jrive

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These are replies submitted by jrive

@Carl Love  --got it,..thank you!

@Carl Love thank you for your response.  What do you mean by nontrivial solution?  I wanted x1 and x2 expressed in terms of DC and T, instead of ton and toff.

@acer thank you for your response, that solution gives me what I am looking for.  But, out of curiosity, how else might I have asked the question to be more clear?  How should, " solve this equation in terms of x,y,z" be phrased?   The intent with that phrase (or phrasing) is that the solution be expressed in terms of those variables, using whatever relationships are available between those variables and the ones in the solution they can "replace"...I wouldn't know how else to say that, so I look forward to your response!

In my my case, I didn't want ton or toff in the solution, but instead wanted it expressed in tems of DC and T which have relationships to ton and toff, and each other, as shown in eqs 3 and 4.....

anyway, you understood what I meant, so much appreciated.

@acer Perfect! thank you.  Thanks for picking up on the typo and proceeding with the solution!

@acer appreciate the help!

Jorge

 

@Kitonum thanks for the assist!

@tomleslie   Thank you!  I appreciate the help!!

@Carl Love  thank you!! Makes sense.  

@Carl Love-- Thank you.  Did you try it on my worksheet?  I just did, and I get, 

restart

````

assum := R1::real, C1::real, omega::real, vab::real

R1::real, C1::real, omega::real, vab::real

(1)

Pab := `assuming`([expand(rationalize(vab/(R1-I/(omega*C1))))], [assum])

vab*omega^2*C1^2*R1/(C1^2*R1^2*omega^2+1)+I*vab*omega*C1/(C1^2*R1^2*omega^2+1)

(2)

Pabmag := `assuming`([abs(Pab)], [assum])

(vab^2*omega^4*C1^4*R1^2/(C1^2*R1^2*omega^2+1)^2+vab^2*omega^2*C1^2/(C1^2*R1^2*omega^2+1)^2)^(1/2)

(3)

Pabang := `assuming`([argument(Pab)], [assum])

argument(vab*omega^2*C1^2*R1/(C1^2*R1^2*omega^2+1)+I*vab*omega*C1/(C1^2*R1^2*omega^2+1))

(4)

``

``

could you show me how to do it?

Download complex_power_RC.mw

btw.....I think it would be extremely helpful to newbies like me if the invlaplace help page explained why the answers are given in trig hyperbolic functions instead of the more traditional (at least for EEs, perhaps not for mathematicians?) and cover how the results can be converted to exponentials via the convert instruction.  The intuitive action is to try simplify --- I had no idea "convert" existed.  Simple documentation like that goes a long way and saves a lot of aggravation!!! 

found it....evalf will evaluate  radicals numerically.

@tomleslie 

How do I simplify the expression to resolve the radicals in one "simplify" command?  right now, I can only do it by selecting "approximate" (from the side panel) on the result from the simplify expression.  I tried simplify(convert(temp2,exp) , radical)  but that didn't work.

@tomleslie thank you.  You misunderstood---I was wondering why the invlaplace resulted in sinh and cosh terms, and why i wouln't simplify to the "other" answer if in fact they are the same.  I had to plot them to find out that they were indeed the same.  I was not surprise of the different form of the answer --- just that I couldn't make them look the same!

@acer 

I'd prefer to have the mantissa greater than or equal to 1, so I'd prefer 6e-10 to be represented as 600e-12.

 

Thank you!

 

@acer   wow..thank you for your response. I've never worked with modules, so I'll have to take some time to understand what you did.

Thanks again!

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