# how can I plot y = (cos(t-2)/4)(rect(t+1)/6 )?

4 views (last 30 days)
SaiRevan Potharaju on 3 Aug 2022
Edited: Jan on 3 Aug 2022
y = [(1-cos(0.25*(t-2).^2))*rect(t+1,6)]
is it correct way to write tht code ?
Jan on 3 Aug 2022
Edited: Jan on 3 Aug 2022
The brackets [ and ] are the concatenation operator in Matlab. What do you concatenate?
The shown code does not allow to plot anything, so how could it be correct?
What is "rect"? Why do you convert "rect(t+1)/6" to "rect(t+1,6)" ?
"cos(t-2)/4" differs from "1-cos(0.25*(t-2).^2)" also. Very strange.

Adam Danz on 3 Aug 2022
is it correct way to write tht code ?
The line of code looks functional without knowing any other details including what rect is.
how can I plot y = (cos(t-2)/4)(rect(t+1)/6 )?
Define t, compute y, and plot(t,y)

Jan on 3 Aug 2022
y = @(t) cos(t-2) / 4 .* rect(t + 1) / 6;
fplot(y)

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