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

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SaiRevan Potharaju
SaiRevan Potharaju 2022 年 8 月 3 日
編集済み: Jan 2022 年 8 月 3 日
y = [(1-cos(0.25*(t-2).^2))*rect(t+1,6)]
is it correct way to write tht code ?
  1 件のコメント
Jan
Jan 2022 年 8 月 3 日
編集済み: Jan 2022 年 8 月 3 日
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.

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回答 (2 件)

Adam Danz
Adam Danz 2022 年 8 月 3 日
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
Jan 2022 年 8 月 3 日
y = @(t) cos(t-2) / 4 .* rect(t + 1) / 6;
fplot(y)

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