Solving for Ellipse Parameters Given a radius and angle (Challenge 2)
Multi tool use
Given an ellipse centered on the origin in an x-y plane expressed as
$$bigg(frac{x}{a} bigg)^2+bigg(frac{y}{b} bigg)^2 = 1$$
In polar coordinates with radius $R$ and angle = $theta$, this can be expressed as:
$$R = sqrt{big(a^2 cos^2thetabig)+big(b^2 sin^2thetabig)}$$
If we set a and b to be related as follows:
$a = 1 - Delta$
$b = 1 + Delta$
The solution for $Delta$ as a function of $R$ and $theta$ is determined from
$$R = sqrt{1 - 2Delta cos(2theta)+ Delta^2}$$
Which as given in Solving for Ellipse Parameters given a radius and angle (Challenge 1)
has the simplest solution as $$Delta = cos(2theta) pm sqrt{R^2-sin^2(2theta)}$$
I am trying to similarly solve for $Delta$ given a modified ellipse with the following relationship:
$$R = (1-Delta)sqrt{1 - 2Delta cos(2theta)+ Delta^2}$$
What is the simplest form of $Delta$ given $R$ and $theta$ from the above equation?
If a simple relationship does not exist, then an approximation will be acceptable given that I can limit $Delta$ to be $0.9<Delta<1$
(Note this is related to my answer at this link on the signal processing site where I had to find the root of the equation to solve but am hoping for a simple closed form equation: https://dsp.stackexchange.com/questions/54006/computation-of-parameter-filter-to-match-a-given-frequency-response/54008#54008)
geometry conic-sections quartic-equations
add a comment |
Given an ellipse centered on the origin in an x-y plane expressed as
$$bigg(frac{x}{a} bigg)^2+bigg(frac{y}{b} bigg)^2 = 1$$
In polar coordinates with radius $R$ and angle = $theta$, this can be expressed as:
$$R = sqrt{big(a^2 cos^2thetabig)+big(b^2 sin^2thetabig)}$$
If we set a and b to be related as follows:
$a = 1 - Delta$
$b = 1 + Delta$
The solution for $Delta$ as a function of $R$ and $theta$ is determined from
$$R = sqrt{1 - 2Delta cos(2theta)+ Delta^2}$$
Which as given in Solving for Ellipse Parameters given a radius and angle (Challenge 1)
has the simplest solution as $$Delta = cos(2theta) pm sqrt{R^2-sin^2(2theta)}$$
I am trying to similarly solve for $Delta$ given a modified ellipse with the following relationship:
$$R = (1-Delta)sqrt{1 - 2Delta cos(2theta)+ Delta^2}$$
What is the simplest form of $Delta$ given $R$ and $theta$ from the above equation?
If a simple relationship does not exist, then an approximation will be acceptable given that I can limit $Delta$ to be $0.9<Delta<1$
(Note this is related to my answer at this link on the signal processing site where I had to find the root of the equation to solve but am hoping for a simple closed form equation: https://dsp.stackexchange.com/questions/54006/computation-of-parameter-filter-to-match-a-given-frequency-response/54008#54008)
geometry conic-sections quartic-equations
add a comment |
Given an ellipse centered on the origin in an x-y plane expressed as
$$bigg(frac{x}{a} bigg)^2+bigg(frac{y}{b} bigg)^2 = 1$$
In polar coordinates with radius $R$ and angle = $theta$, this can be expressed as:
$$R = sqrt{big(a^2 cos^2thetabig)+big(b^2 sin^2thetabig)}$$
If we set a and b to be related as follows:
$a = 1 - Delta$
$b = 1 + Delta$
The solution for $Delta$ as a function of $R$ and $theta$ is determined from
$$R = sqrt{1 - 2Delta cos(2theta)+ Delta^2}$$
Which as given in Solving for Ellipse Parameters given a radius and angle (Challenge 1)
has the simplest solution as $$Delta = cos(2theta) pm sqrt{R^2-sin^2(2theta)}$$
I am trying to similarly solve for $Delta$ given a modified ellipse with the following relationship:
$$R = (1-Delta)sqrt{1 - 2Delta cos(2theta)+ Delta^2}$$
What is the simplest form of $Delta$ given $R$ and $theta$ from the above equation?
If a simple relationship does not exist, then an approximation will be acceptable given that I can limit $Delta$ to be $0.9<Delta<1$
(Note this is related to my answer at this link on the signal processing site where I had to find the root of the equation to solve but am hoping for a simple closed form equation: https://dsp.stackexchange.com/questions/54006/computation-of-parameter-filter-to-match-a-given-frequency-response/54008#54008)
geometry conic-sections quartic-equations
Given an ellipse centered on the origin in an x-y plane expressed as
$$bigg(frac{x}{a} bigg)^2+bigg(frac{y}{b} bigg)^2 = 1$$
In polar coordinates with radius $R$ and angle = $theta$, this can be expressed as:
$$R = sqrt{big(a^2 cos^2thetabig)+big(b^2 sin^2thetabig)}$$
If we set a and b to be related as follows:
$a = 1 - Delta$
$b = 1 + Delta$
The solution for $Delta$ as a function of $R$ and $theta$ is determined from
$$R = sqrt{1 - 2Delta cos(2theta)+ Delta^2}$$
Which as given in Solving for Ellipse Parameters given a radius and angle (Challenge 1)
has the simplest solution as $$Delta = cos(2theta) pm sqrt{R^2-sin^2(2theta)}$$
I am trying to similarly solve for $Delta$ given a modified ellipse with the following relationship:
$$R = (1-Delta)sqrt{1 - 2Delta cos(2theta)+ Delta^2}$$
What is the simplest form of $Delta$ given $R$ and $theta$ from the above equation?
If a simple relationship does not exist, then an approximation will be acceptable given that I can limit $Delta$ to be $0.9<Delta<1$
(Note this is related to my answer at this link on the signal processing site where I had to find the root of the equation to solve but am hoping for a simple closed form equation: https://dsp.stackexchange.com/questions/54006/computation-of-parameter-filter-to-match-a-given-frequency-response/54008#54008)
geometry conic-sections quartic-equations
geometry conic-sections quartic-equations
edited Dec 11 '18 at 16:36
asked Dec 11 '18 at 14:34
Dan Boschen
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Squaring both sides gives
$$R^2 = (1 - Delta)^2 (Delta^2 - 2 Delta cos 2 theta + 1) .$$
This is a quartic equation in $Delta$, so there is a closed form in terms of $R, theta$, but it's too large to reproduce here: For general values of $theta, R$ there won't be a simple form, and given the content of the linked post, the numerical approximations you've already been using are probably the most useful. If you're interested in limiting cases, say, $R ll 1$ or $R gg 1$, one can analyze the above equation to produce good approximate solutions for $Delta$.
Edit If we have the additional condition that $R ll 1$, then substituting gives $Delta approx 1$. To get a next-order approximation, we can set $Delta = 1 + epsilon$ with $epsilon approx 0$ (this is close to the condition specified in the edit to the original question), giving $$epsilon^4 + 4 sin^2 theta , epsilon^3 + 4 sin^2 theta , epsilon^2 = R^2 .$$ Provided that $sin theta$ is not too close to $0$, the $epsilon^2$ term on the left dominates the others, giving $$epsilon approx pm frac{R}{2} csc theta,$$
so
$$Delta approx 1 pm frac{R}{2} csc theta .$$
On the domain $frac{9}{10} < Delta < 1$ of interest, this is a robust approximation:
From innermost to outermost, the red curves are the graphs of the solution functions $Delta(theta)$ for $R = frac{1}{8}, frac{1}{16}, frac{1}{32}, frac{1}{64}, frac{1}{128}$, and the blue curves are the graphs of the above approximations.
On the other hand, for very small $R$, say, $R < frac{1}{64}$, it's possible for to have $frac{9}{10} < Delta < 1$ but $sin theta$ small enough that the above approximation becomes poor: In the limit $sin theta to 0$, the $epsilon^4$ term dominates, giving a zeroth-order approximation $Delta approx 1 - sqrt{R}$. Computing to second order in $sin theta$ gives the approximation $$Delta approx 1 - sqrt{R} + left(1 - frac{1}{sqrt{R}}right) sin^2 theta .$$
This plot for $R = frac{1}{256}$ is typical for $R < frac{1}{100}$; the green curve is the small-angle approximation. (NB the expanded scale of the horizontal axis.)
Thank you Travis. I can limit $Delta$ to be greater than 0.9 and less then 1 if that helps get to an approximate solution. What approach do you suggest to find an approximate in that case? (Or should I post that as another question?)
– Dan Boschen
Dec 11 '18 at 16:18
If we know a priori that $frac{9}{10} < Delta < 1$, one strategy is to change variables to $s = 1 - Delta$, so that we're working in the domain $s ll 1$. Then, the equation becomes $R^2 = 2 (1 - cos 2 theta) s^2 - 2(1 - cos 2 theta) s^3 + s^4$. Provided that $cos 2 theta$ isn't too close to $1$, the $s^2$ term dominates, and ignoring the other terms gives $s approx pm frac{1}{2} R csc theta$ (so, the condition on $Delta$ implies that $R$ must have been small in the first place).
– Travis
Dec 11 '18 at 16:28
1
Super- yes this would be similar to poles for an analog filter on a s plane. In this case it is a digital filter so maps to polar coordinates on a z plane; the vertical imaginary $jomega$ axis becomes the unit circle. Thanks for your help!
– Dan Boschen
Dec 11 '18 at 17:02
1
Works great! I got 0.9696 using your estimate for my example case from the Signal Processing Stack Exchange link, where $theta$ was 0.377 and R=1/44.67. The answer I got by solving for the real root of the equation was 0.9691.
– Dan Boschen
Dec 11 '18 at 17:18
1
(And yes, I agree that for $theta = 0$ we have $Delta = 1 pm sqrt{R}$.)
– Travis
Dec 12 '18 at 1:40
|
show 16 more comments
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Squaring both sides gives
$$R^2 = (1 - Delta)^2 (Delta^2 - 2 Delta cos 2 theta + 1) .$$
This is a quartic equation in $Delta$, so there is a closed form in terms of $R, theta$, but it's too large to reproduce here: For general values of $theta, R$ there won't be a simple form, and given the content of the linked post, the numerical approximations you've already been using are probably the most useful. If you're interested in limiting cases, say, $R ll 1$ or $R gg 1$, one can analyze the above equation to produce good approximate solutions for $Delta$.
Edit If we have the additional condition that $R ll 1$, then substituting gives $Delta approx 1$. To get a next-order approximation, we can set $Delta = 1 + epsilon$ with $epsilon approx 0$ (this is close to the condition specified in the edit to the original question), giving $$epsilon^4 + 4 sin^2 theta , epsilon^3 + 4 sin^2 theta , epsilon^2 = R^2 .$$ Provided that $sin theta$ is not too close to $0$, the $epsilon^2$ term on the left dominates the others, giving $$epsilon approx pm frac{R}{2} csc theta,$$
so
$$Delta approx 1 pm frac{R}{2} csc theta .$$
On the domain $frac{9}{10} < Delta < 1$ of interest, this is a robust approximation:
From innermost to outermost, the red curves are the graphs of the solution functions $Delta(theta)$ for $R = frac{1}{8}, frac{1}{16}, frac{1}{32}, frac{1}{64}, frac{1}{128}$, and the blue curves are the graphs of the above approximations.
On the other hand, for very small $R$, say, $R < frac{1}{64}$, it's possible for to have $frac{9}{10} < Delta < 1$ but $sin theta$ small enough that the above approximation becomes poor: In the limit $sin theta to 0$, the $epsilon^4$ term dominates, giving a zeroth-order approximation $Delta approx 1 - sqrt{R}$. Computing to second order in $sin theta$ gives the approximation $$Delta approx 1 - sqrt{R} + left(1 - frac{1}{sqrt{R}}right) sin^2 theta .$$
This plot for $R = frac{1}{256}$ is typical for $R < frac{1}{100}$; the green curve is the small-angle approximation. (NB the expanded scale of the horizontal axis.)
Thank you Travis. I can limit $Delta$ to be greater than 0.9 and less then 1 if that helps get to an approximate solution. What approach do you suggest to find an approximate in that case? (Or should I post that as another question?)
– Dan Boschen
Dec 11 '18 at 16:18
If we know a priori that $frac{9}{10} < Delta < 1$, one strategy is to change variables to $s = 1 - Delta$, so that we're working in the domain $s ll 1$. Then, the equation becomes $R^2 = 2 (1 - cos 2 theta) s^2 - 2(1 - cos 2 theta) s^3 + s^4$. Provided that $cos 2 theta$ isn't too close to $1$, the $s^2$ term dominates, and ignoring the other terms gives $s approx pm frac{1}{2} R csc theta$ (so, the condition on $Delta$ implies that $R$ must have been small in the first place).
– Travis
Dec 11 '18 at 16:28
1
Super- yes this would be similar to poles for an analog filter on a s plane. In this case it is a digital filter so maps to polar coordinates on a z plane; the vertical imaginary $jomega$ axis becomes the unit circle. Thanks for your help!
– Dan Boschen
Dec 11 '18 at 17:02
1
Works great! I got 0.9696 using your estimate for my example case from the Signal Processing Stack Exchange link, where $theta$ was 0.377 and R=1/44.67. The answer I got by solving for the real root of the equation was 0.9691.
– Dan Boschen
Dec 11 '18 at 17:18
1
(And yes, I agree that for $theta = 0$ we have $Delta = 1 pm sqrt{R}$.)
– Travis
Dec 12 '18 at 1:40
|
show 16 more comments
Squaring both sides gives
$$R^2 = (1 - Delta)^2 (Delta^2 - 2 Delta cos 2 theta + 1) .$$
This is a quartic equation in $Delta$, so there is a closed form in terms of $R, theta$, but it's too large to reproduce here: For general values of $theta, R$ there won't be a simple form, and given the content of the linked post, the numerical approximations you've already been using are probably the most useful. If you're interested in limiting cases, say, $R ll 1$ or $R gg 1$, one can analyze the above equation to produce good approximate solutions for $Delta$.
Edit If we have the additional condition that $R ll 1$, then substituting gives $Delta approx 1$. To get a next-order approximation, we can set $Delta = 1 + epsilon$ with $epsilon approx 0$ (this is close to the condition specified in the edit to the original question), giving $$epsilon^4 + 4 sin^2 theta , epsilon^3 + 4 sin^2 theta , epsilon^2 = R^2 .$$ Provided that $sin theta$ is not too close to $0$, the $epsilon^2$ term on the left dominates the others, giving $$epsilon approx pm frac{R}{2} csc theta,$$
so
$$Delta approx 1 pm frac{R}{2} csc theta .$$
On the domain $frac{9}{10} < Delta < 1$ of interest, this is a robust approximation:
From innermost to outermost, the red curves are the graphs of the solution functions $Delta(theta)$ for $R = frac{1}{8}, frac{1}{16}, frac{1}{32}, frac{1}{64}, frac{1}{128}$, and the blue curves are the graphs of the above approximations.
On the other hand, for very small $R$, say, $R < frac{1}{64}$, it's possible for to have $frac{9}{10} < Delta < 1$ but $sin theta$ small enough that the above approximation becomes poor: In the limit $sin theta to 0$, the $epsilon^4$ term dominates, giving a zeroth-order approximation $Delta approx 1 - sqrt{R}$. Computing to second order in $sin theta$ gives the approximation $$Delta approx 1 - sqrt{R} + left(1 - frac{1}{sqrt{R}}right) sin^2 theta .$$
This plot for $R = frac{1}{256}$ is typical for $R < frac{1}{100}$; the green curve is the small-angle approximation. (NB the expanded scale of the horizontal axis.)
Thank you Travis. I can limit $Delta$ to be greater than 0.9 and less then 1 if that helps get to an approximate solution. What approach do you suggest to find an approximate in that case? (Or should I post that as another question?)
– Dan Boschen
Dec 11 '18 at 16:18
If we know a priori that $frac{9}{10} < Delta < 1$, one strategy is to change variables to $s = 1 - Delta$, so that we're working in the domain $s ll 1$. Then, the equation becomes $R^2 = 2 (1 - cos 2 theta) s^2 - 2(1 - cos 2 theta) s^3 + s^4$. Provided that $cos 2 theta$ isn't too close to $1$, the $s^2$ term dominates, and ignoring the other terms gives $s approx pm frac{1}{2} R csc theta$ (so, the condition on $Delta$ implies that $R$ must have been small in the first place).
– Travis
Dec 11 '18 at 16:28
1
Super- yes this would be similar to poles for an analog filter on a s plane. In this case it is a digital filter so maps to polar coordinates on a z plane; the vertical imaginary $jomega$ axis becomes the unit circle. Thanks for your help!
– Dan Boschen
Dec 11 '18 at 17:02
1
Works great! I got 0.9696 using your estimate for my example case from the Signal Processing Stack Exchange link, where $theta$ was 0.377 and R=1/44.67. The answer I got by solving for the real root of the equation was 0.9691.
– Dan Boschen
Dec 11 '18 at 17:18
1
(And yes, I agree that for $theta = 0$ we have $Delta = 1 pm sqrt{R}$.)
– Travis
Dec 12 '18 at 1:40
|
show 16 more comments
Squaring both sides gives
$$R^2 = (1 - Delta)^2 (Delta^2 - 2 Delta cos 2 theta + 1) .$$
This is a quartic equation in $Delta$, so there is a closed form in terms of $R, theta$, but it's too large to reproduce here: For general values of $theta, R$ there won't be a simple form, and given the content of the linked post, the numerical approximations you've already been using are probably the most useful. If you're interested in limiting cases, say, $R ll 1$ or $R gg 1$, one can analyze the above equation to produce good approximate solutions for $Delta$.
Edit If we have the additional condition that $R ll 1$, then substituting gives $Delta approx 1$. To get a next-order approximation, we can set $Delta = 1 + epsilon$ with $epsilon approx 0$ (this is close to the condition specified in the edit to the original question), giving $$epsilon^4 + 4 sin^2 theta , epsilon^3 + 4 sin^2 theta , epsilon^2 = R^2 .$$ Provided that $sin theta$ is not too close to $0$, the $epsilon^2$ term on the left dominates the others, giving $$epsilon approx pm frac{R}{2} csc theta,$$
so
$$Delta approx 1 pm frac{R}{2} csc theta .$$
On the domain $frac{9}{10} < Delta < 1$ of interest, this is a robust approximation:
From innermost to outermost, the red curves are the graphs of the solution functions $Delta(theta)$ for $R = frac{1}{8}, frac{1}{16}, frac{1}{32}, frac{1}{64}, frac{1}{128}$, and the blue curves are the graphs of the above approximations.
On the other hand, for very small $R$, say, $R < frac{1}{64}$, it's possible for to have $frac{9}{10} < Delta < 1$ but $sin theta$ small enough that the above approximation becomes poor: In the limit $sin theta to 0$, the $epsilon^4$ term dominates, giving a zeroth-order approximation $Delta approx 1 - sqrt{R}$. Computing to second order in $sin theta$ gives the approximation $$Delta approx 1 - sqrt{R} + left(1 - frac{1}{sqrt{R}}right) sin^2 theta .$$
This plot for $R = frac{1}{256}$ is typical for $R < frac{1}{100}$; the green curve is the small-angle approximation. (NB the expanded scale of the horizontal axis.)
Squaring both sides gives
$$R^2 = (1 - Delta)^2 (Delta^2 - 2 Delta cos 2 theta + 1) .$$
This is a quartic equation in $Delta$, so there is a closed form in terms of $R, theta$, but it's too large to reproduce here: For general values of $theta, R$ there won't be a simple form, and given the content of the linked post, the numerical approximations you've already been using are probably the most useful. If you're interested in limiting cases, say, $R ll 1$ or $R gg 1$, one can analyze the above equation to produce good approximate solutions for $Delta$.
Edit If we have the additional condition that $R ll 1$, then substituting gives $Delta approx 1$. To get a next-order approximation, we can set $Delta = 1 + epsilon$ with $epsilon approx 0$ (this is close to the condition specified in the edit to the original question), giving $$epsilon^4 + 4 sin^2 theta , epsilon^3 + 4 sin^2 theta , epsilon^2 = R^2 .$$ Provided that $sin theta$ is not too close to $0$, the $epsilon^2$ term on the left dominates the others, giving $$epsilon approx pm frac{R}{2} csc theta,$$
so
$$Delta approx 1 pm frac{R}{2} csc theta .$$
On the domain $frac{9}{10} < Delta < 1$ of interest, this is a robust approximation:
From innermost to outermost, the red curves are the graphs of the solution functions $Delta(theta)$ for $R = frac{1}{8}, frac{1}{16}, frac{1}{32}, frac{1}{64}, frac{1}{128}$, and the blue curves are the graphs of the above approximations.
On the other hand, for very small $R$, say, $R < frac{1}{64}$, it's possible for to have $frac{9}{10} < Delta < 1$ but $sin theta$ small enough that the above approximation becomes poor: In the limit $sin theta to 0$, the $epsilon^4$ term dominates, giving a zeroth-order approximation $Delta approx 1 - sqrt{R}$. Computing to second order in $sin theta$ gives the approximation $$Delta approx 1 - sqrt{R} + left(1 - frac{1}{sqrt{R}}right) sin^2 theta .$$
This plot for $R = frac{1}{256}$ is typical for $R < frac{1}{100}$; the green curve is the small-angle approximation. (NB the expanded scale of the horizontal axis.)
edited Dec 12 '18 at 2:17
answered Dec 11 '18 at 15:59
Travis
59.8k767146
59.8k767146
Thank you Travis. I can limit $Delta$ to be greater than 0.9 and less then 1 if that helps get to an approximate solution. What approach do you suggest to find an approximate in that case? (Or should I post that as another question?)
– Dan Boschen
Dec 11 '18 at 16:18
If we know a priori that $frac{9}{10} < Delta < 1$, one strategy is to change variables to $s = 1 - Delta$, so that we're working in the domain $s ll 1$. Then, the equation becomes $R^2 = 2 (1 - cos 2 theta) s^2 - 2(1 - cos 2 theta) s^3 + s^4$. Provided that $cos 2 theta$ isn't too close to $1$, the $s^2$ term dominates, and ignoring the other terms gives $s approx pm frac{1}{2} R csc theta$ (so, the condition on $Delta$ implies that $R$ must have been small in the first place).
– Travis
Dec 11 '18 at 16:28
1
Super- yes this would be similar to poles for an analog filter on a s plane. In this case it is a digital filter so maps to polar coordinates on a z plane; the vertical imaginary $jomega$ axis becomes the unit circle. Thanks for your help!
– Dan Boschen
Dec 11 '18 at 17:02
1
Works great! I got 0.9696 using your estimate for my example case from the Signal Processing Stack Exchange link, where $theta$ was 0.377 and R=1/44.67. The answer I got by solving for the real root of the equation was 0.9691.
– Dan Boschen
Dec 11 '18 at 17:18
1
(And yes, I agree that for $theta = 0$ we have $Delta = 1 pm sqrt{R}$.)
– Travis
Dec 12 '18 at 1:40
|
show 16 more comments
Thank you Travis. I can limit $Delta$ to be greater than 0.9 and less then 1 if that helps get to an approximate solution. What approach do you suggest to find an approximate in that case? (Or should I post that as another question?)
– Dan Boschen
Dec 11 '18 at 16:18
If we know a priori that $frac{9}{10} < Delta < 1$, one strategy is to change variables to $s = 1 - Delta$, so that we're working in the domain $s ll 1$. Then, the equation becomes $R^2 = 2 (1 - cos 2 theta) s^2 - 2(1 - cos 2 theta) s^3 + s^4$. Provided that $cos 2 theta$ isn't too close to $1$, the $s^2$ term dominates, and ignoring the other terms gives $s approx pm frac{1}{2} R csc theta$ (so, the condition on $Delta$ implies that $R$ must have been small in the first place).
– Travis
Dec 11 '18 at 16:28
1
Super- yes this would be similar to poles for an analog filter on a s plane. In this case it is a digital filter so maps to polar coordinates on a z plane; the vertical imaginary $jomega$ axis becomes the unit circle. Thanks for your help!
– Dan Boschen
Dec 11 '18 at 17:02
1
Works great! I got 0.9696 using your estimate for my example case from the Signal Processing Stack Exchange link, where $theta$ was 0.377 and R=1/44.67. The answer I got by solving for the real root of the equation was 0.9691.
– Dan Boschen
Dec 11 '18 at 17:18
1
(And yes, I agree that for $theta = 0$ we have $Delta = 1 pm sqrt{R}$.)
– Travis
Dec 12 '18 at 1:40
Thank you Travis. I can limit $Delta$ to be greater than 0.9 and less then 1 if that helps get to an approximate solution. What approach do you suggest to find an approximate in that case? (Or should I post that as another question?)
– Dan Boschen
Dec 11 '18 at 16:18
Thank you Travis. I can limit $Delta$ to be greater than 0.9 and less then 1 if that helps get to an approximate solution. What approach do you suggest to find an approximate in that case? (Or should I post that as another question?)
– Dan Boschen
Dec 11 '18 at 16:18
If we know a priori that $frac{9}{10} < Delta < 1$, one strategy is to change variables to $s = 1 - Delta$, so that we're working in the domain $s ll 1$. Then, the equation becomes $R^2 = 2 (1 - cos 2 theta) s^2 - 2(1 - cos 2 theta) s^3 + s^4$. Provided that $cos 2 theta$ isn't too close to $1$, the $s^2$ term dominates, and ignoring the other terms gives $s approx pm frac{1}{2} R csc theta$ (so, the condition on $Delta$ implies that $R$ must have been small in the first place).
– Travis
Dec 11 '18 at 16:28
If we know a priori that $frac{9}{10} < Delta < 1$, one strategy is to change variables to $s = 1 - Delta$, so that we're working in the domain $s ll 1$. Then, the equation becomes $R^2 = 2 (1 - cos 2 theta) s^2 - 2(1 - cos 2 theta) s^3 + s^4$. Provided that $cos 2 theta$ isn't too close to $1$, the $s^2$ term dominates, and ignoring the other terms gives $s approx pm frac{1}{2} R csc theta$ (so, the condition on $Delta$ implies that $R$ must have been small in the first place).
– Travis
Dec 11 '18 at 16:28
1
1
Super- yes this would be similar to poles for an analog filter on a s plane. In this case it is a digital filter so maps to polar coordinates on a z plane; the vertical imaginary $jomega$ axis becomes the unit circle. Thanks for your help!
– Dan Boschen
Dec 11 '18 at 17:02
Super- yes this would be similar to poles for an analog filter on a s plane. In this case it is a digital filter so maps to polar coordinates on a z plane; the vertical imaginary $jomega$ axis becomes the unit circle. Thanks for your help!
– Dan Boschen
Dec 11 '18 at 17:02
1
1
Works great! I got 0.9696 using your estimate for my example case from the Signal Processing Stack Exchange link, where $theta$ was 0.377 and R=1/44.67. The answer I got by solving for the real root of the equation was 0.9691.
– Dan Boschen
Dec 11 '18 at 17:18
Works great! I got 0.9696 using your estimate for my example case from the Signal Processing Stack Exchange link, where $theta$ was 0.377 and R=1/44.67. The answer I got by solving for the real root of the equation was 0.9691.
– Dan Boschen
Dec 11 '18 at 17:18
1
1
(And yes, I agree that for $theta = 0$ we have $Delta = 1 pm sqrt{R}$.)
– Travis
Dec 12 '18 at 1:40
(And yes, I agree that for $theta = 0$ we have $Delta = 1 pm sqrt{R}$.)
– Travis
Dec 12 '18 at 1:40
|
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