# Solving an Equation for K | Math Help from Experts - Bazza2

• Bazza2
In summary, the conversation discusses the equation W/B=1/Pi*[ln〖(1+R)/(1-R )〗-D/B*ln〖(K+R)/(K-R) 〗 ] and the need to solve for K. The equation is a transmission line analysis and a closed form solution cannot be found. Numerical methods such as root-finding algorithms or bisection may be used to solve for K, with the values of W, B, and D provided by the user. The solution of K will then be used in another equation, Zdbs=293.9/Math.Sqrt(Er)*D/B*0.5* Math.Log[((1+K))/((1-K) )
Bazza2
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.

Cheers,

Bazza2

#### Attachments

• eqn.JPG
7.2 KB · Views: 432
hi your eq. is confusing...could use latex ..?

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

#### Attachments

• eqn.JPG
7.6 KB · Views: 418
Bazza2 said:
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.

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

hi bazza,
i have access to MATLAB (but i am new to it!)...just give some time..i will check and let you know..

Rajini said:
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

Hi Rajini,

Have you been able to solve this equation using Matlab?

I appreciate you help.

Cheers,

Bazza

No explicit solution could be found for the equation:

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

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 :(

You may need to resort to numerical methods to solve for K.

Thanks Glustro and hotvette for your time.

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

Cheers,

Bazza

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.)

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.

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

Bazza2 said:
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?

hotvette said:
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

Hi i am sorry i can't help..
i copy and past the eqn from clustro..i always get undefined variable as error!

Bazza2 said:
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?

Rajini said:
Hi i am sorry i can't help..
i copy and past the eqn from clustro..i always get undefined variable as error!

Thanks Rajini for trying.

hotvette said:
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

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.

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...

Can I ask what aspect of the analysis of a transmission line you're doing? Maybe there's an easier way.

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.

Cheers,

Bazza

Bazza2 said:
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:
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.

#### Attachments

• Formula.bmp
291.3 KB · Views: 451
Mego05 said:
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.

What value of K do you get when W= 5, D=5 & B=10? Does it match what I got via numerical methods?

I'm not quite sure how you would figure that, seeing as how, in order to find K, you need a value for R. The value of R is dependent on K. Wouldn't that make this set of equations recursive? Or, should I have replaced every R in the first equation with the expression in the second equation?

EDIT: Tried doing it that way, now I have an even bigger recursive mess.

$$\frac{W}{B}=\frac{1}{\pi}*[ln(\frac{1+\sqrt{K^{2}B^{2}-K^{3}BD-KBD+K^{3}D}}{1-\sqrt{K^{2}B^{2}-K^{3}BD-KBD+K^{3}D}})*\frac{D}{B}*ln(\frac{K+\sqrt{K^{2}B^{2}-K^{3}BD-KBD+K^{3}D}}{K-\sqrt{K^{2}B^{2}-K^{3}BD-KBD+K^{3}D}})]$$

If anyone can figure out how to get K explicit in that one, I'll get you a Nobel Prize.

Last edited:

## 1. How do I solve an equation for K?

To solve an equation for K, you need to isolate K on one side of the equation. This typically involves using inverse operations to cancel out any numbers or variables that are on the same side as K. Once K is isolated, you can solve for it by performing any necessary operations on both sides of the equation until you are left with K equal to a specific number.

## 2. What are some tips for solving equations for K?

One helpful tip for solving equations for K is to always perform the same operation on both sides of the equation. This ensures that the equation remains balanced and that you are not changing its value. It is also important to carefully distribute any numbers or variables before combining like terms. Lastly, always make sure to check your solution by plugging it back into the original equation to ensure it is correct.

## 3. Can you provide an example of solving an equation for K?

Sure! Let's take the equation 2K + 8 = 20. To solve for K, we first need to isolate it on one side of the equation. We can do this by subtracting 8 from both sides, leaving us with 2K = 12. Next, we can divide both sides by 2 to get K by itself, giving us the solution K = 6. We can check our answer by plugging it back into the original equation: 2(6) + 8 = 12 + 8 = 20. Therefore, our solution is correct.

## 4. What are some common mistakes to avoid when solving equations for K?

One common mistake when solving equations for K is forgetting to perform the same operation on both sides of the equation. This can throw off the balance of the equation and lead to an incorrect solution. Another mistake is not properly distributing any numbers or variables, which can also result in an incorrect solution. Lastly, it is important to always check your solution to ensure it is correct.

## 5. Are there any shortcuts for solving equations for K?

While there are no shortcuts for solving equations for K, there are some strategies that can make the process easier. One strategy is to combine like terms before isolating K, as this can simplify the equation. Another strategy is to use the order of operations to determine which operations to perform first. Additionally, you can use the distributive property to simplify expressions before combining like terms. However, it is always important to follow the standard steps for solving equations to ensure an accurate solution.

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