Laplace transform of a ramp function with a $2$ second delay
$begingroup$
Been trying to solve a ramp function with a $2$ second delay by integrating it from $2$ to infinity, but I end up getting the wrong answer. I know that I can integrate it without looking at the delay then multiply my Laplace transform with $$e^{-2s}$$ but I'd like to know what I am doing wrong when integrating.
$$int_{2}^infty t e^{-st} , mathrm{d}t = frac{-te^{-st}}{s}bigg|_{2}^infty + int_{2}^infty frac{e^{-st}}{s} , mathrm{d}t = frac{2e^{-2s}}{s}+frac{e^{-2s}}{s^2}$$
This is what I get. Where am I going wrong?
laplace-transform
edited Oct 16 '16 at 18:14
Rodrigo de Azevedo
13.1k41960
asked Oct 16 '16 at 18:07
KeilaraKeilara
558
$endgroup$
|
show 1 more comment
$begingroup$
Been trying to solve a ramp function with a $2$ second delay by integrating it from $2$ to infinity, but I end up getting the wrong answer. I know that I can integrate it without looking at the delay then multiply my Laplace transform with $$e^{-2s}$$ but I'd like to know what I am doing wrong when integrating.
$$int_{2}^infty t e^{-st} , mathrm{d}t = frac{-te^{-st}}{s}bigg|_{2}^infty + int_{2}^infty frac{e^{-st}}{s} , mathrm{d}t = frac{2e^{-2s}}{s}+frac{e^{-2s}}{s^2}$$
This is what I get. Where am I going wrong?
laplace-transform
edited Oct 16 '16 at 18:14
Rodrigo de Azevedo
13.1k41960
asked Oct 16 '16 at 18:07
KeilaraKeilara
558
$endgroup$
2
$begingroup$
You're computing the wrong integral. Instead, you should be computing $$int_{2}^infty (t - 2) e^{-st} , mathrm{d}t $$
$endgroup$
– Rodrigo de Azevedo
Oct 16 '16 at 18:13
$begingroup$
Now you have changed from $ds$ to $dt$ which is it?
$endgroup$
– Olba12
Oct 16 '16 at 18:17
$begingroup$
$, mathrm{d}t$
$endgroup$
– Rodrigo de Azevedo
Oct 16 '16 at 18:18
$begingroup$
I saw your comment! @RodrigodeAzevedo
$endgroup$
– Olba12
Oct 16 '16 at 18:18
$begingroup$
Why is it that you integrate (t-2) instead of t? isn't the lower bound (2) enough?
$endgroup$
– Keilara
Oct 16 '16 at 21:29
|
show 1 more comment
$begingroup$
Been trying to solve a ramp function with a $2$ second delay by integrating it from $2$ to infinity, but I end up getting the wrong answer. I know that I can integrate it without looking at the delay then multiply my Laplace transform with $$e^{-2s}$$ but I'd like to know what I am doing wrong when integrating.
$$int_{2}^infty t e^{-st} , mathrm{d}t = frac{-te^{-st}}{s}bigg|_{2}^infty + int_{2}^infty frac{e^{-st}}{s} , mathrm{d}t = frac{2e^{-2s}}{s}+frac{e^{-2s}}{s^2}$$
This is what I get. Where am I going wrong?
laplace-transform
edited Oct 16 '16 at 18:14
Rodrigo de Azevedo
13.1k41960
asked Oct 16 '16 at 18:07
KeilaraKeilara
558
$endgroup$
Been trying to solve a ramp function with a $2$ second delay by integrating it from $2$ to infinity, but I end up getting the wrong answer. I know that I can integrate it without looking at the delay then multiply my Laplace transform with $$e^{-2s}$$ but I'd like to know what I am doing wrong when integrating.
$$int_{2}^infty t e^{-st} , mathrm{d}t = frac{-te^{-st}}{s}bigg|_{2}^infty + int_{2}^infty frac{e^{-st}}{s} , mathrm{d}t = frac{2e^{-2s}}{s}+frac{e^{-2s}}{s^2}$$
This is what I get. Where am I going wrong?
laplace-transform
laplace-transform
edited Oct 16 '16 at 18:14
Rodrigo de Azevedo
13.1k41960
asked Oct 16 '16 at 18:07
KeilaraKeilara
558
edited Oct 16 '16 at 18:14
Rodrigo de Azevedo
13.1k41960
asked Oct 16 '16 at 18:07
KeilaraKeilara
558
edited Oct 16 '16 at 18:14
Rodrigo de Azevedo
13.1k41960
edited Oct 16 '16 at 18:14
Rodrigo de Azevedo
13.1k41960
edited Oct 16 '16 at 18:14
Rodrigo de Azevedo
13.1k41960
13.1k41960
asked Oct 16 '16 at 18:07
KeilaraKeilara
558
asked Oct 16 '16 at 18:07
KeilaraKeilara
558
asked Oct 16 '16 at 18:07
KeilaraKeilara
558
558
2
$begingroup$
You're computing the wrong integral. Instead, you should be computing $$int_{2}^infty (t - 2) e^{-st} , mathrm{d}t $$
$endgroup$
– Rodrigo de Azevedo
Oct 16 '16 at 18:13
$begingroup$
Now you have changed from $ds$ to $dt$ which is it?
$endgroup$
– Olba12
Oct 16 '16 at 18:17
$begingroup$
$, mathrm{d}t$
$endgroup$
– Rodrigo de Azevedo
Oct 16 '16 at 18:18
$begingroup$
I saw your comment! @RodrigodeAzevedo
$endgroup$
– Olba12
Oct 16 '16 at 18:18
$begingroup$
Why is it that you integrate (t-2) instead of t? isn't the lower bound (2) enough?
$endgroup$
– Keilara
Oct 16 '16 at 21:29
|
show 1 more comment
2
$begingroup$
You're computing the wrong integral. Instead, you should be computing $$int_{2}^infty (t - 2) e^{-st} , mathrm{d}t $$
$endgroup$
– Rodrigo de Azevedo
Oct 16 '16 at 18:13
$begingroup$
Now you have changed from $ds$ to $dt$ which is it?
$endgroup$
– Olba12
Oct 16 '16 at 18:17
$begingroup$
$, mathrm{d}t$
$endgroup$
– Rodrigo de Azevedo
Oct 16 '16 at 18:18
$begingroup$
I saw your comment! @RodrigodeAzevedo
$endgroup$
– Olba12
Oct 16 '16 at 18:18
$begingroup$
Why is it that you integrate (t-2) instead of t? isn't the lower bound (2) enough?
$endgroup$
– Keilara
Oct 16 '16 at 21:29
2
2
$begingroup$
You're computing the wrong integral. Instead, you should be computing $$int_{2}^infty (t - 2) e^{-st} , mathrm{d}t $$
$endgroup$
– Rodrigo de Azevedo
Oct 16 '16 at 18:13
$begingroup$
You're computing the wrong integral. Instead, you should be computing $$int_{2}^infty (t - 2) e^{-st} , mathrm{d}t $$
$endgroup$
– Rodrigo de Azevedo
Oct 16 '16 at 18:13
$begingroup$
Now you have changed from $ds$ to $dt$ which is it?
$endgroup$
– Olba12
Oct 16 '16 at 18:17
$begingroup$
Now you have changed from $ds$ to $dt$ which is it?
$endgroup$
– Olba12
Oct 16 '16 at 18:17
$begingroup$
$, mathrm{d}t$
$endgroup$
– Rodrigo de Azevedo
Oct 16 '16 at 18:18
$begingroup$
$, mathrm{d}t$
$endgroup$
– Rodrigo de Azevedo
Oct 16 '16 at 18:18
$begingroup$
I saw your comment! @RodrigodeAzevedo
$endgroup$
– Olba12
Oct 16 '16 at 18:18
$begingroup$
I saw your comment! @RodrigodeAzevedo
$endgroup$
– Olba12
Oct 16 '16 at 18:18
$begingroup$
Why is it that you integrate (t-2) instead of t? isn't the lower bound (2) enough?
$endgroup$
– Keilara
Oct 16 '16 at 21:29
$begingroup$
Why is it that you integrate (t-2) instead of t? isn't the lower bound (2) enough?
$endgroup$
– Keilara
Oct 16 '16 at 21:29
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
The 2 second delayed ramp function with slope 1 is given by
$$ (t - 2) u(t -2) $$
The Laplace transform of this is:
$$ mathcal{L} [f (t- 2) u(t-2)] = int_0^infty f(t-2)u(t-2)e^{-st} dt $$
This can be rewritten as the sum of two segments where the first segment is from 0 to 2 and has an integral of zero and the second segment is from 2 to infinity. Note the $u(t-2)$ no longer contributes anything above 2 (other than one) so we leave it out:
$$ = 0 + int_2^infty f(t-2), e^{-st} dt $$
This is where the mistake is, you can't integrate $t e^{-st}$ between 2 and infinity, you have to integrate $(t-2) e^{-st}$ between 2 and infinity.
To integrate this we make a change of variable were $tau = t - 2$, hence $t = tau + 2$ and $dt = dtau$. Note that the lower limit changes because when $t=2$, $tau = 2 - 2 = 0$, so that
$$ = int_0^infty f(tau) e^{-s(tau + 2)} dtau $$
$$ = e^{-2s} int_0^infty f(tau) e^{-stau} dtau $$
But the integral in the last expression is just $F(s)$ of the orginal function without a delay, therefore:
$$ = e^{-2s} F(s) $$
If $f(t)$ is just the unit slope then $F(s) = 1/s^2$, this leaves us with
$$ mathcal{L} [f (t- 2) u(t-2)] = frac{e^{-2s}}{s^2} $$
answered Jan 17 '17 at 2:40
rhodyrhody
1134
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
The 2 second delayed ramp function with slope 1 is given by
$$ (t - 2) u(t -2) $$
The Laplace transform of this is:
$$ mathcal{L} [f (t- 2) u(t-2)] = int_0^infty f(t-2)u(t-2)e^{-st} dt $$
This can be rewritten as the sum of two segments where the first segment is from 0 to 2 and has an integral of zero and the second segment is from 2 to infinity. Note the $u(t-2)$ no longer contributes anything above 2 (other than one) so we leave it out:
$$ = 0 + int_2^infty f(t-2), e^{-st} dt $$
This is where the mistake is, you can't integrate $t e^{-st}$ between 2 and infinity, you have to integrate $(t-2) e^{-st}$ between 2 and infinity.
To integrate this we make a change of variable were $tau = t - 2$, hence $t = tau + 2$ and $dt = dtau$. Note that the lower limit changes because when $t=2$, $tau = 2 - 2 = 0$, so that
$$ = int_0^infty f(tau) e^{-s(tau + 2)} dtau $$
$$ = e^{-2s} int_0^infty f(tau) e^{-stau} dtau $$
But the integral in the last expression is just $F(s)$ of the orginal function without a delay, therefore:
$$ = e^{-2s} F(s) $$
If $f(t)$ is just the unit slope then $F(s) = 1/s^2$, this leaves us with
$$ mathcal{L} [f (t- 2) u(t-2)] = frac{e^{-2s}}{s^2} $$
answered Jan 17 '17 at 2:40
rhodyrhody
1134
$endgroup$
add a comment |
$begingroup$
The 2 second delayed ramp function with slope 1 is given by
$$ (t - 2) u(t -2) $$
The Laplace transform of this is:
$$ mathcal{L} [f (t- 2) u(t-2)] = int_0^infty f(t-2)u(t-2)e^{-st} dt $$
This can be rewritten as the sum of two segments where the first segment is from 0 to 2 and has an integral of zero and the second segment is from 2 to infinity. Note the $u(t-2)$ no longer contributes anything above 2 (other than one) so we leave it out:
$$ = 0 + int_2^infty f(t-2), e^{-st} dt $$
This is where the mistake is, you can't integrate $t e^{-st}$ between 2 and infinity, you have to integrate $(t-2) e^{-st}$ between 2 and infinity.
To integrate this we make a change of variable were $tau = t - 2$, hence $t = tau + 2$ and $dt = dtau$. Note that the lower limit changes because when $t=2$, $tau = 2 - 2 = 0$, so that
$$ = int_0^infty f(tau) e^{-s(tau + 2)} dtau $$
$$ = e^{-2s} int_0^infty f(tau) e^{-stau} dtau $$
But the integral in the last expression is just $F(s)$ of the orginal function without a delay, therefore:
$$ = e^{-2s} F(s) $$
If $f(t)$ is just the unit slope then $F(s) = 1/s^2$, this leaves us with
$$ mathcal{L} [f (t- 2) u(t-2)] = frac{e^{-2s}}{s^2} $$
answered Jan 17 '17 at 2:40
rhodyrhody
1134
$endgroup$
add a comment |
$begingroup$
The 2 second delayed ramp function with slope 1 is given by
$$ (t - 2) u(t -2) $$
The Laplace transform of this is:
$$ mathcal{L} [f (t- 2) u(t-2)] = int_0^infty f(t-2)u(t-2)e^{-st} dt $$
This can be rewritten as the sum of two segments where the first segment is from 0 to 2 and has an integral of zero and the second segment is from 2 to infinity. Note the $u(t-2)$ no longer contributes anything above 2 (other than one) so we leave it out:
$$ = 0 + int_2^infty f(t-2), e^{-st} dt $$
This is where the mistake is, you can't integrate $t e^{-st}$ between 2 and infinity, you have to integrate $(t-2) e^{-st}$ between 2 and infinity.
To integrate this we make a change of variable were $tau = t - 2$, hence $t = tau + 2$ and $dt = dtau$. Note that the lower limit changes because when $t=2$, $tau = 2 - 2 = 0$, so that
$$ = int_0^infty f(tau) e^{-s(tau + 2)} dtau $$
$$ = e^{-2s} int_0^infty f(tau) e^{-stau} dtau $$
But the integral in the last expression is just $F(s)$ of the orginal function without a delay, therefore:
$$ = e^{-2s} F(s) $$
If $f(t)$ is just the unit slope then $F(s) = 1/s^2$, this leaves us with
$$ mathcal{L} [f (t- 2) u(t-2)] = frac{e^{-2s}}{s^2} $$
answered Jan 17 '17 at 2:40
rhodyrhody
1134
$endgroup$
The 2 second delayed ramp function with slope 1 is given by
$$ (t - 2) u(t -2) $$
The Laplace transform of this is:
$$ mathcal{L} [f (t- 2) u(t-2)] = int_0^infty f(t-2)u(t-2)e^{-st} dt $$
This can be rewritten as the sum of two segments where the first segment is from 0 to 2 and has an integral of zero and the second segment is from 2 to infinity. Note the $u(t-2)$ no longer contributes anything above 2 (other than one) so we leave it out:
$$ = 0 + int_2^infty f(t-2), e^{-st} dt $$
This is where the mistake is, you can't integrate $t e^{-st}$ between 2 and infinity, you have to integrate $(t-2) e^{-st}$ between 2 and infinity.
To integrate this we make a change of variable were $tau = t - 2$, hence $t = tau + 2$ and $dt = dtau$. Note that the lower limit changes because when $t=2$, $tau = 2 - 2 = 0$, so that
$$ = int_0^infty f(tau) e^{-s(tau + 2)} dtau $$
$$ = e^{-2s} int_0^infty f(tau) e^{-stau} dtau $$
But the integral in the last expression is just $F(s)$ of the orginal function without a delay, therefore:
$$ = e^{-2s} F(s) $$
If $f(t)$ is just the unit slope then $F(s) = 1/s^2$, this leaves us with
$$ mathcal{L} [f (t- 2) u(t-2)] = frac{e^{-2s}}{s^2} $$
answered Jan 17 '17 at 2:40
rhodyrhody
1134
answered Jan 17 '17 at 2:40
rhodyrhody
1134
answered Jan 17 '17 at 2:40
rhodyrhody
1134
answered Jan 17 '17 at 2:40
rhodyrhody
1134
1134
add a comment |
add a comment |
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2
$begingroup$
You're computing the wrong integral. Instead, you should be computing $$int_{2}^infty (t - 2) e^{-st} , mathrm{d}t $$
$endgroup$
– Rodrigo de Azevedo
Oct 16 '16 at 18:13
$begingroup$
Now you have changed from $ds$ to $dt$ which is it?
$endgroup$
– Olba12
Oct 16 '16 at 18:17
$begingroup$
$, mathrm{d}t$
$endgroup$
– Rodrigo de Azevedo
Oct 16 '16 at 18:18
$begingroup$
I saw your comment! @RodrigodeAzevedo
$endgroup$
– Olba12
Oct 16 '16 at 18:18
$begingroup$
Why is it that you integrate (t-2) instead of t? isn't the lower bound (2) enough?
$endgroup$
– Keilara
Oct 16 '16 at 21:29