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Laplace Transform of Damped Hyperbolic 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 $x\mathrm{\left ( \mathit{t}\right)}$ is a time domain function, then its Laplace transform is defined as −
$$\mathrm{\mathit{L\mathrm{\left[\mathit{x\mathrm{\left(\mathit{t} \right )}}\right ]}}\mathrm{=}\mathit{X\mathrm{\left(\mathit{s} \right )}}\mathrm{=}\int_{-\infty }^{\infty}\mathit{x\mathrm{\left(\mathit{t} \right )}e^{-st}}\:\mathit{dt}\:\:\:\:\:\:...(1)}$$
Equation (1) gives the bilateral Laplace transform of the function $x\mathrm{\left ( \mathit{t}\right)}$. But for the causal signals, the unilateral Laplace transform is applied, which is defined as,
$$\mathrm{\mathit{L\mathrm{\left[\mathit{x\mathrm{\left(\mathit{t} \right )}}\right ]}}\mathrm{=}\mathit{X\mathrm{\left(\mathit{s} \right )}}\mathrm{=}\int_{\mathrm{0} }^{\infty}\mathit{x\mathrm{\left(\mathit{t} \right )}e^{-st}}\:\mathit{dt}\:\:\:\:\:\:...(2)}$$
Laplace Transform of Damped Hyperbolic Sine Function
The damped hyperbolic sine function is given by,
$$\mathrm{\mathit{x\mathrm{\left(\mathit{t}\right)}}\:\mathrm{=}\:e^{-at}\:\mathrm{sinh\:\mathit{\omega \mathit{t} \:\mathit{u}\mathrm{\left ( \mathit{t} \right )}}}\:\mathrm{=}\:\mathit{e^{-at}}\mathrm{\left ( \frac{\mathit{e^{\omega t}-e^{-\omega t}}}{\mathrm{2}} \right )}\:\mathit{u}\mathrm{\left ( {\mathit{t}}\right)}}$$
Hence, by the definition of the Laplace transform, we have,
$$\mathrm{\mathit{L\mathrm{\left[\mathit{{e}^{-\mathit{at}}} \mathrm{sinh\:\mathit{\omega t}}\:\mathit{u}\mathrm{\left(\mathit{t}\right )}\right]}}\:\mathrm{=}\mathit{L}\mathrm{\left [ \:\mathit{e^{-at}}\mathrm{\left ( \frac{\mathit{e^{\omega t}-e^{-\omega t}}}{\mathrm{2}} \right )}\:\mathit{u}\mathrm{\left ( {\mathit{t}}\right)} \right ]}}$$
$$\mathrm{\Rightarrow \mathit{L\mathrm{\left[\mathit{e^{-at}} \mathrm{sinh\:\mathit{\omega t}}\:\mathit{u}\mathrm{\left(\mathit{t}\right )}\right]}}\:\mathrm{=}\:\frac{1}{2}\mathit{L}\mathrm{\left [\mathit{e^{-at}e^{\omega t}\:u\mathrm{\left ( \mathit{t} \right )}-e^{-at}e^{-\omega t}\:u\mathrm{\left ( \mathit{t} \right )}} \right ]} }$$
$$\mathrm{\Rightarrow \mathit{L\mathrm{\left[\mathit{e^{-at}} \mathrm{sinh\:\mathit{\omega t}}\:\mathit{u}\mathrm{\left(\mathit{t}\right )}\right]}}\:\mathrm{=}\:\frac{1}{2}\mathrm{\left\{\mathit{L}\mathrm{\left [\mathit{ e^{-\mathrm{\left ( a-\omega \right)}t}\:u\mathrm{\left ( \mathit{t} \right )}}\right ]}-\mathit{L}\mathrm{\left [ \mathit{e^{-\mathrm{\left ( a\mathrm{+}\omega \right )}\mathit{t}}\:u \mathrm{\left ( \mathit{t} \right )} } \right ]} \right\}}}$$
$$\mathrm{\Rightarrow \mathit{L\mathrm{\left[\mathit{e^{-at}} \mathrm{sinh\:\mathit{\omega t}}\:\mathit{u}\mathrm{\left(\mathit{t}\right )}\right]}}\:\mathrm{=}\:\frac{1}{2}\mathrm{\left[\frac{1}{\mathit{s\mathrm{+}\mathrm{\left(\mathit{a-\omega}\right)}}}-\frac{1}{\mathit{s}\mathrm{+}\mathrm{\left ( \mathit{a\mathrm{+}\omega } \right )}} \right ]}}$$
$$\mathrm{\Rightarrow \mathit{L\mathrm{\left[\mathit{e^{-at}} \mathrm{sinh\:\mathit{\omega t}}\:\mathit{u}\mathrm{\left(\mathit{t}\right )}\right]}}\:\mathrm{=}\:\frac{1}{2}\mathrm{\left[\frac{1}{\mathrm{\left( \mathit{s\mathrm{+}a} \right)}-\omega }-\frac{1}{\mathrm{\left ( \mathit{s\mathrm{+}a}\right)}+\omega } \right ]}\mathrm{=}\frac{\mathit{\omega}}{\mathrm{\left ( \mathit{s\mathrm{+}a}\right)^{\mathrm{2}}}-\omega ^{^{2}}}}$$
The region of convergence (ROC) of Laplace transform of the damped hyperbolic sine function is Re(s)> -a, which is shown in Figure-1. Therefore, the Laplace transform of damped hyperbolic sine function along with its ROC is given as follows −
$$\mathrm{\mathit{e^{-at}} \mathrm{sinh\:\mathit{\omega t}}\:\mathit{u}\mathrm{\left(\mathit{t}\right)}\overset{LT}{\leftrightarrow}\mathrm{\left [ \frac{\mathit{\omega}}{\mathrm{\left ( \mathit{s\mathrm{+}a}\right)^{\mathrm{2}}}-\omega ^{^{2}}} \right ]} ;\:\mathrm{ROC}\to \mathrm{Re}\mathrm{\left(\mathit{s} \right )}>-\mathit{a}}$$
Laplace Transform of Damped Hyperbolic Cosine Function
The damped hyperbolic cosine function is given by,
$$\mathrm{\mathit{x\mathrm{\left(\mathit{t}\right)}}\:\mathrm{=}\:e^{-\mathit{at}}\:\mathrm{cosh\:\mathit{\omega \mathit{t} \:\mathit{u}\mathrm{\left(\mathit{t} \right )}}}\:\mathrm{=}\:\mathit{e^{-at}}\mathrm{\left ( \frac{\mathit{e^{\omega t}\mathrm{+}e^{-\omega t}}}{\mathrm{2}} \right )}\:\mathit{u}\mathrm{\left ( {\mathit{t}}\right)}}$$
Hence, by the definition of the Laplace transform, we have,
$$\mathrm{\mathit{L\mathrm{\left[\mathit{{e}^{-\mathit{at}}} \mathrm{cosh\:\mathit{\omega t}}\:\mathit{u}\mathrm{\left(\mathit{t}\right )}\right]}}\:\mathrm{=}\mathit{L}\mathrm{\left [ \:\mathit{e^{-at}}\mathrm{\left ( \frac{\mathit{e^{\omega t}\mathrm{+}e^{-\omega t}}}{\mathrm{2}} \right )}\:\mathit{u}\mathrm{\left ( {\mathit{t}}\right)} \right ]}}$$
$$\mathrm{\Rightarrow \mathit{L\mathrm{\left[\mathit{e^{-at}} \mathrm{cosh\:\mathit{\omega t}}\:\mathit{u}\mathrm{\left(\mathit{t}\right )}\right]}}\:\mathrm{=}\:\frac{1}{2}\mathit{L}\mathrm{\left [\mathit{e^{-at}e^{\omega t}\:u\mathrm{\left ( \mathit{t} \right )}\mathrm{+}e^{-at}e^{-\omega t}\:u\mathrm{\left ( \mathit{t} \right )}} \right ]}}$$
$$\mathrm{\Rightarrow \mathit{L\mathrm{\left[\mathit{e^{-at}} \mathrm{cosh\:\mathit{\omega t}}\:\mathit{u}\mathrm{\left(\mathit{t}\right )}\right]}}\:\mathrm{=}\:\frac{1}{2}\mathrm{\left\{\mathit{L}\mathrm{\left [\mathit{ e^{-\mathrm{\left ( a-\omega \right)}t}\:u\mathrm{\left ( \mathit{t} \right )}}\right ]}+\mathit{L}\mathrm{\left [ \mathit{e^{-\mathrm{\left ( a+\omega \right )}\mathit{t}}\:u \mathrm{\left ( \mathit{t} \right )} } \right ]} \right\}}}$$
$$\mathrm{\Rightarrow \mathit{L\mathrm{\left[\mathit{e^{-at}} \mathrm{cosh\:\mathit{\omega t}}\:\mathit{u}\mathrm{\left(\mathit{t}\right )}\right]}}\:\mathrm{=}\:\frac{1}{2}\mathrm{\left[\frac{1}{\mathit{s\mathrm{+}\mathrm{\left(\mathit{a-\omega}\right)}}}+\frac{1}{\mathit{s}+\mathrm{\left ( \mathit{a+\omega } \right )}} \right ]}}$$
$$\mathrm{\Rightarrow \mathit{L\mathrm{\left[\mathit{e^{-at}} \mathrm{cosh\:\mathit{\omega t}}\:\mathit{u}\mathrm{\left(\mathit{t}\right )}\right]}}\:\mathrm{=}\:\frac{1}{2}\mathrm{\left[\frac{1}{\mathrm{\left( \mathit{s\mathrm{+}a} \right)}-\omega }+\frac{1}{\mathrm{\left ( \mathit{s\mathrm{+}a}\right)}+\omega } \right ]}\mathrm{=}\frac{\mathit{s\mathrm{+}a}}{\mathrm{\left ( \mathit{s\mathrm{+}a}\right)^{\mathrm{2}}}-\omega ^{^{2}}}}$$
The ROC of Laplace transform of the damped hyperbolic cosine function is also Re(s)> -a as shown in Figure-1. Therefore, the Laplace transform of damped hyperbolic cosine function along with its ROC is given by,
$$\mathrm{\mathit{e^{-at}} \mathrm{cos\:\mathit{\omega t}}\:\mathit{u}\mathrm{\left(\mathit{t}\right)}\overset{LT}{\leftrightarrow}\mathrm{\left [ \frac{\mathit{s\mathrm{+}a}}{\mathrm{\left ( \mathit{s\mathrm{+}a}\right)^{\mathrm{2}}}-\omega ^{{2}}} \right ]} ;\:\mathrm{ROC}\to \mathrm{Re}\mathrm{\left(\mathit{s} \right )}>-\mathit{a}}$$
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