Laplace transforms

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

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

Notation

• $F=\mathcal{L}(f)$ means $F(s)=\int_0^\infty e^{-st}f(t)dt$
• Heaviside function $u_c(t)=\begin{cases}0 & t\in[0,c) \\ 1& t\geq c\end{cases}$
• Dirac delta $\delta(t-c)$
• Convolution of functions: $h=f*g$ means that $h(t)=\int_0^t f(t-u)g(u)du$

.

---

Question Write the solution using convolution integral. $$y''+y= g(t), \quad y(0)=1,y'(0)=1$$

1. $y(t)=\int_0^t( \cos \tau +\sin \tau) g(t-\tau)d\tau$
2. $y(t)=\cos t +\sin t +\int_0^t g(\tau) \sin(t-\tau)d\tau$
3. $y(t)=\sin t +\cos t+\int_0^t g(t-u) \cos(u)du$
4. $y(t)=\cos t +\int_0^t g(t-u) \sin(u)du $
5. None of these

Show Solution

---

Answer: B.
Solution $s^2 Y(s)-s-1+Y(s)=G(s)$ so $Y(s)=\frac{s}{s^2+1}+\frac{1}{s^2+1}+ \frac{1}{s^2+1}G(s) $ We get $y(t)=\cos(t) +\sin(t)+\int_0^t g(u) \sin(t-u)du= \cos t +\sin t+\int_0^t g(t-u) \sin(u)du$.
Another one?

---

Question If $y^{(4)}+5 y''+4y=g(t) $, $y(0)=0, y'(0)=0,y''(0)=0,y'''(0)=0$ then

1. $y(t)=\frac16\int_0^t \left(2\sin u-\sin(2u)\right) g(t-u) du$
2. $y(t)= \int_0^t g(u)\left(\sin (t)-\sin(u)\right) du$
3. $y(t)= \int_0^t g(t-u)\left(\sin (u)\sin(2 u)\right) g(t-u) du$
4. $y(t)=\frac13 \int_0^t g(u)\left(\sin (t-u)-\sin(2(t-u))\right) du$
5. None of these

Show Solution

---

Answer: A.
Solution $Y(s)=\frac{G(s)}{(s^2+1)(s^2+4)}=\frac13G(s)( \frac{1}{s^2+1}-\frac{1}{s^2+4})=\frac16 G(s) ( 2 \frac{1}{s^2+1}-\frac{2}{s^2+4})$
Another one?

---

Question If $y''-y=\sum_{k=0}^\infty \delta(t-k)$, $y(0)=0,y'(0)=0$ then

1. $y(t)=\int_0^t \sin u \sin (t-u)du$
2. $y(t)=\sum_{k=0}^\infty (e^{t-k}+e^{k-t})u_k(t)$
3. $y(t)=\frac{e^t+e^{-t}}{1-e^{-t}}$
4. $y(t)=\int_0^t \cos \tau \sin (t-\tau)d\tau$
5. None of these

Show Solution

---

Answer: E.
Solution $Y(s)=\sum_{k=0}^\infty \frac{1}{1-s^2}e^{-ks}=\frac12\sum_{k=0}^\infty (\frac{1}{1-s}+\frac{1}{1+s})e^{-ks} $ So $y(t)=\frac12\sum_{k=0}^\infty (e^{t-k}+e^{k-t})u_k(t)$. Answer B is missing $\frac12$
Another one?