Question

Find the laplace transform of the half-wave rectified sine wave.

Answer 1

The laplace transform of the half-wave rectified sine wave is as follows:

$L{f(t)} = \frac{1}{(1-e^{-\pi s} ) (s^{2}+1) }$

• The half wave rectified sine curve is represented as :

        f (t) = {  sint          :(2n+1) π ≤ t ≤ (2n+2) π

                     0             :2nπ ≤ t ≤ (2n+1)π    }

• Thus, the Laplace transform of f(t) will be :

$L{f(t)} = \frac{1}{(1-e^{-\pi s} ) (s^{2}+1) }$

Answer 2

The Laplace transform of the half-wave rectified sine wave is $\frac{1}{\left(1-e^{-\pi s}\right)\left(s^{2}+1\right)}$

Explanation:

Consider the half wave rectified sine curve:

$f(t)=\left\{\begin{array}{ll}\sin t & : 2 n \pi \leq t \leq(2 n+1) \pi \\0 & :(2 n+1) \pi \leq t \leq(2 n+2) \pi\end{array}\right.$

The Laplace transform of f(t) is given by:

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

The f(t) has a period of 2π.

Now, the Laplace Transform of Periodic Function

$\mathcal{L}\{f(t)\}=\frac{1}{1-e^{-2 \pi s}} \int_{0}^{2 \pi} e^{-s t} f(t) \mathrm{d} t$

$\mathcal{L}\{f(t)\}=\frac{1}{1-e^{-2 \pi s}}\left(\int_{0}^{\pi} e^{-s t} \sin t \mathrm{d} t+\int_{\pi}^{2 \pi} e^{-s t} \times 0 \mathrm{d} t\right)$

$\mathcal{L}\{f(t)\}=\frac{1}{1-e^{-2 \pi s}} \int_{0}^{\pi} e^{-s t} \sin t d t$

Now, primitive of $e^{a x} \sin b x$

$\mathcal{L}\{f(t)\}=\frac{1}{1-e^{-2 \pi s}}\left[\frac{e^{-s t}(-s \sin t-\cos t)}{s^{2}+1}\right]_{0}^{\pi}$

$\mathcal{L}\{f(t)\}=\frac{1}{1-e^{-2 \pi s}}\left(\frac{e^{-s \pi}(-s \sin \pi-\cos \pi)}{s^{2}+1}-\frac{e^{-s \times 0}(-s \sin 0-\cos 0)}{s^{2}+1}\right)$

Now, sine of Integer Multiple of π

$\mathcal{L}\{f(t)\}=\frac{1}{1-e^{-2 \pi s}}\left(\frac{e^{-s \pi}(-\cos \pi)}{s^{2}+1}-\frac{e^{-s \times 0}(-\cos 0)}{s^{2}+1}\right)$

Now, cosine of Integer Multiple of π

$\mathcal{L}\{f(t)\}=\frac{1}{1-e^{-2 \pi s}}\left(\frac{e^{-s \pi}(-(-1))}{s^{2}+1}-\frac{e^{-s \times 0}(-1)}{s^{2}+1}\right)$

Now, exponential of Zero

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

On simplifying, we get,

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

Now, difference of Two Squares,

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

On simplifying, we get,

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