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General Solution to nth order linear differential equation w/ x(t)=Asin(wt+phi)
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General Solution to nth order linear differential equation w/ x(t)=Asin(wt+phi)
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Chondrally
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General Solution to nth order linear differential equation w/ x(t)=Asin(wt+phi)
«
on:
03/05/2018 16:24:08 »
let ay''(t)+by'(t)+cy(t) = x(t)
t
If x(t) = Asin(wt+phi)
then y(t) = Bsin(wt+theta)
y'(t) = Bwcos(wt+theta)
y''(t)=-Bw^2*sin(wt+theta)
then the true solution is:
tan(theta)=(tan(phi)*(-aw^2+c)-bw)/(-aw^2+c+bw*tan(phi))
B=(A*sin(phi))/(sin(theta)*(-aw^2+c)+bw*cos(theta))
so y(t)=B*sin(wt+theta)
here is the derivation:
w will not change from these equations as it is a linear time invariant system.....
so we can assume y(t)=B*sin(wt+theta) where only the magnitude and phase change under the transformation.
if the equations are not linear time invariant, ie nonlinear, even the frequency can change nonlinearly or not be preserved at all.
y'(t)=Bw*cos(wt+theta)
y''(t)=-Bw^2*sin(wt+theta)
therefore we can write:
a*(-Bw^2*sin(wt+theta))+b*Bw*cos(wt+theta)+c*B*sin(wt+theta) = A*sin(wt+phi)
since cos(x+y)=cos(x)cos(y)-sin(x)sin(y) and sin(x+y)=sin(x)cos(y)+cos(x)sin(y) we can expand the terms and get
-aBw^2*(sin(wt)*cos(theta)+cos(wt)+sin(theta))+b*Bw*(cos(wt)*cos(theta)-sin(wt)*sin(theta))+cB*(sin(wt)*cos(theta)+cos(wt)*sin(theta)) = A*(sin(wt)*cos(phi)+cos(wt)*sin(phi))
we can now separate this into two equations, one in sin(wt) and one in cos(wt) and eliminate sin(wt) and cos(wt) as they are orthogonal.
-aBw^2*cos(theta)-bBw*sin(theta)+cB*cos(theta)=A*cos(phi) (1)
and
-aBw^2*sin(theta)+bBw*cos(theta)+cB*sin(theta)=A*sin(phi) (2)
divide equation (2) by equation (1) so that cos(phi) is the divisor of sin(phi to give us tan(phi) and eliminate B and A
tan(phi)=(-aw^2*sin(theta)+bw*cos(theta)+c*sin(theta))/(-aw^2*cos(theta)-bw*sin(theta)+c*cos(theta)) (3)
now divide the numerator and denominaotr by cos(theta) to either eliminate cos(theta) or create tan(theta) and this will not affect the tan(phi)
tan(phi)=(-aw^2*tan(theta)+bw+c*tan(theta))/(-aw^2-bw*tan(theta)+c) (4)
multiply the denominator by both sides and separate out tan(theta) and remove it to one side of the equation to give
tan(theta)=(tan(phi)*(-aw^2+c)-bw)/(-aw^2+c+bw*tan(phi)) (5)
now we can solve for theta from a, b,c,w and phi , the givens.
now substitute all these into equation (2) and solve for B to give:
B=(A*sin(phi))/(sin(theta)*(-aw^2+c)+bw*cos(theta)) (6)
and voila, we now have B and theta for any w,a,b,c and phi and A.
similarly from equation (1) we can also say:
B=(A*cos(phi))/(cos(theta)*(-aw^2+c)-bw*sin(theta)) (7)
This does not apply to polynomials, and it is not Fourier or Laplace. Neither Fourier or Laplace can
give you theta, but they will give you B. In fact because both x(t) and y(t) are completely real, there is no complex phase.
if it is third order,
tan(theta)=(tan(phi)*(-bw^2+d)-(aw^3+cw))/(-tan(phi)*(aw^3+cw)+(-bw^2+d))
if it is fourth order
tan(theta)=(tan(phi)*(aw^4-cw^2+e)+(bw^3-dw))/(-tan(phi)*(aw^4-cw^2+e)+(-bw^3-dw))
if a(n)y(n)(t)+a(n-1)y(n-1)(t) +...+a0y(t)=x(t)
and x(t) = Asin(wt+phi)
then y(t) = Bsin(wt+theta)
and there is a general formula for tan(theta) and B depending on n, the nth order linear differential equation.
It is too difficult to express here as the notation is not available...but i will try to express it here
a(n)*y(n)(t)+a(n-1)*y(n-1)(t) +..... + a1y'(t)+a0y(t) = x(t) where x(t) = Asin(wt+phi) and y(t)=B*sin(wt+theta)
then
tan(theta) = ( tan(phi)*(sum k from 0 to n by 2: ak*(jw)^k) - (sum k from 1 to n by 2: ak*w^k*j^(k+1)) )/
(sum k from 0 to n by 2: ak*(jw)^k) +tan(phi)*(sum k from 1 to n by 2: ak*w^k*j^(k+1)) )
B=A*cos(phi)/[ cos(theta)*(sum k from 0 to n by 2: ak*(jw)^k)+sin(theta)*(sum k from 1 to n by 2: ak*w^k*j^(k+1))]
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Home Forums Archive Member Blogs General Solution to nth order linear differential equation w/ x(t)=Asin(wt+phi)
General Solution to nth order linear differential equation w/ x(t)=Asin(wt+phi)
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May 06, 2018 #1
Richard Belshaw
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Last seen: 13 sec ago
Joined: 2007-02-27
General Solution to nth order linear differential equation w/ x(t)=Asin(wt+phi)
et ay''(t)+by'(t)+cy(t) = x(t)
If x(t) = Asin(wt+phi)
then y(t) = Bsin(wt+theta)
y'(t) = Bwcos(wt+theta)
y''(t)=-Bw^2*sin(wt+theta)
then the true solution is:
tan(theta)=(tan(phi)*(-aw^2+c)-bw)/(-aw^2+c+bw*tan(phi))
B=(A*sin(phi))/(sin(theta)*(-aw^2+c)+bw*cos(theta))
so y(t)=B*sin(wt+theta)
here is the derivation:
w will not change from these equations as it is a linear time
invariant system.....
so we can assume y(t)=B*sin(wt+theta) where only the magnitude and
phase change under the transformation.
if the equations are not linear time invariant, ie nonlinear, even
the frequency can change nonlinearly or not be preserved at all.
y'(t)=Bw*cos(wt+theta)
y''(t)=-Bw^2*sin(wt+theta)
therefore we can write:
a*(-Bw^2*sin(wt+theta))+b*Bw*cos(wt+theta)+c*B*sin(wt+theta) = A*sin(wt+phi)
since cos(x+y)=cos(x)cos(y)-sin(x)sin(y) and
sin(x+y)=sin(x)cos(y)+cos(x)sin(y) we can expand the terms and get
-aBw^2*(sin(wt)*cos(theta)+cos(wt)+sin(theta))+b*Bw*(cos(wt)*cos(theta)-sin(wt)*sin(theta))+cB*(sin(wt)*cos(theta)+cos(wt)*sin(theta))
= A*(sin(wt)*cos(phi)+cos(wt)*sin(phi))
we can now separate this into two equations, one in sin(wt) and one
in cos(wt) and eliminate sin(wt) and cos(wt) as they are orthogonal.
-aBw^2*cos(theta)-bBw*sin(theta)+cB*cos(theta)=A*cos(phi) (1)
and
-aBw^2*sin(theta)+bBw*cos(theta)+cB*sin(theta)=A*sin(phi) (2)
divide equation (2) by equation (1) so that cos(phi) is the divisor of
sin(phi to give us tan(phi) and eliminate B and A
tan(phi)=(-aw^2*sin(theta)+bw*cos(theta)+c*sin(theta))/(-aw^2*cos(theta)-bw*sin(theta)+c*cos(theta))
(3)
now divide the numerator and denominaotr by cos(theta) to either
eliminate cos(theta) or create tan(theta) and this will not affect the
tan(phi)
tan(phi)=(-aw^2*tan(theta)+bw+c*tan(theta))/(-aw^2-bw*tan(theta)+c) (4)
multiply the denominator by both sides and separate out tan(theta) and
remove it to one side of the equation to give
tan(theta)=(tan(phi)*(-aw^2+c)-bw)/(-aw^2+c+bw*tan(phi))
(5)
now we can solve for theta from a, b,c,w and phi , the givens.
now substitute all these into equation (2) and solve for B to give:
B=(A*sin(phi))/(sin(theta)*(-aw^2+c)+bw*cos(theta))
(6)
and voila, we now have B and theta for any w,a,b,c and phi and A.
similarly from equation (1) we can also say:
B=(A*cos(phi))/(cos(theta)*(-aw^2+c)-bw*sin(theta))
(7)
This does not apply to polynomials, and it is not Fourier or Laplace.
Neither Fourier or Laplace can
give you theta, but they will give you B. In fact because both x(t)
and y(t) are completely real, there is no complex phase.
if it is third order,
tan(theta)=(tan(phi)*(-bw^2+d)-(aw^3+cw))/(-tan(phi)*(aw^3+cw)+(-bw^2+d))
if it is fourth order
tan(theta)=(tan(phi)*(aw^4-cw^2+e)+(bw^3-dw))/(-tan(phi)*(aw^4-cw^2+e)+(-bw^3-dw))
if a(n)y(n)(t)+a(n-1)y(n-1)(t) +...+a0y(t)=x(t)
and x(t) = Asin(wt+phi)
then y(t) = Bsin(wt+theta)
and there is a general formula for tan(theta) and B depending on n,
the nth order linear differential equation.
It is too difficult to express here as the notation is not
available...but i will try to express it here
a(n)*y(n)(t)+a(n-1)*y(n-1)(t) +..... + a1y'(t)+a0y(t) = x(t) where
x(t) = Asin(wt+phi) and y(t)=B*sin(wt+theta)
then
tan(theta) = ( tan(phi)*(sum k from 0 to n by 2: ak*(jw)^k) - (sum k
from 1 to n by 2: ak*w^k*j^(k+1)) )/
(sum k from 0 to n by 2: ak*(jw)^k)
+tan(phi)*(sum k from 1 to n by 2: ak*w^k*j^(k+1)) )
B=A*cos(phi)/[ cos(theta)*(sum k from 0 to n by 2:
ak*(jw)^k)+sin(theta)*(sum k from 1 to n by 2: ak*w^k*j^(k+1))]
where j=sqrt(-1).
This has broad applications in digital and continuous FIR and IIR
optimal phase filter design
Fluid Companionship –
A reminder about the
limits of Mathematics:
Fluid Companionship – Poem – A Reminder about the limits of Mathematics
Each drop’s splash seen from the window up on high as the edge of the
wind wafts past in wave fronts of rain.
Like wind in the long grass, flowing, alternating shades of green.
Like quicksand collapsing to consume itself, or so it seems.
Like water coursing, etching patterns never to be seen.
Like a plume from an undersea geyser-vent, hidden from view.
Like drifting tendrils of snow on a field, poetry in motion.
Like a lava avalanche from a volcano, beautiful and deadly, scorching.
Like dust devil’s, coalescing, dancing, darting, dispersing.
Like a puff of smoke into a beam of light.
Like a violet, orange, red and pink Sunrise, full of motion and promise!
Like a Commando’s mind when he takes the life of a supposed enemy.
Like a sailor, out on the water, jibing and tacking.
Like a hangglider, floating high on thermals.
Like a Scuba Diver down at 105 feet, getting only a glimpse.
Like a Skier’s mind, racing down the moguls.
Like a Kayaker’s mind going down a series of whitewater rapids.
Like a Fireman entering a burning building, pulling a little girl to safety.
Like a warm shower on a cold winter’s day, full of billoughy steam.
Like a cat purring on your lap in the evening.
Like retro rockets on a space capsule!
Like fire, shifting, shining, reaching for the sky.
The High and the Low colliding, a lightning storm created.
From the window, warm and protected:
Like Dreams…Like no Mathematics can fully describe!
where j=sqrt(-1).
This has broad applications in digital and continuous FIR and IIR optimal phase filter design
«
Last Edit: 07/05/2018 00:01:06 by
Chondrally
»
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Re: General Solution to 2nd order differential equation
«
Reply #1 on:
03/05/2018 18:42:29 »
Quote from: Chondrally on 03/05/2018 16:24:08
it follows that:
if a=b, then s=+/- 1/2
Hello, I really have no idea what you are talking about , but I recognise a few things as the quoted.
If a = b then s = +/- 1/2
does 1/2 = 0.5/0.5 = 1?
Forgive me , I am just interested in what you are doing, I do not know this subject.
0.5 + 0.5 = 1
0.5 / 0.5 = 1
1/2 = 0.5
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