Function modulation and dilatation in a Laplace transform
$begingroup$
My question is about something incomplete in my lesson. I learnt about the modulation and the dilatation of a function in the Laplace transform but what if I have both in the same time?
For the modulation, I have the following relation:
$$
mathcal{L}[e^{-gamma t} f(t)] = mathcal{L}[f](z + gamma)
$$
For the dilatation, I have the following relation:
$$
mathcal{L}[f(lambda t)] = frac{1}{lambda} mathcal{L}[f](frac{z}{lambda})
$$
What if I have the following Laplace transform : $mathcal{L}[e^{gamma t} f(frac{t}{lambda})]$
Is the answer $lambda mathcal{L}[f](lambda z - gamma)$ or $lambda mathcal{L}[f](lambda(z - gamma))$?
laplace-transform
asked Dec 11 '18 at 12:53
WizixWizix
1032
$endgroup$
add a comment |
$begingroup$
My question is about something incomplete in my lesson. I learnt about the modulation and the dilatation of a function in the Laplace transform but what if I have both in the same time?
For the modulation, I have the following relation:
$$
mathcal{L}[e^{-gamma t} f(t)] = mathcal{L}[f](z + gamma)
$$
For the dilatation, I have the following relation:
$$
mathcal{L}[f(lambda t)] = frac{1}{lambda} mathcal{L}[f](frac{z}{lambda})
$$
What if I have the following Laplace transform : $mathcal{L}[e^{gamma t} f(frac{t}{lambda})]$
Is the answer $lambda mathcal{L}[f](lambda z - gamma)$ or $lambda mathcal{L}[f](lambda(z - gamma))$?
laplace-transform
asked Dec 11 '18 at 12:53
WizixWizix
1032
$endgroup$
add a comment |
$begingroup$
My question is about something incomplete in my lesson. I learnt about the modulation and the dilatation of a function in the Laplace transform but what if I have both in the same time?
For the modulation, I have the following relation:
$$
mathcal{L}[e^{-gamma t} f(t)] = mathcal{L}[f](z + gamma)
$$
For the dilatation, I have the following relation:
$$
mathcal{L}[f(lambda t)] = frac{1}{lambda} mathcal{L}[f](frac{z}{lambda})
$$
What if I have the following Laplace transform : $mathcal{L}[e^{gamma t} f(frac{t}{lambda})]$
Is the answer $lambda mathcal{L}[f](lambda z - gamma)$ or $lambda mathcal{L}[f](lambda(z - gamma))$?
laplace-transform
asked Dec 11 '18 at 12:53
WizixWizix
1032
$endgroup$
My question is about something incomplete in my lesson. I learnt about the modulation and the dilatation of a function in the Laplace transform but what if I have both in the same time?
For the modulation, I have the following relation:
$$
mathcal{L}[e^{-gamma t} f(t)] = mathcal{L}[f](z + gamma)
$$
For the dilatation, I have the following relation:
$$
mathcal{L}[f(lambda t)] = frac{1}{lambda} mathcal{L}[f](frac{z}{lambda})
$$
What if I have the following Laplace transform : $mathcal{L}[e^{gamma t} f(frac{t}{lambda})]$
Is the answer $lambda mathcal{L}[f](lambda z - gamma)$ or $lambda mathcal{L}[f](lambda(z - gamma))$?
laplace-transform
laplace-transform
asked Dec 11 '18 at 12:53
WizixWizix
1032
asked Dec 11 '18 at 12:53
WizixWizix
1032
asked Dec 11 '18 at 12:53
WizixWizix
1032
asked Dec 11 '18 at 12:53
WizixWizix
1032
asked Dec 11 '18 at 12:53
WizixWizix
1032
1032
add a comment |
add a comment |
1 Answer
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$begingroup$
Define modulation and dilation operators: $M_{gamma}f(t) = e^{-gamma t} f(t)$ and $D_{lambda}f(t) = f(lambda t)$. Then $t mapsto e^{gamma t} f(t/lambda)$ can be expressed as either $M_{-gamma} (D_{1/lambda} f)$ (dilation then modulation) or $D_{1/lambda}(M_{-gamma lambda} f)$ (modulation then dilation). Using either form, we obtain the same result.
$$mathcal L[M_{-gamma} (D_{1/lambda} f)](z) = mathcal L[D_{1/lambda} f](z gamma) = lambda mathcal L[f] (lambda(z - gamma))$$
$$mathcal L[D_{1/lambda}(M_{-gamma lambda} f)](z) = lambda mathcal L[M_{-gamma lambda} f](lambda z) = lambda mathcal L[f] (lambda (z - gamma))$$
answered Dec 11 '18 at 13:20
Roberto RastapopoulosRoberto Rastapopoulos
950425
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add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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active
oldest
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active
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$begingroup$
Define modulation and dilation operators: $M_{gamma}f(t) = e^{-gamma t} f(t)$ and $D_{lambda}f(t) = f(lambda t)$. Then $t mapsto e^{gamma t} f(t/lambda)$ can be expressed as either $M_{-gamma} (D_{1/lambda} f)$ (dilation then modulation) or $D_{1/lambda}(M_{-gamma lambda} f)$ (modulation then dilation). Using either form, we obtain the same result.
$$mathcal L[M_{-gamma} (D_{1/lambda} f)](z) = mathcal L[D_{1/lambda} f](z gamma) = lambda mathcal L[f] (lambda(z - gamma))$$
$$mathcal L[D_{1/lambda}(M_{-gamma lambda} f)](z) = lambda mathcal L[M_{-gamma lambda} f](lambda z) = lambda mathcal L[f] (lambda (z - gamma))$$
answered Dec 11 '18 at 13:20
Roberto RastapopoulosRoberto Rastapopoulos
950425
$endgroup$
add a comment |
$begingroup$
Define modulation and dilation operators: $M_{gamma}f(t) = e^{-gamma t} f(t)$ and $D_{lambda}f(t) = f(lambda t)$. Then $t mapsto e^{gamma t} f(t/lambda)$ can be expressed as either $M_{-gamma} (D_{1/lambda} f)$ (dilation then modulation) or $D_{1/lambda}(M_{-gamma lambda} f)$ (modulation then dilation). Using either form, we obtain the same result.
$$mathcal L[M_{-gamma} (D_{1/lambda} f)](z) = mathcal L[D_{1/lambda} f](z gamma) = lambda mathcal L[f] (lambda(z - gamma))$$
$$mathcal L[D_{1/lambda}(M_{-gamma lambda} f)](z) = lambda mathcal L[M_{-gamma lambda} f](lambda z) = lambda mathcal L[f] (lambda (z - gamma))$$
answered Dec 11 '18 at 13:20
Roberto RastapopoulosRoberto Rastapopoulos
950425
$endgroup$
add a comment |
$begingroup$
Define modulation and dilation operators: $M_{gamma}f(t) = e^{-gamma t} f(t)$ and $D_{lambda}f(t) = f(lambda t)$. Then $t mapsto e^{gamma t} f(t/lambda)$ can be expressed as either $M_{-gamma} (D_{1/lambda} f)$ (dilation then modulation) or $D_{1/lambda}(M_{-gamma lambda} f)$ (modulation then dilation). Using either form, we obtain the same result.
$$mathcal L[M_{-gamma} (D_{1/lambda} f)](z) = mathcal L[D_{1/lambda} f](z gamma) = lambda mathcal L[f] (lambda(z - gamma))$$
$$mathcal L[D_{1/lambda}(M_{-gamma lambda} f)](z) = lambda mathcal L[M_{-gamma lambda} f](lambda z) = lambda mathcal L[f] (lambda (z - gamma))$$
answered Dec 11 '18 at 13:20
Roberto RastapopoulosRoberto Rastapopoulos
950425
$endgroup$
Define modulation and dilation operators: $M_{gamma}f(t) = e^{-gamma t} f(t)$ and $D_{lambda}f(t) = f(lambda t)$. Then $t mapsto e^{gamma t} f(t/lambda)$ can be expressed as either $M_{-gamma} (D_{1/lambda} f)$ (dilation then modulation) or $D_{1/lambda}(M_{-gamma lambda} f)$ (modulation then dilation). Using either form, we obtain the same result.
$$mathcal L[M_{-gamma} (D_{1/lambda} f)](z) = mathcal L[D_{1/lambda} f](z gamma) = lambda mathcal L[f] (lambda(z - gamma))$$
$$mathcal L[D_{1/lambda}(M_{-gamma lambda} f)](z) = lambda mathcal L[M_{-gamma lambda} f](lambda z) = lambda mathcal L[f] (lambda (z - gamma))$$
answered Dec 11 '18 at 13:20
Roberto RastapopoulosRoberto Rastapopoulos
950425
edited Dec 11 '18 at 14:08
answered Dec 11 '18 at 13:20
Roberto RastapopoulosRoberto Rastapopoulos
950425
answered Dec 11 '18 at 13:20
Roberto RastapopoulosRoberto Rastapopoulos
950425
answered Dec 11 '18 at 13:20
Roberto RastapopoulosRoberto Rastapopoulos
950425
950425
add a comment |
add a comment |
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