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Laplace Transform of Sine and Cosine Functions
Laplace Transform
The Laplace transform is a mathematical tool which is used to convert the differential equation in time domain into the algebraic equations in the frequency domain or s-domain.
Mathematically, if $\mathrm{\mathit{x\left ( \mathit{t} \right )}}$ is a time domain function, then its Laplace transform is defined as −
$$\mathrm{\mathit{L\left [ x\left ( \mathrm{t} \right ) \right ]}= \mathit{X\left ( s \right )}=\int_{-\infty }^{\infty}\mathit{x\left ( \mathrm{t} \right )e^{-st}\; dt}\; \; ...\left ( 1 \right )}$$
Equation (1) gives the bilateral Laplace transform of the function $\mathrm{\mathit{x\left ( \mathit{t} \right )}}$. But for the causal signals, the unilateral Laplace transform is applied, which is defined as,
$$\mathrm{\mathit{L\left [ x\left ( \mathrm{t} \right ) \right ]}\mathrm{=} \mathit{X\left ( s \right )}\mathrm{=}\int_{0 }^{\infty}\mathit{x\left ( \mathrm{t} \right )e^{-st}\; dt}\; \; ...\left ( 2 \right )}$$
Laplace Transform of Sine Function
Let,
$$\mathrm{\mathit{x\left ( t \right )}\mathrm{=}sin\: \mathit{\omega t\: u\left ( t \right )}\mathrm{=}\mathit{\frac{e^{j\:\omega t }-e^{-j\:\omega t}}{\mathrm{2}j}u\left ( t \right )}}$$
Thus, from the definition of the Laplace transform, we have,
$$\mathrm{\mathit{X\left ( s \right )}\mathrm{=}\mathit{L\left [ \mathrm{sin}\: \mathit{\omega t\: u\left ( t \right )} \right ]}\mathrm{=}\mathit{L}\left [ \mathit{\frac{e^{j\:\omega t }-e^{-j\:\omega t}}{\mathrm{2}j}u\left ( t \right )} \right ]}$$
$$\mathrm{\Rightarrow \mathit{L\left [ \mathrm{sin\; }\omega t\; u\left ( t \right ) \right ]\mathrm{=}}\frac{1}{2\mathit{j}}\left\{\mathit{L\left [ e^{j\:\omega t}u\left ( t \right ) \right ]-L\left [ e^{-j\:\omega t}u\left ( t \right ) \right ]} \right\}} $$
$$\mathrm{\Rightarrow \mathit{L\left [ \mathrm{sin\; }\omega t\; u\left ( t \right ) \right ]\mathrm{=}}\frac{1}{2\mathit{j}}\left\{\left [ \frac{1}{\mathit{\left ( s\mathrm{+}j\:\omega \right ) }} \right ]-\left [ \frac{1}{\mathit{\left ( s-j\:\omega \right )}} \right ] \right\}\mathrm{=}\frac{1}{2\mathit{j}}\left [ \frac{\mathit{s-j\:\omega -s-j\:\omega}}{\mathit{s^{\mathrm{2}}}\mathrm{+}\omega ^{\mathrm{2}}} \right ]}$$
$$\mathrm{\therefore \mathit{L\left [ \mathrm{sin\; }\omega t\; u\left ( t \right ) \right ]}\mathrm{=}\mathit{\left (\frac{\omega }{s^{\mathrm{2}}\mathrm{+}\omega ^{\mathrm{2}}} \right )}}$$
The region of convergence (ROC) of Laplace transform of the sine function is 𝑅𝑒(𝑠) > 0, which is shown in Figure-1. Thus, the Laplace transform of the sine function along with its ROC is,
$$\mathrm{sin\: \mathit{\omega t\: u\left ( t \right )\overset{LT}{\leftrightarrow}}\left ( \frac{\omega }{\mathit{s^{\mathrm{2}}}\mathrm{+}\omega ^{\mathrm{2}}} \right )\; \; and\; \; ROC\rightarrow Re\left ( \mathit{s} \right )> 0}$$
Laplace Transform of Cosine Function
Let,
$$\mathrm{\mathit{x\left ( t \right )}\mathrm{=}cos \: \mathit{\omega t\: u\left ( t \right )}\mathrm{=}\mathit{\frac{e^{j\:\omega t }\mathrm{+}e^{-j\:\omega t}}{\mathrm{2}}u\left ( t \right )}}$$
Now, from the definition of Laplace transform, we have,
$$\mathrm{\mathit{X\left ( s \right )}\mathrm{=}\mathit{L\left [ \mathrm{cos}\: \mathit{\omega t\: u\left ( t \right )} \right ]}\mathrm{=}\mathit{L}\left [ \mathit{\frac{e^{j\:\omega t }\mathrm{+}e^{-j\:\omega t}}{\mathrm{2}}u\left ( t \right )} \right ]}$$
$$\mathrm{\Rightarrow \mathit{L\left [ \mathrm{cos\; }\omega t\; u\left ( t \right ) \right ]\mathrm{=}}\frac{1}{2}\left\{\mathit{L\left [ e^{j\:\omega t}u\left ( t \right ) \right ]-L\left [ e^{-j\:\omega t}u\left ( t \right ) \right ]} \right\}}$$
$$\mathrm{\Rightarrow \mathit{L\left [ \mathrm{cos\; }\omega t\; u\left ( t \right ) \right ]\mathrm{=}}\frac{1}{2}\left [ \left ( \frac{1}{\mathit{s-j\omega }} \right ) \right ]\mathrm{+}\left [ \left ( \frac{1}{\mathit{s\mathrm{+}j\omega}} \right ) \right ] \mathrm{=}\frac{1}{2}\left [ \frac{\mathit{s\mathrm{+}j\omega \mathrm{+}s-j\omega}}{\mathit{s^{\mathrm{2}}}\mathrm{+}\omega ^{\mathrm{2}}} \right ]}$$
$$\mathrm{\Rightarrow \mathit{L\left [ \mathrm{cos\; }\omega t\; u\left ( t \right ) \right ]}\mathrm{=}\mathit{\left (\frac{s}{s^{\mathrm{2}}\mathrm{+}\omega ^{\mathrm{2}}} \right )}}$$
The ROC of Laplace transform of the cosine function is also 𝑅𝑒(𝑠) > 0 as shown in Figure-1. Therefore, the Laplace transform of the cosine function along with its ROC is,
$$\mathrm{cos\: \mathit{\omega t\: u\left ( t \right )\overset{LT}{\leftrightarrow}}\left ( \frac{\mathit{s} }{\mathit{s^{\mathrm{2}}}\mathrm{+}\omega ^{\mathrm{2}}} \right )\; \; and\; \; ROC\rightarrow Re\left ( \mathit{s} \right )> 0}$$
Laplace Transform of Hyperbolic Sine Function
Let,
$$\mathrm{\mathit{x\left ( t \right )}\mathrm{=}sinh\: \mathit{\omega t\: u\left ( t \right )}\mathrm{=}\mathit{\frac{e^{\omega t }-e^{-\omega t}}{\mathrm{2}}u\left ( t \right )}} $$
Now, from the definition of the Laplace transform, we have,
$$\mathrm{\mathit{X\left ( s \right )}\mathrm{=}\mathit{L\left [ \mathrm{sinh}\: \mathit{\omega t\: u\left ( t \right )} \right ]}\mathrm{=}\mathit{L}\left [ \mathit{\frac{e^{\omega t }-e^{-\omega t}}{\mathrm{2}}u\left ( t \right )} \right ]} $$
$$\mathrm{\Rightarrow \mathit{L\left [ \mathrm{sinh\; }\omega t\; u\left ( t \right ) \right ]\mathrm{=}}\frac{1}{2}\left\{\mathit{L\left [ e^{\omega t}u\left ( t \right ) \right ]-L\left [ e^{-\omega t}u\left ( t \right ) \right ]} \right\}}$$
$$\mathrm{\Rightarrow \mathit{L\left [ \mathrm{sinh\; }\omega t\; u\left ( t \right ) \right ]\mathrm{=}}\frac{1}{2}\left [ \left ( \frac{1}{\mathit{s-\omega }} \right ) \right ]-\left [ \left ( \frac{1}{\mathit{s\mathrm{+}\omega}} \right ) \right ] \mathrm{=}\frac{1}{2}\left [ \frac{\mathit{s\mathrm{+}\omega -s\mathrm{+}\omega}}{\mathit{s^{\mathrm{2}}}-\omega ^{\mathrm{2}}} \right ]}$$
$$\mathrm{\therefore \mathit{L\left [ \mathrm{sinh\; }\omega t\; u\left ( t \right ) \right ]}\mathrm{=}\mathit{\left (\frac{\omega }{s^{\mathrm{2}}-\omega ^{\mathrm{2}}} \right )}}$$
The ROC of Laplace transform of the hyperbolic sine function is also 𝑅𝑒(𝑠) > 0 as shown in Figure-1 above. Therefore, the Laplace transform of the hyperbolic sine function along with its ROC is,
$$\mathrm{sinh\: \mathit{\omega t\: u\left ( t \right )\overset{LT}{\leftrightarrow}}\left ( \frac{\omega }{\mathit{s^{\mathrm{2}}}-\omega ^{\mathrm{2}}} \right )\; \; and\; \; ROC\rightarrow Re\left ( \mathit{s} \right )> 0}$$
Laplace Transform of Hyperbolic Cosine Function
Let,
$$\mathrm{\mathit{x\left ( t \right )}\mathrm{=}cosh \: \mathit{\omega t\: u\left ( t \right )}\mathrm{=}\mathit{\frac{e^{\omega t }\mathrm{+}e^{-\omega t}}{\mathrm{2}}u\left ( t \right )}}$$
Now, from the definition of the Laplace transform, we have,
$$\mathrm{\mathit{X\left ( s \right )}\mathrm{=}\mathit{L\left [ \mathrm{cosh}\: \mathit{\omega t\: u\left ( t \right )} \right ]}\mathrm{=}\mathit{L}\left [ \mathit{\frac{e^{\:\omega t }\mathrm{+}e^{-\:\omega t}}{\mathrm{2}}u\left ( t \right )} \right ]}$$
$$\mathrm{\Rightarrow \mathit{L\left [ \mathrm{cosh\; }\omega t\; u\left ( t \right ) \right ]\mathrm{=}}\frac{1}{2}\left\{\mathit{L\left [ e^{\omega t}u\left ( t \right ) \right ]\mathrm{+}L\left [ e^{-\omega t}u\left ( t \right ) \right ]} \right\}}$$
$$\mathrm{\Rightarrow \mathit{L\left [ \mathrm{cosh\; }\omega t\; u\left ( t \right ) \right ]\mathrm{=}}\frac{1}{2}\left [ \left ( \frac{1}{\mathit{s-\omega }} \right ) \mathrm{+} \left ( \frac{1}{\mathit{s\mathrm{+}\omega}} \right ) \right ] \mathrm{=}\frac{1}{2}\left [ \frac{\mathit{s\mathrm{+}\omega \mathrm{+}s-\omega}}{\mathit{s^{\mathrm{2}}}-\omega ^{\mathrm{2}}} \right ]}$$
$$\mathrm{\therefore \mathit{L\left [ \mathrm{cosh\; }\omega t\; u\left ( t \right ) \right ]}\mathrm{=}\mathit{\left (\frac{s}{s^{\mathrm{2}}-\omega ^{\mathrm{2}}} \right )}}$$
The ROC of Laplace transform of the hyperbolic cosine function is also 𝑅𝑒(𝑠) > 0 as shown above in Figure-1. Therefore, the Laplace transform of the hyperbolic sine function along with its ROC is,
$$\mathrm{cosh\: \mathit{\omega t\: u\left ( t \right )\overset{LT}{\leftrightarrow}}\left ( \frac{\mathit{s} }{\mathit{s^{\mathrm{2}}}-\omega ^{\mathrm{2}}} \right )\; \; and\; \; ROC\rightarrow Re\left ( \mathit{s} \right )> 0}$$
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