How to use fzero and arrayfun to find roots?

Hi!I got a problem when I ran my code.

This is my code:

f = [1*1e9 2*1e9 3*1e9];
w=2*pi*f;

z0 = 376.73415;
c = 2.997925e8;
L1 = 10*1e-3;
L2 = 20*1e-3;
dL = 0.01*1e-3;
miu0 = z0/c;
miur = 1;

y0 = 1i.*w/c;
ys = @(w,episr) 1i*w./c.*sqrt( miur*episr );

the = @(episr) episr./sqrt(miur);

Zin3_1 = @(w,episr) tanh(ys(w,episr)*L2);

Zin3_2 = @(w,episr) the(episr).*tanh(y0*dL);

Zin3_3 = @(w,episr) the(episr).*( 1+the(episr).*tanh(y0*dL).*tanh(ys(w,episr)*L2) );

Zin3 =@(w,episr) ZL1.*( Zin3_1(w,episr) + Zin3_2(episr) + Zin3_3(w,episr).*tanh(y0*L1) )...
    ./( Zin3_3(w,episr) +( Zin3_1(w,episr) + Zin3_2(w,episr) ).*tanh(y0*L1) );

S11 = @(w,episr) ( Zin3(w,episr)-Z0 )./( Zin3(w,episr)+Z0 );

reS11=@(w,episr) real( S11(w,episr) );

arrayfun( @(i) fzero( @(episr) reS11( w(i),episr ),3 ),1:numel(w) )

After I ran my code, I got a message:Operands to the and && operators must be convertible to logical scalar values.

So I ran my code step by step, I found a problem in Zin3_3.

But I don't know how to fix it.

Does anyone help me? Thanks!

Answer 1

[Walter Roberson]

In Zin3 you call Zin3_2(episr) but that routine needs two input arguments.

[San-Chi Yan]

Thanks! But I still can't run it. I used the following code to test it:

 reS11=@(w,episr) real(Zin3_3(w,episr) );
arrayfun( @(i) fzero( @(episr) reS11( w(i),episr ),4 ),1:numel(w) )

I got the same error message.

[Walter Roberson]

Your f is a vector of length 3. Your w=2*pi*f so your w is length 3. Your y0 = 1i.*w/c so your y0 is length 3. Then you have

Zin3_2 = @(w,episr) the(episr).*tanh(y0*dL);

so your Zin3_2 which uses y0 so Zin3_2 is going to be length 3. You also have

Zin3_3 = @(w,episr) the(episr).*( 1+the(episr).*tanh(y0*dL).*tanh(ys(w,episr)*L2) );

which uses y0 so your Zin3_3 is going to be length 3.

With your reS11 referring to Zin3_3, your reS11 is going to be length 3.

You are using reS11 in your fzero call, so the result of your reS11 is to be compared to 0. However, fzero is strictly for functions of a single variable, so fzero is not expecting to need to test that three different values are all 0 so it fails.

If you intend for reS11 to return a vector of length 3 and you are trying to satisfy that all 3 components are 0, then you need to use fsolve() from the Optimization toolbox.

It is my guess that you need to make y0 a function of the w that has been passed in, instead of being a constant vector.

[San-Chi Yan]

I want to get root of reS11 when I replace different w into reS11.

So I use arratfun that I can replace each w(i) into reS11 to find the root : episr.

This is my reference:

http://stackoverflow.com/questions/16061486/matlab-fzero-with-a-matrix-as-input-to-function

[San-Chi Yan]

So I think I made a mistake in Zin3_3.

Maybe I miss some input.

[Walter Roberson]

The general technique of using arrayfun is fine, but your expression to be solved turns out to be a vector rather than a scalar.

I recommend that you rewrite your expressions so that you distinguish by variable name between the w that you assign in your second line, and the w that is being passed in to the routines, the value that you are trying to solve for. For example call one of them W instead of w. I think when you do that and follow through the logic that you will find that the w that is in y0 should not be the w that you compute originally and should instead by the trial w. Like this:

ZL1 = rand();              %you never defined it

f = [1*1e9 2*1e9 3*1e9];
w=2*pi*f;

z0 = 376.73415;
c = 2.997925e8;
L1 = 10*1e-3;
L2 = 20*1e-3;
dL = 0.01*1e-3;
miu0 = z0/c;
miur = 1;

y0 = @(W) 1i.*W/c;
ys = @(W,episr) 1i*W./c.*sqrt( miur*episr );

the = @(episr) episr./sqrt(miur);

Zin3_1 = @(W,episr) tanh(ys(W,episr)*L2);

Zin3_2 = @(W,episr) the(episr).*tanh(y0(W)*dL);

Zin3_3 = @(W,episr) the(episr).*( 1 + the(episr) .* tanh(y0(W)*dL) .* tanh(ys(W,episr)*L2) );

Zin3 =@(W,episr) ZL1.*( Zin3_1(W,episr) + Zin3_2(W,episr) + Zin3_3(W,episr).*tanh(y0(W)*L1) ) ./ ( Zin3_3(W,episr) + ( Zin3_1(W,episr) + Zin3_2(W,episr) ).*tanh(y0(W)*L1) );

S11 = @(W,episr) ( Zin3(W,episr)-z0 ) ./ ( Zin3(W,episr) + z0 );

reS11 = @(W,episr) real( S11(W,episr) );

arrayfun( @(i) fzero( @(episr) reS11( w(i),episr ),3 ),1:numel(w) )

With those initial w values, two of the three will give results as NaN. I have investigated and found that the range that would need to be searched over for a change in sign can be quite narrow, depending on the initial ZL1 value.

[San-Chi Yan]

Sorry! I missed a error again... I forgot it from posting mt code.

Thank you for your respond! I will remember that.