which one of these is the right sol.? and let me know my mistakes please ..

  xb (t)= r(t ) - r(t-A ) - r(t-B ) + r(t-C ) 
 A = 3, B = 4 and C = 4. 

which one of these is the right sol.
and let me know my mistakes please 

syms Xb t;

y1=t.*heaviside(t);
y2=(t-3).*heaviside(t-3);
y3=(t-4).*heaviside(t-4);
y4=(t-4).*heaviside(t-4);
Xb=y1-y2-y3+y4;
fplot(Xb);
xlabel('t');
ylabel('Xb(t)');
title (' xb(t)= r(t ) - r(t-3 ) - r(t-4 ) + r(t-4 ) ');

***************************************************************


syms t
A=3
B=4
C=4

t= 0: 0.02:5;
u=t>= 0;
u1=t>= 1;


X= t.*u-(t-A).*u1-(t-B).*u1+(t-C).*u1;
plot(t,X)

Answer 1

[Sulaymon Eshkabilov]

Both are accurate and you can get the same results by changing this in the 1st one:

fplot(Xb, [0, 5]);

In the second one, you should not make t symbolic.

Note that the second one is computationally more efficient.

[Rama a]

thanks

so the second code should be like this ?

A=3
B=4
C=4
t= 0: 0.02:5;
u=t>= 0;
u1=t>= 1;
X= t.*u-(t-A).*u1-(t-B).*u1+(t-C).*u1;
plot(X);

and I want you to run the two codes please .. there is  a difference between between the plot of the first code and the second one from 1 to 3 on Y axis

Answer 2

[Sulaymon Eshkabilov]

Both gives the same results, if you fix in the second one the time boundary value for u1.

...
A=3;
B=4;
C=4;
t= 0: 0.02:5;
u=t>= 0;
u1=t>= 3;                               % Start at 3rd second 
X= t.*u-(t-A).*u1-(t-B).*u1+(t-C).*u1;
plot(t,X, 'r*--')