Choosing a complex square root
First off, I'm not sure if this question belongs here so I apologize if it's out of place.
I'm working with a system of coupled linear partial differential equations in space and time with which I'd like to conduct frequency response analysis. I took a LaPlace transform of the system with respect to time, solved a resulting second order ODE for x, and ended up with a complicated transfer function in $x$ and $s$. It looks like this:
$Fleft(sright)=frac{text{Stuff}}{e^{s+sqrt{As^2+Bs+c}}+...}$
Where the exponential term has that form because it is one of the roots of the characteristic equation for the ODE in $x$.
Frequency response analysis techniques then say to replace instances of $s$ with $jomega$, where $omega$ is the frequency of the input sine wave, and then plot the amplitude and angle of $Fleft(jomegaright)$. My question then is is there an "informed" choice to make for which square root solution I should take when computing these values.
The analog that comes to mind for me is when we take roots with real numbers, like when we calculated the roots of the characteristic equation of the above described ODE, we take the positive root and then write plus/minus to indicate roots. Here though since we're rooting a complex number we can have 4 roots and I'm not sure which one to take. When we aren't working in the complex plane, we'll often only consider the positive root because it has physical meaning. I'm just not sure if there is a similar interpretation here.
complex-numbers control-theory
asked Dec 1 '18 at 20:15
Leif Ericson
1248
add a comment |
First off, I'm not sure if this question belongs here so I apologize if it's out of place.
I'm working with a system of coupled linear partial differential equations in space and time with which I'd like to conduct frequency response analysis. I took a LaPlace transform of the system with respect to time, solved a resulting second order ODE for x, and ended up with a complicated transfer function in $x$ and $s$. It looks like this:
$Fleft(sright)=frac{text{Stuff}}{e^{s+sqrt{As^2+Bs+c}}+...}$
Where the exponential term has that form because it is one of the roots of the characteristic equation for the ODE in $x$.
Frequency response analysis techniques then say to replace instances of $s$ with $jomega$, where $omega$ is the frequency of the input sine wave, and then plot the amplitude and angle of $Fleft(jomegaright)$. My question then is is there an "informed" choice to make for which square root solution I should take when computing these values.
The analog that comes to mind for me is when we take roots with real numbers, like when we calculated the roots of the characteristic equation of the above described ODE, we take the positive root and then write plus/minus to indicate roots. Here though since we're rooting a complex number we can have 4 roots and I'm not sure which one to take. When we aren't working in the complex plane, we'll often only consider the positive root because it has physical meaning. I'm just not sure if there is a similar interpretation here.
complex-numbers control-theory
asked Dec 1 '18 at 20:15
Leif Ericson
1248
1
The $dots$ term corresponds to the other root, no? Then every choice of a specific root would yield the same final result.
– Did
Dec 1 '18 at 20:22
I guess I should have been more specific. If you let the roots of the characteristic equation be $r_1$ and $r_2$ then the denominator has the form: $e^{r_2}-e^{r_1}+r_2e^{r_2}-r_1e^{r_1}$
– Leif Ericson
Dec 1 '18 at 20:26
Precisely. Hence the whole denominator does not depend on which root you denote by $r_1$ and which root you denote by $r_2$ except by its sign. Is there a similar sign in "Stuff", which would make the ratio independent of the choice?
– Did
Dec 1 '18 at 20:38
The numerator has this same type of symmetry with the roots. Sorry I've left out all this information. I think I see what you're saying though. When you have this symmetry with the roots, the value will only differ up to a sign, so this shouldn't effect the magnitude of the resulting complex number. It will have an effect on the phase angle though.
– Leif Ericson
Dec 1 '18 at 21:15
No -- in the end the numerator is also affected with a minus sign for the other choice of the roots, hence the ratio stays exactly the same.
– Did
Dec 1 '18 at 23:03
add a comment |
First off, I'm not sure if this question belongs here so I apologize if it's out of place.
I'm working with a system of coupled linear partial differential equations in space and time with which I'd like to conduct frequency response analysis. I took a LaPlace transform of the system with respect to time, solved a resulting second order ODE for x, and ended up with a complicated transfer function in $x$ and $s$. It looks like this:
$Fleft(sright)=frac{text{Stuff}}{e^{s+sqrt{As^2+Bs+c}}+...}$
Where the exponential term has that form because it is one of the roots of the characteristic equation for the ODE in $x$.
Frequency response analysis techniques then say to replace instances of $s$ with $jomega$, where $omega$ is the frequency of the input sine wave, and then plot the amplitude and angle of $Fleft(jomegaright)$. My question then is is there an "informed" choice to make for which square root solution I should take when computing these values.
The analog that comes to mind for me is when we take roots with real numbers, like when we calculated the roots of the characteristic equation of the above described ODE, we take the positive root and then write plus/minus to indicate roots. Here though since we're rooting a complex number we can have 4 roots and I'm not sure which one to take. When we aren't working in the complex plane, we'll often only consider the positive root because it has physical meaning. I'm just not sure if there is a similar interpretation here.
complex-numbers control-theory
asked Dec 1 '18 at 20:15
Leif Ericson
1248
First off, I'm not sure if this question belongs here so I apologize if it's out of place.
I'm working with a system of coupled linear partial differential equations in space and time with which I'd like to conduct frequency response analysis. I took a LaPlace transform of the system with respect to time, solved a resulting second order ODE for x, and ended up with a complicated transfer function in $x$ and $s$. It looks like this:
$Fleft(sright)=frac{text{Stuff}}{e^{s+sqrt{As^2+Bs+c}}+...}$
Where the exponential term has that form because it is one of the roots of the characteristic equation for the ODE in $x$.
Frequency response analysis techniques then say to replace instances of $s$ with $jomega$, where $omega$ is the frequency of the input sine wave, and then plot the amplitude and angle of $Fleft(jomegaright)$. My question then is is there an "informed" choice to make for which square root solution I should take when computing these values.
The analog that comes to mind for me is when we take roots with real numbers, like when we calculated the roots of the characteristic equation of the above described ODE, we take the positive root and then write plus/minus to indicate roots. Here though since we're rooting a complex number we can have 4 roots and I'm not sure which one to take. When we aren't working in the complex plane, we'll often only consider the positive root because it has physical meaning. I'm just not sure if there is a similar interpretation here.
complex-numbers control-theory
complex-numbers control-theory
asked Dec 1 '18 at 20:15
Leif Ericson
1248
asked Dec 1 '18 at 20:15
Leif Ericson
1248
asked Dec 1 '18 at 20:15
Leif Ericson
1248
asked Dec 1 '18 at 20:15
Leif Ericson
1248
asked Dec 1 '18 at 20:15
Leif Ericson
1248
1248
1
The $dots$ term corresponds to the other root, no? Then every choice of a specific root would yield the same final result.
– Did
Dec 1 '18 at 20:22
I guess I should have been more specific. If you let the roots of the characteristic equation be $r_1$ and $r_2$ then the denominator has the form: $e^{r_2}-e^{r_1}+r_2e^{r_2}-r_1e^{r_1}$
– Leif Ericson
Dec 1 '18 at 20:26
Precisely. Hence the whole denominator does not depend on which root you denote by $r_1$ and which root you denote by $r_2$ except by its sign. Is there a similar sign in "Stuff", which would make the ratio independent of the choice?
– Did
Dec 1 '18 at 20:38
The numerator has this same type of symmetry with the roots. Sorry I've left out all this information. I think I see what you're saying though. When you have this symmetry with the roots, the value will only differ up to a sign, so this shouldn't effect the magnitude of the resulting complex number. It will have an effect on the phase angle though.
– Leif Ericson
Dec 1 '18 at 21:15
No -- in the end the numerator is also affected with a minus sign for the other choice of the roots, hence the ratio stays exactly the same.
– Did
Dec 1 '18 at 23:03
add a comment |
1
The $dots$ term corresponds to the other root, no? Then every choice of a specific root would yield the same final result.
– Did
Dec 1 '18 at 20:22
I guess I should have been more specific. If you let the roots of the characteristic equation be $r_1$ and $r_2$ then the denominator has the form: $e^{r_2}-e^{r_1}+r_2e^{r_2}-r_1e^{r_1}$
– Leif Ericson
Dec 1 '18 at 20:26
Precisely. Hence the whole denominator does not depend on which root you denote by $r_1$ and which root you denote by $r_2$ except by its sign. Is there a similar sign in "Stuff", which would make the ratio independent of the choice?
– Did
Dec 1 '18 at 20:38
The numerator has this same type of symmetry with the roots. Sorry I've left out all this information. I think I see what you're saying though. When you have this symmetry with the roots, the value will only differ up to a sign, so this shouldn't effect the magnitude of the resulting complex number. It will have an effect on the phase angle though.
– Leif Ericson
Dec 1 '18 at 21:15
No -- in the end the numerator is also affected with a minus sign for the other choice of the roots, hence the ratio stays exactly the same.
– Did
Dec 1 '18 at 23:03
1
1
The $dots$ term corresponds to the other root, no? Then every choice of a specific root would yield the same final result.
– Did
Dec 1 '18 at 20:22
The $dots$ term corresponds to the other root, no? Then every choice of a specific root would yield the same final result.
– Did
Dec 1 '18 at 20:22
I guess I should have been more specific. If you let the roots of the characteristic equation be $r_1$ and $r_2$ then the denominator has the form: $e^{r_2}-e^{r_1}+r_2e^{r_2}-r_1e^{r_1}$
– Leif Ericson
Dec 1 '18 at 20:26
I guess I should have been more specific. If you let the roots of the characteristic equation be $r_1$ and $r_2$ then the denominator has the form: $e^{r_2}-e^{r_1}+r_2e^{r_2}-r_1e^{r_1}$
– Leif Ericson
Dec 1 '18 at 20:26
Precisely. Hence the whole denominator does not depend on which root you denote by $r_1$ and which root you denote by $r_2$ except by its sign. Is there a similar sign in "Stuff", which would make the ratio independent of the choice?
– Did
Dec 1 '18 at 20:38
Precisely. Hence the whole denominator does not depend on which root you denote by $r_1$ and which root you denote by $r_2$ except by its sign. Is there a similar sign in "Stuff", which would make the ratio independent of the choice?
– Did
Dec 1 '18 at 20:38
The numerator has this same type of symmetry with the roots. Sorry I've left out all this information. I think I see what you're saying though. When you have this symmetry with the roots, the value will only differ up to a sign, so this shouldn't effect the magnitude of the resulting complex number. It will have an effect on the phase angle though.
– Leif Ericson
Dec 1 '18 at 21:15
The numerator has this same type of symmetry with the roots. Sorry I've left out all this information. I think I see what you're saying though. When you have this symmetry with the roots, the value will only differ up to a sign, so this shouldn't effect the magnitude of the resulting complex number. It will have an effect on the phase angle though.
– Leif Ericson
Dec 1 '18 at 21:15
No -- in the end the numerator is also affected with a minus sign for the other choice of the roots, hence the ratio stays exactly the same.
– Did
Dec 1 '18 at 23:03
No -- in the end the numerator is also affected with a minus sign for the other choice of the roots, hence the ratio stays exactly the same.
– Did
Dec 1 '18 at 23:03
add a comment |
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1
The $dots$ term corresponds to the other root, no? Then every choice of a specific root would yield the same final result.
– Did
Dec 1 '18 at 20:22
I guess I should have been more specific. If you let the roots of the characteristic equation be $r_1$ and $r_2$ then the denominator has the form: $e^{r_2}-e^{r_1}+r_2e^{r_2}-r_1e^{r_1}$
– Leif Ericson
Dec 1 '18 at 20:26
Precisely. Hence the whole denominator does not depend on which root you denote by $r_1$ and which root you denote by $r_2$ except by its sign. Is there a similar sign in "Stuff", which would make the ratio independent of the choice?
– Did
Dec 1 '18 at 20:38
The numerator has this same type of symmetry with the roots. Sorry I've left out all this information. I think I see what you're saying though. When you have this symmetry with the roots, the value will only differ up to a sign, so this shouldn't effect the magnitude of the resulting complex number. It will have an effect on the phase angle though.
– Leif Ericson
Dec 1 '18 at 21:15
No -- in the end the numerator is also affected with a minus sign for the other choice of the roots, hence the ratio stays exactly the same.
– Did
Dec 1 '18 at 23:03