Determine the Laplace transform of $f(t) = 1$ using the definition.

$$\mathscr{L}\,\{1\}\;\;\; = \int_{0}^{\infty}1.e^{-s t}\,dt$$

$$= \overset{{\large\lim\;}}{{\small u} {\small\rightarrow \infty}} \int_{0}^{u}e^{-s t}\,dt$$

$$= \overset{{\large\lim\;}}{{\small u} {\small\rightarrow \infty}} \bigg[-\frac{1}{s}e^{-s t}\bigg]_{t=0}^{t=u}$$

$$= - \frac{1}{s}\; \overset{{\large\lim\;}}{{\small u} {\small\rightarrow \infty}}\;(e^{-s u} - e^0)$$

$= - {\Large\frac{1}{s}}\;(0 - 1) $ ( provided $s > 0$ )

$$= \frac{1}{s}.$$

Now watch the following:

📹 Calculating Laplace Transforms from the Definition

Determine the Laplace transform of, $f(t) = t$.

$$\mathscr{L}\{t\} = \int_{0}^{\infty}\;t.e^{-s t}\,dt$$

To evaluate this integral requires integration by parts and it can readily be shown that, provided $s > 0$,

$$\mathscr{L} \{t\} = \frac{1}{s^2}.$$

Very quickly the integrations required to complete the Laplace transformation become difficult and messy. For this reason, we generally work from a table of pre-determined Laplace transforms (see Appendix).

(i). Determine $\mathscr{L}\{t^3\}$.

Now, $t^3$ is not in the table explicitly, but $t^n$ is:

$\mathscr{L}\{t^n\} = {\Large\frac{n!}{s^{n+1}}}$ (#4 in the table)

For $t^3$ we require $n = 3$:

Hence, $\mathscr{L}\{t^3\}\;$ $= {\Large\frac{3!}{s^4}} = {\Large\frac{3 \times 2 \times 1}{s^4}} = {\Large\frac{6}{s^4}}$.

(ii). Determine $\mathscr{L}\{e^{-2t}\}$.

From # 5 in the table:

$\mathscr{L} \{e^{-\alpha t}\} = {\Large\frac{1}{s + \alpha}}$.

Set $\alpha = 2$ to obtain:

$\mathscr{L} \{e^{-2 t}\} =\;$ ${\Large\frac{1}{s + 2}}$.

(iii). Determine $\mathscr{L}\;\{\sin(4t)\}$.

From # 8 in the table:

$\mathscr{L}\;\{\sin(4t)\} = {\Large\frac{\omega}{s^2 + \omega^2}}$.

Set $\omega = 4$ to obtain :

$\mathscr{L} \{\sin(4t)\}\;$ $= {\Large\frac{4}{s^2 + 4^2}} = {\Large\frac{4}{s^2 + 16}}$.

$\mathscr{L}\{2\cos(\omega t) + 3\sin(\omega t)\} = 2{\Large\frac{s}{s^2 + \omega^2}} + 3{\Large\frac{\omega}{s^2 + \omega^2}} = \;$ ${\Large\frac{2s + 3 \omega}{s^2 + \omega^2}}$.

Determine $\mathscr{L}\,\{e^{-2t}t^3\}$.

From tables and Example (3)(i) we have

$\mathscr{L}\;\{t^3\} = {\Large\frac{6}{s^4}}$. $[1]$

Then, by the first shifting property with $\alpha = 2$ we have that

$\mathscr{L}\;\{e^{-2 t}t^3\} = \;$ ${\Large\frac{6}{(s + 2)^4}}$.$[2]$

Note that to obtain $[2]$ from $[1]$ we merely replace $s$ by $(s + 2)$.

Invert the Laplace transform, ${\Large\frac{3}{s^6}}$.

The closest entry in the table is #4 which gives

$$\mathscr{L}^{-1} \bigg\{\frac{n!}{s^{n+1}}\bigg\} = t^n.$$

Setting $n = 5$ we have

$$\mathscr{L}^{-1}\bigg\{\frac{5!}{s^6}\bigg\} = \mathscr{L}^{-1} \bigg\{ \frac{120}{s^6}\bigg\} = t^5.$$

In the given Laplace transform there is a 3 in the numerator and so we must manipulate the expression obtained from the tables which has 120 in the numerator. To do so we multiply by the value we want in the numerator (3) and divide by the value introduced from the tables (120). We then have

$$\frac{3}{s^6} = \frac{3}{120} \frac{120}{s^6} = \frac{1}{40}\bigg[\frac{120}{s^6}\bigg].$$

The term in the square brackets is now exactly the table entry so we can invert that and simply multiply by the fraction in front:

$\mathscr{L}^{-1}\bigg\{{\Large\frac{3}{s^6}}\bigg\} = \;$ ${\Large\frac{1}{40}}t^5$.

Invert the Laplace transform, ${\Large\frac{1}{s^2\, +\, 2s \,+\, 5}}$.

This example requires the technique known as “completing the square” and a little bit of fine tuning to rewrite it in a form that can be inverted directly from the table. Note that the closest entry in the table is #14 giving:

$$\mathscr{L}^{-1}\bigg\{\frac{\omega}{(s + \alpha)^2 + \omega^2}\bigg\} = e^{- \alpha t}\sin(\omega\,t).$$

Now complete the square in the denominator:

$$s^2 + 2s + 5 = (s + 1)^2 - 1 + 5 = (s + 1)^2 + 4.$$

The given transform becomes:

$$\frac{1}{s^2 + 2s + 5} = \frac{1}{(s + 1)^2 + 4} = \frac{1}{(s + 1)^2 + 2^2}\;.$$

This expression is now very close to the table entry with $\alpha = 1$ and $\omega = 2$. We would like there to be a 2 in the numerator so “fine tuning” gives

$$\frac{1}{(s + 1)^2 + 2^2} = \frac{1}{2}\bigg[\frac{2}{(s + 1)^2 + 2^2}\Bigg] ,$$

where the term in the square brackets is exactly the table entry with $\alpha = 1$ and $\omega = 2$.

Inverting this expression and multiplying by the fraction gives

${\Large\frac{1}{2}} \mathscr{L}^{-1}\bigg\{{\Large\frac{2}{(s + 1)^2 + 2^2}}\bigg\} = \;$ ${\Large\frac{1}{2}}e^{- t}\sin(2t)$.

(i). Detemine $\mathscr{L}^{-1}\bigg\{{\Large\frac{s\, +\, 5}{s^2 \,-\, 2s \,-\, 3}}\bigg\}$ using partial fractions.

Factorise the denominator and obtain the partial fraction representation of the resulting expression. Note that the details of the partial fraction expansion have been omitted.

$$\frac{s + 5}{s^2 - 2s - 3} = \frac{s + 5}{(s - 3)(s + 1)} = \frac{2}{s - 3} - \frac{1}{s + 1}.$$

Examination of the tables shows that # 5 is the appropriate choice giving

$$\mathscr{L}^{-1}\bigg\{\frac{1}{s + \alpha}\bigg\} = e^{- \alpha t}.$$

Using this result with $\alpha = - 3$ and $\alpha = + 1$ gives

$\mathscr{L}^{-1} \bigg\{{\Large\frac{2}{s - 3}}\bigg\} - \mathscr{L}^{-1}\bigg\{{\Large\frac{1}{s + 1}}\bigg\} = \;$ $2e^{3t} - e^{-t}$

(ii). Detemine $\mathscr{L}^{-1}\bigg\{{\Large\frac{s \,+\, 5}{s^2 \,-\, 2s \,-\, 3}}\bigg\}$ by completing the square.

If we fail to notice that the denominator can be factorised, then completing the square is still an option:

$$\frac{s + 5}{s^2 - 2s - 3} = \frac{s + 5}{(s - 1)^2 - 1 - 3} = \frac{s + 5}{(s - 1)^2 - 4} = \frac{s + 5}{(s - 1)^2 - 2^2} \;.$$

The minus sign in the denominator is significant since it no longer conforms to the form

$$(s + \alpha)^2 + \omega^2.$$

we had in Example 7.

Instead we need the following results, # 19 and #21 in the tables, i.e.

$$\mathscr{L}^{-1}\{e^{- \alpha t}\sinh \,(\beta t)\} = \frac{\beta}{(s + \alpha)^2 - \beta^2}$$

$$\mathscr{L}^{-1}\{e^{- \alpha t} \cosh\,(\beta t)\} = \frac{s + \alpha}{(s + \alpha)^2 - \beta^2} $$

where sinh (pronounced “shine”) and cosh (pronounced “cosh”) are the so-called hyperbolic functions. Fine tuning the Laplace transform a little bit more gives

$$\frac{s + 5}{(s - 1)^2 - 2^2} = \frac{(s - 1) + 6}{(s - 1)^2 - 2^2} = \bigg[\frac{(s - 1)}{(s - 1)^2 - 2^2}\bigg] + 3\bigg[\frac{2}{(s - 1)^2 - 2^2}\bigg].$$

Inverting the terms in square brackets using # 19 and #21 in the tables gives,

$$\mathscr{L}^{-1} \bigg\{\frac{s + 5}{s^2 - 2s - 3}\bigg\} = \mathscr{L}^{-1} \bigg\{\frac{(s - 1)}{(s - 1)^2 - 2^2}\bigg\} + 3\; \mathscr{L}^{-1}\bigg\{\frac{2}{(s - 1)^2 - 2^2}\bigg\}$$

$$= e^t \cosh(2t) + 3e^t \sinh(2t)$$

$$= e^t [\cosh(2t) + 3 \sinh(2t)].$$

Note: By applying the following mathematical identities:

$\cosh\,(\beta t) = {\Large\frac{1}{2}}[e^{+ \beta t} + e^{- \beta t}]$  and  $\sinh\,(\beta t) = {\Large\frac{1}{2}}[e^{+ \beta t} - e^{- \beta t}]$

we can easily show that the two versions of the answer, in parts (i) and (ii), are equivalent.

Use Laplace transforms to solve the ODE, $\ddot{x} + 2\dot{x} + 5x = 8e^{- 3 t}$ subject to the initial conditions $x(0) = \dot{x}(0) = 0$.

Transform both sides of the ODE ( from left to right use #25, #24, #23 and #5 ):

$$[s^2\,\bar{x} - s\,x(0) - \dot{x}(0)] + 2[s\,\bar{x} - x(0)] + 5\,\bar{x} = \frac{8}{s + 3}$$

Substitute initial values:

$$[s^2\,\bar{x} - s\,0 - 0] + 2[s\,\bar{x} - 0] + 5\,\bar{x} = \frac{8}{s + 3}$$

$$(s^2 + 2s + 5 )\,\bar{x} \;\;\;\;\;\;\;\;\;\;= \frac{8}{s + 3}.$$

Solve for $\bar{x}$:

$$\bar{x} = \frac{8}{(s + 3)(s^2 + 2s + 5)}\;.$$

Manipulate into a form that can be inverted from the tables. In this case we form partial fractions, complete the square on the quadratic denominator and finish up with a little “fine tuning” of the last term:

$$\bar{x} = \bigg[\frac{1}{s + 3}\bigg] - \bigg[\frac{s}{(s + 1)^2 + 2^2}\bigg] + \frac{1}{2}\bigg[\frac{2}{(s + 1) + 2^2}\bigg].$$

Finally invert term-by-term and simplify:

$$x(t) = e^{- 3 t} - e^{- t}\Big[\cos(2t) - \frac{1}{2}\sin(2t)\Big] + \frac{1}{2}e^{-t}\sin(2t)$$

$= e^{- 3 t} - e^{- t} \big[\cos(2t) - \sin(2t)\big]$.

Solve $\ddot{x} + 25x = 1$ subject to $x(0) = 10$ and $\dot{x}(0) = 0$ using Laplace transforms.

Transform the ODE:

$$\Big[s^2\bar{x} - s\,x(0) - \dot{x}(0)\Big] + 25\bar{x} = \frac{1}{s}\;. $$

Substitute initial values:

$$\Big[s^2\bar{x} - 10s - 0\Big] + 25\bar{x} = \frac{1}{s}\;. $$

Solve for $\bar{x}$:

$$s^2\bar{x} + 25\bar{x} = \frac{1}{s} + 10s$$

$$(s^2 + 25)\bar{x} = \frac{1}{s} + 10s$$

$$\bar{x} = \frac{1}{s(s^2 + 25)} + 10\frac{s}{(s^2 + 25)}.$$

Manipulate and invert:

$$\bar{x} = \frac{1}{25}\bigg[\frac{25}{s(s^2 + 25)}\bigg] + 10\bigg[\frac{s}{(s^2 + 25)}\bigg]$$

$x(t) = {\Large\frac{1}{25}}\big[1 - \cos(5t)\big] + 10\cos(5t)$. ( # 10 and # 9 from the tables )

Sketch the graph of the function $f(t) = 2u(t - 5) - 3u(t - 10) + u(t - 15)$.

Here it may be useful to think of the step functions as switches and examine the critical values.

for $t < 5$ all step functions are “off” and are all equal to zero;

for $5 < t < 10 $ the first step function, $u(t - 5)$, is “on” and has the value 1;

for $10 < t < 15$ the second step function, $u(t - 10)$, now switches on to 1 also; and

for $t > 15$ all step functions are now “on” and equal to 1.

We now evaluate $f(t)$ noting that it is a linear combination of the three step functions.

$$ \begin{align*} f(t) = \left\{ \begin{matrix} \;\;0 & , & 0 < t < 5\; \\ \;\;2 & , & 5 < t < 10 \\ -1 & , & 10 < t < 15 \\ \;\;0 & , & t > 15 \end{matrix}\right. \end{align*}$$

The graph of $f(t)$ is shown below

Graph the function, $f(t) = (2 - t)(t - 4)[u(t - 2) - u(t - 4)]$.

We shall do this in two parts. Without the step functions we would have the quadratic function

$$g(t) = (2 - t)(t - 4) = -t^2 + 6t - 8.$$

This graph crosses the $x$-axis at $t = 2$ and $t = 4$ and has a maximum turning point at $(3, 1)$ .

The step functions take the following values:

$$ \begin{align*} u(t - 2) - u(t - 4) = \left\{ \begin{matrix} 0 - 0 = 0\ & , & t < 2 \\ 1 - 0 = 1 & , & 2 < t < 4 \\ 1 - 1 = 0 & , & t > 4 \end{matrix}\right. \end{align*}.$$

When the two parts ( quadratic and step functions ) are multiplied together we obtain

$$ \begin{align*} f(t) = \left\{ \begin{matrix} 0 & , & t < 2 \\-t^2 + 6t - 8 & , & 2 < t < 4 \\ 0 & , & t > 4 \end{matrix}\right. \end{align*} $$

resulting in the graph below

We can think of a difference of two step functions

$$u(t - a) - u(t - b)$$

like a “mathematical window” whose frame obscures the graph of any multiplying function to the sides, i.e. for $t < a$ and $t > b$.

In the previous example by including the “mathematical window”, $u(t - 2) - u(t - 4)$, the quadratic function $g(t) = -t^2 + 6t - 8$ was obscured outside the window, i.e. set to zero for $t < 2$ and $t > 4$.

Determine $f(t)$ from the following graph:

The jumps in the graph can be represented by step functions. The location of the jump is determined by the critical value of the step function, the size of the jump can be achieved by multiplying the step function by a suitably sized constant, the direction of the jump is obtained from the sign of the constant.

Working from left to right:

1st jump: $4u(t - 1)$ up to $4$ at $T = 1$

2nd jump: $-6u(t - 2)$ down by $6$ at $T = 2$

3rd jump: $4u(t - 3)$ up by $4$ at $T = 3$.

Adding these together gives us the function:

$f(t) = 4u(t - 1) - 6u(t - 2) + 4u(t - 3)$.

Determine $f(t)$ from the following graph:

Using basic coordinate geometry the equation of the oblique straight line is given by

$$y = 3t - 3.$$

The straight line is only visible between $t = 1$ and $t = 2$ with the function taking the value zero outside this interval, i.e. for $t < 1$ and $t > 2$. Hence, we have the “mathematical window”,

$$u(t - 1) - u(t - 2).$$

Multiplying the two parts together gives us our function:

$f(t) = (3t - 3)[u(t - 1) - u(t - 2)]$.

Write down the following function in terms of step functions:

$$ \begin{align*} f(t) = \left\{ \begin{matrix} t & , & 0 <t <1 \\ 2 - t & , & 1 < t < 2 \\ 0 & , & t > 2 \end{matrix}\right. \end{align*}.$$

Examine the function $f(t)$ over each of the given intervals.

On the interval $0 < t < 1$, $f(t) = t$ giving, $t\big[u(t - 0) - u(t - 1)\big]$.

On the interval $1 < t < 2$, $f(t) = 2 - t$ giving, $(2 - t)\big[u(t - 1) - u(t - 2)\big]$.

For $t > 2,\; f(t) = 0$.

Combining the above gives

$$f(t) = t\big[u(t - 0) - u(t - 1)\big] + (2 - t)\big[u(t - 1) - u(t - 2)\big].$$

As far as Laplace transforms are concerned $u(t - 0) \equiv 1$ so the above can be rewritten as

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

Determine $\mathscr{L}\,\{t^2u(t - 10)\}$.

To calculate LTs using the SST we start at the bottom-left of the diagram and move round in a clockwise direction ending up in the bottom-right and obtaining the result:

$$\class{outbox}{\matrix{f(t) & \rightarrow & F(s) \\ \uparrow & & \downarrow \\ f(t - T)u(t - T) & & F(s)e^{-sT}}}$$

However, we have a slight problem in that the given function of $t$ does not quite conform to the structure in the bottom-left. We need to rewrite the multiplying function $t^2$ so that any occurrence of the variable $t$ always includes a “minus $T$ ”. For this example $T = 10$ so we introduce it as follows:

$$t^2 = \big[(t - 10) + 10\big]^2 = (t - 10)^2 + 20(t - 10) + 100 .$$

The original function becomes $\big[(t - 10)^2 + 20(t - 10) + 100\big]u(t - 10)$ which now does conform. Next we use the SST to transform:

Step 1: Cover up the step function and the “minus 10s” to leave

$$f(t) = t^2 + 20t + 100.$$

Step 2: Transform to give $F(s) = {\Large\frac{2}{s^3}} + {\Large\frac{20}{s^2}} + {\Large\frac{100}{s}}$.

Step 3: Everything we covered up in Step 1 is now accounted for by multiplying $F(s)$ by $e^{-10s}$ and so completes the transformation,

$\mathscr{L}\,\{t^2u(t - 10)\} = \;$ $\bigg[{\Large\frac{2}{s^3}} + {\Large\frac{20}{s^2}} + {\Large\frac{100}{s}}\bigg]e^{-10s}$.

Invert the Laplace transform ${\Large\frac{1}{s(s + 2)}}e^{-4s}$.

To invert LTs, using the SST, we start at the bottom-right of the diagram and move round in an anticlockwise direction ending up in the bottom-left and obtaining the result:

$$\class{outbox}{\matrix{f(t) &\leftarrow & F(s) \\ \downarrow & & \uparrow \\f(t - T)u(t - T) & & F(s)e^{-sT}}}$$

Step 1: Ignore exponential to give $F(s) = {\Large\frac{1}{s(s + 2)}} = {\Large\frac{1}{2}}\bigg[{\Large\frac{2}{s(s + 2)}}\bigg]$.

Step 2: Invert $F(s)$ to give $f(t) = {\Large\frac{1}{2}}\big[1 - e^{-2t}\big]$.

Step 3: Shift and truncate with $T = 4$, i.e. replace $t$ in $f(t)$ by $(t - 4)$ and multiply by $u(t - 4)$, to give the completed inversion $f(t - T)u(t - T)$. Hence,

$\mathscr{L}^{-1} \bigg\{{\Large\frac{1}{s(s + 2)}}e^{-4s}\bigg\} = \;$ ${\Large\frac{1}{2}}\big[1 - e^{-2(t - 4)}\big]u(t - 4)$.

Solve the differential equation $\ddot{x} + 4x = u(t - 3)$ subject to $x(0) = \dot{x}(0) = 0$.

Transform: $\big[s^2\bar{x} - s\,x(0) - \dot{x}(0)\big] + 4\bar{x} = {\Large\frac{1}{s}}e^{-3s}$

Substitute initial conditions and solve for $\bar{x}$:

$$(s^2 + 4)\bar{x} = \frac{1}{s}e^{-3s}$$

so

$$\bar{x} = \frac{1}{s(s^2 + 4)}e^{-3s}.$$

Invert with the help of the Second Shifting Theorem:

Again, we are starting at the bottom-right of the diagram and moving round anti-clockwise:

$$\class{outbox}{\matrix{f(t) & \leftarrow & F(s) \\ \downarrow & & \uparrow \\ f(t - T)u(t - T) & & F(s)e^{-sT}}} $$

Step 1: Ignore exponential to give $F(s) = {\Large\frac{1}{s(s^2 + 4)}} = {\Large\frac{1}{4}}\bigg[{\Large\frac{2^2}{s(s^2 + 2^2)}}\bigg]$.

Step 2: Invert $F(s)$ to give $f(t) = {\Large\frac{1}{4}}\big[1 - \cos(2t)\big]$.

Step 3: Shift and truncate with $T = 3$, i.e. in $f(t)\,$ replace $t$ by $(t - 3)$ and multiply by $u(t - 3)$ to give $f(t - T)u(t - T)$ and the completed solution,

$x(t) = {\Large\frac{1}{4}}\big[1 - \cos(2(t - 3))\big]u(t - 3)$.

Solve the differential equation $\dot{x} + 4x = 3\,\delta(t - 2)\;,\; x(0) = 5$.

Transform: $\big[s\bar{x} - x(0)\big] + 4\bar{x} = 3e^{-2s}$.

Substitute initial condition and solve for $\bar{x}$:

$$(s + 4)\;\bar{x} = 3e^{-2s} + 5$$

$$\bar{x} = \frac{3}{(s + 4)}e^{-2s} + \frac{5}{(s + 4)}.$$

$[A] \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;[B]$

Invert to give solution:

$$[A]: \;\;\;\;\;\mathscr{L}^{-1}\bigg\{\frac{3}{(s + 4)}\bigg\} = 3\,e^{-4t}.$$

By the Second Shifting Theorem,

$$\mathscr{L}^{-1} \bigg\{\frac{3}{(s + 4)}e^{-2s}\bigg\} = 3\,e^{-4(t - 2)}u(t - 2)$$

$$[B]: \;\;\;\;\;\mathscr{L}^{-1}\bigg\{\frac{5}{(s + 4)}\bigg\} = 5\,e^{-4t}$$

Combine $[A]$ and $[B]$ above to obtain the solution:

$x(t) = 3e^{-4(t - 2)}u(t - 2) + 5\,e^{-4t}$.

In a branch of an electronic circuit the variation of current with time is modelled by the differential equation

$$\frac{d^2i}{dt^2} + 36i = \frac{dV}{dt},$$

where $V(t)$ is an input voltage. Suppose $V(t) = 240[u(t - 5) - u(t - 10)]$. Assuming zero initial conditions, determine the current, $i$, as a function of time, $t$ .

Solution

Recalling that the Dirac delta function is the derivative of the unit step function we can write

$$\frac{dV}{dt} = 240[\delta(t - 5) - \delta(t - 10)]$$

and so the differential equation becomes

$$\frac{d^2i}{dt^2} + 36i = 240[\delta(t - 5) - \delta(t - 10)].$$

The Laplace transformation process results in

$$\bar{i} = \frac{240}{s^2 + 36}[e^{-5s} - e^{-10s}],$$

which can be inverted with the help of the Second Shifting Theorem to give the solution:

$i(t) = 40\,\sin[6(t - 5)]u(t - 5) - 40\,\sin[6(t - 10)]u(t - 10)$.

Invert the Laplace transform ${\Large\frac{1}{s(s + 4)^2}}$.

The solution could be obtained using partial fractions (try it!), but convolution can be used as an alternative:

Set $F(s) = {\Large\frac{1}{s}}$ and $G(s) = {\Large\frac{1}{(s + 4)^2}}$ and invert individually from the tables to obtain:

$f(t) = 1$ and$g(t) = te^{-4t}$.

Form the convolution of $f$ and $g$ using form $[a]$ above :

$f(t - z) = 1$, $g(z) = ze^{-4z}$

$$f * g = \int_{0}^{t}\, f(t - z)\,g(z)\,dz$$

$$= \int_{0}^{t}\,z\,e^{-4z}\,dz. $$

Omitting the details, integration by parts gives

$f * g = - {\Large\frac{1}{4}}t\,e^{-4t} - {\Large\frac{1}{16}}e^{-4t} + {\Large\frac{1}{16}}$.