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

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Rama a on 5 Jun 2021
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)

Sulaymon Eshkabilov on 5 Jun 2021
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 on 5 Jun 2021
Edited: Rama a on 5 Jun 2021
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

Sulaymon Eshkabilov on 15 Jun 2021
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*--')