Examples

Let

$\displaystyle f(t)$	$\displaystyle =$	$\begin{displaymath}\begin{cases} 1 & \text{if $-T \leq t \leq T$.} \\ 0 & \text{otherwise} \end{cases}\end{displaymath}$
	$\displaystyle =$	$\displaystyle [u(t+T) - u(t-T)]$

Then

$\displaystyle {\cal L}[f(t)] = \frac{e^{sT}}{s} - \frac{e^{-sT}}{s} = \frac{e^{sT}-e^{-sT}}{s}, s \neq 0$

$\displaystyle F(i\omega) = {\cal L}[f(t)] \Bigl\lvert_{s = i\omega} = \frac{e^{... ...-e^{-i\omega T}}{i\omega} = \frac{2\text{sin}(\omega T)}{\omega}, \omega \neq 0$

and

$\displaystyle F(i\cdot 0) = \int_{-\infty}^\infty f(t) dt = \int_{-T}^T 1\cdot dt = 2T$

Problem 5.3 from Text.

$\displaystyle f(t)$	$\displaystyle =$	$\begin{displaymath}\begin{cases} 1-\frac{\vert t\vert}{T} & \text{if $-T \leq t \leq T$.} \\ 0 & \text{otherwise} \end{cases}\end{displaymath}$

Solution: Using

$\displaystyle \vert t\vert$

$\displaystyle =$

$\begin{displaymath}\begin{cases} - t & \text{if $t < 0$} \\ t & \text{if $t > 0$} \end{cases}\end{displaymath}$

and expressing

in terms of step functions, we get

$\displaystyle f(t)$	$\displaystyle =$	$\displaystyle \left(1+\frac{t}{T} \right) [u(t+T) - u(t)] + \left( 1-\frac{t}{T} \right)[u(t) - u(t-T)]$
	$\displaystyle =$	$\displaystyle \frac{(t+T)}{T}[u(t+T) - u(t)] - \frac{(t-T)}{T}[u(t) - u(t-T)]$
	$\displaystyle =$	$\displaystyle \frac{1}{T}\left[ (t+T)u(t+T) - 2tu(t) + (t-T)u(t-T) \right]$

Hence

$\displaystyle F(s) = \frac{1}{T}\left[\frac{e^{sT}}{s^2} -\frac{1}{s^2} + \frac{e^{-sT}}{s^2}\right] = \frac{e^{sT}-2+e^{-sT}}{s^2 T}, s\neq 0$

$\displaystyle F(i\omega) = {\cal L}[f(t)] \Bigl\lvert_{s = i\omega} = \frac{e^{... ...}}{(i\omega)^2 T} = \frac{2(1-\text{cos}(\omega T))}{\omega^2 T}, \omega \neq 0$

and

$\displaystyle F(i\cdot 0) = \int_{-\infty}^\infty f(t) dt = \int_{-T}^T f(t) dt = T$

Problem 5.1 (iii) from Text.

$\displaystyle f(t)$	$\displaystyle =$	$\begin{displaymath}\begin{cases} -\text{sin }t & \text{if } \pi \leq t \leq 2\pi \\ 0 & \text{otherwise} \end{cases}\end{displaymath}$

Solution:

$\displaystyle f(t)$	$\displaystyle =$	$\displaystyle (-$ sin $\displaystyle t)[u(t-\pi) - u(t-2\pi)]$
	$\displaystyle =$	$\displaystyle -$ sin $\displaystyle t u(t-\pi) +$ sin $\displaystyle t u(t-2\pi)$
	$\displaystyle =$	$\displaystyle -$ sin $\displaystyle (t-\pi+\pi) u(t-\pi) +$ sin $\displaystyle (t-2\pi+2\pi) u(t-2\pi)$
	$\displaystyle =$	sin $\displaystyle (t-\pi) u(t-\pi) +$ sin $\displaystyle (t-2\pi) u(t-2\pi)$

Hence

$\displaystyle F(s) = \frac{e^{-s\pi}\cdot 1}{s^2+1} + \frac{e^{-s2\pi}\cdot 1}{s^2+1^2} = \frac{e^{-s\pi}+e^{-s2\pi}}{s^2+1}, s \neq \pm i$

$\displaystyle F(i\omega) = {\cal L}[f(t)] \Bigl\lvert_{s = i\omega} = \frac{e^{... ...\pi}{2}}\cdot \frac{ 2\text{cos}(\omega\pi/2)}{1- \omega^2}, \omega \neq \pm 1$

For $\omega = \pm 1$ , we could first find $F(i\cdot 1)$ using $\omega = 1$ in the Fourier Transform integral:

$\displaystyle F(i\cdot 1)$	$\displaystyle =$	$\displaystyle \int_\pi^{2\pi} (-$ sin $\displaystyle t)e^{-i t} dt = (-1) \int_\pi^{2\pi} \left(\frac{e^{i t}-e^{-i t}}{2i}\right) e^{-i t} dt$
	$\displaystyle =$	$\displaystyle \left(\frac{-1}{2i}\right) \int_\pi^{2\pi} (1-e^{-2i t}) dt$
	$\displaystyle =$	$\displaystyle \frac{i\pi}{2}$

Then, since

is real,

$\displaystyle F(i\cdot (-1)) = \overline{F(i\cdot 1) } = -\frac{i\pi}{2}$

Alternately, we could also find $F(i\cdot 1)$ as follows:

$\displaystyle F(i\cdot 1)$	$\displaystyle =$	$\displaystyle \lim_{\omega\rightarrow 1} e^{-i\frac{\omega3\pi}{2}}\cdot \frac{... ...}\cdot \lim_{\omega\rightarrow 1} \frac{ 2\text{cos}(\omega\pi/2)}{1- \omega^2}$
	$\displaystyle =$	$\displaystyle i \cdot \lim_{\omega\rightarrow 1} \frac{ -\pi\text{sin}(\omega\pi/2)}{-2\omega} \text{ (using L'Hospital's rule)}$
	$\displaystyle =$	$\displaystyle \frac{i\pi}{2}$

EE102 Systems and Signals (Discussion)

Examples