Solving non-linear trigonometric equations with two unknowns.

We have two solve two non-linear equations as follows:

B1=1/(Z1*tan(x))

B2=1/(Z1*tan(r*x))

Here, 'B1', 'B2', and 'r' are known quantities (e.g. B1 = 0.002, B2 = 0.004 and r = 1.8)

We need to find out the value of Z1 and x.

Thanks in advance.

Answer 1

[John D'Errico]

Too easy?

Eliminate Z1, by taking the ratio.

B1=1/(Z1*tan(x))
B2=1/(Z1*tan(r*x))

Therefore,

B2/B1 = tan(x)/tan(r*x)

The solve:

B1 = 0.002;
B2 = 0.004;
r = 1.8;

% First, plot things:
ezplot(@(x) tan(r*x)./tan(x))
grid on
H = refline(0,B2/B1);
H.Color = 'r';

So it looks like a root exists a little under 4. (These are radians of course!)

syms x
xs = vpasolve(B2/B1 == tan(x)/tan(r*x),x,4)
xs =
3.6373857978313951727197712680783

It is symmetric around zero of course, and there will be infinitely many roots, but the primary root seems to be roughly around 4. This would probably change for different values of r.

Solve for Z1 for the root.

Z1s = 1./(B1*tan(xs))
Z1s =
924.46628620202999172157648998132

I could have used fzero for this too.

As a test,

B2
B2 =
      0.004

1/(Z1s*tan(r*xs))
ans =
0.004

If you wanted to do this without the graphic, that too can be accomplished, at the cost of slightly more effort. Not too much though.

[Antra Saxena]

Thanks for the answer.

[Antra Saxena]

Hello,

Graphical approach is very interesting, but our scenario is as follows.

We have to fabricate our design, so fabricable limits are:

Z1 should lie between 20 to 160.

x should lie between 10 to 150.

Now, there is little difference in the above equations as follows:

B1=tan(x)/(Z1)

B2=tan(r*x)/(Z1)

Also, value of r = 4.2.

So, can you please tell if its possible to get the values between defined limits.

Thanks in advance.

[John D'Errico]

I would not call this a graphical approach. Merely a suggestion of how to learn if a solution exists at all.

I would also note that your equations have changed in a significant way. But the solution approach does not change that much. I'd just automate it to look for all solutions in that interval. Now we have the ratio of B2/B1 as...

% B2/B1 = tan(r*x)/tan(x)

So lets solve this.

B1 = 0.002;
B2 = 0.004;
r = 4.2;

ezplot(@(x) tan(r*x)./tan(x),[10 150])
H = refline(0,B2/B1);
set(H,'Color','r')

The larger value of r means there are literally hundreds of solutions between 10 and 150 in x.

To find them, you need to use an automatic scheme, instead of a graphical one where you can see the solution.

xmin = 10;
xmax = 150;
x0 = linspace(xmin,xmax,10001);
fun = @(x) tan(r*x)./tan(x) - B2/B1;

fsign = sign(fun(x0));
crossind = find(fsign(1:end-1).*fsign(2:end) <= 0);
xbrac = [x0(crossind);x0(crossind + 1)];

There are 338 candidate brackets for a solution.

size(xbrac)
ans =
     2   338

Some of them are not truly brackets however. We can try to exclude those that are not by an extra test. If our function at the midpoint of the interval does not lie between the function values at the endpoints, then we will choose to ignore that bracket. You can get the idea of what I am saying from the first 10 brackets:

xbrac(:,1:10)
ans =
       10.084       10.812        10.84       11.582        11.68       12.338       12.562       12.828        13.08       13.794
       10.098       10.826       10.854       11.596       11.694       12.352       12.576       12.842       13.094       13.808

fun(xbrac(:,1:10))
ans =
        19.96      -0.7077        4.249      -15.201       0.1593      -259.94       170.74     -0.10049        40.31    -0.087157
       -12645     0.036114      -6.2269       78.427     -0.12048       106.98      -71.211      0.06843      -103.36      0.69965

fun(mean(xbrac(:,1:10)))
ans =
       41.402     -0.43179      -36.756      -34.291    0.0096058       347.26      -271.95    -0.015843       137.83      0.23244

So brackets {1,3,4,6,7,9} will not actually contain a root. Instead, those brackets are locations where the function transitions from +inf to -inf, or -inf to +inf. How can we find them?

f1 = fun(xbrac(1,:));
f2 = fun(xbrac(2,:));
fm = fun(mean(xbrac,1));

keepbrac = ((f1 <= fm) & (f2 >= fm)) | ((f2 <= fm) & (f1 >= fm));

xbrac = xbrac(:,keepbrac);

size(xbrac)
ans =
     2   124

Much better. There are now 124 possible brackets we have identified. Just throw them at fzero.

nb = size(xbrac,2);
xroot = zeros(1,nb);
fval = zeros(1,nb);
for i = 1:nb
  [xroot(i),fval(i)] = fzero(fun,xbrac(:,i)');
end

xroot
xroot =
  Columns 1 through 14
       10.826       11.687       12.836       13.796       14.652       16.764        17.62        18.58       19.728        20.59       21.496       22.685       24.439       25.628
  Columns 15 through 28
       26.534       27.395       28.544       29.504       30.359       32.472       33.328       34.288       35.436       36.298       37.204       38.393       40.147       41.336
  Columns 29 through 42
       42.242       43.103       44.252       45.212       46.067        48.18       49.036       49.996       51.144       52.006       52.912       54.101       55.855       57.044
  Columns 43 through 56
       57.949       58.811        59.96        60.92       61.775       63.888       64.744       65.704       66.852       67.714        68.62       69.809       71.563       72.752
  Columns 57 through 70
       73.657       74.519       75.668       76.628       77.483       79.596       80.452       81.411        82.56       83.422       84.328       85.517       87.271        88.46
  Columns 71 through 84
       89.365       90.227       91.376       92.336       93.191       95.304        96.16       97.119       98.268        99.13       100.04       101.22       102.98       104.17
  Columns 85 through 98
       105.07       105.94       107.08       108.04        108.9       111.01       111.87       112.83       113.98       114.84       115.74       116.93       118.69       119.88
  Columns 99 through 112
       120.78       121.64       122.79       123.75       124.61       126.72       127.58       128.54       129.68       130.55       131.45       132.64       134.39       135.58
  Columns 113 through 124
       136.49       137.35        138.5       139.46       140.32       142.43       143.28       144.24       145.39       146.25       147.16       148.35
>> fval
fval =
  Columns 1 through 14
    7.816e-14   4.8406e-14   -1.199e-14   1.9096e-14   1.2879e-14  -2.0872e-14  -8.2823e-14  -3.6859e-14  -1.5543e-15   1.6298e-13  -2.4203e-14  -2.9088e-14            0  -3.3307e-15
  Columns 15 through 28
  -3.6859e-14  -2.5313e-14  -3.6415e-14  -3.6859e-14   3.9968e-14   1.7364e-13  -7.5273e-14   4.3965e-14   7.2387e-14   1.7542e-13   6.6169e-14   2.7534e-14   1.0747e-13   3.1086e-15
  Columns 29 through 42
  -2.9754e-13  -1.1413e-13   3.9524e-14  -2.1538e-14   1.3856e-13   2.0073e-13  -1.1324e-13   6.2172e-14  -6.5947e-14   1.0214e-13  -3.7748e-14  -7.9714e-14  -1.3767e-14  -8.7041e-14
  Columns 43 through 56
   2.7223e-13   2.4425e-14    2.176e-14  -3.1397e-13   1.4433e-13  -2.0672e-13  -1.5121e-13  -7.7049e-14  -4.3077e-14   1.1458e-13   1.3234e-13   1.4122e-13  -1.0614e-13   1.6875e-14
  Columns 57 through 70
    3.455e-13   1.6209e-13  -1.9984e-14  -6.0707e-13  -2.8355e-13   2.2116e-13  -1.4411e-13  -1.6942e-13  -4.9738e-14   1.2657e-13   1.4255e-13  -6.5947e-14  -2.9843e-13   7.1054e-15
  Columns 71 through 84
   3.3307e-13   1.6875e-13  -1.9629e-13   4.6629e-14   9.0594e-14  -1.5277e-13   4.3388e-13   6.6613e-15  -5.6399e-14  -3.1308e-14  -2.3537e-13  -2.7311e-13   2.5047e-13  -3.3307e-15
  Columns 85 through 98
   3.2063e-13   1.1724e-13   1.0125e-13   6.1018e-13   9.6811e-14  -1.5943e-13   4.4142e-13  -1.3278e-13   -4.885e-15  -1.9318e-14   9.4591e-14  -2.3315e-14  -3.4106e-13   5.4623e-14
  Columns 99 through 112
   4.7917e-13    1.239e-13    2.407e-13  -5.3846e-13   1.0347e-13   2.6912e-13  -7.8293e-13   2.0117e-13   1.9451e-13   1.1569e-12  -2.8311e-13   3.6904e-13  -1.3389e-13  -1.8319e-13
  Columns 113 through 124
  -6.9722e-13  -4.5475e-13   1.7319e-14   6.8612e-13  -6.9322e-13  -5.3979e-13   4.5564e-13  -9.6589e-14   3.0376e-13   1.1693e-12   4.6629e-14  -3.8192e-14

Now z1 is simple. z1 is just tan(xroot)/B1.

Z1 = tan(xroot)/B1
Z1 =
Columns 1 through 14
     2913.1      -603.46       138.35       1408.9       -884.8        884.8      -1408.9      -138.35       603.46      -2913.1       -270.2       416.01      -416.01        270.2
Columns 15 through 28
     2913.1      -603.46       138.35       1408.9       -884.8        884.8      -1408.9      -138.35       603.46      -2913.1       -270.2       416.01      -416.01        270.2
Columns 29 through 42
     2913.1      -603.46       138.35       1408.9       -884.8        884.8      -1408.9      -138.35       603.46      -2913.1       -270.2       416.01      -416.01        270.2
Columns 43 through 56
     2913.1      -603.46       138.35       1408.9       -884.8        884.8      -1408.9      -138.35       603.46      -2913.1       -270.2       416.01      -416.01        270.2
Columns 57 through 70
     2913.1      -603.46       138.35       1408.9       -884.8        884.8      -1408.9      -138.35       603.46      -2913.1       -270.2       416.01      -416.01        270.2
Columns 71 through 84
     2913.1      -603.46       138.35       1408.9       -884.8        884.8      -1408.9      -138.35       603.46      -2913.1       -270.2       416.01      -416.01        270.2
Columns 85 through 98
     2913.1      -603.46       138.35       1408.9       -884.8        884.8      -1408.9      -138.35       603.46      -2913.1       -270.2       416.01      -416.01        270.2
Columns 99 through 112
     2913.1      -603.46       138.35       1408.9       -884.8        884.8      -1408.9      -138.35       603.46      -2913.1       -270.2       416.01      -416.01        270.2
Columns 113 through 124
     2913.1      -603.46       138.35       1408.9       -884.8        884.8      -1408.9      -138.35       603.46      -2913.1       -270.2       416.01

What are the roots of interest? You said that Z1 must lie between 20 and 160.

validind = find((Z1 >= 20) & (Z1 <= 160))
validind =
     3    17    31    45    59    73    87   101   115

So there are infinitely many roots, as I suggested. All have the same value for Z1. They are effectively the same periodic root.

[xroot(validind);Z1(validind)]
ans =
     12.836       28.544       44.252        59.96       75.668       91.376       107.08       122.79        138.5
     138.35       138.35       138.35       138.35       138.35       138.35       138.35       138.35       138.35

diff(xroot(validind))
ans =
     15.708       15.708       15.708       15.708       15.708       15.708       15.708       15.708

diff(xroot(validind))/pi
ans =
          5            5            5            5            5            5            5            5

So the period is 5*pi.

Essentially, that means there is only one solution of interest. I'd call it the principal solution. It occurs at x=12.836, Z1 = 138.35.

format long g
xroot(3)
ans =
          12.8363147171853

Z1(3)
ans =
          138.348979092173

fun(xroot(3))
ans =
     -1.19904086659517e-14

The general solution of interest to you is therefore

12.8363147171853 + n*5*pi

for integer n. As I said, you will only care about the first, i.e., principal solution.

[Antra Saxena]

Thanks a lot. Its working for my design.

Answer 2

[Birdman]

syms Z1 x
B1 = 0.002;
B2 = 0.004; 
r = 1.8;
eqns=[B1==1/(Z1*tan(x));
B2==1/(Z1*tan(r*x))];
sol=solve(eqns,[Z1,x])
Z1=double(sol.Z1)
x=double(sol.x)

[Antra Saxena]

Hi Birdman,

Thanks for the answer, unfortunately it shows this error.

Snapshot is shared.

Please Comment.

[Birdman]

It worked well for me. What is your MATLAB version?

[Antra Saxena]

Thanks for sharing the solution.

I am using MATLAB 2014.

[Birdman]

Try to use vpasolve instead of solve.

[Antra Saxena]

Its working on 2015 version.

Thanks alot.

[Antra Saxena]

We have to fabricate our design, so fabricable limits are:

Z1 should lie between 20 to 160.

x should lie between 10 to 150.

So, I guess this needs another solution.

Given answer not working for our design.