Help please to solve equation

  • Thread starter Bazza2
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  • #1
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Hi Guys,

Its been a while since I studied math and I would appreciate if someone could be of assistance in solving an equation for me.

I need to solve for K in the following:

W/B=1/Pi*[ln〖(1+R)/(1-R )〗-D/B*ln〖(K+R)/(K-R) 〗 ]

where

R=√(K*(K*B-D)/(B-K*D ))

I have attached a jpeg of the equations in scientific format.

Thanks for your help.

Cheers,

Bazza2
 

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  • #2
611
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hi your eq. is confusing...could use latex ..?
 
  • #3
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Hi Rajini,

Unfortunately I am not familiar with latex. However, I have redone the equations in Word's Equation Editor so it should be clear now. Please find it attached.

W/B=1/Pi*{ln[(1+R)/(1-R )]-D/B*ln[(K+R)/(K-R )] }

R=√((K*(K*B-D ))/(B-K*D ))

Thanks,

Bazza2
 

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  • #4
berkeman
Mentor
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Hi Rajini,

Unfortunately I am not familiar with latex. However, I have redone the equations in Word's Equation Editor so it should be clear now. Please find it attached.

W/B=1/Pi*{ln[(1+R)/(1-R )]-D/B*ln[(K+R)/(K-R )] }

R=√((K*(K*B-D ))/(B-K*D ))

Thanks,

Bazza2
Wow, what is that equation from? Do you know that there is a closed form solution? Do you have Mathematica? You might also try Wolfram|Alpha to see if it's able to solve it for you.
 
  • #5
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Hi Berkeman,

It is transmission line analysis. And I'm afraid it's been too long since I was last at Uni to remember how to solve such equations. I tried Wolfram Alpha - quite interesting but just kept changing things around.

I would appreciate if anyone could help me out here.

Thanks,

Bazza2
 
  • #6
611
2
hi bazza,
i have access to matlab (but i am new to it!)...just give some time..i will check and let you know..
 
  • #7
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hi bazza,
i have access to matlab (but i am new to it!)...just give some time..i will check and let you know..
Thanks Rajini. Much appreciated.

Cheers,

Bazza
 
  • #8
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Hi Rajini,

Have you been able to solve this equation using Matlab?

I appreciate you help.

Cheers,

Bazza
 
  • #9
79
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No explicit solution could be found for the equation:

[tex]\frac {w} {b} = \frac{1} {\pi} \ln\!\left(-\frac{\sqrt{-\frac{k\, \left(d - b\, k\right)}{b - d\, k}} + 1}{\sqrt{-\frac{k\, \left(d - b\, k\right)}{b - d\, k}} - 1}\right)\, \ln\!\left(\frac{k + \sqrt{-\frac{k\, \left(d - b\, k\right)}{b - d\, k}}}{k - \sqrt{-\frac{k\, \left(d - b\, k\right)}{b - d\, k}}}\right)}[/tex]

Here is the equation in string form I used in the "solve()" function:

eqn = 'w/b = (1 / pi) * ( log( (1 + (-(k*(d - b*k))/(b - d*k))^(1/2)) / (1 - (-(k*(d - b*k))/(b - d*k))^(1/2)) ) * log( (k + (-(k*(d - b*k))/(b - d*k))^(1/2))/(k -(-(k*(d - b*k))/(b - d*k))^(1/2))))'

Here is the solve() output:

>> solve(eqn,'k')

Warning: Explicit solution could not be found.
> In solve at 170

ans =

[ empty sym ]

MATLAB is a fickle mistress :(
 
  • #10
hotvette
Homework Helper
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You may need to resort to numerical methods to solve for K.
 
  • #11
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Thanks Glustro and hotvette for your time.

Excuse my ignorance but what do you mean by numerical methods?

Cheers,

Bazza
 
  • #12
79
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Root-finding algorithms.

Assume values for all variables except K, and then use something like Newton-Raphson, Wegstein, or substitution method to iterate on a value of K until it converges.

(This is not done by hand - you use a computer.)
 
  • #13
hotvette
Homework Helper
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Since you are dealing with a single unknown (i.e. K), you could even use something as simple as bisection if you knew a range that brackets the value you want. Converges slowly but very simple to implement.
 
  • #14
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Unfortunaely, the K is a variable in an even more complex equation so i have no idea of what sort of value it should be.

Cheers,

Bazza
 
  • #15
hotvette
Homework Helper
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Unfortunaely, the K is a variable in an even more complex equation so i have no idea of what sort of value it should be.
Sorry, don't understand. I thought K was the solution to the equation in your first post, meaning you know the values of all other quantities (i.e. W, B, D) and K is the only unknown. Not true?
 
  • #16
10
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Sorry, don't understand. I thought K was the solution to the equation in your first post, meaning you know the values of all other quantities (i.e. W, B, D) and K is the only unknown. Not true?
Hi hotvette,

Yes I am looking for the solution of K.

Then K goes in this equation:

Zdbs=293.9/Math.Sqrt(Er)*D/B*0.5* Math.Log[((1+K))/((1-K) )]

and it is actually Zdbs that I want at the end of the day.

Cheers,

Bazza
 
  • #17
611
2
Hi i am sorry i cant help..
i copy and past the eqn from clustro..i always get undefined variable as error!
 
  • #18
hotvette
Homework Helper
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Yes I am looking for the solution of K
Thanks for clarifying. If finding the value of K by numerical methods is acceptable, then you should be able to obtain a solution. Here is a suggestion. Tell me the values of W, B, and D and I'll play around with it and let you know what may be a reasonable approach.

Question: is this a one time only solution, or do you need to repeatedly find the value of K based on multiple values of W, B, D?
 
  • #19
10
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Hi i am sorry i cant help..
i copy and past the eqn from clustro..i always get undefined variable as error!
Thanks Rajini for trying.

Thanks for clarifying. If finding the value of K by numerical methods is acceptable, then you should be able to obtain a solution. Here is a suggestion. Tell me the values of W, B, and D and I'll play around with it and let you know what may be a reasonable approach.

Question: is this a one time only solution, or do you need to repeatedly find the value of K based on multiple values of W, B, D?
It will be calculated using different values. Typical values are:
W= 5, D=5 & B=10

Thanks hotvette.

Cheers,

Bazza
 
  • #20
79
0
I tried it again and it worked.

Rajni, are you sure you have the Symbolic Math Toolbox? solve() is contained in that toolbox.

Bazza: If you know all of those values you can easily extract a root for K.
 
  • #21
611
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Hi clustro,
i have to say you..that i really a amateur in matlab..never used..but i notice many people use and strongly recommend it..But at least i would like to sue it to solve this problem..
At present only one thing i can do for baza..if he could tell me hot to execute and send me the codes i can execute and let you know..
PS: i login to linux pc..then go to mat lab from command by typing matlab then a window comes...there f(x):...here i typed your codes..after entering no error but when i type solve(..)..like k not define invalid char...
 
  • #22
Jmf
49
0
Can I ask what aspect of the analysis of a transmission line you're doing? Maybe there's an easier way.
 
  • #23
10
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I thought this may be easy to solve but it seems not. I have another way to solve this which is less accurate but it will have to do.

Thanks guys for your time.

Cheers,

Bazza
 
  • #24
hotvette
Homework Helper
994
4
Typical values are:
W= 5, D=5 & B=10
For the above values of W, D, B, K = 0.978619036085362. Newton-Raphson worked well but because of the logs, there is a relatively narrow range of starting points that can be used (~0.95 to ~0.99).

I'm curious about something. Does K represent some sort of efficiency?

Just for fun I varied the values for W, D, and B. The result follows:

Code:
W	   D	   B	   K
1	   5	   10	   0.756276579421452
2	   5	   10	   0.865849758609811
3	   5	   10	   0.926653501309252
4	   5	   10	   0.960270604440607
5	   5	   10	   0.978619036085362
6	   5	   10	   0.988538243178375
7	   5	   10	   0.993869188156523
8	   5	   10	   0.996724645149870
9	   5	   10	   0.998251307920962
5	   4	   10	   0.953551108282295
5	   3	   10	   0.915051241605172
5	   2	   10	   0.860069503637580
5	   1	   10	   0.782715023759189
5	   5	   11	   0.959554866316289
5	   5	   12	   0.935843903716909
5	   5	   13	   0.909035721120116
5	   5	   14	   0.880465274020010
5	   5	   15	   0.851159374048799
 
Last edited:
  • #25
5
0
Well, after a couple of hours of work, I figured out how to get K explicit in the first equation. I came up with the following:

K = (-R[1-e^([W*pi]/[D*ln([1+R]/[1-R])])])/(1+3^[(W*pi)/(D*ln[(1+R)/(1-R)])])

It should be fairly easy to subsititute the second equation in for R, and then substitute that entire thing into the third equation.

Sorry for the sloppy text, LaTeX really doesn't like me. I'll attach a picture to make it easier to read.
 

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