Chapter 3 · Laplace Transform Applications

Step Functions & Convolution

Handling Discontinuous Inputs & System Response

Dr. Mohamed Mabrok · Qatar University

The Big Picture: Chapter 3 Roadmap

Definition
✓ Done

Inverse
✓ Done

Laplace Transform

Solving ODEs
✓ Done

Step Functions &
Convolution
← THIS LECTURE

Where We Are

      We've mastered the Laplace transform. Now we unlock two powerful tools: handling discontinuous inputs with step functions and modeling system impulse response with convolution.

The Unit Step Function

Definition

      $$u(t - \tau) = \begin{cases} 0, & t < \tau \\ 1, & t \geq \tau \end{cases}$$
    
      A switch that turns ON at time $\tau$. Special case: $u(t)$ turns on at $t = 0$.

Physical Interpretation

      The step function models a sudden change: a light switch flipping on, a constant force suddenly applied to a system, or a voltage step in a circuit.

Using Step Functions to Build Piecewise Functions

      $$f(t) \cdot u(t-a) = \begin{cases} 0, & t < a \\ f(t), & t \geq a \end{cases}$$
    
      Multiply any function by $u(t-a)$ to "turn it on" at $t = a$.

Converting Piecewise Functions to Step Functions

General Conversion Formula

      $$f(t) = \begin{cases} f_1(t), & 0 \leq t < a \\ f_2(t), & t \geq a \end{cases} = f_1(t) + [f_2(t) - f_1(t)]u(t-a)$$

Worked Example

      $$f(t) = \begin{cases} 2, & 0 \leq t < 3 \\ 5, & t \geq 3 \end{cases}$$
    
      Start at $2$, then add $(5 - 2) = 3$ at $t = 3$:
    
      $$f(t) = 2 + 3u(t-3)$$

Key insight: We can express ANY piecewise function as a linear combination of step functions. This is crucial for applying the Laplace transform!

Laplace Transform of the Step Function

The Transform

      $$\mathcal{L}\{u(t-\tau)\} = \frac{e^{-\tau s}}{s}$$
    
      Special case: $\mathcal{L}\{u(t)\} = \frac{1}{s}$ (delay of 0).

Intuition

      The factor $e^{-\tau s}$ encodes the delay $\tau$ in the s-domain. The $\frac{1}{s}$ is the transform of a constant function (1) in the time domain.

Verification

      $$\mathcal{L}\{u(t-\tau)\} = \int_0^\infty e^{-st}u(t-\tau)\,dt = \int_\tau^\infty e^{-st}\,dt = \left[-\frac{e^{-st}}{s}\right]_\tau^\infty = \frac{e^{-\tau s}}{s}$$

Second Shifting Theorem: Time Delay

The Theorem

      $$\mathcal{L}\{f(t-\tau)u(t-\tau)\} = e^{-\tau s}F(s)$$
    
      where $F(s) = \mathcal{L}\{f(t)\}$.

Inverse Form

      $$\mathcal{L}^{-1}\{e^{-\tau s}F(s)\} = f(t-\tau)u(t-\tau)$$

The Big Picture

      Multiplication by $e^{-\tau s}$ in the s-domain = delay by $\tau$ in the time domain. This is one of the most powerful tools in control theory and signal processing!

First vs Second Shifting: Side by Side

First Shifting (s-shift)

        Exponential multiplication in time
      
        $$e^{at}f(t) \leftrightarrow F(s-a)$$
      
        Shift the entire s-domain graph left or right by $a$.

Second Shifting (t-shift)

        Time delay in time domain
      
        $$f(t-\tau)u(t-\tau) \leftrightarrow e^{-\tau s}F(s)$$
      
        Multiply by exponential $e^{-\tau s}$ in s-domain.

Remember

      Both are essential. First shifting handles exponential growth/decay. Second shifting handles sudden changes and delays in real-world systems.

Worked Example

Piecewise Function & Its Laplace Transform

STEP 1 Set up the piecewise function

f(t) = \begin{cases} t, & 0 \leq t < 2 \\ 4, & t \geq 2 \end{cases}

STEP 2 Convert to step functions

$$f(t) = t + (4 - t)u(t-2) = t - (t-2)u(t-2) - 2u(t-2) + 4u(t-2)$$

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

STEP 3 Apply the Laplace transform term by term

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

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

Worked Examples

Applying the Second Shifting Theorem

Example 1: Find $\mathcal{L}\{(t-3)^2 u(t-3)\}$

\mathcal{L}\{t^2\} = \frac{2}{s^3}

\mathcal{L}\{(t-3)^2 u(t-3)\} = e^{-3s} \cdot \frac{2}{s^3}

Example 2: Find $\mathcal{L}\{e^{2(t-1)}u(t-1)\}$

\mathcal{L}\{e^{2t}\} = \frac{1}{s-2}

\mathcal{L}\{e^{2(t-1)}u(t-1)\} = e^{-s} \cdot \frac{1}{s-2}

Worked Example

Finding the Inverse Laplace Transform with Delay

STEP 1 Identify the exponential delay

\mathcal{L}^{-1}\left\{\frac{e^{-2s}}{s^2+4}\right\}

STEP 2 Find the inverse of $F(s)$ (without the exponential)

\mathcal{L}^{-1}\left\{\frac{1}{s^2+4}\right\} = \frac{1}{2}\sin(2t)

STEP 3 Apply the second shifting theorem: replace $t$ with $(t-2)$ and multiply by $u(t-2)$

f(t) = \frac{1}{2}\sin(2(t-2))u(t-2)

Interpretation

      The sine wave is delayed by 2 seconds and doesn't start until $t = 2$.

Worked Example

Solving an ODE with a Step Input

SETUP IVP: $y'' + y = u(t-\pi)$, $y(0)=0$, $y'(0)=1$

STEP 1 Transform both sides (using shifting theorem on the right)

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

STEP 2 Solve for Y(s)

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

STEP 3 Use partial fractions on $\frac{1}{s(s^2+1)}$

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

STEP 4 Inverse transform

y(t) = \sin(t) + [1 - \cos(t-\pi)]u(t-\pi) = \sin(t) + [1 + \cos(t)]u(t-\pi)

Convolution: Definition & Intuition

Convolution Integral

      $$(f * g)(t) = \int_0^t f(\tau)g(t-\tau)\,d\tau$$
    
      Integration from 0 to $t$ (causal systems: nothing happens before $t = 0$).

Properties of Convolution

      Commutative: $f * g = g * f$
    
      Associative: $(f * g) * h = f * (g * h)$
    
      Distributive: $f * (g + h) = f * g + f * h$

Intuitive Meaning

      The result at time $t$ is a weighted average of the input $f(\tau)$ at all past times $\tau$, weighted by how the system $g$ responds to impulses that happened at time $\tau$.

The Convolution Theorem: Time to s-Domain Magic

The Theorem

      $$\mathcal{L}\{(f * g)(t)\} = F(s) \cdot G(s)$$
    
      where $F(s) = \mathcal{L}\{f(t)\}$ and $G(s) = \mathcal{L}\{g(t)\}$.

Inverse Form

      $$\mathcal{L}^{-1}\{F(s) \cdot G(s)\} = (f * g)(t)$$

The Power

      Product in s-domain = Convolution in time domain. This transforms difficult inverse Laplace transforms (products) into manageable integrals!

⚠️ Important

Don't confuse: $\mathcal{L}^{-1}\{F \cdot G\}$ is NOT $f \cdot g$ (pointwise product). It's $(f * g)$ (convolution)!
Convolution is an integral, not a simple multiplication.

Impulse Response & Linear Systems

The Impulse Function

      $$\mathcal{L}\{\delta(t)\} = 1, \quad \mathcal{L}\{\delta(t-a)\} = e^{-as}$$
    
      The Dirac delta $\delta(t)$ is an idealized instantaneous impulse.

Impulse Response

      A system with transfer function $G(s)$ (or differential equation coefficients) has impulse response:
    
      $$g(t) = \mathcal{L}^{-1}\{G(s)\}$$
    
      This is what comes out when you give the system a $\delta(t)$ impulse as input.

System Output Formula

      $$\text{Output} = y(t) = (f * g)(t)$$
    
      Input $f(t)$ convolved with impulse response $g(t)$ gives the output. This is the foundation of signal processing!

Worked Example

Computing a Convolution Directly

Problem: Compute $(1 * e^{2t})(t)$

STEP 1 Set up the integral

(1 * e^{2t})(t) = \int_0^t 1 \cdot e^{2(t-\tau)}\,d\tau

STEP 2 Factor out $e^{2t}$

= e^{2t} \int_0^t e^{-2\tau}\,d\tau

STEP 3 Integrate

= e^{2t} \left[-\frac{e^{-2\tau}}{2}\right]_0^t = e^{2t} \left(-\frac{e^{-2t}}{2} + \frac{1}{2}\right)

= \frac{1}{2}(e^{2t} - 1)

Verification via Convolution Theorem

      $$\mathcal{L}\{1\} \cdot \mathcal{L}\{e^{2t}\} = \frac{1}{s} \cdot \frac{1}{s-2} = \frac{1}{s(s-2)}$$
      $$\mathcal{L}^{-1}\left\{\frac{1}{s(s-2)}\right\} = \frac{1}{2}(e^{2t} - 1) \,\checkmark$$
    

Worked Example

Inverse Laplace Transform via Convolution

Problem: Find $\mathcal{L}^{-1}\left\{\dfrac{1}{(s+1)(s^2+1)}\right\}$

STEP 1 Split the product: $\frac{1}{s+1} \cdot \frac{1}{s^2+1}$

STEP 2 Find the inverse of each factor

f(t) = \mathcal{L}^{-1}\left\{\frac{1}{s+1}\right\} = e^{-t}$$ $$g(t) = \mathcal{L}^{-1}\left\{\frac{1}{s^2+1}\right\} = \sin(t)

STEP 3 Compute the convolution

(e^{-t} * \sin(t)) = \int_0^t e^{-\tau}\sin(t-\tau)\,d\tau

STEP 4 Use integration by parts twice (or a table)

y(t) = \frac{1}{2}(e^{-t} - \cos t + \sin t)

Key Takeaways & Summary

Step Functions

$u(t-\tau)$ represents a switch turning on at $t = \tau$. Piecewise functions → linear combinations of steps.

Second Shifting

$e^{-\tau s}F(s) \leftrightarrow f(t-\tau)u(t-\tau)$. Time delay in s-domain is exponential multiplication.

Convolution

$(f*g)(t) = \int_0^t f(\tau)g(t-\tau)d\tau$. Product in s-domain maps to time-domain convolution.

Convolution Theorem

$\mathcal{L}^{-1}\{F(s) \cdot G(s)\} = (f*g)(t)$. Powerful for inverse transforms of products.

Impulse Response

$g(t) = \mathcal{L}^{-1}\{G(s)\}$ tells us how a system responds to an instantaneous impulse.

System Output

$y(t) = (f*g)(t)$ is the output when input is $f(t)$ and impulse response is $g(t)$.

⚠️ Common Mistakes to Avoid

Forgetting the $u(t-\tau)$ factor: Shifted functions MUST include the step function. Otherwise they're active for all $t < 0$ too!
Confusing second shifting with first shifting: First = exponential in $t$. Second = delay in $t$.
Forgetting the limits of integration: Convolution is $\int_0^t$, not $\int_0^\infty$. We only "look back" to $t = 0$.
Thinking convolution is multiplication: $\mathcal{L}^{-1}\{F \cdot G\} = (f*g)$, NOT $f(t) \cdot g(t)$!

Next Up: Chapter Review & Exam Prep

      We've now mastered the Laplace transform from start to finish: definition, inverse, ODEs, step functions, and convolution. Time to review and prepare for assessments!