Exercício 7.1. a) Use a igualdade (7.13) para calcular L[\operatorname{sen} \omega t] e L[\cos \omega t]. Sugestão: (\operatorname{sen} \omega t)^{\prime \prime}=-\omega^{2} \operatorname{sen} \omega t e(\cos \omega t)^{\prime \prime}=-\omega^{2} \cos \omega t. b) Calcule as seguintes transformadas:
i) L[7 \cos 3 t-4 \operatorname{sen} 3 t]
ii) L\left[5 e^{-3 t}\right]
c) (i) Mostre que, se f é T-periódica (isto é, f(t+T)=f(t), para todo t \geq 0 ), então L[f(t)]=\frac{1}{1-e^{-s T}} \int_{0}^{T} e^{-s t} f(t) d t.
(ii) Mostre que, se f é 2 -periódica e f(t)=\left\{\begin{aligned} 1, & \text { se } 0<t<1 \\\\ -1, & \text { se } 1<t<2,\end{aligned}\right. então L[f(t)]=\frac{1}{s} \operatorname{tg} \mathrm{~h} \frac{s}{2}.
iii) L\left[e^{-3 t} \operatorname{sen} 4 t\right]
iv) L\left[e^{-3 t} \cos 2 t\right]
v) L\left[t^{3} e^{-5 t}\right]
vi) L\left[e^{-2 t}-4 t^{3}\right]
vii) L[\operatorname{cosh} 2 t-3 \operatorname{senh} 2 t]
viii) L\left[t^{2} \cos t\right]
ix) L\left[t e^{3 t}-\operatorname{sen} 2 t\right]
x) L[4 \cos 2 t-t \cos 2 t]
xi) L\left[t e^{3 t} \operatorname{sen} 4 t\right]
xii) L\left[3 \cos 2 t-t e^{5 t}\right]
xiii) L[\operatorname{sen} t-t \cos t]
xiv) L\left[\left(e^{t}-2 t\right)^{2}\right]
xv) L\left[t e^{-3 t} \cos 4 t\right]
(iii) Mostre que, se f é 2 \pi-periódica e f(t)=\left\{\begin{array}{cl}\operatorname{sen} t, & \text { se } 0 \leq t \leq \pi \\\\ 0, & \text { se } \pi \leq t \leq 2 \pi,\end{array}\right. então L[f(t)]=\frac{1}{\left(1-e^{-\pi s}\right)\left(s^{2}+1\right)}\.

Passo 1

O enunciado pede várias transformadas de Laplace. Vamos usar:
  1. A igualdade (7.13), que provém de
L\{f''(t)\}(s)=s^{2}F(s)-s f(0)-f'(0),
aplicada a f(t)=\sin\omega t e f(t)=\cos\omega t.
2. A linearidade de L.
3. A transformada de e^{at}:
L\{e^{at}\}(s)=\frac{1}{s-a},
e de \sin bt,\cos bt.
4. A propriedade de funções periódicas.
5. As fórmulas de deslocamento em s para e^{-at}f(t).
6. A fórmula L\{t^n f(t)\}=(-1)^n \frac{d^n}{ds^n}F(s).

Passo 2

Calcular L\{\sin\omega t\} e L\{\cos\omega t\} usando f''(t)=-\omega^2 f(t).
Para f(t)=\sin\omega t, temos f(0)=0 e f'(0)=\omega. Aplicando
L\{f''\}=s^2F(s)-s f(0)-f'(0),
e sabendo f''=-\omega^2f, obtemos
-\omega^2F(s)=s^2F(s)-\omega \quad\Longrightarrow\quad F(s)=\frac{\omega}{s^2+\omega^2}.
Logo
L\{\sin\omega t\}(s)=\frac{\omega}{s^2+\omega^2}.
Para g(t)=\cos\omega t, g(0)=1, g'(0)=0, e g''=-\omega^2g. Assim
-\omega^2G(s)=s^2G(s)-s \quad\Longrightarrow\quad G(s)=\frac{s}{s^2+\omega^2}.
Portanto
L\{\cos\omega t\}(s)=\frac{s}{s^2+\omega^2}.

Passo 3

b.i) Usando linearidade e substituindo \omega=3:
L\{7\cos3t-4\sin3t\} =7\cdot\frac{s}{s^2+9}-4\cdot\frac{3}{s^2+9} =\frac{7s-12}{s^2+9}.
b.ii)
L\{5e^{-3t}\} =5\cdot\frac{1}{s+3} =\frac{5}{s+3}.

Passo 4

c.i) Se f é T-periódica,
L\{f\}(s) =\int_{0}^{\infty}e^{-st}f(t)\,dt =\sum_{k=0}^{\infty}\int_{kT}^{(k+1)T}e^{-st}f(t)\,dt.
Fazendo u=t-kT em cada integral e somando a série geométrica em e^{-sT}, obtém-se
L\{f\}(s) =\frac{1}{1-e^{-sT}}\int_{0}^{T}e^{-st}f(t)\,dt.

Passo 5

c.ii) Aqui T=2 e
f(t)=\begin{cases} 1,&0<t<1,\\\\ -1,&1<t<2, \end{cases}
logo
\int_{0}^{2}e^{-st}f(t)\,dt =\int_{0}^{1}e^{-st}dt-\int_{1}^{2}e^{-st}dt =\frac{1-e^{-s}}{s}-\frac{e^{-s}-e^{-2s}}{s} =\frac{1-2e^{-s}+e^{-2s}}{s}.
Aplicando a fórmula periódica:
L\{f\}(s) =\frac{1}{1-e^{-2s}}\cdot\frac{(1-e^{-s})^2}{s} =\frac{(1-e^{-s})^2}{s(1-e^{-s})(1+e^{-s})} =\frac{1-e^{-s}}{s(1+e^{-s})} =\frac{1}{s}\operatorname{tanh}\frac{s}{2}.

Passo 6

c.iii–xv) Usamos:
– Deslocamento: L\{e^{-at}f(t)\}(s)=F(s+a).
– Derivação em s: L\{t^n f(t)\}=(-1)^n\dfrac{d^n}{ds^n}F(s).
L\{\operatorname{sinh} at\}=\dfrac{a}{s^2-a^2}, L\{\operatorname{cosh} at\}=\dfrac{s}{s^2-a^2}.
Aplicando cada caso, encontra-se:
iii) L\{e^{-3t}\sin4t\}=\dfrac{4}{(s+3)^2+16}.
iv) L\{e^{-3t}\cos2t\}=\dfrac{s+3}{(s+3)^2+4}.
v) L\{t^3e^{-5t}\}=(-1)^3\frac{d^3}{ds^3}\frac{1}{s+5}=\frac{6}{(s+5)^4}.
vi) L\{e^{-2t}-4t^3\}=\frac{1}{s+2}-4\cdot\frac{6}{s^4}.
vii) L\{\cosh2t-3\sinh2t\}=\frac{s}{s^2-4}-3\cdot\frac{2}{s^2-4}=\frac{s-6}{s^2-4}.
viii) L\{t^2\cos t\}= \frac{d^2}{ds^2}\!\Bigl(\frac{s}{s^2+1}\Bigr) =\frac{2s(s^2-3)}{(s^2+1)^3}.
ix) L\{t e^{3t}-\sin2t\} =-\frac{d}{ds}\frac{1}{s-3}-\frac{2}{s^2+4} =\frac{1}{(s-3)^2}-\frac{2}{s^2+4}.
x) L\{4\cos2t-t\cos2t\} =4\frac{s}{s^2+4}-\bigl(-\frac{d}{ds}\tfrac{s}{s^2+4}\bigr) =\frac{4s(s^2+4)- (s^2-4)}{(s^2+4)^2}.
xi) L\{t e^{3t}\sin4t\} =-\frac{d}{ds}\!\Bigl(\tfrac{4}{(s-3)^2+16}\Bigr) =\frac{8(s-3)}{\bigl((s-3)^2+16\bigr)^2}.
xii) L\{3\cos2t-t e^{5t}\} =3\frac{s}{s^2+4}-\bigl(-\tfrac{d}{ds}\tfrac{1}{s-5}\bigr) =3\frac{s}{s^2+4}+\frac{1}{(s-5)^2}.
xiii) L\{\sin t - t\cos t\} =\frac{1}{s^2+1}-\bigl(-\tfrac{d}{ds}\tfrac{s}{s^2+1}\bigr) =\frac{1}{s^2+1}+\frac{s^2-1}{(s^2+1)^2}.
xiv) L\{(e^t-2t)^2\} =L\{e^{2t}-4te^t+4t^2\} =\frac{1}{s-2}-4\bigl(-\tfrac{d}{ds}\tfrac{1}{s-1}\bigr) +4\bigl(\tfrac{d^2}{ds^2}\tfrac{1}{s}\bigr).
xv) L\{t e^{-3t}\cos4t\} =-\frac{d}{ds}\!\Bigl(\tfrac{s+3}{(s+3)^2+16}\Bigr) =\frac{16-(s+3)^2}{\bigl((s+3)^2+16\bigr)^2}.

Passo 7

c.iii) Agora T=2\pi e
f(t)=\begin{cases} \sin t,&0\le t\le\pi,\\\\ 0,&\pi\le t\le2\pi. \end{cases}
Calculamos
\int_{0}^{2\pi}e^{-st}f(t)\,dt =\int_{0}^{\pi}e^{-st}\sin t\,dt =\frac{1+e^{-s\pi}(s\sin\pi-\cos\pi)}{s^2+1} =\frac{1+e^{-s\pi}}{s^2+1}.
Usando a fórmula periódica:
L\{f\}(s) =\frac{1}{1-e^{-2\pi s}}\cdot\frac{1+e^{-s\pi}}{s^2+1} =\frac{1}{(1-e^{-s\pi})(s^2+1)}.

Resposta

a)
\displaystyle L\{\sin\omega t\}=\frac{\omega}{s^2+\omega^2},\quad L\{\cos\omega t\}=\frac{s}{s^2+\omega^2}.
b.i) \displaystyle \frac{7s-12}{s^2+9}.
b.ii) \displaystyle \frac{5}{s+3}.
c.i) \displaystyle L\{f\}=\frac{1}{1-e^{-sT}}\int_{0}^{T}e^{-st}f(t)\,dt.
c.ii) \displaystyle \frac{1}{s}\operatorname{tanh}\frac{s}{2}.
c.iii) \displaystyle \frac{4}{(s+3)^2+16}.
iv) \displaystyle \frac{s+3}{(s+3)^2+4}.
v) \displaystyle \frac{6}{(s+5)^4}.
vi) \displaystyle \frac{1}{s+2}-\frac{24}{s^4}.
vii) \displaystyle \frac{s-6}{s^2-4}.
viii) \displaystyle \frac{2s(s^2-3)}{(s^2+1)^3}.
ix) \displaystyle \frac{1}{(s-3)^2}-\frac{2}{s^2+4}.
x) \displaystyle \frac{4s(s^2+4)-(s^2-4)}{(s^2+4)^2}.
xi) \displaystyle \frac{8(s-3)}{\bigl((s-3)^2+16\bigr)^2}.
xii) \displaystyle 3\frac{s}{s^2+4}+\frac{1}{(s-5)^2}.
xiii) \displaystyle \frac{1}{s^2+1}+\frac{s^2-1}{(s^2+1)^2}.
xiv) \displaystyle \frac{1}{s-2}+ \frac{4}{(s-1)^2}+ \frac{8}{s^3}.
xv) \displaystyle \frac{16-(s+3)^2}{\bigl((s+3)^2+16\bigr)^2}.
d) Para f de período 2\pi dado,
\displaystyle L\{f\}=\frac{1}{(1-e^{-\pi s})(s^2+1)}.

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