how to plot the imaginary part of an array numbers as a minus y-axis data

in developing a matlab code I need to plot the imaginary part of the frequency range in the negative y-axis range . the real part ( in the positive y-axis) and the imaginary part ( in the negative y- axis range)

here is the function in which the real part is ploted successfully, but the imag part is stuck

function []=plot_dispersion(Nk,k_norm,name_special_k,f)
f=real(f);
f1=imag(f);
max_nu=max(max(f));
plot(ones(100,1)*k_norm(1),linspace(0,max_nu*1.1,100),'k-','linewidth',2);
hold on;
for n=1:size(name_special_k,2)-1
    plot(linspace(k_norm(n),k_norm(n+1),Nk),f(:,(n-1)*Nk+1:n*Nk),'b.');
    plot(ones(100,1)*k_norm(n+1),linspace(0,max_nu*1.1,100),'k-','linewidth',2);
end
set(gca,'xtick',[],'fontsize',12);
ylabel('\nu (THz)','fontsize',12);
 xlabel ( ' (wavevector) ' )
axis tight;
for n=1:size(name_special_k,2)
    text(k_norm(n),-max_nu*0.05,name_special_k(n),...
        'interpreter','latex','fontsize',12);
end

[Mathieu NOE]

hello

it would help if :

1/ you could add the data to display and maybe the main code that calls this function

2/ show what the expected result should look like (not everybody knows your application)

tx

[zina shadidi]

Hello Mathieu NOE.

thank you very very much , that is very kind of you.

In fact I solved it with another function , but let me ask you please

if we can draw a link between the plotted data in the negative part of the folowing plot .

( the bold one ) in fact i have a good data curve, but my data lines seems like broken .

the figure attached in below.

I can not making fitting because every point here in this figure regarded as a set of data. on the other hand ( lines or arrows) iwas not a good choice.

thank you in advance

[Mathieu NOE]

hello

do you mean the bold dots in the negative Y ? because wesee bold dots both in positive and negative Y ...

yes the lines seems broken, and there are gaps . if you want to fill those gaps you have to interpolate the data (using interp1) , then you have one data set and you can also then do fitting.

[zina shadidi]

Thanks a lot , Mathieu NOE

interp1 was a good choice, and my problem solved.

[Mathieu NOE]

my pleasure