\magnification 2200
\parindent 0pt
\parskip 10pt
\pageno=1
\hsize 7truein
\vsize 9.8truein
\def\u{\vskip -10pt}
\def\v{\vskip -6pt}
\def\s{\vskip -1pt}

6.5:  Impulse functions

Unit impulse function = Dirac delta function is a generalized function with the properties

\vskip 5 pt
\centerline{$\delta(t) =
 0, ~~~~t \not= 0$}
\vskip 8 pt

\centerline{$\int_{-\infty}^\infty \delta (t) dt = 1$}


Let $d_\tau (t) = \cases{{1 \over 2 \tau} & $ -\tau < t < \tau $
\cr 0 & $t \leq -\tau$ or $t \geq \tau$}$

Note $lim_{\tau \rightarrow 0} d_\tau (t) = 0$ if $t \not= 0$

and $lim_{\tau \rightarrow 0}\int_{-\infty}^\infty  d_\tau (t) = lim_{\tau \rightarrow 0}= 1 = \int_{-\infty}^\infty \delta 
(t) dt$


${\cal L}(\delta(t - t_0)) = lim_{\tau \rightarrow 0}{\cal L}(d_\tau(t - t_0))$

$~~~~~~~~~~~~~~~~= lim_{\tau \rightarrow 0}\int_0^\infty e^{-st}d_\tau(t - t_0)dt$

$~~~~~~~~~~~~~~~~= lim_{\tau \rightarrow 0}{1 \over 2 \tau} \int_{t_0 - \tau}^{t_0 + \tau} e^{-st}dt$

$~~~~~~~~~~~~~~~~= lim_{\tau \rightarrow 0}{-1 \over 2s \tau} e^{-st} |_{t_0 - \tau}^{t_0 + \tau} $

$~~~~~~~~~~~~~~~~= lim_{\tau \rightarrow 0}{1 \over 2s \tau} e^{-st_0} ( e^{s\tau} - e^{-s\tau}) $

$~~~~~~~~~~~~~~~~= lim_{\tau \rightarrow 0}{sinh s \tau \over s \tau} e^{-st_0 } $
$= lim_{\tau \rightarrow 0}{ s cosh s \tau \over s } e^{-st_0 } $

$~~~~~~~~~~~~~~~~= e^{-st_0 } $

\end
6.6:  The Convolution Integral

Defn: The convolution of $f$ and $g$ is the function $f * g$ defined by

\vskip 5pt

\centerline{$(f * g)(t) = \int_0^t f(t - s)g(s)ds = \int_0^t f(x)g(t - x)dx  $}

Note $*$ is

1.)  commutative: $f * g = g * f$

2.) associative:  $(f*g)*h = f*(g*h)$

3.)  distributive w.r.t $+$:  $f*(g_1 + g_2) = f*g_1 + f* g_2$

4.)  $f*0 = 0*f = 0$

Example:  $cos(t) * 1 =$

\vskip 30pt

Example:  $sin(t) * sin(t) \not\geq 0$

\eject
Thm: ${\cal L}((f*g)(t)) = {\cal L}(f(t)){\cal L}(g(t))$
\vskip -5pt

Proof:
\vskip -5pt



${\cal L}(f(t)){\cal L}(g(t)) 
= \int_0^\infty e^{-sy}f(y)dy \int_0^\infty e^{-sx}g(x)dx$
\s
$= \int_0^\infty [\int_0^\infty e^{-sy}f(y)dy] e^{-sx}g(x)  dx$
\s

$= \int_0^\infty [\int_0^\infty e^{-sy}f(y) e^{-sx}g(x) dy] dx$
\s


$= \int_0^\infty [\int_0^\infty e^{-s(y + x)}f(y) g(x) dy] dx$
\s

$= \int_0^\infty [\int_0^\infty e^{-s(y + x)}f(y) g(x) dx] dy$
\s


Let $t = x + y$, $dt = dx$
\s

$= \int_0^\infty [\int_y^\infty e^{-s(y + t - y)}f(y) g(t-y) dt] dy$
\s

$= \int_0^\infty [\int_y^\infty e^{-st}f(y) g(t-y) dt] dy$
\s

$= \int_0^\infty [\int_0^t e^{-st}f(y) g(t-y) dy] dt$
\s

$= \int_0^\infty e^{-st} [\int_0^t f(y) g(t-y) dy] dt$
\s

$= \int_0^\infty e^{-st} (f*g)(t) dt$
\s

$= {\cal L}(f*g)$

Example:  ${\cal L}^{-1}({1 \over s (s - a)}) = $


\end