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johnzfrebbies@ wrote: hey Thanks! The code you posted works great and gave me a better understanding by example. Thats what I need to fully understand what I'm doing.
I'll post my full code later today with the modifications, just so and if anyone else needs an example they can be directed to. Thanks again for your help.
Okay, great. Glad it worked out.
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hello a question about desktops if you don't mind...do you have any recommendations other than what I use (yod'm 3D) I'd like to have something like beryl, but I'm not using the installed version of Ubuntu 8.04 (it runs atop xp as beta on 1 of the desktops) I like the switching desktops, I just need more functions. Do you know any other functions or maybe just another program?
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johnzfrebbies@ wrote: hello a question about desktops if you don't mind...do you have any recommendations other than what I use (yod'm 3D) I'd like to have something like beryl, but I'm not using the installed version of Ubuntu 8.04 (it runs atop xp as beta on 1 of the desktops) I like the switching desktops, I just need more functions. Do you know any other functions or maybe just another program?
Oh geez, no idea. That's a little out of my realm of experience. I used Linux back in the late 90's early 2000's. I think it was Enlightenment that I used to run and occasionally Gnome. Now I mostly use XP on a Dell laptop. I also have a Mac that I do some cross-testing on. I just go with the standard desktops out of the box. Hmmm, you could try asking in the Lounge and see if someone recommends something.
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Problem:
<br />
Let w = 2000<br />
V = 6V<br />
L = 0.2H<br />
R = 400<br />
<br />
Given the differential equation L*(di/dt) + R*i = V*cos(w*t)<br />
<br />
Find i(t). - The above equation is true for t = 0<br />
<br />
This is what I have come up with. - NOTE: When im integrating ive used the '!' symbol
I solved the above using the steady state and transient method.
For the transient part:
<br />
-> 0.2*(di/dt) + 400*i = 6*cos(w*t) <br />
-> (di/dt) + 2000*i = 30*cos(w*t) <br />
<br />
-> di/dt + 2000*i = 0 <br />
<br />
-> di = -2000*i*dt<br />
<br />
-> !(1/i)*di = !-2000*dt ('!' means integrate)<br />
<br />
-> ln(i) = -2000t + c<br />
<br />
-> i(transient) = e^(-2000t)*e^c<br />
<br />
Let e^c = A (a constant)<br />
<br />
-> i(transient) = A*e^-2000t<br />
For the steady state part:
Ohms law states that the current i = V/Z, where V is voltage, and Z is impedance.
So, V = 6>0 V The '>' is the polar form (or phasor some might know it as) representation of the complex angle associated with the voltage, so 6V, 0 angle.
<br />
Z = R + j*w*L<br />
Z = 400 + j*2000*0.2<br />
Z = 400 + j400<br />
Z = 565.685>0.785 (all angles are in radians)<br />
<br />
-> i = V/Z<br />
<br />
-> i = 6>0 / 565.685>0.785<br />
<br />
-> i = 10.61>-0.785 mA<br />
<br />
-> i(steady state) = 10.61*sin(w*t - 0.785) mA<br />
Now to find the final equation for i(t)
<br />
To derrive i(t):<br />
<br />
-> i(t) = i(steady state) + i(transient)<br />
<br />
-> i(t) = A*e^-2000t + 10.61*sin(w*t - 0.785) mA <br />
<br />
To find A make t = 0<br />
<br />
-> i(0) = A(1) + (10.61*10^-3)*sin(-0.785)<br />
<br />
-> 0 = A - 7.5*10^-3<br />
-> A = 7.5*10^-3<br />
<br />
So the final equation for i(t) is:<br />
<br />
-7.5*e^-2000t + 10.61*sin(w*t - 0.785) mA<br />
<br />
However this is not working out - I should be able to substitute this equation back into the original:
<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)) + 400((7.5*10^-3)*e^-2000t + (10.61*10^-3)*sin(2000*t - 0.785)) = 6*cos(2000*t)<br />
<br />
Let t = 1<br />
<br />
-> 0.2(0 + 8.434) + 400(0 + (9.8677*10^-3)) = 6cos(2000)<br />
<br />
-> 1.6869 + 3.947 = -2.205<br />
<br />
This does not satisfy the equation!!!<br />
<br />
Can anyone see where I went wrong in my working?
[Edit] - Fixed a typo.
Cheers,
modified on Tuesday, April 22, 2008 10:28 AM
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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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