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Laplace Transform of $f(t) = u(sin(2t))$?
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What is the Laplace Transform of $f(t) = u(sin(2t))$?
I'm confused how to proceed further? because $u(t)=0$ for t<0. What does having $sin(2t)$ in argument mean?
Or $u$ has to be taken as unknown function?
laplace-transform
asked Jul 30 at 7:25
Akash Kishore
33
add a comment |Â
up vote
-2
down vote
favorite
What is the Laplace Transform of $f(t) = u(sin(2t))$?
I'm confused how to proceed further? because $u(t)=0$ for t<0. What does having $sin(2t)$ in argument mean?
Or $u$ has to be taken as unknown function?
laplace-transform
asked Jul 30 at 7:25
Akash Kishore
33
2
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 30 at 7:32
1
I've always seen $u$ defined as $1$ for $t>0$, and $0$ for $t<0$. But this (or another) definition should be given to you, not guessed at.
– mr_e_man
Jul 30 at 7:34
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
What is the Laplace Transform of $f(t) = u(sin(2t))$?
I'm confused how to proceed further? because $u(t)=0$ for t<0. What does having $sin(2t)$ in argument mean?
Or $u$ has to be taken as unknown function?
laplace-transform
asked Jul 30 at 7:25
Akash Kishore
33
What is the Laplace Transform of $f(t) = u(sin(2t))$?
I'm confused how to proceed further? because $u(t)=0$ for t<0. What does having $sin(2t)$ in argument mean?
Or $u$ has to be taken as unknown function?
laplace-transform
asked Jul 30 at 7:25
Akash Kishore
33
edited Jul 30 at 8:44
asked Jul 30 at 7:25
Akash Kishore
33
asked Jul 30 at 7:25
Akash Kishore
33
asked Jul 30 at 7:25
Akash Kishore
33
33
2
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 30 at 7:32
1
I've always seen $u$ defined as $1$ for $t>0$, and $0$ for $t<0$. But this (or another) definition should be given to you, not guessed at.
– mr_e_man
Jul 30 at 7:34
add a comment |Â
2
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 30 at 7:32
1
I've always seen $u$ defined as $1$ for $t>0$, and $0$ for $t<0$. But this (or another) definition should be given to you, not guessed at.
– mr_e_man
Jul 30 at 7:34
2
2
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 30 at 7:32
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 30 at 7:32
1
1
I've always seen $u$ defined as $1$ for $t>0$, and $0$ for $t<0$. But this (or another) definition should be given to you, not guessed at.
– mr_e_man
Jul 30 at 7:34
I've always seen $u$ defined as $1$ for $t>0$, and $0$ for $t<0$. But this (or another) definition should be given to you, not guessed at.
– mr_e_man
Jul 30 at 7:34
add a comment |Â
1 Answer
1
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oldest
votes
up vote
1
down vote
accepted
If $u(t) = 1$ for positive $t$, and $0$ otherwise, then $u(sin(2t)) = 1$ whenever $sin(2t)$ is positive.
That is, $f(t) = 1$ whenever $2t$ is in one of the intervals $(0,pi),;(2pi,3pi),;(4pi,5pi),;(-2pi,-pi),;cdots$
meaning $t$ is in one of the intervals $(0,fracpi2),;(pi,frac3pi2),;(2pi,frac5pi2),;(-pi,-fracpi2),;cdots$
and $f(t) = 0$ otherwise. This is a square wave.
answered Jul 30 at 7:46
mr_e_man
748219
Yes i wrote the defn of step function wrongly. And Thanks for Solution
– Akash Kishore
Jul 30 at 8:44
The Laplace transform of $f$ is not a square wave; I just set up the problem for you to finish. Or did you already? If so, click on the check mark to "accept" this answer.
– mr_e_man
Jul 30 at 8:46
Yes i understood that f(t) is a square wave
– Akash Kishore
Jul 30 at 8:50
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If $u(t) = 1$ for positive $t$, and $0$ otherwise, then $u(sin(2t)) = 1$ whenever $sin(2t)$ is positive.
That is, $f(t) = 1$ whenever $2t$ is in one of the intervals $(0,pi),;(2pi,3pi),;(4pi,5pi),;(-2pi,-pi),;cdots$
meaning $t$ is in one of the intervals $(0,fracpi2),;(pi,frac3pi2),;(2pi,frac5pi2),;(-pi,-fracpi2),;cdots$
and $f(t) = 0$ otherwise. This is a square wave.
answered Jul 30 at 7:46
mr_e_man
748219
Yes i wrote the defn of step function wrongly. And Thanks for Solution
– Akash Kishore
Jul 30 at 8:44
The Laplace transform of $f$ is not a square wave; I just set up the problem for you to finish. Or did you already? If so, click on the check mark to "accept" this answer.
– mr_e_man
Jul 30 at 8:46
Yes i understood that f(t) is a square wave
– Akash Kishore
Jul 30 at 8:50
add a comment |Â
up vote
1
down vote
accepted
If $u(t) = 1$ for positive $t$, and $0$ otherwise, then $u(sin(2t)) = 1$ whenever $sin(2t)$ is positive.
That is, $f(t) = 1$ whenever $2t$ is in one of the intervals $(0,pi),;(2pi,3pi),;(4pi,5pi),;(-2pi,-pi),;cdots$
meaning $t$ is in one of the intervals $(0,fracpi2),;(pi,frac3pi2),;(2pi,frac5pi2),;(-pi,-fracpi2),;cdots$
and $f(t) = 0$ otherwise. This is a square wave.
answered Jul 30 at 7:46
mr_e_man
748219
Yes i wrote the defn of step function wrongly. And Thanks for Solution
– Akash Kishore
Jul 30 at 8:44
The Laplace transform of $f$ is not a square wave; I just set up the problem for you to finish. Or did you already? If so, click on the check mark to "accept" this answer.
– mr_e_man
Jul 30 at 8:46
Yes i understood that f(t) is a square wave
– Akash Kishore
Jul 30 at 8:50
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If $u(t) = 1$ for positive $t$, and $0$ otherwise, then $u(sin(2t)) = 1$ whenever $sin(2t)$ is positive.
That is, $f(t) = 1$ whenever $2t$ is in one of the intervals $(0,pi),;(2pi,3pi),;(4pi,5pi),;(-2pi,-pi),;cdots$
meaning $t$ is in one of the intervals $(0,fracpi2),;(pi,frac3pi2),;(2pi,frac5pi2),;(-pi,-fracpi2),;cdots$
and $f(t) = 0$ otherwise. This is a square wave.
answered Jul 30 at 7:46
mr_e_man
748219
If $u(t) = 1$ for positive $t$, and $0$ otherwise, then $u(sin(2t)) = 1$ whenever $sin(2t)$ is positive.
That is, $f(t) = 1$ whenever $2t$ is in one of the intervals $(0,pi),;(2pi,3pi),;(4pi,5pi),;(-2pi,-pi),;cdots$
meaning $t$ is in one of the intervals $(0,fracpi2),;(pi,frac3pi2),;(2pi,frac5pi2),;(-pi,-fracpi2),;cdots$
and $f(t) = 0$ otherwise. This is a square wave.
answered Jul 30 at 7:46
mr_e_man
748219
answered Jul 30 at 7:46
mr_e_man
748219
answered Jul 30 at 7:46
mr_e_man
748219
answered Jul 30 at 7:46
mr_e_man
748219
748219
Yes i wrote the defn of step function wrongly. And Thanks for Solution
– Akash Kishore
Jul 30 at 8:44
The Laplace transform of $f$ is not a square wave; I just set up the problem for you to finish. Or did you already? If so, click on the check mark to "accept" this answer.
– mr_e_man
Jul 30 at 8:46
Yes i understood that f(t) is a square wave
– Akash Kishore
Jul 30 at 8:50
add a comment |Â
Yes i wrote the defn of step function wrongly. And Thanks for Solution
– Akash Kishore
Jul 30 at 8:44
The Laplace transform of $f$ is not a square wave; I just set up the problem for you to finish. Or did you already? If so, click on the check mark to "accept" this answer.
– mr_e_man
Jul 30 at 8:46
Yes i understood that f(t) is a square wave
– Akash Kishore
Jul 30 at 8:50
Yes i wrote the defn of step function wrongly. And Thanks for Solution
– Akash Kishore
Jul 30 at 8:44
Yes i wrote the defn of step function wrongly. And Thanks for Solution
– Akash Kishore
Jul 30 at 8:44
The Laplace transform of $f$ is not a square wave; I just set up the problem for you to finish. Or did you already? If so, click on the check mark to "accept" this answer.
– mr_e_man
Jul 30 at 8:46
The Laplace transform of $f$ is not a square wave; I just set up the problem for you to finish. Or did you already? If so, click on the check mark to "accept" this answer.
– mr_e_man
Jul 30 at 8:46
Yes i understood that f(t) is a square wave
– Akash Kishore
Jul 30 at 8:50
Yes i understood that f(t) is a square wave
– Akash Kishore
Jul 30 at 8:50
add a comment |Â
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2
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 30 at 7:32
1
I've always seen $u$ defined as $1$ for $t>0$, and $0$ for $t<0$. But this (or another) definition should be given to you, not guessed at.
– mr_e_man
Jul 30 at 7:34