Download original notes
PAGE 1

6.1 (continued)

For \( f(t), t > 0 \), the Laplace transform is defined as:

\[ \mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} f(t) e^{-st} dt \]

\( f(t) \) must be piecewise continuous.

Last time: \( \mathcal{L}\{1\} = \frac{1}{s}, s > 0 \)

Now, let's try \( f(t) = t \)

\[ \mathcal{L}\{t\} = \int_{0}^{\infty} t e^{-st} dt \]

Using integration by parts (LIATE):

\( u = t \)
\( du = dt \)
\( dv = e^{-st} dt \)
\( v = -\frac{1}{s} e^{-st} \)

Formula: \( uv - \int v du \)

\[ = \lim_{a \to \infty} \int_{0}^{a} t e^{-st} dt \]\[ = \lim_{a \to \infty} \left( -\frac{t}{s} e^{-st} \Big|_{0}^{a} + \int_{0}^{a} \frac{1}{s} e^{-st} dt \right) \]\[ = \lim_{a \to \infty} \left( -\frac{t}{s} e^{-st} \Big|_{0}^{a} - \frac{1}{s^2} e^{-st} \Big|_{0}^{a} \right) \]
PAGE 2
\[ = \lim_{a \to \infty} \left( -\frac{a}{s} e^{-sa} - \frac{1}{s^2} e^{-sa} \right) + \frac{1}{s^2} \]

The limit exists only if \( s > 0 \) and the limit is 0.

\( \mathcal{L}\{t\} = \frac{1}{s^2}, s > 0 \)
\( \mathcal{L}\{1\} = \frac{1}{s}, s > 0 \)

Repeat for \( t^2, t^3, \dots \)

We found:

\( \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}, s > 0 \)

Linearity Property

If \( \mathcal{L}\{1\} = \frac{1}{s} \), then what is \( \mathcal{L}\{3\} = ? \)

\[ \mathcal{L}\{k\} = \int_{0}^{\infty} k \cdot e^{-st} dt = k \cdot \int_{0}^{\infty} 1 \cdot e^{-st} dt = k \cdot \mathcal{L}\{1\} = \frac{k}{s}, s > 0 \]

Same thing goes for \( t, t^n, \text{etc.} \)

\[ \mathcal{L}\{k \cdot f(t)\} = k \cdot \mathcal{L}\{f(t)\} \]
PAGE 3

Linearity of the Laplace Transform

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

Laplace Transform is linear

\[ \mathcal{L} \{ f(t) g(t) \} \neq \mathcal{L} \{ f(t) \} \mathcal{L} \{ g(t) \} \]

Example: Polynomial

\[ \begin{aligned} \mathcal{L} \{ 5t^2 - 3t + 10 \} &= 5 \mathcal{L} \{ t^2 \} - 3 \mathcal{L} \{ t \} + 10 \mathcal{L} \{ 1 \} \\ &= 5 \cdot \frac{2}{s^3} - 3 \cdot \frac{1}{s^2} + 10 \cdot \frac{1}{s} \quad (s > 0) \end{aligned} \]

Exponential Function

Exponential: \( f(t) = e^{at} \) where \( a \) is constant.

\[ \begin{aligned} \mathcal{L} \{ e^{at} \} &= \int_{0}^{\infty} e^{at} e^{-st} dt = \lim_{b \to \infty} \int_{0}^{b} e^{(a-s)t} dt \\ &= \lim_{b \to \infty} \left. \frac{1}{a-s} e^{(a-s)t} \right|_{0}^{b} \end{aligned} \]
PAGE 4
\[ \begin{aligned} &= \lim_{b \to \infty} \frac{1}{a-s} e^{(a-s)b} - \frac{1}{a-s} = \frac{1}{s-a} \end{aligned} \]

\[ \mathcal{L} \{ e^{at} \} = \frac{1}{s-a} \quad s > a \]

exists if \( a-s < 0 \)

\( s > a \)

if \( a=0 \) it reduces to:

\[ \mathcal{L} \{ 1 \} = \frac{1}{s} \quad s > 0 \]

Trigonometric Function

now \( f(t) = \cos(t) \)

\[ \mathcal{L} \{ \cos(t) \} = \int_{0}^{\infty} \cos(t) e^{-st} dt \]

one option: integration by parts

\( u = \cos(t) \), \( dv = e^{-st} dt \)

Euler's Identity Approach

Euler's identity:

  • \( e^{it} = \cos(t) + i \sin(t) \)
  • \( e^{-it} = \cos(t) - i \sin(t) \)

add:

\[ \begin{aligned} 2 \cos(t) &= e^{it} + e^{-it} \\ \cos(t) &= \frac{1}{2} e^{it} + \frac{1}{2} e^{-it} \end{aligned} \]
PAGE 5

Laplace Transform of Cosine

\[ \begin{aligned} \mathcal{L} \{ \cos(t) \} &= \frac{1}{2} \mathcal{L} \{ e^{it} \} + \frac{1}{2} \mathcal{L} \{ e^{-it} \} \\ &= \frac{1}{2} \frac{1}{s-i} + \frac{1}{2} \frac{1}{s+i} \\ &= \frac{1}{2} \left( \frac{1}{s-i} + \frac{1}{s+i} \right) = \frac{1}{2} \left( \frac{2s}{s^2+1} \right) = \frac{s}{s^2+1} \end{aligned} \]

Domain Restriction?

\[ \begin{aligned} \mathcal{L} \{ e^{it} \} &= \int_{0}^{\infty} e^{it} e^{-st} dt \\ &= \int_{0}^{\infty} e^{(i-s)t} dt \end{aligned} \]
\[ \int_{0}^{\infty} [\cos(t) + i \sin(t)] e^{-st} dt \]

The integral must be bounded:

  • \(\cos(t)\), \(\sin(t)\) are bounded
  • \(e^{-st}\) bounded only if \(s > 0\)
\[ \mathcal{L} \{ \cos(t) \} = \frac{s}{s^2+1} \quad s > 0 \]
PAGE 6
\[ \begin{aligned} \mathcal{L} \{ \cos(at) \} &= \frac{1}{2} \mathcal{L} \{ e^{(ia)t} \} + \frac{1}{2} \mathcal{L} \{ e^{-(ia)t} \} \\ &= \dots = \frac{s}{s^2+a^2} \quad s > 0 \end{aligned} \]

Similarly, we can find:

\[ \mathcal{L} \{ \sin(at) \} = \frac{a}{s^2+a^2} \quad s > 0 \]

What is Laplace transform doing?

\(f(t) \to F(s)\)

time domain \(\to\) frequency domain

There is no point to point correspondence between \(t\) and \(s\).

\(t = 10 \to s = ?\) meaningless question

Connection to Taylor Series

We've seen a similar thing in the case of Taylor series:

\[ 1 + x + x^2 + x^3 + \dots = \frac{1}{1-x} \quad |x| < 1 \]

\(\to\) takes the sequence \(\{1, 1, 1, 1, \dots\}\) and transforms it by attaching to \(x^n\) to turn into \(\frac{1}{1-x}\).

\[ \left\{ 1, 1, \frac{1}{2!}, \frac{1}{3!}, \frac{1}{4!}, \frac{1}{5!}, \dots \right\} = e^x \]
PAGE 7

What if we don't attach them to \( x^n \) but \( e^{-st} \) where \( t \) is a continuous variable

and the coefficients are \( f(t) \) at various \( t \)?

\[ f(0)e^{-s(0)} + f(t_1)e^{-st_1} + f(t_2)e^{-st_2} + \dots \]

\( \sum \) (sum) \( \rightarrow \) \( \int \) (integral) because \( t \) is continuous and not discrete.

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

So, take \( f(t) \), evaluate at each \( t \), attach to \( e^{-st} \) at same \( t \), add them all up