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Looks to me like you're missing an exponential term. You only have one. The generalized solution to an equation of your type is:
/ R t 2 R t 2 2 \
|exp(- ---) C1 R + exp(- ---) C1 w L + V sin(w t) w L + V cos(w t) R|
\ L L /
/ 2 2 2
/ (R + w L )
/
You should check your transient solution and make sure you've solved for the integration properly. Did you omit a term?
And when the sunlight hits the olive oil, don't hesitate.
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Oh geez mate i've never seen that equation before!
I got my answer by following a worked solution:
<br />
<br />
(di/dt) + 1000i = 10cos(wt)<br />
<br />
What is the transient part of the solution?<br />
<br />
The transient part satisfies the DE di/dt + 1000i = 0.<br />
<br />
Let i(transient) = A*e^(theta *t)<br />
<br />
di = -1000i*dt<br />
!(1/i)*di = !-1000*dt where ! means integral<br />
ln(i) = -1000t + c<br />
i = e^(-1000*t)*e^c<br />
<br />
Let e^c = A<br />
<br />
-> i = A*e^(-1000*t).<br />
<br />
This example was given to me in class, so I assume that it is the method they want me to follow.
Cheers,
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Your final line of your transient solution:
-> i(transient) = e^-2000t + e^c
-> i(transient) = A*e^-2000t
ln i = -Rt/L + C
i = exp(-Rt/L + C)
which does not equal: exp(-2000t) + exp(C)
it equals exp(-2000t + C)
Try fixing that...
Sorry, I see you've got A*exp(-2000t) there...okay.
And when the sunlight hits the olive oil, don't hesitate.
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Yea sorry that was a typo on my part. I was meant to write that as:
<br />
i(transient) = e^(-2000t+c)<br />
i(transient) = e^(-2000t)*e^c<br />
<br />
Let e^c = a constant A<br />
<br />
i(transient) = A*e^(-2000t)<br />
That typo does not effect any working following it, as the next line corrected the error.
Cheers,
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MarkBrock wrote: -> (di/dt) + 2000*i = 30*cos(w*t)
-> di/dt + 2000*i = 0
Uhm.............
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
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From my notes, given the DE -> (di/dt) + 1000i = 10cos(wt)
What is the transient part of the solution?<br />
<br />
The transient part satisfies the DE di/dt + 1000i = 0
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Ah, then, it's OK. I didn't remember the definition.
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
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MarkBrock wrote: -> i = 6>0 / 565.685>0.785
I think the V phase is wrong. It should be PI/2 instead of 0.
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
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Yes I think so too...
And when the sunlight hits the olive oil, don't hesitate.
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It's comforting: I don't remember a lot of that stuff...
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
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Yup that was it!
t=1 and t=0.1 both worked so im satisfied with that.
<br />
<br />
-7.5*e^-2000t + 10.61*sin(w*t + 0.785) mA<br />
<br />
<br />
-> L*(di/dt) + R*i = V*cos(w*t)<br />
<br />
L = 0.2<br />
R = 400<br />
i(t) = -7.5*e^-2000t + 10.61*sin(2000*t + 0.785) mA<br />
i'(t) = 15*e^-2000t + 21.22*cos(2000*t + 0.785)<br />
V = 6<br />
<br />
-> 0.2*(15*e^-2000t + 21.22*cos(2000*t + 0.785)) <br />
+ 400((-7.5*10^-3)*e^-2000t + (10.61*10^-3)*sin(2000*t + 0.785))<br />
= 6*cos(2000*t)<br />
<br />
Let t = 1<br />
<br />
-> 0.2(0 -19.46) + 400(0 + (9.8677*10^-3)) = 6cos(2000)<br />
<br />
-> -3.893 + 1.6898 = -2.2<br />
-> -2.2 = -2.2<br />
<br />
Horray! <br />
<br />
Thankyou so much for your help!
Cheers,
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You're welcome.
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
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Did you work it out?
And when the sunlight hits the olive oil, don't hesitate.
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Yup that was it!!!
t=1 and t=0.1 both worked so im satisfied with that.
<br />
<br />
-7.5*e^-2000t + 10.61*sin(w*t + 0.785) mA<br />
<br />
<br />
-> L*(di/dt) + R*i = V*cos(w*t)<br />
<br />
L = 0.2<br />
R = 400<br />
i(t) = -7.5*e^-2000t + 10.61*sin(2000*t + 0.785) mA<br />
i'(t) = 15*e^-2000t + 21.22*cos(2000*t + 0.785)<br />
V = 6<br />
<br />
-> 0.2*(15*e^-2000t + 21.22*cos(2000*t + 0.785)) <br />
+ 400((-7.5*10^-3)*e^-2000t + (10.61*10^-3)*sin(2000*t + 0.785))<br />
= 6*cos(2000*t)<br />
<br />
Let t = 1<br />
<br />
-> 0.2(0 -19.46) + 400(0 + (9.8677*10^-3)) = 6cos(2000)<br />
<br />
-> -3.893 + 1.6898 = -2.2<br />
-> -2.2 = -2.2<br />
<br />
Horray! <br />
<br />
Thankyou so much for your help!
Cheers,
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Good!
MarkBrock wrote: Thankyou so much for your help!
No problem.
And when the sunlight hits the olive oil, don't hesitate.
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MarkBrock wrote: ...and Z is impedance.
Back when I took electronics courses, we used R for resistance.
"Love people and use things, not love things and use people." - Unknown
"To have a respect for ourselves guides our morals; to have deference for others governs our manners." - Laurence Sterne
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DavidCrow wrote: Back when I took electronics courses, we used R for resistance.
We still do; impedance is complex: Z = R + jX. 
"A Journey of a Thousand Rest Stops Begins with a Single Movement"
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Thanks for the refresher (AC vs. DC). It's been a long time.
"Love people and use things, not love things and use people." - Unknown
"To have a respect for ourselves guides our morals; to have deference for others governs our manners." - Laurence Sterne
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Hi all,
I appologize if this is an inappropriate place to ask,
but I would appreciate if someone could help me with
the following issue.
I need to define an algebraic group (with prime
order), and define a method to transform strings of
bits, into elements of this group.
Could anyone, target me to some literature, or
programming libraries, that illustrate how to achieve
this?
Thank you in advance.
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I've created a polygon component for a small vector graphics section in my program. I have a function where the user can convert a line section to a cubic bezier.
My question is: I would like the control points to initially bend the curve to the outside of the polygon. Mathematically speaking, how do I determine what coordinates are inside the shape or outside the shape?
If you could rent a programmer a hour and a half for a dollar and a half, how much would he charge for 10 hours?
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If you want to determine points inside and outside a polygon, this[^] should do the trick! Make sure you handle points that lie directly on a vertex.
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This is a homework question. If your against helping students, I understand.
<br />
<br />
Given the differential equation :<br />
<br />
dy/dx + 2y = -5, y(0) = 3<br />
<br />
Solve using Newton's first order numerical method, taking sufficient steps to estimate y when x = 2.<br />
<br />
I can't find any examples on how to solve the equation using this method.
It would be appreciated if someone showed me how to do this, or pointed me to a useful example.
Cheers,
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The numerical recipes book(s) has been a God-send for many of us. It was originally for Fortran, then translated to (a rather clunky, but functional) C. The really great thing is that they do a good job of explaining the algorithms. I have, for the most part, just used their explanations and written my own code.
Here's a link to their older versions that you can read on-line.
http://www.nrbook.com/a/[^]
You can also find their most recent versions at www.nr.com
The book will explain the Newton method, and have examples of it's use.
David
---------
Empirical studies indicate that 20% of the people drink 80% of the beer. With C++ developers, the rule is that 80% of the developers understand at most 20% of the language. It is not the same 20% for different people, so don't count on them to understand each other's code.
http://yosefk.com/c++fqa/picture.html#fqa-6.6
---------
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Thanks! 
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MarkBrock wrote: I can't find any examples on how to solve the equation using this method.
You obviously haven't looked. 
Doing my part to piss off the religious right.
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