Linearity Property of The Laplace Transform

Electrical Circuit Analysis > The Laplace Transform > Properties of The Laplace Transform

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If $ F_{1}(s) $ and $ F_{2}(s) $ are, respectively, the Laplace transforms of $ f_{1}(t) $ and $ f_{2}(t) $, then

$$\mathcal{L}\left[a_{1} f_{1}(t)+a_{2} f_{2}(t)\right]=a_{1} F_{1}(s)+a_{2} F_{2}(s) \tag{1}$$

$$\mathcal{L}[f(t)]=F(s)=\int_{0-}^{\infty} f(t) e^{-s t} d t \tag{2}$$

where $ a_{1} $ and $ a_{2} $ are constants. Equation $ 1 $ expresses the linearity property of the Laplace transform. The proof of Eq. (1) follows readily from the definition of the Laplace transform in Eq. (2). For example, by the linearity property in Eq. (1), we may write

$$\mathcal{L}[\cos w t]=\mathcal{L}\left[\frac{1}{2}\left(e^{j \omega t}+e^{-j \omega t}\right)\right]=\frac{1}{2} \mathcal{L}\left[e^{j \omega t}\right]+\frac{1}{2} \mathcal{L}\left[e^{-j \omega t}\right]$$

But from Example 1, $ \mathcal{L}\left[e^{-a t}\right]=1 /(s+a) $. Hence,

$$\mathcal{L}[\cos w t]=\frac{1}{2}\left(\frac{1}{s-j \omega}+\frac{1}{s+j \omega}\right)=\frac{s}{s^{2}+\omega^{2}}$$

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