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## Q. 2.6.2

The Rectangular Pulse Function

The rectangular pulse function P(t) is shown in Figure 2.6.2a. Derive the Laplace transform of this function (a) from the basic definition of the transform and (b) from the time-shifting property.

## Verified Solution

a. From the definition of the transform,

$\mathcal{L}[P(t)]=\int_{0-}^{\infty }P(t)e^{-st} dt=\int_{0}^{D} 1e^{-st} dt + \int_{D}^{\infty} 0e^{-st} dt= \int_{0}^{D} 1e^{-st} dt \\ = \frac{e^{-st}}{-s} |_0^D = \frac{1}{s} (1-e^{-sD})$

b. Figure 2.6.2b shows that the pulse can be considered to be composed of the sum of a unit-step function and a shifted, negative unit-step function. Thus, $P(t) = u_s(t) − u_s(t − D)$ and from the time-shifting property,

$P(s)=\mathcal{L}[u_s(t)]-\mathcal{L}[u_s(t-D)]=\frac{1}{s}-e^{-sD}\frac{1}{s}=\frac{1}{s}(1-e^{-sD})$

which is the same result obtained in part (a).