Properties of the Shifting Theorem The Next CEO of Stack OverflowFind the Laplace transform of $f(t) = begincases 0, & textif $t<5$ \ t^2−10t+31, & textif $tge 5$ \ endcases $How do I go about performing the following Laplace transform?Heaviside function in the function whose Laplace transformation is $e^-(gamma+s)/[(s+gamma)^2+b^2]$Finding Laplace Transform of $(t-4)h(t-4)$Solve differential equation using Laplace Transform and Second Shifting TheoremLaplace transform second shifting ruleConvolution of $te^2t$ and $delta_1-delta_2$?Bilateral Laplace transformation of $sin(t)*(H(t)-H(t -pi))$Find the inverse Laplace transform of $e^-3s frac 3s+1s^2-s-6$Find the solution of $x''-x=f(t)$ using Heaviside Functions
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Properties of the Shifting Theorem
The Next CEO of Stack OverflowFind the Laplace transform of $f(t) = begincases 0, & textif $t<5$ \ t^2−10t+31, & textif $tge 5$ \ endcases $How do I go about performing the following Laplace transform?Heaviside function in the function whose Laplace transformation is $e^-(gamma+s)/[(s+gamma)^2+b^2]$Finding Laplace Transform of $(t-4)h(t-4)$Solve differential equation using Laplace Transform and Second Shifting TheoremLaplace transform second shifting ruleConvolution of $te^2t$ and $delta_1-delta_2$?Bilateral Laplace transformation of $sin(t)*(H(t)-H(t -pi))$Find the inverse Laplace transform of $e^-3s frac 3s+1s^2-s-6$Find the solution of $x''-x=f(t)$ using Heaviside Functions
$begingroup$
The Shifting Theorem states that
$$textIf H(t)=begincases
0 & t<0 \
1 & t>0 \
endcases, textthen mathcalL(f(t-c)H(t-c))=e^-csF(s).$$
Does this mean that
$$mathcalL((-1+t)H(t-1))equiv -mathcalL((t-1)H(t-1))?$$ I'm unsure if the Heaviside function, $H(t-1)$, is affected by this change in sign.
laplace-transform
asked Mar 20 at 9:09
BettyBetty
636
$endgroup$
|
show 1 more comment
$begingroup$
The Shifting Theorem states that
$$textIf H(t)=begincases
0 & t<0 \
1 & t>0 \
endcases, textthen mathcalL(f(t-c)H(t-c))=e^-csF(s).$$
Does this mean that
$$mathcalL((-1+t)H(t-1))equiv -mathcalL((t-1)H(t-1))?$$ I'm unsure if the Heaviside function, $H(t-1)$, is affected by this change in sign.
laplace-transform
asked Mar 20 at 9:09
BettyBetty
636
$endgroup$
$begingroup$
Your last equation says $f(t)equiv -f(t)$, where $f(t):= (t-1)H(t-1)$.
$endgroup$
– Minus One-Twelfth
Mar 20 at 9:13
$begingroup$
@Minus One-Twelfth did my edit clear things up?
$endgroup$
– Betty
Mar 20 at 9:13
$begingroup$
$newcommandLmathcalL$Yes, it did. In that case though, you can note that $L(-f(t))=-L(f(t))$ just by linearity of the Laplace transform -- no need for the sifting theorem.
$endgroup$
– Minus One-Twelfth
Mar 20 at 9:17
$begingroup$
Yes of course. I just wasn't sure if the Heaviside function was affected. So my statement is true?
$endgroup$
– Betty
Mar 20 at 9:18
$begingroup$
Should be all good. If in doubt, you could always write down the integral expressions for both Laplace transforms.
$endgroup$
– Minus One-Twelfth
Mar 20 at 9:21
|
show 1 more comment
$begingroup$
The Shifting Theorem states that
$$textIf H(t)=begincases
0 & t<0 \
1 & t>0 \
endcases, textthen mathcalL(f(t-c)H(t-c))=e^-csF(s).$$
Does this mean that
$$mathcalL((-1+t)H(t-1))equiv -mathcalL((t-1)H(t-1))?$$ I'm unsure if the Heaviside function, $H(t-1)$, is affected by this change in sign.
laplace-transform
asked Mar 20 at 9:09
BettyBetty
636
$endgroup$
The Shifting Theorem states that
$$textIf H(t)=begincases
0 & t<0 \
1 & t>0 \
endcases, textthen mathcalL(f(t-c)H(t-c))=e^-csF(s).$$
Does this mean that
$$mathcalL((-1+t)H(t-1))equiv -mathcalL((t-1)H(t-1))?$$ I'm unsure if the Heaviside function, $H(t-1)$, is affected by this change in sign.
laplace-transform
laplace-transform
asked Mar 20 at 9:09
BettyBetty
636
asked Mar 20 at 9:09
BettyBetty
636
asked Mar 20 at 9:09
BettyBetty
636
asked Mar 20 at 9:09
BettyBetty
636
asked Mar 20 at 9:09
BettyBetty
636
636
$begingroup$
Your last equation says $f(t)equiv -f(t)$, where $f(t):= (t-1)H(t-1)$.
$endgroup$
– Minus One-Twelfth
Mar 20 at 9:13
$begingroup$
@Minus One-Twelfth did my edit clear things up?
$endgroup$
– Betty
Mar 20 at 9:13
$begingroup$
$newcommandLmathcalL$Yes, it did. In that case though, you can note that $L(-f(t))=-L(f(t))$ just by linearity of the Laplace transform -- no need for the sifting theorem.
$endgroup$
– Minus One-Twelfth
Mar 20 at 9:17
$begingroup$
Yes of course. I just wasn't sure if the Heaviside function was affected. So my statement is true?
$endgroup$
– Betty
Mar 20 at 9:18
$begingroup$
Should be all good. If in doubt, you could always write down the integral expressions for both Laplace transforms.
$endgroup$
– Minus One-Twelfth
Mar 20 at 9:21
|
show 1 more comment
$begingroup$
Your last equation says $f(t)equiv -f(t)$, where $f(t):= (t-1)H(t-1)$.
$endgroup$
– Minus One-Twelfth
Mar 20 at 9:13
$begingroup$
@Minus One-Twelfth did my edit clear things up?
$endgroup$
– Betty
Mar 20 at 9:13
$begingroup$
$newcommandLmathcalL$Yes, it did. In that case though, you can note that $L(-f(t))=-L(f(t))$ just by linearity of the Laplace transform -- no need for the sifting theorem.
$endgroup$
– Minus One-Twelfth
Mar 20 at 9:17
$begingroup$
Yes of course. I just wasn't sure if the Heaviside function was affected. So my statement is true?
$endgroup$
– Betty
Mar 20 at 9:18
$begingroup$
Should be all good. If in doubt, you could always write down the integral expressions for both Laplace transforms.
$endgroup$
– Minus One-Twelfth
Mar 20 at 9:21
$begingroup$
Your last equation says $f(t)equiv -f(t)$, where $f(t):= (t-1)H(t-1)$.
$endgroup$
– Minus One-Twelfth
Mar 20 at 9:13
$begingroup$
Your last equation says $f(t)equiv -f(t)$, where $f(t):= (t-1)H(t-1)$.
$endgroup$
– Minus One-Twelfth
Mar 20 at 9:13
$begingroup$
@Minus One-Twelfth did my edit clear things up?
$endgroup$
– Betty
Mar 20 at 9:13
$begingroup$
@Minus One-Twelfth did my edit clear things up?
$endgroup$
– Betty
Mar 20 at 9:13
$begingroup$
$newcommandLmathcalL$Yes, it did. In that case though, you can note that $L(-f(t))=-L(f(t))$ just by linearity of the Laplace transform -- no need for the sifting theorem.
$endgroup$
– Minus One-Twelfth
Mar 20 at 9:17
$begingroup$
$newcommandLmathcalL$Yes, it did. In that case though, you can note that $L(-f(t))=-L(f(t))$ just by linearity of the Laplace transform -- no need for the sifting theorem.
$endgroup$
– Minus One-Twelfth
Mar 20 at 9:17
$begingroup$
Yes of course. I just wasn't sure if the Heaviside function was affected. So my statement is true?
$endgroup$
– Betty
Mar 20 at 9:18
$begingroup$
Yes of course. I just wasn't sure if the Heaviside function was affected. So my statement is true?
$endgroup$
– Betty
Mar 20 at 9:18
$begingroup$
Should be all good. If in doubt, you could always write down the integral expressions for both Laplace transforms.
$endgroup$
– Minus One-Twelfth
Mar 20 at 9:21
$begingroup$
Should be all good. If in doubt, you could always write down the integral expressions for both Laplace transforms.
$endgroup$
– Minus One-Twelfth
Mar 20 at 9:21
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
$newcommandLmathcalL$Yes, the last equation is correct, since $L(-f(t))=-L(f(t))$, just using linearity of the Laplace transform.
answered Mar 20 at 9:27
Minus One-TwelfthMinus One-Twelfth
2,933413
$endgroup$
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$begingroup$
$newcommandLmathcalL$Yes, the last equation is correct, since $L(-f(t))=-L(f(t))$, just using linearity of the Laplace transform.
answered Mar 20 at 9:27
Minus One-TwelfthMinus One-Twelfth
2,933413
$endgroup$
add a comment |
$begingroup$
$newcommandLmathcalL$Yes, the last equation is correct, since $L(-f(t))=-L(f(t))$, just using linearity of the Laplace transform.
answered Mar 20 at 9:27
Minus One-TwelfthMinus One-Twelfth
2,933413
$endgroup$
add a comment |
$begingroup$
$newcommandLmathcalL$Yes, the last equation is correct, since $L(-f(t))=-L(f(t))$, just using linearity of the Laplace transform.
answered Mar 20 at 9:27
Minus One-TwelfthMinus One-Twelfth
2,933413
$endgroup$
$newcommandLmathcalL$Yes, the last equation is correct, since $L(-f(t))=-L(f(t))$, just using linearity of the Laplace transform.
answered Mar 20 at 9:27
Minus One-TwelfthMinus One-Twelfth
2,933413
answered Mar 20 at 9:27
Minus One-TwelfthMinus One-Twelfth
2,933413
answered Mar 20 at 9:27
Minus One-TwelfthMinus One-Twelfth
2,933413
answered Mar 20 at 9:27
Minus One-TwelfthMinus One-Twelfth
2,933413
2,933413
add a comment |
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$begingroup$
Your last equation says $f(t)equiv -f(t)$, where $f(t):= (t-1)H(t-1)$.
$endgroup$
– Minus One-Twelfth
Mar 20 at 9:13
$begingroup$
@Minus One-Twelfth did my edit clear things up?
$endgroup$
– Betty
Mar 20 at 9:13
$begingroup$
$newcommandLmathcalL$Yes, it did. In that case though, you can note that $L(-f(t))=-L(f(t))$ just by linearity of the Laplace transform -- no need for the sifting theorem.
$endgroup$
– Minus One-Twelfth
Mar 20 at 9:17
$begingroup$
Yes of course. I just wasn't sure if the Heaviside function was affected. So my statement is true?
$endgroup$
– Betty
Mar 20 at 9:18
$begingroup$
Should be all good. If in doubt, you could always write down the integral expressions for both Laplace transforms.
$endgroup$
– Minus One-Twelfth
Mar 20 at 9:21