How to use interp1 command?

I have the code to draw each of these 6 figures as follows:

ph1 = 1;
ph2 = 2;
ph3 = 3;
ph4 = 4;
ph5 = 5;
ph6 = 6;

Vi=30;
Vo=0:0.1:30;
d= Vo./Vi ;

m1= d*ph1; 
m2= d*ph2;
m3= d*ph3;
m4= d*ph4;
m5= d*ph5;
m6= d*ph6;
floor1= floor(m1);
floor2= floor(m2);
floor3= floor(m3);
floor4= floor(m4);
floor5= floor(m5);
floor6= floor(m6);

Kcm1 = (1-(floor1./(ph1*d))).*(1+floor1-ph1*d) 
Kcm2 = (1-(floor2./(ph2*d))).*(1+floor2-ph2*d) 
Kcm3 = (1-(floor3./(ph3*d))).*(1+floor3-ph3*d) 
Kcm4 = (1-(floor4./(ph4*d))).*(1+floor4-ph4*d) 
Kcm5 = (1-(floor5./(ph5*d))).*(1+floor5-ph5*d) 
Kcm6 = (1-(floor6./(ph6*d))).*(1+floor6-ph6*d)

figure (1)
 hold on
 title('for the output current ripple') 
xlabel('Duty Cycle') 
ylabel('Kcm')
plot(d,Kcm1)
plot(d,Kcm2)
plot(d,Kcm3)
plot(d,Kcm4)
plot(d,Kcm5)
plot(d,Kcm6)

legend('Phase1','Phase2','Phase3','Phase4','Phase5','Phase6');
grid on
hold off

What I want to do is that based on an input value (which is going to be a value on the X-axis), I want the output to choose the figures out of these 6 figures that give the minimum Y-axis value.

For example,

If I put the input value to be 0.5, which is a duty cycle value on the X-axis, then the output should display and recommend (phase-2, phase-4 and figure 6) as they give the minimum value of Y-axis (exactly 0) at this specific input value of 0.5

Another example,

If the input value is 0.25, then the recommend figure to be displayed is (phase-4 the yellow one) as this also gives the minimum value on the Y-axis, and so on.

I was told that I could use the "interp1" command for each curve and pick the minimum but I don't know how to use it.

Please help me out?

[Stephen23]

https://web.archive.org/web/20210927033125/https://www.mathworks.com/matlabcentral/answers/1461144-how-to-use-interp1-command

Answer 1

[Stephen23]

Numbering your variables like that is a red-herring that makes this task more complex.

MATLAB is designed to work efficiently with arrays. Using arrays makes your code much simpler:

ph = 1:6;
Vi = 30;
Vo = 0:0.1:30;
d  = Vo./Vi;
m  = d(:).*ph; % note the orientation!
Kcm = (1-(floor(m)./m)).*(1+floor(m)-m);
figure (1)
title('for the output current ripple') 
xlabel('Duty Cycle') 
ylabel('Kcm')
plot(d,Kcm)
legend(sprintfc('Phase%d',ph));
grid on

Using arrays also makes your task much simpler:

inp = 0.5;
tmp = interp1(d,Kcm,inp);
idx = find(tmp==min(tmp))

idx = 1×3

2 4 6

inp = 0.75;
tmp = interp1(d,Kcm,inp);
idx = find(tmp==min(tmp))

idx = 4

Note that the orientation of the data in m is significant, because INTERP1 treats each column as its own dataset.

[Mohammed Alharbi]

Thanks you so much Stephen. That was very helpful!

[Mohammed Alharbi]

Hi Stephen,

why would I get only one "idx" for some cases when two different "idx" are expected. for example for inp = 0.6667, idx should be 6 and 3 as both of them will give a minumim value of 0 as per the graph, but I am only getting the result for 6

inp = 0.6667;
tmp = interp1(d,Kcm,inp);
idx = find(tmp==min(tmp))

idx =

     6

[Stephen23]

I was wondering if you would ask about this.

The answer is that those values are actually different, due to the accumulated floating point error.

Lets have a look:

ph = 1:6;
Vi = 30;
Vo = 0:0.1:30;
d  = Vo./Vi;
m  = d(:).*ph; % note the orientation!
Kcm = (1-(floor(m)./m)).*(1+floor(m)-m);
inp = 0.6667;
tmp = interp1(d,Kcm,inp)

tmp = 1×6

0.3333 0.1667 0.0000 0.0833 0.0667 0.0000

fprintf('%.40f\n',tmp([3,6])) % clearly not the same!

0.0000492537313432778867247283094421561600
0.0000487562189054669805677623417583532728

In you want, you could compare the absolute difference against a tolerance:

tol = 1e-5; % pick this value to suit YOUR definition of how different values can be.
idx = find(abs(tmp-min(tmp))<tol)

idx = 1×2

3 6

Read more about binary floating point numbers:

https://www.mathworks.com/matlabcentral/answers/57444-faq-why-is-0-3-0-2-0-1-not-equal-to-zero

https://www.mathworks.com/matlabcentral/answers/316889-why-two-equal-numbers-are-not-equal

http://www.mathworks.com/matlabcentral/answers/102419-how-do-i-determine-if-the-error-in-my-answer-is-the-result-of-round-off-error-or-a-bug

http://www.mathworks.com/matlabcentral/answers/140656-about-numerical-difficulty-of-equality-comparison-operation-between-two-double-real-numbers-in-matlab

This is worth reading as well:

https://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html

http://faculty.tarleton.edu/agapie/documents/cs_343_arch/papers/1991_Goldberg_FloatingPoint.pdf

[Mohammed Alharbi]

Super!

I can't thank you enogh Stephen. Thanks for your time and the info

Mohammed

[Mohammed Alharbi]

Hi Stephen,

Sorry to keep bothering you with my questions and I really appreciate it. Just last question:

is there a way I can plot "ph against d" using the minimum data of Kcm of all the different Ph variables.

for example according to the graph above, when d is from 0 to 0.183, 6-phase should be selected as this will give the minimum values of Kcm, then from around 0.184 up to 0.224, 5-phase will be slected, and so on. in some cases, as discussed, for example when d=0.5, then we can select 2-phase, or 4-phase or 6-phase.

A similar graph of how I want it is shown in the bottom graph below comparing phase-3 and phase-4 to get the lowest value of Y-axis of the other graph

[Stephen23]

Perhaps something like this will get you started:

ph = 1:6;
Vi = 30;
Vo = 0:0.1:30;
d  = Vo./Vi;
m  = d(:).*ph; % note the orientation!
Kcm = (1-(floor(m)./m)).*(1+floor(m)-m);
% new code:
X = Kcm==min(Kcm,[],2);
Y = 7-sum(cumsum(X,2)>0,2);
plot(d,Y)

[Mohammed Alharbi]

YOU ARE GENOIUS!

Thanks a lot mate!

is just that I am thinking of a way how I can represent that for some duty cycles values like for example when D=0.5, then Phase-2, Phase-4, and Phase-6, can all give the lowest Kcm value. The above figure shows only the selction of phase-2 for this case. would it be possible to overlap it with other figures likewise the orignal figure?

Thanks again!

Mohammed

[Stephen23]

"how I can represent that for some duty cycles values like for example when D=0.5, then Phase-2, Phase-4, and Phase-6, can all give the lowest Kcm value. "

That is a bit more tricky because there are varying numbers of possible results. Some possible approaches:

• use a loop, calculate and plot each duty-cycle separately.
• use a scatter plot, for example:

ph = 1:6;
Vi = 30;
Vo = 0:0.01:30; % <- smaller step!
d  = Vo./Vi;
m  = d(:).*ph; % note the orientation!
Kcm = (1-(floor(m)./m)).*(1+floor(m)-m);
% new code:
tol = 1e-5; % pick this value to suit your needs. 
idx = abs(Kcm-min(Kcm,[],2))<tol;
[idr,idc] = find(idx);
scatter(d(idr),ph(idc),'.')

[Mohammed Alharbi]

Hi Stephen,

When I am using the MATLAB Function in Simulink I get the below error:

Althoug I predfined the dimensions of what the output is going to be, as follows:

function [PWM_signals,phase_delay,Num_phases]  = PWM_FCN(vin,vout,ref) %inputs: referance current and sensed current in each phase
Num_phases = zeros(3,1);
   phase_delay = zeros(6,1); %output phase shift
x=0;
ph=zeros(1,6);
tmp=zeros(1,6);
d=zeros(1,1001);
m=zeros(1001,6);
Kcm=zeros(1001,6);
Vo=zeros(1,1001);
tol=0;

  ph = 1:6;
  Vi = 1;
Vo = 0:0.001:1;
d  = Vo./Vi;
m  = d(:).*ph; % note the orientation!
Kcm = (1-(floor(m)./m)).*(1+floor(m)-m);
tol = 1e-2;

inp = vout/vin;
tmp = interp1(d,Kcm,inp);
Num_phases = find(abs(tmp-min(tmp))<tol);
PWM_signals = zeros(6,1); %output state 1 or 0

     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
     % Chose the phases according to the refernce current:
      if (ref > 200 && ref <= 240)
          x=6;
      elseif (ref > 165 && ref <= 200)
          x=5;
      elseif (ref > 130 && ref <= 165)
          x=4;
      elseif (ref > 90 && ref <= 130)
          x=3;
      elseif (ref > 0 && ref <= 90)
          x=2;
      elseif ref==0
          x=0;
      end
     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
     % for 6 phases
     %%%%%%%%%%%%%%5%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
     if x==6 || max(Num_phases)==6 && x >=3   % for very high current switch on all phases, Or switch the 6 phases if the current ripple is excellent to switch 6 phases yet the current should not be very low
     PWM_signals = ones (6,1);  % all phases are 1s
     phase_delay(1) =0 ; phase_delay(2) =60*((1/100e3)/360) ; phase_delay(3) =120*((1/100e3)/360) ; phase_delay(4) =180*((1/100e3)/360) ; phase_delay(5) =240*((1/100e3)/360) ; phase_delay(6) =300*((1/100e3)/360) ;
     end
     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
     % for 5 phases
%    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
     if (x > 2 && x <= 5) && (max(Num_phases) ==5) || (x == 5) && (max(Num_phases) < 5 )    % if the current is to hight to handel less the number of phases, then always swithc 5 phases 
     PWM_signals(1) =1 ; PWM_signals(2) =1 ; PWM_signals(3) =1 ; PWM_signals(4) =1 ; PWM_signals(5) =1 ; PWM_signals(6) =0 ; %then switch ON all the first 5 phases and switch OFF the last phase
     phase_delay(1) =0 ; phase_delay(2) =72*((1/100e3)/360) ; phase_delay(3) =144*((1/100e3)/360) ; phase_delay(4) =216*((1/100e3)/360) ; phase_delay(5) =288*((1/100e3)/360) ; phase_delay(6) =0 ;
     end
     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
     % for 4 phases
     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
     if ((x > 2 && x <= 4) && (max(Num_phases) ==4 || sum(Num_phases)/3 ==4 )) || ((x == 4) && (max(Num_phases) < 4 )) % if the  
     PWM_signals(1) =1 ; PWM_signals(2) =1 ; PWM_signals(3) =1 ; PWM_signals(4) =1 ; PWM_signals(5) =0 ; PWM_signals(6) =0 ;    
     phase_delay(1) =0 ; phase_delay(2) =90*((1/100e3)/360) ; phase_delay(3) =180*((1/100e3)/360) ; phase_delay(4) =270*((1/100e3)/360) ; phase_delay(5) =0 ; phase_delay(6) =0 ;
     end
     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
     % for 3 phases
     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
     if (x > 0 && x <= 3) && (sum(Num_phases)/3 ==3) || (x == 3) && (max(Num_phases) < 3 )
     PWM_signals(1) =1 ; PWM_signals(2) =1 ; PWM_signals(3) =1 ; PWM_signals(4) =0 ; PWM_signals(5) =0 ; PWM_signals(6) =0 ;     phase_delay(1) =0 ; phase_delay(2) =120*((1/100e3)/360) ; phase_delay(3) =240*((1/100e3)/360) ; phase_delay(4) =0 ; phase_delay(5) =0 ; phase_delay(6) =0 ;
     end
     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
     % for 2 phases 
     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
     if (x > 0 && x <= 2) && (sum(Num_phases)/6 == 2 || max(Num_phases) >= 4) % 2+4+6=12/2=2 
     PWM_signals(1) =1 ; PWM_signals(2) =1 ; PWM_signals(3) =0 ; PWM_signals(4) =0 ; PWM_signals(5) =0 ; PWM_signals(6) =0 ;
     phase_delay(1) =0 ; phase_delay(2) =180*((1/100e3)/360) ; phase_delay(3) =0 ; phase_delay(4) =0 ; phase_delay(5) =0 ; phase_delay(6) =0 ;
     end
     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
     % for 0 phases 
     %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
     if ref == 0
     PWM_signals = zeros (6,1);
     phase_delay(1) =0 ; phase_delay(2) =0; phase_delay(3) =0; phase_delay(4) =0; phase_delay(5) =0 ; phase_delay(6) =0 ;
     end 
     

I removed the function and includded the inputs variables (vin,vout,ref) and put them all in a normal m.file and I don't get any error and get exactly what I want.

So can you please let me know what is the issue?

I am including both the Simulink file and the m.file

thanks

[Stephen23]

@Mohammed Alharbi: you should ask this in a new thread.

[Mohammed Alharbi]

Hi Stephen,

Thanks for your response.

I already did but still waiting for a help and the thought of asking you here. This is the link for the new post:

https://uk.mathworks.com/matlabcentral/answers/1465904-matlab-function-simulink-simulink-does-not-have-enough-information-to-determine-output-sizes-for-th?s_tid=srchtitle

Thanks a lot!

[Mohammed Alharbi]

Hi Stephen,

I sloved the previous issue by putting

m  = d'*ph;

instead of:

m  = d(:).*ph;

But now I have another issue with using the "find" command in Simulink function block as it always returns a variable-length vector. Please have a look at this thread if you have time, I am sure you will be the one who can hlep me out with this

Thannks in advance!

Mohammed