Example. Restore the original $f(t)$ from its image

$$
F(p)=\frac{p^{2}+p+1}{(p-1)(p+1)^{2}}
$$

using the decomposition theorem.
Solution.

1. Find the singular points of the function

$$
F_{1}(p)=e^{p t} F(p)=e^{p t} \frac{p^{2}+p+1}{(p-1)(p+1)^{2}}
$$

$F_{1}(p)$ has two poles: $p=1-$ a pole of the 1st order and $p=-1-$ a pole of the 2nd order.

2. Calculate the residues at these points

$$
\operatorname{res}_{p=1} \frac{\left(p^{2}+p+1\right) e^{p t}}{(p-1)(p+1)^{2}}=\lim _{p \rightarrow 1} \frac{\left(p^{2}+p+1\right)(p-1) e^{p t}}{(p-1)(p+1)^{2}}=\frac{3}{4} e^{t}
$$

$$
\begin{aligned}
& \operatorname{res}_{p=-1} \frac{\left(p^{2}+p+1\right) e^{p t}}{(p-1)(p+1)^{2}}=\lim _{p \rightarrow-1} \frac{d}{d p}\left[\frac{\left(p^{2}+p+1\right)(p+1)^{2} e^{p t}}{(p-1)(p+1)^{2}}\right]= \\
& =\lim _{p \rightarrow-1} \frac{\left[(2 p+1) e^{p t}+t e^{p t}\left(p^{2}+p+1\right)\right](p-1)-\left(p^{2}+p+1\right) e^{p t}}{(p-1)^{2}}= \\
& =\frac{1}{4} e^{-t}-\frac{1}{2} t e^{-t}
\end{aligned}
$$

3. Calculate the original $f(t)$ using formula (1)

$$
\begin{aligned}
f(t)=\operatorname{res}_{p=1} \frac{\left(p^{2}+p+1\right) e^{p t}}{(p-1)(p+1)^{2}}+\operatorname{res}_{p=-1} \frac{\left(p^{2}+p+1\right) e^{p t}}{(p-1)(p+1)^{2}} & = \\
= & \frac{3}{4} e^{t}+\frac{1}{4} e^{-t}-\frac{1}{2} t e^{-t}
\end{aligned}
$$

Answer. $f(t)=\frac{3}{4} e^{t}+\frac{1}{4} e^{-t}-\frac{1}{2} t e^{-t}$.

Conditions of the Problems. Restore the original $f(t)$ from the image $F(p)$ using the expansion theorem.

1. $F(p)=\frac{1}{(p-1)^{3}}$.
2. $F(p)=\frac{p+1}{p(p-1)(p-2)(p-3)}$.
3. $F(p)=\frac{1}{(p-1)\left(p^{2}-4\right)}$.
4. $F(p)=\frac{p+3}{p^{3}-4 p^{2}+3 p}$.
5. $F(p)=\frac{1}{p\left(p^{2}+1\right)\left(p^{2}+4\right)}$.
6. $F(p)=\frac{p}{(p+1)(p+2)(p+3)(p+4)}$.
7. $F(p)=\frac{4-p-p^{2}}{p^{3}-p^{2}}$.
8. $F(p)=\frac{1}{p^{4}-6 p^{3}+11 p-6 p}$.
9. $F(p)=\frac{1}{p^{2}-2 p-3}$.
10. $F(p)=\frac{1}{p^{3}+2 p^{2}+p}$.

Answers. 1. $f(t)=t^{2} e^{t} / 2$. 2. $f(t)=-1 / 6+e^{t}-3 e^{2 t} / 2+2 e^{3 t} / 3$. 3. $f(t)=-e^{t} / 3+e^{2 t} / 4+e^{-2 t} / 12$. 4. $f(t)=1-2 e^{t}+e^{3 t}$. 5. $f(t)=(3-4 \cos t+\cos 2 t) / 12$. 6. $f(t)=-1 / 6-3 e^{-3 t} / 2+2 e^{-4 t} / 3+e^{-2 t}$. 7. $f(t)=2 e^{t}-4 t-3$. 8. $f(t)=-1 / 6+e^{t} / 2-e^{2 t} / 2+e^{3 t} / 6$. 9. $f(t)=\left(e^{3 t}-e^{-t}\right) / 4$. 10. $f(t)=1-e^{-t}-t e^{-t}$.

### 2.10. Restoration of the Original from the Image $F(p) \cdot G(p)$

Problem Statement. Restore the original from the image $F(p) \cdot G(p)$, if $F(p)$ is the image of the function $f(t)$ and $G(p)$ is the image of the function $g(t)$.

Solution Plan.

If $F(p)$ is the image of the original $f(t)$ and $G(p)$ is the image of the original $g(t)$, then by the multiplication theorem of images (convolution theorem)

$$
f * g \longleftrightarrow F(p) \cdot G(p)
$$

where the original

$$
f * g=\int_{0}^{\tau} f(\tau) \cdot g(t-\tau) d \tau
$$

is called the convolution of the originals $f(t)$ and $g(t)$.

1. Restore the originals $f(t)$ and $g(t)$ from their images $F(p)$ and $G(p)$.
2. Compute the convolution

$$
f * g=\int_{0}^{\tau} f(\tau) \cdot g(t-\tau) d \tau
$$

3. Write the answer using formula (1).