Biomedical engineers in Kenya frequently encounter dynamic systems such as physiological signals and medical devices whose behaviour is best analysed in the frequency domain. The Laplace transform is a powerful mathematical tool that converts complex time-domain differential equations into simpler algebraic equations in the s-domain. Mastering Laplace transforms enables biomedical engineers to model, analyse, and design control systems for biomedical instrumentation, signal processing, and system dynamics effectively.

4.1 Meaning of Laplace transforms deriving Laplace transforms from first principles

Laplace transforms convert a time-domain function \( f(t) \) defined for \( t \geq 0 \) into a complex frequency-domain function \( F(s) \). This transformation simplifies solving differential equations encountered in biomedical systems such as heart rate variability analysis and drug delivery dynamics. The Laplace transform is defined by the integral

$$ F(s) = \int_0^\infty f(t) e^{-st} dt $$

where \( s \) is a complex number, \( s = \sigma + j\omega \). This integral transforms \( f(t) \) into a function of \( s \), capturing the system’s frequency characteristics. Deriving Laplace transforms from first principles involves evaluating this integral for specific functions \( f(t) \).

4.1.1 Definition and fundamental formula of Laplace transform

The Laplace transform of a function \( f(t) \) is defined as

$$ F(s) = \int_0^\infty f(t) e^{-st} dt $$

where \( t \) represents time (seconds), and \( s \) is a complex frequency parameter with units of reciprocal seconds. \( F(s) \) is the image of \( f(t) \) in the complex frequency domain. This transformation assumes \( f(t) = 0 \) for \( t < 0 \), consistent with causal biomedical signals such as ECG or EEG.

Worked Examples

Example 1: Find the Laplace transform of \( f(t) = 1 \) for \( t \geq 0 \).

Given: \( f(t) = 1 \), \( s > 0 \)

$$ F(s) = \int_0^\infty 1 \times e^{-st} dt $$

$$ = [ \frac{-1}{s} e^{-st} ]_0^\infty $$

Evaluate the limits:

At \( t = \infty \), \( e^{-s \times \infty} = 0 \) (since \( s > 0 \))

At \( t = 0 \), \( e^{0} = 1 \)

$$ F(s) = 0 - ( \frac{-1}{s} \times 1 ) = \frac{1}{s} $$

Answer: \( F(s) = \frac{1}{s} \), \( s > 0 \)

Example 2: Find the Laplace transform of \( f(t) = t \), \( t \geq 0 \).

Given: \( f(t) = t \), \( s > 0 \)

$$ F(s) = \int_0^\infty t e^{-st} dt $$

Use integration by parts:

Let \( u = t \Rightarrow du = dt \)

Let \( dv = e^{-st} dt \Rightarrow v = \frac{-1}{s} e^{-st} \)

$$ F(s) = uv \Big|_0^\infty - \int_0^\infty v du = ( t \times \frac{-1}{s} e^{-st} )_0^\infty + \frac{1}{s} \int_0^\infty e^{-st} dt $$

Evaluate the boundary term:

At \( t = \infty \), \( t e^{-st} \to 0 \) (exponential dominates)

At \( t = 0 \), term is zero

So boundary term is 0.

Now evaluate the integral:

$$ \frac{1}{s} \times \frac{1}{s} = \frac{1}{s^2} $$

Answer: \( F(s) = \frac{1}{s^2} \), \( s > 0 \)

Example 3: Find the Laplace transform of \( f(t) = e^{at} \), \( t \geq 0 \), where \( a \) is a constant.

Given: \( f(t) = e^{at} \), \( s > a \)

$$ F(s) = \int_0^\infty e^{at} e^{-st} dt = \int_0^\infty e^{-(s, a)t} dt $$

$$ = [ \frac{-1}{s - a} e^{-(s, a)t} ]_0^\infty $$

At \( t = \infty \), \( e^{-(s, a)\infty} = 0 \) (since \( s > a \))

At \( t = 0 \), \( e^0 = 1 \)

$$ F(s) = 0 - ( \frac{-1}{s - a} \times 1 ) = \frac{1}{s - a} $$

Answer: \( F(s) = \frac{1}{s - a} \), \( s > a \)

4.1.2 Derivation of Laplace transform for piecewise functions relevant to biomedical signals

Biomedical signals such as heartbeats or respiratory signals often exhibit piecewise behaviour. The Laplace transform for piecewise functions requires splitting the integral at points where the function changes. For a function defined as

$$ f(t) = \begin{cases} f_1(t), & 0 \leq t < T \\ f_2(t), & t \geq T \end{cases} $$

the Laplace transform is

$$ F(s) = \int_0^T f_1(t) e^{-st} dt + \int_T^\infty f_2(t) e^{-st} dt $$

Worked Examples

Example 1: Find the Laplace transform of

$$ f(t) = \begin{cases} 0, & 0 \leq t < 2 \\ 1, & t \geq 2 \end{cases} $$

Given: \( f(t) \) as above, \( s > 0 \)

$$ F(s) = \int_0^2 0 \times e^{-st} dt + \int_2^\infty 1 \times e^{-st} dt = 0 + \int_2^\infty e^{-st} dt $$

$$ = [ \frac{-1}{s} e^{-st} ]_2^\infty = 0 - ( \frac{-1}{s} e^{-2s} ) = \frac{e^{-2s}}{s} $$

Answer: \( F(s) = \frac{e^{-2s}}{s} \), \( s > 0 \)

Example 2: Find the Laplace transform of

$$ f(t) = \begin{cases} t, & 0 \leq t < 1 \\ 1, & t \geq 1 \end{cases} $$

Given: \( f(t) \) as above, \( s > 0 \)

$$ F(s) = \int_0^1 t e^{-st} dt + \int_1^\infty 1 \times e^{-st} dt $$

First integral (using integration by parts as in 4.1.1 Example 2):

$$ \int_0^1 t e^{-st} dt = [ \frac{-t}{s} e^{-st} ]_0^1 + \frac{1}{s} \int_0^1 e^{-st} dt $$

$$ = \frac{-1}{s} e^{-s} + \frac{1}{s} [ \frac{-1}{s} e^{-st} ]_0^1 = \frac{-1}{s} e^{-s} + \frac{1}{s} ( \frac{-1}{s} e^{-s} + \frac{1}{s} ) $$

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

Second integral:

$$ \int_1^\infty e^{-st} dt = [ \frac{-1}{s} e^{-st} ]_1^\infty = 0 - ( \frac{-1}{s} e^{-s} ) = \frac{e^{-s}}{s} $$

Add both:

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

Answer: \( F(s) = \frac{1}{s^2} ( 1 - e^{-s} ) \), \( s > 0 \)

Example 3: Find the Laplace transform of a biomedical pulse signal modeled as

$$ f(t) = \begin{cases} 5, & 0 \leq t < 0.5 \\ 0, & t \geq 0.5 \end{cases} $$

Given: \( f(t) \), \( s > 0 \)

$$ F(s) = \int_0^{0.5} 5 e^{-st} dt + \int_{0.5}^\infty 0 dt = 5 \int_0^{0.5} e^{-st} dt $$

$$ = 5 [ \frac{-1}{s} e^{-st} ]_0^{0.5} = 5 ( \frac{-1}{s} e^{-0.5 s} + \frac{1}{s} ) = \frac{5}{s} ( 1 - e^{-0.5 s} ) $$

Answer: \( F(s) = \frac{5}{s} ( 1 - e^{-0.5 s} ) \), \( s > 0 \)

4.1.3 Deriving Laplace transforms of derivatives and integrals from first principles

Many biomedical engineering problems involve differential equations describing system dynamics. The Laplace transform of derivatives and integrals converts these into algebraic equations. From the definition,

$$ L \{ f'(t) \} = \int_0^\infty f'(t) e^{-st} dt $$

Using integration by parts and assuming \( f(t) \) is of exponential order,

$$ L \{ f'(t) \} = s F(s) - f(0) $$

Similarly for integrals,

$$ L \{ \int_0^t f(\tau) d\tau \} = \frac{1}{s} F(s) $$

Worked Examples

Example 1: Find the Laplace transform of \( f'(t) \) where \( f(t) = t^2 \), \( f(0) = 0 \).

Given: \( f(t) = t^2 \), so \( f'(t) = 2t \), \( f(0) = 0 \)

First find \( F(s) \):

$$ F(s) = \int_0^\infty t^2 e^{-st} dt $$

Using standard Laplace transform table, \( L \{ t^n \} = \frac{n!}{s^{n+1}} \), so

$$ F(s) = \frac{2!}{s^{3}} = \frac{2}{s^{3}} $$

Then,

$$ L \{ f'(t) \} = s F(s) - f(0) = s \times \frac{2}{s^3} - 0 = \frac{2}{s^2} $$

Directly, \( f'(t) = 2t \), and \( L \{ 2t \} = 2 \times \frac{1}{s^2} = \frac{2}{s^2} \)

Answer: \( L \{ f'(t) \} = \frac{2}{s^2} \), \( s > 0 \)

Example 2: Find the Laplace transform of the integral \( g(t) = \int_0^t e^{2 \tau} d\tau \).

Given: \( f(\tau) = e^{2 \tau} \)

First find \( F(s) = L \{ e^{2 t} \} = \frac{1}{s - 2} \), \( s > 2 \)

Then,

$$ L \{ g(t) \} = \frac{1}{s} F(s) = \frac{1}{s} \times \frac{1}{s - 2} = \frac{1}{s(s - 2)} $$

Answer: \( L \{ g(t) \} = \frac{1}{s(s - 2)} \), \( s > 2 \)

Example 3: Given \( f(t) = \sin(\omega t) \), find \( L \{ f'(t) \} \) knowing \( f(0) = 0 \).

First,

$$ F(s) = L \{ \sin(\omega t) \} = \frac{\omega}{s^2 + \omega^2} $$

Then,

$$ L \{ f'(t) \} = s F(s) - f(0) = s \times \frac{\omega}{s^2 + \omega^2} - 0 = \frac{s \omega}{s^2 + \omega^2} $$

Answer: \( L \{ f'(t) \} = \frac{s \omega}{s^2 + \omega^2} \), \( s > 0 \)

4.1.4 Using Laplace transform properties to derive transforms of common biomedical functions

The Laplace transform has properties such as linearity, frequency shifting, and time shifting that simplify derivations for biomedical signals like step inputs, impulses, and exponentially decaying signals. These properties follow from the integral definition and help build transforms for complex signals encountered in medical instrumentation.

Key properties:

• Linearity: \( L \{ a f(t) + b g(t) \} = a F(s) + b G(s) \)
• Time shifting: \( L \{ f(t, a) u(t, a) \} = e^{-a s} F(s) \) where \( u(t) \) is the unit step
• Frequency shifting: \( L \{ e^{a t} f(t) \} = F(s, a) \)

Worked Examples

Example 1: Using linearity, find the Laplace transform of \( f(t) = 3 + 2t \).

Given: \( L \{ 1 \} = \frac{1}{s} \), \( L \{ t \} = \frac{1}{s^2} \)

$$ F(s) = 3 \times \frac{1}{s} + 2 \times \frac{1}{s^2} = \frac{3}{s} + \frac{2}{s^2} $$

Answer: \( F(s) = \frac{3}{s} + \frac{2}{s^2} \), \( s > 0 \)

Example 2: Using time shifting, find the Laplace transform of \( f(t) = u(t - 3) \) where \( u(t) \) is the unit step function.

Given: \( L \{ u(t) \} = \frac{1}{s} \)

By time shifting,

$$ F(s) = e^{-3 s} \times \frac{1}{s} $$

Answer: \( F(s) = \frac{e^{-3 s}}{s} \), \( s > 0 \)

Example 3: Using frequency shifting, find the Laplace transform of \( f(t) = e^{4 t} \sin(2 t) \).

Given: \( L \{ \sin(2 t) \} = \frac{2}{s^2 + 4} \)

Frequency shift by \( a = 4 \):

$$ F(s) = \frac{2}{(s - 4)^2 + 4} $$

Answer: \( F(s) = \frac{2}{(s - 4)^2 + 4} \), \( s > 4 \)

Example 4: Find the Laplace transform of \( f(t) = t e^{3 t} \).

Using frequency shift property on \( L \{ t \} = \frac{1}{s^2} \):

$$ F(s) = \frac{1}{(s - 3)^2} $$

Answer: \( F(s) = \frac{1}{(s - 3)^2} \), \( s > 3 \)

Chapter Summary

This chapter introduced the concept of Laplace transforms, explaining their meaning and demonstrating how they can be derived from first principles. The fundamental properties of Laplace transforms were outlined to provide a basis for manipulating and simplifying transforms in various contexts. Methods for determining the inverse Laplace transform were discussed, including the use of partial fraction decomposition to handle complex expressions. The chapter then showed how Laplace transforms can be applied to solve ordinary differential equations efficiently. Extending this, the solution of simultaneous differential equations using given initial conditions was explored, emphasizing practical problem-solving techniques. Finally, the chapter highlighted the application of Laplace transforms in electrical circuits, illustrating their value in analyzing circuit behavior and system responses.