Review: The Unit Step Function

Recall from Section 7.1, the shifted unit step function \(u_a(t)\) is defined as:

\[ u_a(t) = u(t-a) = \begin{cases} 0 & t < a \\ 1 & t \ge a \end{cases} \]

If \(a = 0\), we have the standard unit step function \(u(t)\). The function \(u_a(t)\) is a piecewise function with a jump at \(t=a\).

Any piecewise continuous function \(f(t)\) can be written in terms of \(u_a(t)\) with different values of \(a\).

Example 1: Sketching a Piecewise Function

Sketch the graph of \( f(t) = u_2(t) - 2u_3(t) \) for \( t \ge 0 \).

Solution: We analyze the value of \(f(t)\) across different intervals based on the jumps at \(t=2\) and \(t=3\):

  • For \(0 \le t < 2\): \( f(t) = 0 - 2(0) = 0 \)
  • For \(2 \le t < 3\): \( f(t) = 1 - 2(0) = 1 \) (jump up 1 unit at \(t=2\))
  • For \(t \ge 3\): \( f(t) = 1 - 2(1) = -1 \) (jump down 2 units at \(t=3\))

The function \(f(t)\) has two jumps: at \(t=2\) and \(t=3\).

A piecewise step function graph of f(t) = u2(t) - 2u3(t). The function is zero from t=0 to 2. At t=2, it jumps up to +1 and remains constant until t=3. At t=3, it jumps down two units to -1 and stays at -1 for all t greater than 3. The jumps are indicated by dashed vertical lines and open/closed circles at the points of discontinuity.
Figure 1: Graph of the unit step function \(f(t) = u_2(t) - 2u_3(t)\). Note the jump up of 1 unit at \(t=2\) and the jump down of 2 units at \(t=3\).

Recall the Laplace transform of the unit step function:

\[ \mathcal{L}\{u(t-a)\} = \frac{e^{-as}}{s} \quad \text{and} \quad \mathcal{L}\{u(t)\} = \frac{1}{s} \]

Multiplication of the transform of \(u(t)\) by \(e^{-as}\) corresponds to a translation \(t \to t-a\).

(I) Translation on the t-axis

Theorem 1: Translation on the t-axis

If \( \mathcal{L}\{f(t)\} = F(s) \) exists for \( s > c \), then:

\[ \mathcal{L}\{u(t-a)f(t-a)\} = e^{-as}F(s) \]

And inversely:

\[ \mathcal{L}^{-1}\{e^{-as}F(s)\} = u(t-a)f(t-a) \]
Example 2

Find \( \mathcal{L}\{g(t)\} \) if \( g(t) = \begin{cases} 0 & \text{if } t < 3 \\ t^2 & \text{if } t \ge 3 \end{cases} \).

Solution: We can write \( g(t) = u(t-3)f(t-3) \). Since \( f(t-3) = t^2 \), we must find \( f(t) \):

\( \begin{aligned} f(t) &= (t+3)^2 \\ F(s) &= \mathcal{L}\{(t+3)^2\} = \mathcal{L}\{t^2 + 6t + 9\} \\ &= \frac{2}{s^3} + \frac{6}{s^2} + \frac{9}{s} \end{aligned} \)

Applying Theorem 1:

\[ \mathcal{L}\{g(t)\} = \mathcal{L}\{u(t-3)f(t-3)\} = e^{-3s} \left( \frac{2}{s^3} + \frac{6}{s^2} + \frac{9}{s} \right) \]
Example 3

Find \( \mathcal{L}\{f(t)\} \) if \( f(t) = \begin{cases} \sin t & \text{if } 0 \le t \le 3\pi \\ 0 & \text{if } t > 3\pi \end{cases} \).

Solution: Express \( f(t) \) using unit step functions:

\( \begin{aligned} f(t) &= \sin t - \sin t \cdot u(t-3\pi) \\ &= \sin t + \sin(t-3\pi) \cdot u(t-3\pi) \end{aligned} \)

Now, take the Laplace transform:

\( \begin{aligned} \mathcal{L}\{f(t)\} &= \mathcal{L}\{\sin t\} + e^{-3\pi s}\mathcal{L}\{\sin t\} \\ &= \frac{1}{s^2+1} + e^{-3\pi s}\left(\frac{1}{s^2+1}\right) \end{aligned} \)
\[ \mathcal{L}\{f(t)\} = \frac{1 + e^{-3\pi s}}{s^2+1} \]
Example 4

Find \( \mathcal{L}\{f(t)\} \) if \( f(t) = \begin{cases} \sin(2t) & \text{if } \pi \le t \le 2\pi \\ 0 & \text{if } t < \pi \text{ or } t > 2\pi \end{cases} \).

Solution: Express \( f(t) \) using unit step functions:

\( \begin{aligned} f(t) &= 0 + \sin(2t)u(t-\pi) - \sin(2t)u(t-2\pi) \\ &= \sin(2(t-\pi))u(t-\pi) - \sin(2(t-2\pi))u(t-2\pi) \end{aligned} \)

Now, take the Laplace transform:

\( \begin{aligned} \mathcal{L}\{f(t)\} &= e^{-\pi s}\mathcal{L}\{\sin(2t)\} - e^{-2\pi s}\mathcal{L}\{\sin(2t)\} \\ &= e^{-\pi s}\left(\frac{2}{s^2+4}\right) - e^{-2\pi s}\left(\frac{2}{s^2+4}\right) \end{aligned} \)
\[ \mathcal{L}\{f(t)\} = \frac{2(e^{-\pi s} - e^{-2\pi s})}{s^2+4} \]
Example 5

Find \( \mathcal{L}\{f(t)\} \) for \( f(t) = \begin{cases} t-1 & \text{if } t < 2 \\ 3-t & \text{if } 2 \le t \le 3 \\ 0 & \text{if } t > 3 \end{cases} \).

Solution: Express \( f(t) \) using unit step functions. The function starts as \(t-1\). At \(t=2\), we need to add a term to make it \(3-t\). At \(t=3\), we need to add a term to make it \(0\).

\( \begin{aligned} f(t) &= (t-1) + [(3-t) - (t-1)]u(t-2) + [0 - (3-t)]u(t-3) \\ &= (t-1) - 2(t-2)u(t-2) + (t-3)u(t-3) \end{aligned} \)

Now, take the Laplace transform:

\( \begin{aligned} \mathcal{L}\{f(t)\} &= \mathcal{L}\{t-1\} - 2e^{-2s}\mathcal{L}\{t\} + e^{-3s}\mathcal{L}\{t\} \\ &= \frac{1}{s^2} - \frac{1}{s} - \frac{2e^{-2s}}{s^2} + \frac{e^{-3s}}{s^2} \end{aligned} \)
\[\mathcal{L}\{f(t)\} = \frac{1 - s - 2e^{-2s} + e^{-3s}}{s^2} \]
Example 6: Inverse Transform with Translation

Find \( \mathcal{L}^{-1}\{F(s)\} \) for \( F(s) = e^{-3s}\frac{s+1}{s^2-8s+20} \).

Solution: First, complete the square in the denominator: \(s^2 - 8s + 20 = (s-4)^2 + 4 = (s-4)^2 + 2^2\).

\( \begin{aligned} F(s) &= e^{-3s}\frac{s-4+5}{(s-4)^2+2^2} \\ &= e^{-3s}\left[ \frac{s-4}{(s-4)^2+2^2} \right] + e^{-3s}\left[ \frac{5}{2} \cdot \frac{2}{(s-4)^2+2^2} \right] \end{aligned} \)

The inverse transform of the bracketed terms are \(e^{4t}\cos(2t)\) and \(\frac{5}{2}e^{4t}\sin(2t)\). The \(e^{-3s}\) factor means we apply the translation \(t \to t-3\) and multiply by \(u(t-3)\).

\[ f(t) = u(t-3)e^{4(t-3)}\left[ \cos(2(t-3)) + \frac{5}{2}\sin(2(t-3)) \right] \]
Example 7

Find the Laplace transform of \( f(t) = \begin{cases} 0 & t < 1 \\ t^2 e^{2t} & t \ge 1 \end{cases} \).

Solution: Express \( f(t) \) using a unit step function:

\( \begin{aligned} f(t) &= u(t-1)t^2 e^{2t} \\ &= u(t-1)[(t-1)^2 + 2(t-1) + 1]e^{2(t-1)} \cdot e^2 \\ &= e^2 [ u(t-1)(t-1)^2 e^{2(t-1)} + 2u(t-1)(t-1)e^{2(t-1)} + u(t-1)e^{2(t-1)} ] \end{aligned} \)

Now take the Laplace transform of each term using the translation theorem:

\[ \mathcal{L}\{f(t)\} = e^{2}e^{-s}\left[ \frac{2}{(s-2)^3} + \frac{2}{(s-2)^2} + \frac{1}{s-2} \right] \]

(II) Transforms of Periodic Functions

\( f(t) \) is defined to be periodic if there is a number \( P > 0 \) such that \( f(t+P) = f(t) \) for all \( t \ge 0 \). The least positive value of \( P \) is called the period of \( f \).

Theorem 2: Transforms of Periodic Functions

Let \( f(t) \) be periodic with period \( P \) and piecewise continuous for \( t \ge 0 \). Then the transform \( F(s) = \mathcal{L}\{f(t)\} \) exists for \( s > 0 \) and is given by:

\[ F(s) = \frac{1}{1 - e^{-Ps}} \int_{0}^{P} e^{-st} f(t) \, dt \]