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6.3 Step Functions

The unit step function \( u_c(t) \) is defined as:

\[ u_c(t) = \begin{cases} 0 & t < c \\ 1 & t \geq c \end{cases} \]
Graph of the unit step function u_c(t) which is 0 for t < c and jumps to 1 at t = c.

We can use that to model many discontinuous functions.

For example,

Graph of a rectangular pulse function f(t) that is 1 between t=0 and t=3, and 0 otherwise.
\[ f(t) = \begin{cases} 1 & 0 \leq t < 3 \\ 0 & t \geq 3 \end{cases} \]
\[ f(t) = 1 - u_3(t) \]
  • \( 0 \) before \( t = 3 \)
  • \( 1 \) at and after \( t = 3 \)
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Let's try

\[ f(t) = \begin{cases} 10 & 0 \leq t < 1 \\ -20 & 1 \leq t < 2 \\ 30 & 2 \leq t < 3 \\ -40 & t \geq 3 \end{cases} \]
Graph of a piecewise constant function with steps at t=1, 2, and 3, reaching values 10, -20, 30, and -40.
\[ f(t) = 10 + u_1(t)(-30) + u_2(t)(50) + u_3(t)(-70) \]

brings down by 30 at \( t = 1 \)

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Piecewise Functions and Unit Step Functions

Example: Trapezoidal Pulse

How about a function defined as follows:

\[ f(t) = \begin{cases} t & 0 < t < 1 \\ 1 & 1 \le t < 2 \\ 3-t & 2 \le t < 3 \\ 0 & t \ge 3 \end{cases} \]
Graph of a trapezoidal pulse function f(t) starting at (0,0), rising to (1,1), flat until (2,1), and falling to (3,0).

Representation using Unit Step Functions

We can express \( f(t) \) using unit step functions \( u_c(t) \). The logic involves adding terms to "reset" the previous behavior and "go to where I want" for the next interval.

\[ f(t) = t + u_1(t)(-t + 1) + u_2(t)(-1 + 3 - t) + u_3(t)(-3 + t) \]

Note: The term \( (-t+1) \) resets the initial slope to 0, and subsequent terms adjust the function to the desired piecewise segments.

Simplifying the expression:

\[ f(t) = t + u_1(t)(1 - t) + u_2(t)(2 - t) + u_3(t)(t - 3) \]

Verification

Check at \( t = 2.5 \):

\[ f(2.5) = 2.5 + (1 - 2.5) + (2 - 2.5) + 0(2.5 - 3) \]\[ = 0.5 \text{ (agrees w/ graph)} \]
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Laplace Transform of the Unit Step Function

Derivation

Definition of the unit step function:

\[ u_c(t) = \begin{cases} 0 & t < c \\ 1 & t \ge c \end{cases} \]

Calculating the Laplace transform \( \mathcal{L}\{u_c(t)\} \):

\[ \mathcal{L}\{u_c(t)\} = \int_{0}^{\infty} u_c(t) e^{-st} dt \]\[ = \int_{c}^{\infty} e^{-st} dt = \lim_{a \to \infty} \int_{c}^{a} e^{-st} dt \]\[ = \lim_{a \to \infty} \left. -\frac{1}{s} e^{-st} \right|_c^a = \lim_{a \to \infty} -\frac{1}{s} e^{-sa} + \frac{1}{s} e^{-cs} \]

Since \( \lim_{a \to \infty} e^{-sa} \to 0 \) if \( s > 0 \), we obtain the final result:

\[ \mathcal{L}\{u_c(t)\} = e^{-cs} \frac{1}{s} \]

Example Application

Given a piecewise constant function:

\[ f(t) = 10 + u_1(t)(-30) + u_2(t)(50) + u_3(t)(-70) \]\[ f(t) = 10 - 30 u_1(t) + 50 u_2(t) - 70 u_3(t) \]

The Laplace transform \( F(s) \) is:

\[ F(s) = \frac{10}{s} - 30 e^{-s} \frac{1}{s} + 50 e^{-2s} \frac{1}{s} - 70 e^{-3s} \frac{1}{s} \]
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Delayed Functions and Unit Step

The ramp one is more complicated. Let's try a simpler one first.

Coordinate graph showing a function that is zero until t=1, then increases linearly with slope 1.
\[ y = \begin{cases} 0 & t < 1 \\ t-1 & t \ge 1 \end{cases} \]

This is equivalent to delaying \( f(t) = t \) by one second. The delayed function has its \( t \) changed to \( t - 1 \) (shift to the RIGHT by 1: \( t \to t - 1 \)).

\[ y = u_c(t) f(t-c) \]

delay by \( c \)

the function activated shifted RIGHT by \( c \): \( t \to t - c \)

Laplace Transform of a Delayed Function

\[ \mathcal{L} \{ u_c(t) f(t-c) \} = ? \]
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Derivation of the Transform

\[ = \int_{0}^{\infty} u_c(t) f(t-c) e^{-st} dt = \int_{c}^{\infty} f(t-c) e^{-st} dt \]

Let \( T = t - c \), then \( dT = dt \).

\[ = \int_{0}^{\infty} f(T) e^{-s(T+c)} dT = e^{-cs} \underbrace{\int_{0}^{\infty} f(T) e^{-sT} dT}_{F(s)} \]
\[ \mathcal{L} \{ u_c(t) f(t-c) \} = e^{-cs} F(s) \]

\( \mathcal{L} \{ f(t) \} \) NOT \( \mathcal{L} \{ f(t-c) \} \)

Shift back to origin: \( t \to t + c \) THEN transform.

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Laplace Transform of a Shifted Function

Consider a piecewise function defined by a ramp starting at \( t = 1 \). We want to find its Laplace transform.

\[ f(t) = \begin{cases} 0 & t < 1 \\ t - 1 & t \geq 1 \end{cases} \]
Graph of f(t) showing a value of 0 for t less than 1, and a linear ramp with slope 1 starting at t=1.

Calculating the Transform

To transform this, we use the shifting property. The function is shifted by 1 unit to the right.

\[ F(s) = e^{-s} \mathcal{L} \{ t - 1 + 1 \} \]
Shifted back to origin
\[ = e^{-s} \mathcal{L} \{ t \} \]
\[ = e^{-s} \frac{1}{s^2} \]
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Example: Piecewise Linear Function

Let's try another example.

\[ y = \begin{cases} t & 0 < t < 2 \\ 2 & t \geq 2 \end{cases} \]

We can rewrite this using the unit step function \( u_2(t) \):

\[ y = t + u_2(t)(-t + 2) \]
Graph of y(t) which is a ramp y=t from t=0 to t=2, then stays constant at y=2 for t greater than 2.

Applying the Laplace Transform

\[ Y = \mathcal{L} \{ t \} + \mathcal{L} \{ u_2(t)(-t + 2) \} \]
Shift to origin: \( t \to t + 2 \)
\[ = \frac{1}{s^2} + e^{-2s} \mathcal{L} \{ -(t + 2) + 2 \} \]
Delay of 2 sec in s domain
\[ = \frac{1}{s^2} + e^{-2s} \mathcal{L} \{ -t \} = \frac{1}{s^2} - e^{-2s} \frac{1}{s^2} \]
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Laplace Transform Shifting Properties

Transformation Rules

  • t to s:
    \[ e^{-cs} \mathcal{L} \{ \text{function shifted LEFT by } c : t \to t+c \} \]
  • s to t: reverse the steps

    inverse transform, THEN shift RIGHT by: \( t \to t-c \)

Example Problem

\[ Y = \frac{1}{s^2} + e^{-2s} \frac{1}{s^2} \]
\[ y = \mathcal{L}^{-1} \left\{ \frac{1}{s^2} \right\} + \mathcal{L}^{-1} \left\{ e^{-2s} \frac{1}{s^2} \right\} \]

Annotation for the second term:

Inverse is \( t \), then shift RIGHT by 2: \( t \to t-2 \)

\[ = t + u_2(t)(t-2) \]
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Shifting in the s-domain

shift in \( t \) by \( c \) \( \to \) \( e^{-cs} \) something in \( s \)

shift in \( s \) by \( c \) \( \to \) ? in \( t \)

\( F(s-c) \to \) ? in \( t \)

Derivation using the Definition

\[ F(s) = \int_{0}^{\infty} f(t) e^{-st} dt \]
\[ \begin{aligned} F(s-c) &= \int_{0}^{\infty} f(t) e^{-(s-c)t} dt \\ &= \int_{0}^{\infty} [e^{ct} f(t)] e^{-st} dt \\ &= \mathcal{L} \{ e^{ct} f(t) \} \end{aligned} \]