How do I plot this function graphically?
up vote
0
down vote
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Let $F(t)$ be a function of $t$, given by
$$F(t) = t U(t)-(t-1) U(t-1) + (t-2) U(t-2) - (t-3) U(t-3)$$
where $U$ is the Heaviside step function.
I am getting confused regarding its graphical representation. Any help would be highly appreciated.
graphing-functions
edited Nov 18 at 18:55
Fakemistake
1,682815
asked Nov 18 at 18:48
Jasmine
313
add a comment |
up vote
0
down vote
favorite
Let $F(t)$ be a function of $t$, given by
$$F(t) = t U(t)-(t-1) U(t-1) + (t-2) U(t-2) - (t-3) U(t-3)$$
where $U$ is the Heaviside step function.
I am getting confused regarding its graphical representation. Any help would be highly appreciated.
graphing-functions
edited Nov 18 at 18:55
Fakemistake
1,682815
asked Nov 18 at 18:48
Jasmine
313
What is $F(t)$ for $t<0$.
– hamam_Abdallah
Nov 18 at 18:50
1
Draw each term on the RHS alone. For example draw $tU(t)$, $(t-1)U(t-1)$, and so on and add/subtract them graphically
– Fakemistake
Nov 18 at 18:55
I added the solution in my answer below. I think now it's better not to think about how each term looks like, it's better to think piecewise from $0$ to $1$, $1$ to $2$ and so on because $U(x)$ is zero for $x<0$.
– Fakemistake
Nov 25 at 9:18
1
Notice $tU(t) = max(0,t)$. throwing the commandPlot[Max[t,0] - Max[t-1,0] + Max[t-2,0] - Max[t-3,0],{t,-2,5}]to WA, you will get this graph.
– achille hui
Nov 25 at 9:24
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $F(t)$ be a function of $t$, given by
$$F(t) = t U(t)-(t-1) U(t-1) + (t-2) U(t-2) - (t-3) U(t-3)$$
where $U$ is the Heaviside step function.
I am getting confused regarding its graphical representation. Any help would be highly appreciated.
graphing-functions
edited Nov 18 at 18:55
Fakemistake
1,682815
asked Nov 18 at 18:48
Jasmine
313
Let $F(t)$ be a function of $t$, given by
$$F(t) = t U(t)-(t-1) U(t-1) + (t-2) U(t-2) - (t-3) U(t-3)$$
where $U$ is the Heaviside step function.
I am getting confused regarding its graphical representation. Any help would be highly appreciated.
graphing-functions
graphing-functions
edited Nov 18 at 18:55
Fakemistake
1,682815
asked Nov 18 at 18:48
Jasmine
313
edited Nov 18 at 18:55
Fakemistake
1,682815
asked Nov 18 at 18:48
Jasmine
313
edited Nov 18 at 18:55
Fakemistake
1,682815
edited Nov 18 at 18:55
Fakemistake
1,682815
edited Nov 18 at 18:55
Fakemistake
1,682815
1,682815
asked Nov 18 at 18:48
Jasmine
313
asked Nov 18 at 18:48
Jasmine
313
asked Nov 18 at 18:48
Jasmine
313
313
What is $F(t)$ for $t<0$.
– hamam_Abdallah
Nov 18 at 18:50
1
Draw each term on the RHS alone. For example draw $tU(t)$, $(t-1)U(t-1)$, and so on and add/subtract them graphically
– Fakemistake
Nov 18 at 18:55
I added the solution in my answer below. I think now it's better not to think about how each term looks like, it's better to think piecewise from $0$ to $1$, $1$ to $2$ and so on because $U(x)$ is zero for $x<0$.
– Fakemistake
Nov 25 at 9:18
1
Notice $tU(t) = max(0,t)$. throwing the commandPlot[Max[t,0] - Max[t-1,0] + Max[t-2,0] - Max[t-3,0],{t,-2,5}]to WA, you will get this graph.
– achille hui
Nov 25 at 9:24
add a comment |
What is $F(t)$ for $t<0$.
– hamam_Abdallah
Nov 18 at 18:50
1
Draw each term on the RHS alone. For example draw $tU(t)$, $(t-1)U(t-1)$, and so on and add/subtract them graphically
– Fakemistake
Nov 18 at 18:55
I added the solution in my answer below. I think now it's better not to think about how each term looks like, it's better to think piecewise from $0$ to $1$, $1$ to $2$ and so on because $U(x)$ is zero for $x<0$.
– Fakemistake
Nov 25 at 9:18
1
Notice $tU(t) = max(0,t)$. throwing the commandPlot[Max[t,0] - Max[t-1,0] + Max[t-2,0] - Max[t-3,0],{t,-2,5}]to WA, you will get this graph.
– achille hui
Nov 25 at 9:24
What is $F(t)$ for $t<0$.
– hamam_Abdallah
Nov 18 at 18:50
What is $F(t)$ for $t<0$.
– hamam_Abdallah
Nov 18 at 18:50
1
1
Draw each term on the RHS alone. For example draw $tU(t)$, $(t-1)U(t-1)$, and so on and add/subtract them graphically
– Fakemistake
Nov 18 at 18:55
Draw each term on the RHS alone. For example draw $tU(t)$, $(t-1)U(t-1)$, and so on and add/subtract them graphically
– Fakemistake
Nov 18 at 18:55
I added the solution in my answer below. I think now it's better not to think about how each term looks like, it's better to think piecewise from $0$ to $1$, $1$ to $2$ and so on because $U(x)$ is zero for $x<0$.
– Fakemistake
Nov 25 at 9:18
I added the solution in my answer below. I think now it's better not to think about how each term looks like, it's better to think piecewise from $0$ to $1$, $1$ to $2$ and so on because $U(x)$ is zero for $x<0$.
– Fakemistake
Nov 25 at 9:18
1
1
Notice $tU(t) = max(0,t)$. throwing the command
Plot[Max[t,0] - Max[t-1,0] + Max[t-2,0] - Max[t-3,0],{t,-2,5}] to WA, you will get this graph.– achille hui
Nov 25 at 9:24
Notice $tU(t) = max(0,t)$. throwing the command
Plot[Max[t,0] - Max[t-1,0] + Max[t-2,0] - Max[t-3,0],{t,-2,5}] to WA, you will get this graph.– achille hui
Nov 25 at 9:24
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Hint: The first term is
$$tU(t)=
begin{cases}
0 & t<0\
t & tgeq 0
end{cases}$$
the second
$$(t-1)U(t-1)=begin{cases}
0 &t<1\
t-1 & tgeq 1
end{cases}$$
and so on... Can you take it from here?
Edit:
$$F(t)=
begin{cases}
0, & t<0\
t, & 0leq t<1\
1, & 1leq t<2\
t-1,& 2leq t<3\
2, &tgeq 3
end{cases}$$
answered Nov 18 at 19:01
Fakemistake
1,682815
I'm afraid I cannot. I am able to draw functions like U(t)+2U(t-1)+4U(t-2) - 3U(t-3), whose graph is a combination of step-like figures. But my professor said that the first term of the given function would give a straight line with 45 degree slope, which will continue till t=1 on the t axis, and so on. I, on the other hand, always thought that these functions yield some horizontal lines at different steps. Perhaps the graph could clarify everything.
– Jasmine
Nov 18 at 19:43
Maybe you can do the next step when you subtract these two functions. What value holds for negative $t$, for $tgeq 1$ and for $0leq t<1$?
– Fakemistake
Nov 18 at 21:20
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint: The first term is
$$tU(t)=
begin{cases}
0 & t<0\
t & tgeq 0
end{cases}$$
the second
$$(t-1)U(t-1)=begin{cases}
0 &t<1\
t-1 & tgeq 1
end{cases}$$
and so on... Can you take it from here?
Edit:
$$F(t)=
begin{cases}
0, & t<0\
t, & 0leq t<1\
1, & 1leq t<2\
t-1,& 2leq t<3\
2, &tgeq 3
end{cases}$$
answered Nov 18 at 19:01
Fakemistake
1,682815
I'm afraid I cannot. I am able to draw functions like U(t)+2U(t-1)+4U(t-2) - 3U(t-3), whose graph is a combination of step-like figures. But my professor said that the first term of the given function would give a straight line with 45 degree slope, which will continue till t=1 on the t axis, and so on. I, on the other hand, always thought that these functions yield some horizontal lines at different steps. Perhaps the graph could clarify everything.
– Jasmine
Nov 18 at 19:43
Maybe you can do the next step when you subtract these two functions. What value holds for negative $t$, for $tgeq 1$ and for $0leq t<1$?
– Fakemistake
Nov 18 at 21:20
add a comment |
up vote
2
down vote
accepted
Hint: The first term is
$$tU(t)=
begin{cases}
0 & t<0\
t & tgeq 0
end{cases}$$
the second
$$(t-1)U(t-1)=begin{cases}
0 &t<1\
t-1 & tgeq 1
end{cases}$$
and so on... Can you take it from here?
Edit:
$$F(t)=
begin{cases}
0, & t<0\
t, & 0leq t<1\
1, & 1leq t<2\
t-1,& 2leq t<3\
2, &tgeq 3
end{cases}$$
answered Nov 18 at 19:01
Fakemistake
1,682815
I'm afraid I cannot. I am able to draw functions like U(t)+2U(t-1)+4U(t-2) - 3U(t-3), whose graph is a combination of step-like figures. But my professor said that the first term of the given function would give a straight line with 45 degree slope, which will continue till t=1 on the t axis, and so on. I, on the other hand, always thought that these functions yield some horizontal lines at different steps. Perhaps the graph could clarify everything.
– Jasmine
Nov 18 at 19:43
Maybe you can do the next step when you subtract these two functions. What value holds for negative $t$, for $tgeq 1$ and for $0leq t<1$?
– Fakemistake
Nov 18 at 21:20
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint: The first term is
$$tU(t)=
begin{cases}
0 & t<0\
t & tgeq 0
end{cases}$$
the second
$$(t-1)U(t-1)=begin{cases}
0 &t<1\
t-1 & tgeq 1
end{cases}$$
and so on... Can you take it from here?
Edit:
$$F(t)=
begin{cases}
0, & t<0\
t, & 0leq t<1\
1, & 1leq t<2\
t-1,& 2leq t<3\
2, &tgeq 3
end{cases}$$
answered Nov 18 at 19:01
Fakemistake
1,682815
Hint: The first term is
$$tU(t)=
begin{cases}
0 & t<0\
t & tgeq 0
end{cases}$$
the second
$$(t-1)U(t-1)=begin{cases}
0 &t<1\
t-1 & tgeq 1
end{cases}$$
and so on... Can you take it from here?
Edit:
$$F(t)=
begin{cases}
0, & t<0\
t, & 0leq t<1\
1, & 1leq t<2\
t-1,& 2leq t<3\
2, &tgeq 3
end{cases}$$
answered Nov 18 at 19:01
Fakemistake
1,682815
edited Nov 25 at 9:14
answered Nov 18 at 19:01
Fakemistake
1,682815
answered Nov 18 at 19:01
Fakemistake
1,682815
answered Nov 18 at 19:01
Fakemistake
1,682815
1,682815
I'm afraid I cannot. I am able to draw functions like U(t)+2U(t-1)+4U(t-2) - 3U(t-3), whose graph is a combination of step-like figures. But my professor said that the first term of the given function would give a straight line with 45 degree slope, which will continue till t=1 on the t axis, and so on. I, on the other hand, always thought that these functions yield some horizontal lines at different steps. Perhaps the graph could clarify everything.
– Jasmine
Nov 18 at 19:43
Maybe you can do the next step when you subtract these two functions. What value holds for negative $t$, for $tgeq 1$ and for $0leq t<1$?
– Fakemistake
Nov 18 at 21:20
add a comment |
I'm afraid I cannot. I am able to draw functions like U(t)+2U(t-1)+4U(t-2) - 3U(t-3), whose graph is a combination of step-like figures. But my professor said that the first term of the given function would give a straight line with 45 degree slope, which will continue till t=1 on the t axis, and so on. I, on the other hand, always thought that these functions yield some horizontal lines at different steps. Perhaps the graph could clarify everything.
– Jasmine
Nov 18 at 19:43
Maybe you can do the next step when you subtract these two functions. What value holds for negative $t$, for $tgeq 1$ and for $0leq t<1$?
– Fakemistake
Nov 18 at 21:20
I'm afraid I cannot. I am able to draw functions like U(t)+2U(t-1)+4U(t-2) - 3U(t-3), whose graph is a combination of step-like figures. But my professor said that the first term of the given function would give a straight line with 45 degree slope, which will continue till t=1 on the t axis, and so on. I, on the other hand, always thought that these functions yield some horizontal lines at different steps. Perhaps the graph could clarify everything.
– Jasmine
Nov 18 at 19:43
I'm afraid I cannot. I am able to draw functions like U(t)+2U(t-1)+4U(t-2) - 3U(t-3), whose graph is a combination of step-like figures. But my professor said that the first term of the given function would give a straight line with 45 degree slope, which will continue till t=1 on the t axis, and so on. I, on the other hand, always thought that these functions yield some horizontal lines at different steps. Perhaps the graph could clarify everything.
– Jasmine
Nov 18 at 19:43
Maybe you can do the next step when you subtract these two functions. What value holds for negative $t$, for $tgeq 1$ and for $0leq t<1$?
– Fakemistake
Nov 18 at 21:20
Maybe you can do the next step when you subtract these two functions. What value holds for negative $t$, for $tgeq 1$ and for $0leq t<1$?
– Fakemistake
Nov 18 at 21:20
add a comment |
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What is $F(t)$ for $t<0$.
– hamam_Abdallah
Nov 18 at 18:50
1
Draw each term on the RHS alone. For example draw $tU(t)$, $(t-1)U(t-1)$, and so on and add/subtract them graphically
– Fakemistake
Nov 18 at 18:55
I added the solution in my answer below. I think now it's better not to think about how each term looks like, it's better to think piecewise from $0$ to $1$, $1$ to $2$ and so on because $U(x)$ is zero for $x<0$.
– Fakemistake
Nov 25 at 9:18
1
Notice $tU(t) = max(0,t)$. throwing the command
Plot[Max[t,0] - Max[t-1,0] + Max[t-2,0] - Max[t-3,0],{t,-2,5}]to WA, you will get this graph.– achille hui
Nov 25 at 9:24