Laplace transforms

$f(t)$	$F(s)$
$e^{at}$	$\frac{1}{s-a}$
$t^n$	$\frac{n!}{s^{n+1}}$
$\cos (at)$	$\frac{s}{s^2+a^2}$
$\sin(at)$	$\frac{a}{s^2+a^2}$
$u_c(t)$	$\frac{e^{-cs}}{s}$
$\delta(t-c)$	$e^{-cs}$

$$\mathcal{L}(c_1f_1+c_2f_2)=c_1F_1(s)+c_2F_2(s)$$ $$ \mathcal{L}(f')=s F(s)-f(0)$$ $$ \mathcal{L}(f'')=s^2 F(s)-s f(0)-f'(0)$$ $$ \mathcal{L}(f(t-c)u_c(t))=e^{-cs} F(s)$$

Notation $F=\mathcal{L}(f)$ means $F(s)=\int_0^\infty e^{-st}f(t)dt$. Heaviside function $u_c(t)=\begin{cases}0 & t\in[0,c) \\ 1& t\geq c\end{cases}$

---

Solve the differential equation $y''+4y=4u_{\pi}(t),\; y(0)=1, y'(0)=0$.

1. $y(t)=\cos (2t) + u_\pi(t)(1-\cos (2t))$
2. $y(t)=\cos (2t) + u_\pi(t)(1-\cos (2t-\pi))$
3. $y(t)=\sin (2t) + u_\pi(t)(1-\sin (2t))$
4. $y(t)=\sin (2t) + u_\pi(t)(1-\sin (2t-\pi))$
5. None of the above

Show Solution

---

Answer: A.
Solution The Laplace transform is $Y(s)=\frac{s}{s^2+4}+\frac{4e^{-\pi s}}{s(s^2+4)}= \frac{s}{s^2+4}+ e^{-\pi s}\left(\frac{1}{s}-\frac{s}{s^2+4}\right) $. Therefore the solution is $y(t)=\cos 2t + u_\pi(t)(1-\cos 2(t-\pi))=\cos (2t) + u_\pi(t)(1-\cos (2t))$. Thus $$y(t)=\begin{cases} \cos 2 t & t<\pi \\ 1& t>\pi \end{cases}.$$

Another one?

---

Solve $$y''+y= u_{\pi/2}(t)-u_{\pi}(t), \quad y(0)=0,y'(0)=1$$

1. $y=\sin t +(1- \cos(t) )u_{\pi/2}(t)-(1-\cos (t))u_\pi(t)$
2. $y=\cos t +(1- \cos(t) )u_{\pi/2}(t)-(1+\sin (t))u_\pi(t)$
3. $y=\sin t +(1- \sin(t) )u_{\pi/2}(t)-(1+\cos (t))u_\pi(t)$
4. $y=\sin t +\cos(t-\pi/2) u_{\pi/2}(t)-\cos (t-\pi)u_\pi(t)$
5. None of the above

Show Solution

---

Answer: C.
Solution $s^2 Y(s)-1+Y(s)=\frac{e^{-\pi/2 s}}{s}-\frac{e^{-\pi s}}{s} $ so $Y(s)=\frac{1}{s^2+1}+ \frac{e^{-\pi/2 s}}{s(s^2+1)}-\frac{e^{-\pi s}}{s(s^2+1)} $ Noting that $\frac{1}{s(s^2+1)}=\frac{1}{s}-\frac{s}{s^2+1}=\mathcal{L}(1-\cos t)$ we get $\begin{multline*}y(t)=\sin t +(1- \cos(t-\pi/2) )u_{\pi/2}(t)-(1-\cos (t-\pi))u_\pi(t) \\=\sin t +(1- \sin(t) )u_{\pi/2}(t)-(1+\cos (t))u_\pi(t) =\begin{cases} \sin t & t\in(0,\pi/2]\\ 1 & t\in (\pi/2,\pi]\\ -\cos t & t>\pi \end{cases}\end{multline*}$
Another one?

---

Express $$f(t)=\begin{cases} 0 & t\in[0,1) \\ \\ 3 & t\in[1,2)\\ \\ 0 & t\geq 2 \end{cases}$$ in terms of functions $u_c$.

1. $f(t)= u_1(t)+2u_2(t)$
2. $f(t)=2 u_1(t)-u_2(t)$
3. $f(t)= 3 u_2(t)-3 u_1(t)$
4. None of the above

Show Solution

---

Answer: C.
Solution $f(t)=3(u_1-u_2)$
One more?

---

Determine the Laplace transform of $ (1-t)u_1(t)+ 2(t-2) u_2(t)+(3-t)u_3(t)$

1. $\frac{e^{-s}+2 e^{-2s}+e^{-3s}}{s}$
2. $\frac{-e^{-s}+2 e^{-2s}-e^{-3s}}{s^2}$
3. $\frac{e^{-s}+2 e^{-2s}+e^{-3s}}{s^2}$
4. None of these

Show Solution

---

Answer: B.
Solution
We need to re-write the function as $-(t-1)u_1(t)+2(t-2)u_2(t)-(t-3)u_3(t)$
One more?

---

Determine the Laplace transform of $ t u_1(t)+ 2 t u_2(t)$

1. $\frac{e^{-s}+2e^{-2s}}{s^2}+\frac{e^{-s}+4 e^{-2s}}{s}$
2. $\frac{e^{-s}+2e^{-2s}}{s^2}$
3. $\frac{e^{-s}+2 e^{-2s}}{s}$
4. $\frac{e^{-s}+2e^{-2s}}{s^2+s}$
5. None of these

Show Solution

---

Answer: A.
Solution We need to rewrite the function: $ t u_1(t)+ 2 t u_2(t)=(t-1)u_1(t)+2(t-2)u_2(t)+u_1(t)+4 u_2(t)$
One more?

---

Determine the Laplace transform of $(1-u_\pi(t))\sin t $

1. $ \frac{e^{-\pi s}}{s^2+1}$
2. $ \frac{1-u_\pi(s)}{s^2+1}$
3. $\frac{1-e^{-\pi s}}{s^2+1}$
4. $\frac{1+e^{-\pi s}}{s^2+1}$
5. None of these

Show Solution

---

Answer: D.
Solution The function is $\sin t + \sin(t-\pi)u_\pi(t)$
One more?