Laplace transforms are powerful tools for simplifying complex expressions and solving differential equations. They convert time-domain functions into s-domain representations, making it easier to analyze and manipulate systems.

By applying properties like linearity, shifting, and differentiation, we can break down intricate problems into manageable parts. This approach streamlines the process of solving initial value problems and analyzing system behavior, including periodic functions and convolution.

Laplace Transform Properties

Simplifying Complex Expressions

• The linearity property simplifies the Laplace transform of a sum of functions
  • Expresses the Laplace transform of a sum as the sum of the individual Laplace transforms: $ℒ{𝑓(𝑡)+𝑔(𝑡)}=ℒ{𝑓(𝑡)}+ℒ{𝑔(𝑡)}$
  • Enables the Laplace transform of complex expressions to be broken down into simpler components
  • Example: $ℒ{3𝑒^(−2𝑡)+4𝑠𝑖𝑛(5𝑡)}=3ℒ{𝑒^(−2𝑡)}+4ℒ{𝑠𝑖𝑛(5𝑡)}$
• The shifting property relates the Laplace transform of a time-shifted function to the original function's Laplace transform
  • Introduces a complex exponential factor: $ℒ{𝑓(𝑡−𝑎)𝑢(𝑡−𝑎)}=𝑒^(−𝑎𝑠)ℒ{𝑓(𝑡)}$
  • Allows for the analysis of systems with time delays or shifted inputs
  • Example: $ℒ{𝑒^(−3(𝑡−2))𝑢(𝑡−2)}=𝑒^(−2𝑠)ℒ{𝑒^(−3𝑡)}$
• The scaling property relates the Laplace transform of a scaled function to the original function's Laplace transform
  • Involves a scaling factor in both the time and frequency domains: $ℒ{𝑓(𝑎𝑡)}=(1/𝑎)ℒ{𝑓(𝑡)}|_(𝑠/𝑎)$
  • Useful for analyzing systems with scaled inputs or time constants
  • Example: $ℒ{𝑐𝑜𝑠(2𝑡)}=(1/2)ℒ{𝑐𝑜𝑠(𝑡)}|_(𝑠/2)$
• The differentiation property expresses the Laplace transform of a function's derivative in terms of the original function's Laplace transform
  • Introduces the initial value of the function: $ℒ{𝑓'(𝑡)}=𝑠ℒ{𝑓(𝑡)}−𝑓(0^(−))$
  • Enables the solution of differential equations using algebraic methods in the s-domain
  • Example: $ℒ{𝑡𝑒^(−𝑡)}=ℒ{𝑡𝑒^(−𝑡)}'−𝑡𝑒^(−𝑡)|_(𝑡=0)=(−𝑑/𝑑𝑠)ℒ{𝑒^(−𝑡)}−0$

Combining Properties for Simplification

• These properties can be used in combination to simplify complex expressions involving Laplace transforms
  • Break down complex expressions into simpler components using linearity
  • Account for time shifts, scaling, and derivatives using the appropriate properties
  • Simplify the resulting expression in the s-domain
• Simplifying complex expressions makes it easier to solve differential equations and analyze system behavior
  • Convert differential equations into algebraic equations in the s-domain
  • Solve the algebraic equations using standard techniques
  • Interpret the results in the context of the original system or problem
• Example: Simplify $ℒ{3𝑒^(−2𝑡)𝑢(𝑡−1)+4𝑡𝑐𝑜𝑠(5𝑡)}$
  • Using linearity: $3ℒ{𝑒^(−2𝑡)𝑢(𝑡−1)}+4ℒ{𝑡𝑐𝑜𝑠(5𝑡)}$
  • Applying the shifting property: $3𝑒^(−𝑠)ℒ{𝑒^(−2𝑡)}+4ℒ{𝑡𝑐𝑜𝑠(5𝑡)}$
  • Using the differentiation property: $3𝑒^(−𝑠)/(𝑠+2)+4(−𝑑/𝑑𝑠)ℒ{𝑐𝑜𝑠(5𝑡)}$

Solving Initial Value Problems

Using Laplace Transforms

• Initial value problems involve solving differential equations with given initial conditions
  • Differential equations describe the behavior of a system over time
  • Initial conditions specify the state of the system at a particular time (usually t=0)
• The Laplace transform converts a differential equation into an algebraic equation in the complex frequency domain (s-domain)
  • Transforms the derivative terms into algebraic expressions involving s
  • Simplifies the process of solving the differential equation
• The initial conditions are incorporated into the s-domain equation using the differentiation property of Laplace transforms
  • The differentiation property introduces the initial values of the function and its derivatives
  • These initial values appear as additional terms in the s-domain equation
• Solving the resulting algebraic equation in the s-domain yields the Laplace transform of the solution to the differential equation
  • Use algebraic manipulation and partial fraction decomposition to solve for the Laplace transform of the solution
  • The solution in the s-domain represents the system's behavior in the frequency domain
• The inverse Laplace transform is then applied to obtain the time-domain solution to the initial value problem
  • The inverse Laplace transform converts the s-domain solution back to the time domain
  • The resulting time-domain function represents the system's behavior over time, satisfying the initial conditions

Applying the Initial Value Theorem

• The initial value theorem provides a shortcut for finding the initial value of a function without explicitly computing the inverse Laplace transform
  • Useful when only the initial value of the solution is required
  • Avoids the need for partial fraction decomposition and inverse Laplace transform tables
• The initial value theorem states that the initial value of a function can be found by evaluating the limit of the product of s and the Laplace transform of the function as s approaches infinity: $lim_(𝑠→∞)⁡𝑠ℒ{𝑓(𝑡)}=𝑓(0^(+))$
  • Multiply the Laplace transform of the function by s
  • Take the limit as s approaches infinity
  • The result is the initial value of the function
• Example: Find the initial value of the solution to the differential equation $𝑦''+3𝑦'+2𝑦=𝑒^(−𝑡)$, given $𝑦(0)=1$ and $𝑦'(0)=0$
  • Take the Laplace transform of the differential equation: $𝑠^2𝑌(𝑠)−𝑠𝑦(0)−𝑦'(0)+3[𝑠𝑌(𝑠)−𝑦(0)]+2𝑌(𝑠)=1/(𝑠+1)$
  • Substitute the initial conditions and simplify: $(𝑠^2+3𝑠+2)𝑌(𝑠)=1/(𝑠+1)+𝑠+3$
  • Apply the initial value theorem: $𝑦(0^(+))=lim_(𝑠→∞)⁡𝑠𝑌(𝑠)=lim_(𝑠→∞)⁡𝑠[(1/(𝑠+1)+𝑠+3)/(𝑠^2+3𝑠+2)]=1$

System Behavior Analysis

Utilizing the Convolution Property

• The convolution property relates the Laplace transform of the convolution of two functions to the product of their Laplace transforms: $ℒ{𝑓(𝑡)∗𝑔(𝑡)}=ℒ{𝑓(𝑡)}ℒ{𝑔(𝑡)}$
  • Convolution is a mathematical operation that combines two functions to produce a third function
  • In the context of systems, convolution represents the output of a linear time-invariant (LTI) system as a function of the input and the system's impulse response
• The convolution property allows for the analysis of LTI systems by simplifying the calculation of the system's output given its input and impulse response
  • The impulse response characterizes the system's behavior
  • The input function represents the signal or disturbance applied to the system
  • Convolving the input with the impulse response yields the system's output
• To analyze an LTI system using the convolution property:
  • Determine the impulse response of the system (usually denoted as h(t))
  • Take the Laplace transform of the impulse response to obtain H(s)
  • Multiply H(s) by the Laplace transform of the input function (X(s)) to find the Laplace transform of the output (Y(s))
  • Apply the inverse Laplace transform to Y(s) to obtain the time-domain output y(t)
• Example: Find the output of an LTI system with impulse response $ℎ(𝑡)=𝑒^(−2𝑡)𝑢(𝑡)$ when the input is $𝑥(𝑡)=𝑢(𝑡)$
  • Take the Laplace transform of the impulse response: $ℒ{𝑒^(−2𝑡)𝑢(𝑡)}=𝐻(𝑠)=1/(𝑠+2)$
  • Multiply H(s) by the Laplace transform of the input: $𝑌(𝑠)=𝐻(𝑠)𝑋(𝑠)=(1/(𝑠+2))(1/𝑠)=1/[𝑠(𝑠+2)]$
  • Apply the inverse Laplace transform to find the output: $𝑦(𝑡)=ℒ^(−1){𝑌(𝑠)}=1−𝑒^(−2𝑡)$

Analyzing Periodic Functions

• Periodic functions can be represented as a sum of sinusoids using Fourier series
  • Fourier series decompose a periodic function into a sum of sine and cosine functions with different frequencies and amplitudes
  • The Fourier series coefficients determine the contribution of each sinusoidal component to the overall function
• The Laplace transform of a periodic function can be expressed in terms of the Fourier series coefficients and the complex frequency s
  • The Laplace transform of a sine or cosine function is a rational function in s
  • The Laplace transform of a periodic function is the sum of the Laplace transforms of its Fourier series components
• The Laplace transform of a periodic function is useful in analyzing the steady-state behavior of systems subjected to periodic inputs
  • Steady-state behavior refers to the long-term response of a system after transient effects have died out
  • Periodic inputs, such as sinusoidal signals, result in periodic steady-state outputs
• To analyze the steady-state behavior of a system with periodic input:
  • Represent the periodic input as a Fourier series
  • Take the Laplace transform of the Fourier series to obtain the input's representation in the s-domain
  • Multiply the input's Laplace transform by the system's transfer function (the Laplace transform of the impulse response) to find the output's Laplace transform
  • Apply the inverse Laplace transform to the output's Laplace transform to obtain the time-domain steady-state output
• Example: Find the steady-state output of an LTI system with transfer function $𝐻(𝑠)=1/(𝑠+3)$ when the input is a square wave with amplitude 1 and period 2
  • The Fourier series of a square wave with amplitude 1 and period 2 is: $𝑥(𝑡)=(4/𝜋)∑_(𝑛=1,3,5,…)^∞▒[𝑠𝑖𝑛(𝑛𝜋𝑡)/𝑛]$
  • Take the Laplace transform of the Fourier series: $𝑋(𝑠)=(4/𝜋)∑_(𝑛=1,3,5,…)^∞▒[𝑛𝜋/(𝑠^2+𝑛^2 𝜋^2 )]$
  • Multiply X(s) by the transfer function to find the output's Laplace transform: $𝑌(𝑠)=𝐻(𝑠)𝑋(𝑠)=(4/𝜋)∑_(𝑛=1,3,5,…)^∞▒[𝑛𝜋/[(𝑠+3)(𝑠^2+𝑛^2 𝜋^2 )]]$
  • Apply the inverse Laplace transform to obtain the steady-state output: $𝑦(𝑡)=(4/𝜋)∑_(𝑛=1,3,5,…)^∞▒[(𝑛𝜋/(9+𝑛^2 𝜋^2 ))𝑠𝑖𝑛(𝑛𝜋𝑡)𝑒^(−3𝑡)]$