unsure about using the hold on funciton

hi, i'm still new to matlab and im unsure of how to use the hold on command.

After running the code, i'd like to produce a graph with multiple results from my code but im unsure how to use the hold on function (ie; where to put within the code)

(mulitple results as in, i'm using the limits from 0-9, 9-15 etc as listed in the above code as t and the equations x1 thru 10.

t   = 0:0.1:9;
N   =   length(t)-1;
y   =  (-(1.37)*t.^2)+6000;     %eqn 1  


%Spline Equations
%The equations are substituted into the code.

%y1 (-(1.37)*t.^2)+6000;                            0<=t<=9      0:0.1:9;
%y2 (0.943*(t-9).^2)-(24.99*t+6113.91);             9<=t<=15     9:0.1:15;
%y3 (-(0.6054)*(t-15).^2)-((13.34*t)+5975.1);       15<=t<=39    15:0.1:39;
%y4 (-(6.1557)*(t-39).^2)-((960.74*t)+7254.97);     39<=t<=49    39:0.1:49;
%y5 (5.622*(t-49).^2)-((183.8*t)+12673.95);         49<=t<=57    49:0.1:57;
%y6 (2.4008*(t-57).^2)-((93.89*t)+7903.9);          57<=t<=69    57:0.1:69;
%y7 (-(12.91)*(t-69).^2)-((36.27*t)+4273.63);       69<=t<=74    69:0.1:74;
%y8 (3.091*(t-74).^2)-((92.92*t)+8324.598);         74<=t<=81    74:0.1:81;
%y9 (0.2117*(t-81).^2)-((49.64*t)+4969.84);         81<=t<=96    81:0.1:96;
%y10 (-(5.0625)*(t-96).^2)-((43.28*t)+4407.744);    96<=t<=100   96:0.1:100

% The unknowns are 3*N with ao=0 "Linear Spline"
% Filling Matrix from point matching


V   =   [0;zeros(2*N,1);zeros(N-1,1)];
Z   =   zeros(length(V),length(V));
j=1;
f=1;
for i=2:2:2*N    
    Z(i,f:f+2)          =   [t(j)^2 t(j) 1];
    V(i)                =   y(j);
    j                   =   j+1;
    Z(i+1,f:f+2)        =   [t(j)^2 t(j) 1];  
    V(i+1)              =   y(j);
    f                   =   f+3;
 end
% Filling Matrix from smoothing condition
j=1;
l=2;

for i=2*N+2:3*N
    
    Z(i,j:j+1)            =   [2*t(l) 1];
    Z(i,j+3:j+4)          =   [-2*t(l) -1];
    j                     =   j+3;
    l                     =   l+1;
  end
% Adjusting the value of a1 to be zero "Linear Spline"
Z(1,1)=1;
% Inverting and obtaining the coeffiecients, Plotting
Coeff       =       Z\V;
j=1;

for i=1:N
    curve=@(l) Coeff(j)*l.^2+Coeff(j+1)*l+Coeff(j+2);
    ezplot(curve,[t(i),t(i+1)]);
    hndl=get(gca,'Children');
    set(hndl,'LineWidth',2);
    j=j+3;
end
scatter(t,y,50,'r','filled')



figure (1)
title('Quadratic Spline')
xlim([min(t)-2 max(t)+2]);
ylim([min(y)-2 max(y)+2]);
xlabel('x');
ylabel('y');
grid on 
grid minor
hold on

i placed a hold on at the end but i am not able to collate the results on one graph.

any help would be appreciated!

thank you!!

[Adam]

You should get into the habit of always using the full syntax for hold and plotting instructions, with an explicit axes handle, e.g.

figure; hAxes = gca;
plot( hAxes,... )
hold( hAxes, 'on' )
scatter( hAxes,...)

Answer 1

[Sajeer Modavan]

clear
clc
close all
t   = 0:0.1:9;
N   =   length(t)-1;
y   =  (-(1.37)*t.^2)+6000;     %eqn 1  
%Spline Equations
%The equations are substituted into the code.
%y1 (-(1.37)*t.^2)+6000;                            0<=t<=9      0:0.1:9;
%y2 (0.943*(t-9).^2)-(24.99*t+6113.91);             9<=t<=15     9:0.1:15;
%y3 (-(0.6054)*(t-15).^2)-((13.34*t)+5975.1);       15<=t<=39    15:0.1:39;
%y4 (-(6.1557)*(t-39).^2)-((960.74*t)+7254.97);     39<=t<=49    39:0.1:49;
%y5 (5.622*(t-49).^2)-((183.8*t)+12673.95);         49<=t<=57    49:0.1:57;
%y6 (2.4008*(t-57).^2)-((93.89*t)+7903.9);          57<=t<=69    57:0.1:69;
%y7 (-(12.91)*(t-69).^2)-((36.27*t)+4273.63);       69<=t<=74    69:0.1:74;
%y8 (3.091*(t-74).^2)-((92.92*t)+8324.598);         74<=t<=81    74:0.1:81;
%y9 (0.2117*(t-81).^2)-((49.64*t)+4969.84);         81<=t<=96    81:0.1:96;
%y10 (-(5.0625)*(t-96).^2)-((43.28*t)+4407.744);    96<=t<=100   96:0.1:100
% The unknowns are 3*N with ao=0 "Linear Spline"
% Filling Matrix from point matching
V   =   [0;zeros(2*N,1);zeros(N-1,1)];
Z   =   zeros(length(V),length(V));
j=1;
f=1;
for i=2:2:2*N    
    Z(i,f:f+2)          =   [t(j)^2 t(j) 1];
    V(i)                =   y(j);
    j                   =   j+1;
    Z(i+1,f:f+2)        =   [t(j)^2 t(j) 1];  
    V(i+1)              =   y(j);
    f                   =   f+3;
 end
% Filling Matrix from smoothing condition
j=1;
l=2;
for i=2*N+2:3*N
    
    Z(i,j:j+1)            =   [2*t(l) 1];
    Z(i,j+3:j+4)          =   [-2*t(l) -1];
    j                     =   j+3;
    l                     =   l+1;
  end
% Adjusting the value of a1 to be zero "Linear Spline"
Z(1,1)=1;
% Inverting and obtaining the coeffiecients, Plotting
Coeff       =       Z\V;
j=1;
fig = figure; 
scatter(t,y,50,'r','filled'),hold on
for i=1:N
    curve=@(l) Coeff(j)*l.^2+Coeff(j+1)*l+Coeff(j+2);
    ezplot(curve,[t(i),t(i+1)]);
    hndl=get(gca,'Children');
    set(hndl,'LineWidth',2);
    j=j+3;
end
title('Quadratic Spline')
xlim([min(t)-2 max(t)+2]);
ylim([min(y)-2 max(y)+2]);
xlabel('x');
ylabel('y');
grid on 
grid minor

Please let me know if you are looking different, if you are using latest matlab you can replaceezplot with fplot and adjust color of line to single color

eg:

replace: ezplot(curve,[t(i),t(i+1)]);

fplot(curve,[t(i),t(i+1)],'color','k');

[Nevin Pang]

i am using the r2017b version

yes i am able to replace ezplot with fplot and adjust color of line to single color

[Sajeer Modavan]

ok. you can accept answer and close this issue if you don't need any further help.

[Nevin Pang]

thank you for the help!