Question

Irreducible Quadratics II (4) Consider the Laplace transform pair $\cos t \leftrightarrow \frac{s}{s^2 + 1}$. (a) Use the dilation principle to show that $\cos(\omega t) \leftrightarrow \frac{s}{s^2 + \omega^2}$. (b) Use Part (a) and the output translation principle to show $e^{at}\cos(\omega t) \leftrightarrow \frac{s - a}{(s - a)^2 + \omega^2}$. (c) Use Part (b) and linearity to find $\mathcal{L}^{-1} \left(\frac{2s + 4}{s^2 + 4s + 20}\right)$.

          Irreducible Quadratics II
(4) Consider the Laplace transform pair
$\cos t \leftrightarrow \frac{s}{s^2 + 1}$.
(a) Use the dilation principle to show that $\cos(\omega t) \leftrightarrow \frac{s}{s^2 + \omega^2}$.
(b) Use Part (a) and the output translation principle to show $e^{at}\cos(\omega t) \leftrightarrow \frac{s - a}{(s - a)^2 + \omega^2}$.
(c) Use Part (b) and linearity to find $\mathcal{L}^{-1} \left(\frac{2s + 4}{s^2 + 4s + 20}\right)$.

Irreducible Quadratics II
(4) Consider the Laplace transform pair
cos t ↔(s)/(s^2 + 1).
(a) Use the dilation principle to show that cos(ω t) ↔(s)/(s^2 + ω^2).
(b) Use Part (a) and the output translation principle to show e^atcos(ω t) ↔(s - a)/((s - a)^2 + ω^2).
(c) Use Part (b) and linearity to find ℒ^-1((2s + 4)/(s^2 + 4s + 20)).

Added by Ricardo S.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Charles Machakwa

Step 1

The dilation principle states that if F(s) is the Laplace transform of f(t), then F(as) is the Laplace transform of f(at), where a is a positive constant. In this case, we want to show that cos(wt) is the Laplace transform of cos(2t+1). To do this, we need to Show more…

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Reducible Quadratics Consider the Laplace transform pair: cos(2t+1) a) Use the dilation principle to show that cos(wt) is the Laplace transform of cos(2t+1).