long as the degree of the polynomial is no so large as to hit a computer precision problem.
– herb steinberg
Nov 18 at 19:47
@herbsteinberg It's been a while, but doesn't Sturm only address the real roots in an interval? Here the OP wants the roots with real part in the specified range.
– lulu
Nov 18 at 19:49
@herbsteinberg It's been a while, but doesn't Sturm only address the real roots in an interval? Here the OP wants the roots with real part in the specified range.
– lulu
Nov 18 at 19:49
@lulu The interval is specified by the user. It is not applicable for finding complex roots.
– herb steinberg
Nov 18 at 19:53
@lulu The interval is specified by the user. It is not applicable for finding complex roots.
– herb steinberg
Nov 18 at 19:53
@herbsteinberg As I say, though, the OP is not trying to count the negative real roots. For that, I agree Sturm would get the job done. The OP is looking for complex roots with real parts in a specified interval. Unless I have misunderstood, of course. Anyway, I'm not aware of a good way to do the desired count (which, of course, doesn't mean there isn't one)
– lulu
Nov 18 at 19:59
@herbsteinberg As I say, though, the OP is not trying to count the negative real roots. For that, I agree Sturm would get the job done. The OP is looking for complex roots with real parts in a specified interval. Unless I have misunderstood, of course. Anyway, I'm not aware of a good way to do the desired count (which, of course, doesn't mean there isn't one)
– lulu
Nov 18 at 19:59
@lulu actually also counting the number of roots with negative real parts would also be useful since total number of roots is capped by degree of the polynomial.Then I could ascertain that there are no roots with positive real parts. But strictly real roots would not be useful in general
– TheCoolDrop
Nov 18 at 20:17
@lulu actually also counting the number of roots with negative real parts would also be useful since total number of roots is capped by degree of the polynomial.Then I could ascertain that there are no roots with positive real parts. But strictly real roots would not be useful in general
– TheCoolDrop
Nov 18 at 20:17
|
show 2 more comments
1 Answer
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You could use a Nyquist plot and instead look at encirclements of the origin instead of the minus one point. Starting with a rational polynomial transfer function
$$
G(s) = frac{N(s)}{D(s)} = frac{b_0 + b_1,s + cdots b_m,s^m}{a_0 + a_1,s + cdots a_n,s^n},
$$
with $b_m,a_n neq 0$ and $n geq m$. Since the zeros of $G(s)$ are also not known (and currently not of interest) and one would prefer that the considered transfer function in the Nyquist plot to be bounded (for ease of encirclement counting) you could use
$$
H(s) = frac{(1 + s)^n}{a_0 + a_1,s + cdots a_n,s^n}.
$$
It can also be noted that if $a_0=0$ then the Nyquist plot of $H(s)$ will still blow up near $s=0$. This however can easily be fixed since $a_0=0$ implies that $D(s)$ has a root at $s=0$ which can be factored out, such that one can continu as stated above with the residual of $D(s)$.
Now if $G(s)$ would not have any unstable poles (all roots of $D(s)$ have negative real part) then the Nyquist plot of $H(s)$ would have zero encirclements of the origin.
answered Nov 18 at 21:55
Kwin van der Veen
5,2202826
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You could use a Nyquist plot and instead look at encirclements of the origin instead of the minus one point. Starting with a rational polynomial transfer function
$$
G(s) = frac{N(s)}{D(s)} = frac{b_0 + b_1,s + cdots b_m,s^m}{a_0 + a_1,s + cdots a_n,s^n},
$$
with $b_m,a_n neq 0$ and $n geq m$. Since the zeros of $G(s)$ are also not known (and currently not of interest) and one would prefer that the considered transfer function in the Nyquist plot to be bounded (for ease of encirclement counting) you could use
$$
H(s) = frac{(1 + s)^n}{a_0 + a_1,s + cdots a_n,s^n}.
$$
It can also be noted that if $a_0=0$ then the Nyquist plot of $H(s)$ will still blow up near $s=0$. This however can easily be fixed since $a_0=0$ implies that $D(s)$ has a root at $s=0$ which can be factored out, such that one can continu as stated above with the residual of $D(s)$.
Now if $G(s)$ would not have any unstable poles (all roots of $D(s)$ have negative real part) then the Nyquist plot of $H(s)$ would have zero encirclements of the origin.
answered Nov 18 at 21:55
Kwin van der Veen
5,2202826
add a comment |
up vote
0
down vote
You could use a Nyquist plot and instead look at encirclements of the origin instead of the minus one point. Starting with a rational polynomial transfer function
$$
G(s) = frac{N(s)}{D(s)} = frac{b_0 + b_1,s + cdots b_m,s^m}{a_0 + a_1,s + cdots a_n,s^n},
$$
with $b_m,a_n neq 0$ and $n geq m$. Since the zeros of $G(s)$ are also not known (and currently not of interest) and one would prefer that the considered transfer function in the Nyquist plot to be bounded (for ease of encirclement counting) you could use
$$
H(s) = frac{(1 + s)^n}{a_0 + a_1,s + cdots a_n,s^n}.
$$
It can also be noted that if $a_0=0$ then the Nyquist plot of $H(s)$ will still blow up near $s=0$. This however can easily be fixed since $a_0=0$ implies that $D(s)$ has a root at $s=0$ which can be factored out, such that one can continu as stated above with the residual of $D(s)$.
Now if $G(s)$ would not have any unstable poles (all roots of $D(s)$ have negative real part) then the Nyquist plot of $H(s)$ would have zero encirclements of the origin.
answered Nov 18 at 21:55
Kwin van der Veen
5,2202826
add a comment |
up vote
0
down vote
up vote
0
down vote
You could use a Nyquist plot and instead look at encirclements of the origin instead of the minus one point. Starting with a rational polynomial transfer function
$$
G(s) = frac{N(s)}{D(s)} = frac{b_0 + b_1,s + cdots b_m,s^m}{a_0 + a_1,s + cdots a_n,s^n},
$$
with $b_m,a_n neq 0$ and $n geq m$. Since the zeros of $G(s)$ are also not known (and currently not of interest) and one would prefer that the considered transfer function in the Nyquist plot to be bounded (for ease of encirclement counting) you could use
$$
H(s) = frac{(1 + s)^n}{a_0 + a_1,s + cdots a_n,s^n}.
$$
It can also be noted that if $a_0=0$ then the Nyquist plot of $H(s)$ will still blow up near $s=0$. This however can easily be fixed since $a_0=0$ implies that $D(s)$ has a root at $s=0$ which can be factored out, such that one can continu as stated above with the residual of $D(s)$.
Now if $G(s)$ would not have any unstable poles (all roots of $D(s)$ have negative real part) then the Nyquist plot of $H(s)$ would have zero encirclements of the origin.
answered Nov 18 at 21:55
Kwin van der Veen
5,2202826
You could use a Nyquist plot and instead look at encirclements of the origin instead of the minus one point. Starting with a rational polynomial transfer function
$$
G(s) = frac{N(s)}{D(s)} = frac{b_0 + b_1,s + cdots b_m,s^m}{a_0 + a_1,s + cdots a_n,s^n},
$$
with $b_m,a_n neq 0$ and $n geq m$. Since the zeros of $G(s)$ are also not known (and currently not of interest) and one would prefer that the considered transfer function in the Nyquist plot to be bounded (for ease of encirclement counting) you could use
$$
H(s) = frac{(1 + s)^n}{a_0 + a_1,s + cdots a_n,s^n}.
$$
It can also be noted that if $a_0=0$ then the Nyquist plot of $H(s)$ will still blow up near $s=0$. This however can easily be fixed since $a_0=0$ implies that $D(s)$ has a root at $s=0$ which can be factored out, such that one can continu as stated above with the residual of $D(s)$.
Now if $G(s)$ would not have any unstable poles (all roots of $D(s)$ have negative real part) then the Nyquist plot of $H(s)$ would have zero encirclements of the origin.
answered Nov 18 at 21:55
Kwin van der Veen
5,2202826
answered Nov 18 at 21:55
Kwin van der Veen
5,2202826
answered Nov 18 at 21:55
Kwin van der Veen
5,2202826
answered Nov 18 at 21:55
Kwin van der Veen
5,2202826
5,2202826
add a comment |
add a comment |
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en.wikipedia.org/wiki/Sturm%27s_theorem Ive used this method many times to locate real roots in intervals. It will work as long as the degree of the polynomial is no so large as to hit a computer precision problem.
– herb steinberg
Nov 18 at 19:47
@herbsteinberg It's been a while, but doesn't Sturm only address the real roots in an interval? Here the OP wants the roots with real part in the specified range.
– lulu
Nov 18 at 19:49
@lulu The interval is specified by the user. It is not applicable for finding complex roots.
– herb steinberg
Nov 18 at 19:53
@herbsteinberg As I say, though, the OP is not trying to count the negative real roots. For that, I agree Sturm would get the job done. The OP is looking for complex roots with real parts in a specified interval. Unless I have misunderstood, of course. Anyway, I'm not aware of a good way to do the desired count (which, of course, doesn't mean there isn't one)
– lulu
Nov 18 at 19:59
@lulu actually also counting the number of roots with negative real parts would also be useful since total number of roots is capped by degree of the polynomial.Then I could ascertain that there are no roots with positive real parts. But strictly real roots would not be useful in general
– TheCoolDrop
Nov 18 at 20:17