Laplace Transform of Damped Hyperbolic Sine and Cosine Functions

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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 $\mathrm{x(t)}$ is a time domain function, then its Laplace transform is defined as −

$$\mathrm{L\left[x(t)\right]\:=\:X(s)\:=\:\int_{-\infty}^{\infty}\:x(t)e^{-st}\:dt \:\:\dotso\: (1)}$$

Equation (1) gives the bilateral Laplace transform of the function $\mathrm{x(t)}$. But for the causal signals, the unilateral Laplace transform is applied, which is defined as,

$$\mathrm{L\left[x(t)\right]=X(s)=\int_{0}^{\infty}x\left(t \right )e^{-st}\:dt \:\:\dotso\: (2)}$$

Laplace Transform of Damped Hyperbolic Sine Function

The damped hyperbolic sine function is given by,

$$\mathrm{x(t)\:=\:e^{-at}\:sinh\:\omega t \:u(t)\:=\:e^{-at}\left (\frac{e^{\omega t}\:-\:e^{-\omega t}}{2} \right )\: u(t)}$$

Hence, by the definition of the Laplace transform, we have,

$$\mathrm{L\left[{e}^{-at} sinh\: \omega t\:u(t)\right]\:=\:L\left [e^{-at}\left (\frac{e^{\omega t}\:-\:e^{-\omega t}}{2} \right )\:u(t) \right ]}$$

$$\mathrm{\Rightarrow \: L\left[e^{-at} sinh\:\omega t\:u(t)\right]\:=\:\frac{1}{2}L\left [e^{-at}e^{\omega t}\:u(t) \:-\: e^{-at}e^{-\omega t}\:u(t) \right ] }$$

$$\mathrm{\Rightarrow\: L\left[e^{-at} sin\:h\:\omega t\:u(t)\right]\:=\:\frac{1}{2}\left\{L\left [e^{-(a\:-\:\omega)t} \:u(t)\right ]\:-\:L\left [ e^{-( a\:+\:\omega)t}\:u(t) \right ] \right\}}$$

$$\mathrm{\Rightarrow\: L\left[e^{-at} sinh\:\omega t\:u(t)\right]\:=\:\frac{1}{2}\left[\frac{1}{s\:+\:(a\:-\:\omega)} \:-\: \frac{1}{s\:+\:(a\:+\:\omega)} \right ]}$$

$$\mathrm{\Rightarrow\: L\left[e^{-at} sinh\:\omega t\:u(t)\right]\:=\:\frac{1}{2}\left[\frac{1}{(s\:+\:a)\:-\:\omega } \:-\:\frac{1}{(s\:+\:a)\:+\:\omega } \right ]\:=\:\frac{\omega}{(s\:+\:a)^{2}\:-\:\omega^{2}}}$$

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 −

$$\mathrm{e^{-at}sinh\:\omega t\:u(t)\overset{LT}{\leftrightarrow}\left[ \frac{\omega}{(s\:+\:a)^{2}\:-\:\omega^{2}} \right ] \:;\:ROC \:\to\: Re (s)\: \gt-\: a}$$

Laplace Transform of Damped Hyperbolic Cosine Function

The damped hyperbolic cosine function is given by,

$$\mathrm{x(t)\:=\:e^{-at}\:cosh\:\omega t \:u(t)\:=\:e^{-at}\left (\frac{e^{\omega t}\:+\:e^{-\omega t}}{2} \right )\:u(t)}$$

Hence, by the definition of the Laplace transform, we have,

$$\mathrm{L\left[{e}^{-at} cosh\:\omega t\:u(t)\right]\:=\:L\left [e^{-at}\left ( \frac{e^{\omega t}\:+\:e^{-\omega t}}{2} \right )\:us(t) \right ]}$$

$$\mathrm{\Rightarrow\: L\left[e^{-at} cosh\:\omega t\:u(t)\right]\:=\:\frac{1}{2}L\left [e^{-at}e^{\omega t}\:u(t) \:+\: e^{-at}e^{-\omega t}\:u(t) \right ]}$$

$$\mathrm{\Rightarrow\:L\left[e^{-at} cosh\:\omega t\:u(t)\right]\:=\:\frac{1}{2}\left\{L\left[ e^{-(a-\omega)t}\:u(t) \right ]\:+\:L\left [e^{-(a\:+\:\omega)t}\:u(t) \right ] \right\}}$$

$$\mathrm{\Rightarrow\:L\left[e^{-at} cosh\:\omega t\:u(t)\right]\:=\:\frac{1}{2}\left[\frac{1}{s\:+\:(a\:-\:\omega)} \:+\: \frac{1}{s\:+\:(a\:+\:\omega)} \right ]}$$

$$\mathrm{\Rightarrow\: L\left[e^{-at} cosh\:\omega t\:u(t)\right]\:=\:\frac{1}{2}\left[\frac{1}{(s\:+\:a)\:-\:\omega }\: +\:\frac{1}{(s\:+\:a)\:+\:\omega} \right ]\:=\:\frac{s\:+\:a}{(s\:+\:a)^{2}\:-\:\omega^{^{2}}}}$$

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,

$$\mathrm{e^{-at} cos\:\omega t\:u(t)\overset{LT}{\leftrightarrow}\left [ \frac{s\:+\:a}{(s\:+\:a)^{2}-\omega^{2}}\right ] \:;\:ROC\:\to\: Re(s) \:\gt\:-a}$$