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## Q. 2.6.1

The Shifted Step Function

If the discontinuity in the unit-step function occurs at t = D, Figure 2.6.1, then the function x(t)= $Mu_s(t−D)$ is 0 for t < D and M for t > D. The function $u_s(t−D)$ is called the shifted step function. Determine X(s).

## Verified Solution

From the transform definition, we have

$\mathcal{L}[x(t)]=\underset{T\rightarrow \infty }{\text{lim}}[\int_{0-}^{T} Mu_s(t-D)e^{-st}dt ]=\underset{T\rightarrow \infty }{\text{lim}}(\int_{0-}^{D} 0e^{-st}dt + \int\limits_{D}^{T} Me^{-st}dt)$

or

$\mathcal{L}[x(t)]= 0 + M \underset{T\rightarrow \infty }{\text{lim}} (\frac{1}{-s}e^{-st} |_D^T)=M \underset{T\rightarrow \infty }{\text{lim}} (\frac{1}{-s}e^{-sT}+\frac{1}{s}e^{-sD})= \frac{M}{s}e^{-sD}$

Thus, the transform of the shifted unit-step function $u_s(t − D)$ is $e^{−sD}/s$.