Laplace transforms and applications notes — Unit 3

The Laplace Transform is an integral transform that converts a function of time $f(t)$ into a function of a complex variable $s$.

It is an incredibly powerful tool in engineering because it transforms difficult differential equations into simple algebraic equations.

You must memorize these standard formulas:

The Laplace transform is a linear operator. This means you can break apart sums and pull out constants, exactly like standard integration.

$$L\\{a f(t) + b g(t)\\} = a L\\{f(t)\\} + b L\\{g(t)\\}$$

Example: Find $L\\{4e^{3t} - 2\sin(5t) + 3\\}$

$$= 4L\\{e^{3t}\\} - 2L\\{\sin(5t)\\} + 3L\\{1\\}$$
$$= 4\left(\frac{1}{s-3}\right) - 2\left(\frac{5}{s^2 + 25}\right) + 3\left(\frac{1}{s}\right)$$

If a function is multiplied by an exponential $e^{at}$, the transform $F(s)$ simply shifts to $F(s-a)$.

Example: Find $L\\{e^{-2t} \cos(4t)\\}.$

• Here, $f(t) = \cos(4t)$.
• We know $L\\{\cos(4t)\\} = \frac{s}{s^2 + 16}$.
• By First Shifting Theorem, replace every $s$ with $(s - (-2)) = (s+2)$.
• Answer: $\frac{s+2}{(s+2)^2 + 16}$

The Unit Step Function $u(t-a)$ (or $H(t-a)$) acts like a switch that turns 'ON' at time $t=a$.

$$u(t-a) = \begin{cases} 0 & \text{for } t < a \\\ 1 & \text{for } t \geq a \end{cases}$$

The Laplace transform of the basic unit step function is:
$$L\\{u(t-a)\\} = \frac{e^{-as}}{s}$$

It is heavily used in electrical engineering to model circuits being switched on at a specific time.

This theorem handles functions that are delayed in time.

This means if a function is delayed by $a$ seconds, its Laplace transform is simply multiplied by $e^{-as}$.

Alternative Form: If you just have $L\\{f(t) \cdot u(t-a)\\}$ (where the function itself is not shifted), the formula is:
$$= e^{-as} L\\{f(t+a)\\}$$

If a function is multiplied by $t$, you differentiate its transform with respect to $s$.

Example: Find $L\\{t \sin(at)\\}$

• We know $L\\{\sin(at)\\} = \frac{a}{s^2 + a^2}$.
• By the rule (for $n=1$): $(-1)^1 \frac{d}{ds} \left(\frac{a}{s^2 + a^2}\right)$.
• Using chain rule: $-a \left( -1(s^2 + a^2)^{-2} (2s) \right)$
• Answer: $\frac{2as}{(s^2 + a^2)^2}$

If a function is divided by $t$, you integrate its transform with respect to $s$ from $s$ to $\infty$.

Provided the limit $\lim_{t \to 0} \frac{f(t)}{t}$ exists (is finite). If the limit does not exist, the transform cannot be found this way.

This property is the entire reason Laplace Transforms are useful for solving Differential Equations. It converts derivatives in the time domain ($t$) into simple algebraic multiplication by $s$ in the frequency domain ($s$).

• First Derivative: $L\\{f'(t)\\} = sF(s) - f(0)$
• Second Derivative: $L\\{f''(t)\\} = s^2F(s) - sf(0) - f'(0)$
• n-th Derivative: $L\\{f^n(t)\\} = s^nF(s) - s^{n-1}f(0) - s^{n-2}f'(0) - ... - f^{n-1}(0)$

Notice how the initial conditions ($f(0), f'(0)$) are built directly into the formula! No need to find arbitrary constants $C_1, C_2$ later.

If taking a derivative means multiplying by $s$, it makes sense that taking an integral means dividing by $s$.

Example: Find $L\\{\int_0^t \sin(2x) \\, dx\\}$

• $L\\{\sin(2t)\\} = \frac{2}{s^2 + 4}$
• Therefore, $L\\{\text{integral}\\} = \frac{1}{s} \left( \frac{2}{s^2 + 4} \right)$

The inverse Laplace transform operator $L^{-1}$ takes a function from the $s$-domain back to the time domain $t$.

If $L\\{f(t)\\} = F(s)$, then $L^{-1}\\{F(s)\\} = f(t)$.

⇒Standard Inverse Formulas:

• $L^{-1}\\{\frac{1}{s}\\} = 1$
• $L^{-1}\\{\frac{1}{s^n}\\} = \frac{t^{n-1}}{(n-1)!}$
• $L^{-1}\\{\frac{1}{s-a}\\} = e^{at}$
• $L^{-1}\\{\frac{s}{s^2+a^2}\\} = \cos(at)$

Just like the forward theorem, if you see an $(s-a)$ shift in the denominator, you can pull it out as an exponential $e^{at}$ in the time domain.

$$L^{-1}\\{F(s-a)\\} = e^{at} L^{-1}\\{F(s)\\}$$

Example: Find L^{-1}\left\\{ \frac{1}{(s-3)^2 + 16} \right\\}

• We see a shift of $(s-3)$, so pull out $e^{3t}$.
• This leaves L^{-1}\left\\{ \frac{1}{s^2 + 16} \right\\}.
• Multiply and divide by 4 to get standard sine form: \frac{1}{4} L^{-1}\left\\{ \frac{4}{s^2 + 4^2} \right\\}.
• Answer: $\frac{1}{4} e^{3t} \sin(4t)$.

Most $F(s)$ expressions are complicated rational functions. You must break them down using Partial Fractions before applying inverse rules.

• Linear Factors: $\frac{1}{(s-a)(s-b)} = \frac{A}{s-a} + \frac{B}{s-b}$
• Repeated Factors: $\frac{1}{(s-a)^2} = \frac{A}{s-a} + \frac{B}{(s-a)^2}$
• Quadratic Factors: $\frac{1}{(s-a)(s^2+b)} = \frac{A}{s-a} + \frac{Bs+C}{s^2+b}$

Convolution is a mathematical way of combining two functions to form a third. It allows us to find the inverse transform of the product of two functions.

It is very useful when partial fractions are too difficult (e.g., finding L^{-1}\left\\{ \frac{1}{(s^2+a^2)^2} \right\\}).

Solve: L^{-1}\left\\{ \frac{1}{(s-1)(s-2)} \right\\} using Convolution.

• Let $F(s) = \frac{1}{s-1} \Rightarrow f(t) = e^t$
• Let $G(s) = \frac{1}{s-2} \Rightarrow g(t) = e^{2t}$
• By Convolution: $\int_0^t f(u) g(t-u) \\, du = \int_0^t e^u \cdot e^{2(t-u)} \\, du$
• $= \int_0^t e^u \cdot e^{2t} \cdot e^{-2u} \\, du = e^{2t} \int_0^t e^{-u} \\, du$
• $= e^{2t} [-e^{-u}]_0^t = e^{2t} [-e^{-t} - (-1)]$
• $= e^{2t}(1 - e^{-t}) = e^{2t} - e^t$

(You would get the exact same answer using partial fractions: $\frac{1}{s-2} - \frac{1}{s-1}$).

This is the ultimate application of the Laplace transform.

1. 1.Take the Laplace Transform of both sides of the differential equation.
2. 2.Use the derivative formulas: replace $y''$ with $s^2Y(s) - sy(0) - y'(0)$ and $y'$ with $sY(s) - y(0)$.
3. 3.Substitute the given Initial Conditions (e.g., $y(0)=1, y'(0)=0$).
4. 4.Solve algebraically for $Y(s)$.
5. 5.Take the Inverse Laplace Transform of $Y(s)$ using partial fractions to find $y(t)$.

Solve: $y'' + y = t$, given $y(0)=1, y'(0)=-2$

• Step 1 (Transform): $L\\{y''\\} + L\\{y\\} = L\\{t\\}$
• Step 2 (Formulas): $[s^2Y(s) - sy(0) - y'(0)] + Y(s) = \frac{1}{s^2}$
• Step 3 (Substitute ICs): $s^2Y(s) - s(1) - (-2) + Y(s) = \frac{1}{s^2}$
• Step 4 (Solve for Y): $Y(s)(s^2 + 1) = \frac{1}{s^2} + s - 2$
  $Y(s) = \frac{1}{s^2(s^2+1)} + \frac{s}{s^2+1} - \frac{2}{s^2+1}$
• Step 5 (Inverse): Use partial fractions on the first term: $\frac{1}{s^2} - \frac{1}{s^2+1}$.
  $Y(s) = \frac{1}{s^2} - \frac{3}{s^2+1} + \frac{s}{s^2+1}$
• Final Answer: $y(t) = t - 3\sin(t) + \cos(t)$

A function is periodic if $f(t) = f(t+T)$ for all $t$. The constant $T$ is the period.

This formula is crucial for analyzing square waves and sawtooth waves in AC circuits.

The Dirac Delta function $\delta(t-a)$ represents an infinitely large force acting for an infinitely short time (like a hammer strike).

• $\delta(t-a) = \infty$ at $t=a$, and $0$ everywhere else.
• The area under the curve is exactly 1: $\int_0^\infty \delta(t-a) \\, dt = 1$.

Laplace Transform of Dirac Delta:

$$L\\{\delta(t-a)\\} = e^{-as}$$

If the impulse is at $t=0$, then $L\\{\delta(t)\\} = e^0 = 1$.

⇒Before the exam, ensure you can:

• Write the Laplace transform for $\sin(at)$ and $\cosh(at)$.
• Apply the First Shifting theorem to $e^{at}t^2$.
• State the formula for Convolution theorem.
• Perform partial fraction decomposition for repeated and quadratic roots.
• Write the $L\\{f''(t)\\}$ formula including initial conditions.
• Apply the periodic function formula to a basic square wave.