incorrect number of input arguments in imagesc

Hi guys,

Matlab says:

Error using image

Incorrect number of arguments.

Error in imagesc (line 52)

hh = image(varargin{:}, 'CDataMapping', 'scaled');

Error in test4 (line 12)

imagesc(w, kx, ky, absolute);

And my code is on the photo. Could someone please help me correct the errors.

[Rena Berman]

(Answers Dev) Restored edit

Answer 1

[Walter Roberson]

You are calling

imagesc(w, kx, ky, absolute);

Your w and kx are vectors and would be interpreted as being in the x and y position slots of the call to image() that is made. Your ky is a vector and would be interpreted as being in the C (color information) slot in the call to image(). Your absolute is a 3D array and would be interpreted as being in the clim slot in the call to imagesc() . However, imagesc() notices that the value there is not a vector of 2 elements and so does not recognize it as being intensity limits, and so just passes it as data to image(). image() then complains because it has been passed too many numeric inputs.

This is not just a matter of you needing to drop one of the inputs: image() and imagesc() and imshow() can never handle 3D volumes such as your absolute array is.

To visualize 3D arrays, you can use one of:

• implay (view the slices as time)
• volumeViewer() or volshow() (3D viewer)
• slice()
• isosurface()
• vol3d v2 from File Exchange

[J K]

What I did is I created a third value kperp which I know is equals to sqrt(kx^2+ky^2) from initial conditions. Is this a valid solution? (see below)

w_o = 1;
 w = -3:0.1:3;
 kx = -3:0.1:3;
ky = kx;
kperp = sqrt(kx.^2+ky.^2);
delta =0.5;
[W, Kperp] = ndgrid(w, kperp);
functione = exp(-((w-w_o).^2)/delta.^2).*exp(-(Kperp.^2)/(delta.^2));
fourier = fftn(functione);
shifted = fftshift(fourier)
absolute = abs(shifted);
figure(1);
imagesc(w, kperp, absolute);
figure(2);
mesh(W, Kperp, functione);
nw = length(w);
dt = pi/w(end);
 t =[(-nw/2+1:nw/2)]*dt;
nkx = length(kx);
  dx = pi/kx(end);
%  t =[(-nw/2+1:-1) (0:nw/2)]*dt;
  x =[(-nkx/2+1:nkx/2)]*dx;
  y = x
  perp = sqrt(x.^2+y.^2);
  [T, PERP] = ndgrid(t, perp);
  mesh(T, PERP, absolute);

[Walter Roberson]

I guess it is okay. Note that the range of values produced is much larger than where the "interesting" information is. If you use imcontrast and ask it to eliminate 1% outliers you will see that most of the range is chopped off.

[J K]

Thank you so much! I will try imcontrast.

[J K]

May I ask if I decide to do 2D slices of 3D function instead how would I go about doing it?

[Walter Roberson]

If you have a 3D array of computed values, then slice() can extract for display. slice() uses interp3() internally

[Walter Roberson]

Go back to your original code that created the 3D arrays and slice() the result.