Laplace transform

Example

Show the Laplace transform of DE $y'' -y = e^{-t},\ y(0)=1,\ y'(0)=0$.

$$ s^2 Y -s - Y = \frac{1}{s+1} $$

Note: $Y= Y(s)$ is Laplace transform of $y(t)$.

Example

Show the Laplace transform of unit box $u_{ab}$

$$ u_{ab} = u(t-a) - u(t-b) \\ {\scr L}\left[u(t-a)\right] = e^{-as}\frac 1 s \\ {\scr L}\left[u(t-b)\right] = e^{-bs}\frac 1 s\\ \underline{\underline{{\scr L}\left[u_{ab}\right] = \frac{e^{-as}}{s} - \frac{e^{-bs}}{s}}}\\ $$

Example

Show the Laplace transform of $t^2$ when the function until $t=1$ is required to be erased to zero.

We have to erase whatever would be before $t=1$, i.e. the function $f(t)$ have to be multiplied by unit step function with the jump from 0 to 1 at $a=1$: $u_a(t) = u(t-1)$. In order to multiply we have to employ the rule $(\ref{ref:lapl_4})$ for multiplication of the function with function of the type unit step.

$$ {\scr L}\left[u(t-1)\ t^2\right] = e^{-as}{\scr L}\left[(t+1)^2\right] = e^{-as}{\scr L}\left[t^2 + 2t + 1\right] = \underline{\underline{e^{-s}\left(\frac {2}{s^3} + \frac{2}{s^2} + \frac{1}{s}\right) }} $$

Example

1. At $t=0$ the function looks like the very basic unit step function $(\ref{ref:lapl_unit_step})$. But unit function knows only about $0$ and $1$, here we have $f(t)=2$. That means we have to use $2u(t)$.
2. Then in time $t=2$ its value changes from $2$ to $-1$ (i.e. $3$ down at $t=2$) which means we have to add $-3u(t-2)$.
3. Finally the value at $t=3$ jumps $1$ higher, which brings member $u(t-3)$.

$$ f(t) = 2u(t) - 3u(t-2) + u(t-3) $$

So far we collected unit step functions to express function from the graph. Now we have to use the rule $(\ref{ref:lapl_3})$ to evaluate their Laplace transform.

$$ \begin{align} {\scr L}\left[f(t)\right] &= {\scr L}\left[2u(t)\right] - {\scr L}\left[3u(t-2)\right] + {\scr L}\left[u(t-3)\right] = \nonumber \\ &= \underline{\underline{\frac 2 s -3e^{-2s}\frac 1 s +e^{-3s}\frac 1 s}} \nonumber \end{align} $$

Example

Find $f(t)$ if $F(s) = e^{-2s}/(s^2+1)$.

We are given result in the domain $s$ and we have to use inverse Laplace transform. Since member $\boldsymbol{e^{-2s}}$ is present, it is known that either a shift or intervals are involved. In this case it can be identified that the shift was made to $a=2$. The other member $1/(s^2 + 1)$ is—according to the table of transforms—the product of $\sin kt$ when $k=1$.

So we have

$$ \underline{\underline{f(t) = u(t-2) \sin(t-2)}} \hskip2em \text{or} \hskip2em \underline{\underline{f(t)=u_2(t) \sin(t-2)}}. $$

It is $\sin()$ moved on $t$ axis from zero to $1$ and in such case the unit step function has to be involved to erase whatever was before the beginning.

Example

We are given result in the domain $s$ and we have to use inverse Laplace transform. Since member $\boldsymbol{e^{-\pi s}}$ is present, it is known that either a shift or interval is involved. In this case it can be identified that the shift was made to $a=\pi$. Now let us split the fraction into two and solve them separately.

$$ F(s) = \frac{1}{s^2+1} + \frac{e^{-\pi s}}{s^2+1} $$

It comes from observation and from the table of transform that the first fraction is a product of $\sin kt$ when $k=1$.

The second fraction involves shift to $a=\pi$ and $\sin()$ as well. You will find that the second fraction corresponds to $f(t) = u(t-\pi)\sin(t-\pi)$.

So we have piecewise function, let us combine both members together.

$$ f(t) = \sin t + u_{\pi}(t)\sin(t-\pi) $$

What we see is that the first part $(\sin t)$ is valid for the whole interval, while the second part is erased by $u_{\pi}$ for $t \lt \pi$ and for $t\geq \pi:\ \sin(t-\pi)$ comes into life.

We have enough information to transform $f(t)$ into the case form:

$$ f(t) = \left\{ \begin{aligned} \sin t &\hskip2em 0\leq t\lt \pi \\ \sin t + 1\cdot \sin(t-\pi) &\hskip2em t \geq \pi \end{aligned} \right. $$

The second case deserves a note: it starts with $\sin t$ because this member is valid from $t\gt 0$. The number $1$ is value of $u_{\pi}(t)$ when $t\geq \pi$, because that is the purpose of the unit step function (to be one or zero). Finally $\sin(t-\pi)$ is taken.

Now you can read from graph of $\sin()$ that $\sin(t) = -\sin(t-\pi)$. The consequence is that in the second case both sines cancel each other:

$$ f(t) = \left\{ \begin{aligned} \sin t &\hskip2em 0\leq t\lt \pi \\ 0 &\hskip2em t \geq \pi \end{aligned} \right. $$

And that is the final answer.

Example

Solve DE $y''+y =t$ with initial values $y(\pi) = 0,\ y'(\pi) = 0$

There is an inconvenience that the inital values are described at $t=\pi$ but we expect them at $t=0$. We will convert DE into other DE which has origin at $\pi$:

$$ \begin{equation} w'' + w = t+ \pi,\hskip2em w(0) = 0,\ w'(0) = 0 \label{ref:lapla_ex_move} \end{equation} $$

We have moved with the origin of $t$ axis to $\pi$, so the nonhomogenous term of new DE $(\ref{ref:lapla_ex_move})$ has to have value of $t+\pi$. You can check that now at $t=0$ within $w$ everything behaves like at $t=\pi$ within $y$. We used substitution $\boldsymbol{w(t) = y(t+\pi)}$. Now the DE is ready to be transformed and solved:

$$ \begin{align} (s^2 W - s\cdot 0 - 0) + (s\cdot 0 - 0) + W &= \frac{1}{s^2} + \frac{\pi}{s} \nonumber \\ W(s^2 + 1) &= \frac{1}{s^2} + \frac{\pi}{s} \nonumber \\ W &= \left(\frac{1}{s^2} + \frac{\pi}{s}\right) \frac{1}{s^2+1} \nonumber \\ W &= \frac{1+\pi s}{s^2(s^2+1)} \nonumber \end{align} $$

We have Laplace transform of solution $w(t)$. The inverse transform of $W(s)$ is made in example solved before.

$$ \begin{align} w(t) &= \pi + t - \pi \cos t - \sin t \nonumber \\ \underline{\underline{y(t)}} &= \pi + (t-\pi) -\pi \cos(t-\pi) - \sin(t-\pi) = \underline{\underline{t + \pi \cos t+ \sin t}} \nonumber \end{align} $$

Example

Solve DE $\color{red}{y''}\color{blue}{-5y'}+\color{green}{6y} = e^t(2t-3)$ with initial values $y(0) = 1,\ y'(0) = -1$

We have to transform each term to obtain equation in domain of $Y(s)$, where it is easy to solve. We will use $(\ref{ref:laplac_der1}),\ (\ref{ref:laplac_der2})$ and table of transforms of basic functions (for the nonhomogenous members on the right side of DE).

$$ \begin{align} \color{red}{s^2Y - s - (-1)} \color{blue}{-5(sY-1)} + \color{green}{6Y} &= 2\frac{1}{(s-1)^2} - 3\frac{1}{s-1} \nonumber \\ s^2Y -s +1 -5sY + 5 + 6Y &= 2\frac{1}{(s-1)^2} - 3\frac{1}{s-1} \nonumber \\ Y(s^2-5s+6) &= 2\frac{1}{(s-1)^2} - 3\frac{1}{s-1} - 6 + s \nonumber \\ Y &= \frac{2-3(s-1) -6(s-1)^2 + s(s-1)^2}{(s-1)^2}\cdot \frac{1}{s^2-5s+6} \nonumber \\ Y &= \frac{2-3s+3-6(s^2-2s+1) + s(s^2-2s+1)}{(s-1)^2(s-2)(s-3)} \nonumber \\ Y &= \frac{5-3s-6s^2+12s-6+s^3-2s^2+s}{(s-1)^2(s-2)(s-3)} \nonumber \\ Y &= \frac{-1+10s-8s^2+s^3}{(s-1)^2(s-2)(s-3)} \nonumber \\ \end{align} $$

That was basic algebra in $s$ domain. We have to continue with decomposition into partial fractions.

$$ \begin{align} \frac{A}{s-1} + \frac{B}{(s-1)^2} + \frac{C}{s-2} + \frac{D}{s-3} &= \frac{-1+10s-8s^2+s^3}{(s-1)^2(s-2)(s-3)} \nonumber \\[2mm] A(s-1)(s-2)(s-3) + B(s-2)(s-3) + &\nonumber \\ + C(s-3)(s-1)^2 + D(s-2)(s-1)^2 &= -1 + 10s -8s^2 +s^3 \nonumber \\[2mm] A(s^3-6s^2+11s-6) + B(s^2-5s+6) + & \nonumber \\ +C(s^3-5s^2+7s-3) + D(s^3-4s^2+5s-2) &=-1 + 10s -8s^2 +s^3 \nonumber \\[2mm] s^3(A+C+D) + s^2(-6A+B-5C-4D)+ &\nonumber \\ + s(11A-5B+7C+5D) + (-6A+6B-3C-2D) &= -1 + 10s -8s^2+s^3 \nonumber \\ \end{align} $$

When system of 4 linear equations is solved we have three fractions, each of them represents one term of solution of DE.

$$ \left. \begin{aligned} A+C+D &= 1 \\ -6A+B-5C-4D &= -8 \\ 11A-5B+7C+5D&=10 \\ -6A + 6B -3C-2D&=-1 \end{aligned} \right\} \hskip2em \left. \begin{aligned} A&=0 \\ B&=1 \\ C&=5\\ D&=-4 \end{aligned} \right. \\ Y(s) = \frac{1}{(s-1)^2} + \frac{5}{s-2} - \frac{4}{s-3} \\ \underline{\underline{y(t) = e^{-t}t + 5e^{-2t} -4e^{-3t}}} $$

Example

Solve DE $y''+y' = \pi$ with initial values $y(0) = \pi,\ y'(0) = 0$

We have to transform each term to obtain equation in domain of $Y(s)$, where it is easy to solve. We will use $(\ref{ref:laplac_der1}),\ (\ref{ref:laplac_der2})$ and table of transforms of basic functions (for the nonhomogenous members on the right side of DE).

$$ \begin{align} (s^2Y -s\pi -0) +(sY-\pi) &= \pi\frac1 s \nonumber \\ s^2Y - s\pi +sY -\pi &= \frac{\pi}{s} \nonumber \\ Y(s^2+s) &= \frac{\pi}{s} + \pi +s\pi \nonumber \\ Y &= \frac{\pi+\pi s + s^2\pi}{s(s^2+s)} \nonumber \\ Y &= \frac{\pi (1+s+s^2)}{s\cdot s (s+1)} \nonumber \\ \end{align} $$

The fraction is too difficult for inverse transform. We have to split it into pieces by means of decomposition to partial fractions.

$$ \begin{align} \frac{\pi(1+s+s^2)}{s\cdot s (s+1)} &= \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+1} \nonumber \\ \pi(1+s+s^2) &= A(s)(s+1) + B(s+1) + Cs^2 \nonumber \\ \pi + \pi s + \pi s^2 &= As^2 +As +Bs +B + Cs^2 \nonumber \\ \pi + \pi s + \pi s^2 &= s^2 (A+C) + s(A+B) +1(B) \implies \nonumber \\ \implies A&=0,\ B=\pi,\ C=\pi \nonumber \\ \end{align} $$

The solution is splitted into simple fractions:

$$ Y(s) = \frac{\pi}{s^2}+\frac{\pi}{s+1} $$

The first fraction is Laplace transform of $\pi t$, the second fraction can be identified as a Laplace transform of $\pi e^{-t}$.

$$ \underline{\underline{y(t) = \pi t + \pi e^{-t}}} $$

Example

Solve DE $y''+y' = e^t$ with initial values $y(0) = 1,\ y'(0) = 0$

We have to transform each term to obtain equation in domain of $Y(s)$, where it is easy to solve. We will use $(\ref{ref:laplac_der1}),\ (\ref{ref:laplac_der2})$ and table of transforms of basic functions (for the nonhomogenous members on the right side of DE).

$$ \begin{align} (s^2Y -s -0) +(sY-1) &= \frac1 {s-1} \nonumber \\ s^2Y - s +sY - 1&= \frac{1}{s-1} \nonumber \\ Y(s^2+s) &= \frac{1}{s-1} + 1 + s\nonumber \\ Y &= \frac{1+(s-1)+s(s-1)}{(s-1)(s^2+s)} \nonumber \\ Y &= \frac{\color{silver}{1+s-1+}s^2\color{silver}{-s}}{(s-1)s(1+s)} \nonumber \\ \end{align} $$

The fraction is too difficult for inverse transform. We have to split it into pieces by means of decomposition to partial fractions.

$$ \begin{align} \frac{s^2}{(s-1)s(1+s)} &= \frac{A}{s-1} + \frac{B}{s} + \frac{C}{s+1} \nonumber \\ s^2 &= A(s)(s+1) + B(s-1)(s+1) + C(s-1)(s)\nonumber \\ s^2 &= As^2 +As +Bs +Bs^2-B + Cs^2 -Cs\nonumber \\ s^2 &= s^2 (A+B+C) + s(A-C) -1(B) \implies \nonumber \\ \implies A&=1/2,\ B=0,\ C=1/2 \nonumber \\ \end{align} $$

The solution is splitted into simple fractions:

$$ Y(s) = \frac 1 2 \cdot \frac{1}{s-1} + \frac 1 2 \cdot \frac{1}{s+1} $$

The first fraction is Laplace transform of $e^{-t}$, the second fraction can be identified as a Laplace transform of $e^{t}$.

$$ \underline{\underline{y(t) = \frac 1 2 e^{-t} + \frac 1 2 e^{t}}} $$

Example

Solve DE $y''-y = \sin t$ with initial values $y(0) = -1,\ y'(0) = 0$

We have to transform each term to obtain equation in domain of $Y(s)$, where it is easy to solve. We will use $(\ref{ref:laplac_der1}),\ (\ref{ref:laplac_der2})$ and table of transforms of basic functions (for the nonhomogenous members on the right side of DE).

$$ \begin{align} (s^2Y+s-0)-(Y) &= \frac{1}{s^2+1} \nonumber \\ s^2Y - Y &= \frac{1}{s^2+1} - s \nonumber \\ Y(s^2-1) &= \frac{1-s^3-s}{s^2+1} \nonumber \\ Y &= \frac{1-s^3-s}{(s^2+1)(s-1)(s+1)} \nonumber \\ \end{align} $$

Such fraction is too difficult for inverse transform. We have to split it into pieces by means of decomposition to partial fractions.

$$ \begin{align} \frac{1-s^3 -s}{(s^2+1)(s-1)(s+1)} &= \frac{As+B}{s^2+1} + \frac{C}{s-1} + \frac{D}{s+1} \nonumber \\ \end{align} $$

Let us multiply both sides by $(s^2+1)(s-1)(s+1)$:

$$ \begin{align} 1-s^3-s &= (As+B)(s-1)(s+1) + C(s^2+1)(s+1) + D(s^2+1)(s-1) \nonumber \\ 1-s^3-s &= As^3 +Bs^2-As-B+Cs^3+Cs^2+Cs + C +Ds^3 -Ds2 +Ds -D \nonumber \\ 1-s^3-s &= s^3(A+C+D) + s^2(B+C-D) + s(-A+C+D) +1(-B+C-D) \nonumber \\ \end{align} $$

We are employing method of undetermined coefficients to find values $A,\ B,\ C,\ D$. That means there is a system of linear equations (the first one is $-1 = A + C+ D$) which is not shown and solved in here. Once the system is solved, the results are

$$ \begin{align} A&=0,\ B=-\frac 1 2,\ C= -\frac 1 4 ,\ D=-\frac 3 4 \implies \nonumber \\ \implies Y(s)&= -\frac 1 2 \cdot \frac 1 {s^2+1} - \frac 1 4 \cdot \frac{1}{s-1} - \frac 3 4 \frac 1 {s+1} \nonumber \\ \end{align} $$

We can use table of transforms of basic function to identify that the first fraction is transform of $\sin t$, the second $e^{-t}$ and the last one is $e^t$.

$$ \underline{\underline{ y(t) = -\frac 1 2 \sin t - \frac 1 4 e^{-t} - \frac 3 4 e^{t} }} $$

Example

No initial values are given, we have to provide arbitrary values $x(0) = x_0,\ y(0)=y_0$. First we transform both equations by Laplace and then we solve in domain $s$.

$$ \begin{align} (sX - x_0) &= Y \label{ref:lapl_set1}\\ (sY-y_0) &= -X + 2Y \label{ref:lapl_set2} \end{align} $$

From $(\ref{ref:lapl_set1})$ we can express $X$ and substitute into $(\ref{ref:lapl_set2})$.

$$ \begin{align} X &= (Y+x_0) / s \nonumber \\ (sY + y_0) &= -\left( \frac{Y+x_0}{s}\right) + 2Y \nonumber \\ sY - y_0 &= -\frac{Y}{s} - \frac{x_0}{s} + 2Y \nonumber \\ sY+ \frac Y s - 2Y &= y_0 - \frac{x_0}{s} \nonumber \\ Y(s+\frac 1 s -2) &= \frac{y_0s-x_0}{s} \nonumber \\ Y(s^2 + 1 -2s) &= y_0s - x_0\nonumber \\ Y &= \frac{y_0 s - x_0}{(s-1)^2} = \frac{A}{s-1} + \frac{B}{(s-1)^2} \label{ref:lapl_set3} \\ y_0s-x_0 &= A(s-1) + B \nonumber \\ y_0s-x_0 &= As -A+ B \nonumber \\ y_0s-x_0 &= s(A) + 1( -A+ B) \implies \nonumber\\ \implies A &= y_0,\ B = -x_0+y_0 \nonumber \\ Y &= \frac{y_0}{s-1} + \frac{-x_0+y_0}{(s-1)^2} \nonumber \\ \end{align} $$

The method of undetermined coefficients was used to find $A,\ B$. The solution $y(t)$ is inverse transform of $Y(s)$:

$$ \underline{\underline{y(t) = e^ty_0 + te^t(y_0-x_0)}}. $$

There are more ways how to find $x(t)$. This time $(\ref{ref:lapl_set3})$ is substituted into $(\ref{ref:lapl_set1})$:

$$ \begin{align} sX-x_0 &= \frac{y_0s-x_0}{(s-1)^2} \nonumber \\ sX &= \frac{y_0s-x_0}{(s-1)^2} + x_0 \frac{s^2-2s+1}{(s-1)^2} \nonumber \\ sX &= \frac{y_0s -x_0 +x_0s^2 -x_0\cdot2s + x_0}{(s-1)^2} \nonumber \\ sX &= \frac{y_0s +x_0s^2 -x_0\cdot2s}{(s-1)^2} \nonumber \\ X &= \frac{y_0 +x_0s -2x_0}{(s-1)^2} = \frac{A}{s-1} + \frac{B}{(s-1)^2}\nonumber \\ y_0+x_0s-2x_0 &= A(s-1)+B \nonumber \\ y_0+x_0s-2x_0 &= As - A+B \nonumber \\ y_0+x_0s-2x_0 &= s(A) + 1( - A+B) \nonumber \implies \\ \implies A &=x_0,\ B=y_0-x_0 \nonumber \end{align} $$

Again the complicated fraction was decomposited into two simpler fractions, method of undetermined coefficients was used to determine $A,\ B$.

$$ \underline{\underline{x(t) = e^tx_0 +te^t (y_0-x_0)}} $$

Example

We can borrow general description of spring motion from chapter modeling: $M\ddot{u} + C\dot{u} + Ku = f$. In this example is no dumping involved, no external force. We are used to use variable $y$ instead of $u$.

$$ \begin{align} my'' + ky &= 0 \nonumber \hskip2em \text{(general form)}\\ m_1y_1'' +k_1y_1 + k_2(y_1-y_2) &= 0 \nonumber \\ m_2y_2'' + k_2(y_2-y_1)+k_3y_2 &= 0 \nonumber \\ \end{align} $$

When the masses and spring constant are substituted:

$$ \begin{align} y_1'' + y_1 + y_1 - y_2 &= 0 \nonumber \\ y_2'' +y_2 -y_1 +y_2 &= 0 \nonumber \\[2mm] y_1'' + 2y_1 - y_2 &= 0 \nonumber \\ y_2'' +2y_2 -y_1 &= 0 \nonumber \\[2mm] \end{align} $$

We have a system of DE which will be solved by Laplace. Let us transform each member and solve in the domain $s$.

$$ \begin{align} s^2Y_1 -s\cdot 0 -(-1) +2Y_1 - Y_2 &= 0 \nonumber \\ s^2Y_2 -s\cdot 0 -1 +2Y_2 - Y_1 &= 0 \nonumber \\[2mm] s^2Y_1 + 1+2Y_1 -Y_2 &= 0 \nonumber \\ s^2Y_2 -1 +2Y_2-Y_1 &= 0 \nonumber \\[2mm] Y_1(s^2+2)+1-Y_2 &= 0 \nonumber \\ Y_2(s^2+2)-1-Y_1 &= 0 \label{ref:lapl_set6}\\[2mm] Y_1 &= \frac{Y_2-1}{s^2 + 2} \label{ref:lapl_set4} \\ Y_2 &= \frac{Y_1+1}{s^2+2} \label{ref:lapl_set5}\\ \end{align} $$

Now $(\ref{ref:lapl_set4})$ is substituted into $(\ref{ref:lapl_set6})$.

$$ \begin{align} Y_2(s^2+2)-1-\frac{Y_2-1}{s^2+2} &= 0 \nonumber \\ Y_2s^2 + 2Y_2 -1 -\frac{Y_2 - 1}{s^2 + 2} &= 0\nonumber \\ Y_2s^4 + 2Y_2s^2 - s^2+2Y_2s^2 + 4Y_2 - 2 -Y_2+1 &= 0 \nonumber \\ Y_2s^4 + 4Y_2s^2 +3Y_2 -s^2-1&=0 \nonumber \\ Y_2(s^4+4s^2+3) &= 1+s^2 \nonumber \\ Y_2 &= \frac{1+s^2}{(s^2+1)(s^2+3)} \nonumber \\ Y_2 &= \frac{1}{s^2+3}\nonumber \end{align} $$

The last line can be substituted into $(\ref{ref:lapl_set4})$:

$$ Y_1 = \frac{\frac{1}{s^2+3}-1}{s^2+2} = \frac{\frac{1-s^2-3}{s^2+3}}{s^2+2} = \frac{\frac{-s^2-2}{s^2+3}}{s^2+2} = \frac{-1}{s^2+3} $$

The solution of the system of DE is done by inverse Laplace transform of $Y_1(s)$ and $Y_2(s)$.

$$ \underline{\underline{y_1(t) = -\sin \sqrt3 t \cdot \frac{1}{\sqrt 3}}} \\ \underline{\underline{y_2(t) = \sin \sqrt3 t \cdot \frac{1}{\sqrt 3}}} \\ $$

Example

The first step is to describe the given function $f(t)$ by means of unit step function, otherwise we can not make Laplace transform of $f(t)$.

$$ \begin{align} f(t) &= u_5\left(\frac{t-5}{5}\right) - u_{10}\left(\frac{t-5}{5}\right) +u_{10}(1) = \nonumber \\ &= \frac 1 5 u_5(t-5) + u_{10}\left(-\frac{t-5}{5}+ \frac{5}{5}\right) = \nonumber \\ &= \frac{1}{5}u_5(t-5) - \frac 1 5 u_{10}(t-10) \nonumber \end{align} $$

The next step is usual: we have to transform each term of DE. The members of function $f(t)$, which involves unit step time functions will be transformed the using shifting theorem $(\ref{ref:lapl_4})$.

$$ \begin{align} (s^2Y - s\cdot 0 - 0) + 4Y & = \frac 1 5 \frac 1 {s^2}e^{-5s} - \frac 1 5 \frac{1}{s^2}e^{-10s} \nonumber \\ Y(s^2 +4) &= \frac{\frac 1 5 e^{-5s} - \frac{1}{5}e^{-10s}}{s^2} \nonumber \\ Y &= \frac{\frac 1 5 e^{-5s} - \frac{1}{5}e^{-10s}}{s^2(s^2 +4)} \nonumber \\ \end{align} $$

Both the terms in numerator express time shift, not the functions themselves. Thus we can bring time shift outside of the fraction and bring them back later.

$$ \begin{align} Y &= \left(\frac 1 5 e^{-5s} - \frac{1}{5} e^{-10s}\right)\frac{1}{s^2(s^2 +4)} \nonumber \\ \end{align} $$

We have to do partial fraction decomposition of $\frac{1}{s^2(s^2 +4)}$:

$$ \begin{align} \frac{1}{s^2(s^2 +4)} &= \frac{A}{s} + \frac{B}{s^2} + \frac{Cs+D}{s^2+4} \nonumber \\ 1 &= A(s)(s^2 + 4) + B (s^2+4) + Cs(s^2) + D(s^2) \nonumber \\ 1 &= s^3 (A+C) + s^2(B+D) + s(4A) + 1(4B) \implies \nonumber \\ B &= \frac 1 4,\ A= 0,\ C = 0,\ D =-\frac 1 4 \nonumber \\ \end{align} $$

Then we have to do inverse transform of $Y(s)$:

$$ \begin{align} Y &= \left(\frac 1 5 e^{-5s} - \frac{1}{5} e^{-10s}\right)\left(\frac 1 4\frac{1}{s^2} -\frac 1 8 \frac{2}{s^2 + 4} \right) \nonumber \\ \end{align} $$

The inverse terms inside the second pair of parentheses are $\frac{1}{4} t - \frac{1}{8}\sin 2t$ and the inverse transform of $Y(s)$ is

$$ \begin{align} y(t) &= \left(\frac 1 5 u_5(t) - \frac 1 5 u_{10}(t)\right) \left( \frac 1 4t -\frac 1 8\sin 2t \right) = \nonumber \\ &= \frac 1 5 u_5(t) \left( \frac 1 4 t - \frac 1 8 \sin 2t \right) - \frac 1 5 u_{10}(t) \left( \frac 1 4 t -\frac 1 8 \sin 2t \right) \nonumber \end{align} $$

• Note that both $\frac{1}{4}t$ and $\frac{1}{8}\sin 2t$ are erased by unit step function: until 5 s there is a zero response.
• At time $t=5$ both members $\frac{1}{4} t$ and $\frac{1}{8}\sin 2t$ become active (we also have to shift the functions according to the shifting theorem): $y(5 \leq t \lt 10) = \frac{1}{5}\cdot\frac 1 4 (t-5) - \frac{1}{5}\cdot \frac{1}{8}\sin2(t-5)$.
• At time $t=10$ we have to subtract $\frac{1}{5}(\frac 1 4 (t-10) - \frac{1}{8}\sin2(t-10))$

$$ y(t) = \left\{ \begin{aligned} 0, &\hskip2em t\lt5, \\ \frac 1 {20} (t-5) - \frac 1 {40} \sin 2(t-5), &\hskip2em 5 \leq t \lt 10\\ \frac 5 {20} - \frac 1 {40} \sin 2(t-5) + \frac{1}{40} \sin 2(t-10), & \hskip2em t \geq 10. \end{aligned} \right. $$

That is the final answer.

Example

An event is happening at time $t=1$. According to the shifting theorem, it can be observed that each function is time-shifted into $t=1$. Since they are shifted, they are also erased.

$$ y(t) = u_1(t)\left((t-1)^2 + 2(t-1) + 1\right) = u_1(t)(t^2 - 2t +1 +2t-2 +1) = u_1(t)\cdot t^2. $$

What seemed as a time shift is finally $t^2$ erased until $t=1$. Indeed, such example has been computed above already.

Example

We can recognize some terms which are active from $t=0$ and some other terms shifted and active from time $t=3$:

$$ y(t) = 2e^t - 1 -2u_3(t) - u_3(t)\cdot (t-3) + \frac 3 2 e^{t-3} u_3(t) + \frac 1 2 e^{-(t-3)}u_3(t) $$

Such solution is correct but not convenient and it is expected we provide also the case form:

$$ y(t) = \left\{ \begin{aligned} -1+2e^t & \hskip2em 0 \leq t \lt 3, \\ -1 + 2e^t - 2 -(t-3) + \frac 3 2 e^{t-3} + \frac 1 2 e^{-(t-3)} & \hskip2em t \geq 3, \end{aligned} \right. $$

which can be simplified to

$$ y(t) = \left\{ \begin{aligned} -1+2e^t & \hskip2em 0 \leq t \lt 3, \\ 2e^t - t + \frac 3 2 e^{t-3} + \frac 1 2 e^{3-t} & \hskip2em t \geq 3. \end{aligned} \right. $$

Example

It can be observed that sine is involved, more precisely $\sin 3t$ which has been shifted to $t=2$.

$$ y(t) = \frac 1 3 \sin 3(t-2)\cdot u_2(t) $$

So the solution is $1/3 \sin 3t$ moved to $t=2$. The case form is more convenient:

$$ y(t) = \left\{ \begin{aligned} 0 & \hskip1em t \leq 2, \\ \frac 1 3 \sin3(t-2) &\hskip1em t \gt 2. \end{aligned} \right. $$

Example

We can identify two functions $f(t) = t$ which were moved to time $t=1$ and $t=2$

$$ f(t) = (t-1) u(t-1) - (t-2) u(t-2) $$

Such result, written with unit step function, is formally correct, but for convenience its case form is expected:

$$ f(t) = \left\{ \begin{aligned} 0 &\hskip2em t \lt 1, \\ t - 1 & \hskip2em 1 \leq t \lt 2, \\ 1 & \hskip2em t\geq 2. \end{aligned} \right. $$