Transform of Unit Step Function

DEFINITION 10.3.17 (Unit Step Function)   The Unit-Step function is defined by

$$U_a(t) = \left\{\begin{array}{ll} 0 & {\mbox { if }} \;\; 0 \leq t < a \\ 1 & {\mbox{ if }} \;\; t \geq a \end{array}. \right.$$

EXAMPLE 10.3.18   ${\mathcal L}\bigl(U_a(t)\bigr) = \displaystyle \int\limits_a^\infty e^{-st} dt = \displaystyle\frac{e^{-sa}}{s}, \; s > 0.$

Figure 10.3: Graphs of $f(t)$ and $U_a(t) f(t-a)$

LEMMA 10.3.19 ($t$ -Shifting)   Let ${\mathcal L}(f(t)) = F(s).$ Define $g(t)$ by

$$g(t) = \left\{\begin{array}{ll} 0 & {\mbox { if }} \;\; 0 \leq t < a \\ f(t-a) & {\mbox{ if }} \;\; t \geq a \end{array}. \right.$$

Then $g(t) = U_a(t) f(t-a)$ and

$${\mathcal L} \bigl(g(t)\bigr) = e^{-a s} F(s).$$

Proof. Let $0 \leq t < a$ . Then $U_a(t) = 0$ and so, $U_a(t) f(t-a) = 0 = g(t).$
If $t \geq a$ , then $U_a(t) = 1$ and $U_a(t) f(t-a) = f(t-a) = g(t).$ Since the functions $g(t)$ and $U_a(t) f(t-a)$ take the same value for all $t \geq 0,$ we have $g(t) = U_a(t) f(t-a).$ Thus,

$${\mathcal L}(g(t)) = \int\limits_0^\infty e^{-st} g(t)dt = \int\limits_a^\infty e^{-st} f(t-a) dt$$

$$= \int\limits_0^\infty e^{-s(t+a)} f(t) dt = e^{-as} \int\limits_0^\infty e^{-st} f(t) dt$$

$$= e^{-as} F(s).$$

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EXAMPLE 10.3.20   Find ${\mathcal L}^{-1} \left(\frac{e^{-5s}}{s^2 - 4 s - 5}\right).$
Solution: Let $G(s) = \frac{e^{-5s}}{s^2 - 4 s - 5}= e^{-5 s} F(s),$ with $F(s) = \frac{1}{s^2 - 4 s - 5}.$ Since $s^2 - 4 s - 5 = (s-2)^2 - 3^2$

$${\mathcal L}^{-1}(F(s)) = {\mathcal L}^{-1}\left(\frac{1}{3} \cdot \frac{3}{(s-2)^2 - 3^2} \right)= \frac{1}{3} \sinh (3 t) e^{2t}.$$

Hence, by Lemma 10.3.19

$${\mathcal L}^{-1}(G(s)) = \displaystyle\frac{1}{3} \; U_5(t) \sinh\bigl( 3(t-5)\bigr) e^{2(t-5)}.$$

EXAMPLE 10.3.21   Find ${\mathcal L}(f(t))$ , where $f(t) = \left\{\begin{array}{ll} 0 & t < 2 \pi \\ t \cos t & t > 2 \pi. \end{array}\right.$
Solution: Note that

$$f(t) = \left\{\begin{array}{ll} 0 & t < 2 \pi \\ (t- 2 \pi) \cos (t- 2 \pi) + 2 \pi \cos (t - 2 \pi) & t > 2 \pi. \end{array}\right.$$

Thus, ${\mathcal L}(f(t))= \displaystyle e^{- 2 \pi s} \left( \frac{s^2 -1}{(s^2+ 1)^2} + 2 \pi \frac{s}{s^2 + 1}\right)$

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