EE102 Systems and Signals (Discussion)

Course description
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Examples

1. Problem 4.5 from Text:
   $$f(t) = \begin{cases} e^{-t} & \text{if $0 \leq t \leq 1$} \\ 0 & \text{if $1 \leq t \leq 2$.} \end{cases}$$
   and $f(t+2) = f(t) \forall t$.

   Solution: $T = 2$ and $\omega_0 = \frac{2\pi}{T} = \pi$
   

   $$f_1(t) = \begin{cases} f(t) & \text{if $0 \leq t \leq T$.} \\ 0 & \text{otherwise} \end{cases}$$

   $$= e^{-t}[u(t) - u(t-1)]$$

   $$= e^{-t}u(t) - e^{-t}u(t-1)$$

   $$= e^{-t}u(t) - e^{-1}.e^{-(t-1)}u(t-1)$$

   
   
   
   

   $$F_1(s) = \frac{1}{s+1} - e^{-1}.\frac{e^{-s}}{s+1}$$

   $$= \frac{1-e^{-(s+1)}}{s+1}$$

   
   
   Hence,
   

   $$F_n = \frac{1}{T} F_1(s)\Bigl\lvert_{s = in\omega_0}$$

   $$= \frac{1}{2}. \frac{1-e^{-(s+1)}}{s+1}\Bigl\lvert_{s = in\pi}$$

   $$= \frac{1}{2}. \frac{1-e^{-1}.e^{-in\pi}}{in\pi+1}$$

   $$= \frac{1}{2}. \frac{1-(-1)^n.e^{-1}}{in\pi+1}$$

   
   

2. Problem 4.13 from Text:

   Solution: $T = 2\pi$ and $\omega_0 = \frac{2\pi}{2\pi} = 1$

   $$f_1(t) = [u(t) - u(t-\pi)]$$
   $$F_1(s) = \frac{1-e^{-s\pi}}{s}$$
   Hence,
   

   $$F_n = \frac{1}{T} F_1(s)\Bigl\lvert_{s = in\omega_0}$$

   $$= \frac{1}{2\pi}.\frac{1-e^{-s}}{s}\Bigl\lvert_{s = in}$$

   $$= \frac{1}{2\pi}.\frac{1-e^{-in}}{in},  n \neq 0$$

   $$= \frac{1-(-1)^n}{in2\pi},  n \neq 0$$

   
   
   where we have added the condition $n \neq 0$, to avoid division by zero. For $n = 0$, we have to go back to the integral:
   

   $$F_0 = \frac{1}{T} \int_{0}^{T} f(t) dt = \frac{1}{2\pi} \int_{0}^{\pi} 1. dt$$

   $$= \frac{1}{2}$$

   
   

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