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- Laplace Transform
- Laplace Transforms
- Common Laplace Transform Pairs
- Laplace Transform of Unit Impulse Function and Unit Step Function
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- Laplace Transforms Properties
- Linearity Property of Laplace Transform
- Time Shifting Property of Laplace Transform
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- Difference between Laplace Transform and Fourier Transform
- Difference between Z Transform and Laplace Transform
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- Relation between Laplace Transform and Fourier Transform
- Laplace Transform – Time Reversal, Conjugation, and Conjugate Symmetry Properties
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- Z Transform
- Z-Transforms (ZT)
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- What is Inverse Z Transform?
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- System Realization
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- Discrete Fourier Transform
- Discrete-Time Fourier Transform
- Properties of Discrete Time Fourier Transform
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- Inverse Discrete-Time Fourier Transform
- Time Convolution and Frequency Convolution Properties of Discrete-Time Fourier Transform
- Differentiation in Frequency Domain Property of Discrete Time Fourier Transform
- Parseval’s Power Theorem
- Miscellaneous Concepts
- What is Mean Square Error?
- What is Fourier Spectrum?
- Region of Convergence
- Hilbert Transform
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- Filter Characteristics of Linear Systems
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- Zero Order Hold and its Transfer Function
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- What is the Frequency Response of Discrete Time Systems?
- Basic Elements to Construct the Block Diagram of Continuous Time Systems
- BIBO Stability Criterion
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- Rayleigh’s Energy Theorem
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 $\mathrm{x\left ( t \right )}$ is a time domain function, then its Laplace transform is defined as −
$$\mathrm{L\left [ x\left (t \right ) \right ]\:=\: X\left ( s \right )\:=\:\int_{-\infty }^{\infty}\:x\left ( t \right )e^{-st}\: dt\:\: \dotso\:(1)}$$
Equation (1) gives the bilateral Laplace transform of the function $\mathrm{x\left (t \right )}$. But for the causal signals, the unilateral Laplace transform is applied, which is defined as,
$$\mathrm{L\left [ x\left (t \right ) \right ]\:=\: X\left ( s \right )\:=\:\int_{0}^{\infty}x\left ( t \right )e^{-st}\: dt\:\: \dotso\: (2)}$$
Laplace Transform of Sine Function
Let,
$$\mathrm{x\left ( t \right )\:=\:sin\: \omega t\: u\left ( t \right )\:=\: \frac{e^{j\:\omega t }\:-\:e^{-j\:\omega t}}{2j}u\left ( t \right )}$$
Thus, from the definition of the Laplace transform, we have,
$$\mathrm{X\left ( s \right )\:=\:L\left [ sin\: \omega t\: u\left ( t \right ) \right ]\:=\:L\left [ \frac{e^{j\:\omega t }\:-\:e^{-j\:\omega t}}{2j}u\left ( t \right ) \right ]}$$
$$\mathrm{\Rightarrow\: L\left [ sin\: \omega t\: u\left ( t \right ) \right ]\:=\:\frac{1}{2j}\left\{L\left [ e^{j\:\omega t}u\left ( t \right ) \right ]\:-\:L\left [ e^{-j\:\omega t}u\left ( t \right ) \right ] \right\}} $$
$$\mathrm{\Rightarrow\: L\left [ sin\: \omega t\: u\left ( t \right ) \right ]\:=\:\frac{1}{2j}\left\{\left [ \frac{1}{\left ( s\:+\:j\:\omega \right ) } \right ]\:-\:\left [ \frac{1}{\left ( s\:-\:j\:\omega \right )} \right ] \right\}\:=\:\frac{1}{2j}\left [ \frac{s\:-\:j\:\omega \:-\:s\:-\:j\:\omega}{s^{2}\:+\:\omega^{2}} \right ]}$$
$$\mathrm{\therefore\:L\left [sin\: \omega t\: u\left ( t \right ) \right ]\:=\: \left (\frac{\omega }{s^{2}\:+\:\omega ^{2}} \right )}$$
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,
$$\mathrm{sin\: \omega t\: u\left ( t \right )\overset{LT}{\leftrightarrow}\left ( \frac{\omega }{s^{2}\:+\:\omega^{2}} \right )\:\: and\:\: ROC\:\rightarrow\: Re\left ( s \right )\:\gt\: 0}$$
Laplace Transform of Cosine Function
Let,
$$\mathrm{x\left ( t \right )\:=\:cos \: \omega t\: u\left ( t \right )\:=\:\frac{e^{j\:\omega t }\:+\:e^{-j\:\omega t}}{2}u\left ( t \right )}$$
Now, from the definition of Laplace transform, we have,
$$\mathrm{X\left ( s \right )\:=\:L\left [ cos\: \omega t\: u\left ( t \right ) \right ]\:=\:L\left [\frac{e^{j\:\omega t }\:+\:e^{-j\:\omega t}}{2}u\left ( t \right ) \right ]}$$
$$\mathrm{\Rightarrow\: L\left [ cos\: \omega t\: u\left ( t \right ) \right ]\:=\:\frac{1}{2}\left\{L\left [ e^{j\:\omega t}u\left ( t \right ) \right ]\:-\:L\left [ e^{-j\:\omega t}u\left ( t \right ) \right ] \right\}}$$
$$\mathrm{\Rightarrow\: L\left [ cos\: \omega t\: u\left ( t \right ) \right ]\:=\:\frac{1}{2}\left [ \left ( \frac{1}{s\:-\:j\omega } \right ) \right ]\:+\:\left [ \left ( \frac{1}{s\:+\:j\omega} \right ) \right ] \:=\:\frac{1}{2}\left [ \frac{s\:+\:j\omega \:+\:s\:-\:j\omega}{s^{2}\:+\:\omega^{2}} \right ]}$$
$$\mathrm{\Rightarrow\: L\left [ cos\:\omega t\:u\left ( t \right ) \right ]\:=\: \left (\frac{s}{s^{2}\:+\:\omega^{2}} \right )}$$
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,
$$\mathrm{cos\: \omega t\: u\left ( t \right )\overset{LT}{\leftrightarrow}\left ( \frac{s}{s^{2}\:+\:\omega^{2}} \right )\: \: and\: \: ROC\:\rightarrow\: Re\left (s \right )\:\gt \:0}$$
Laplace Transform of Hyperbolic Sine Function
Let,
$$\mathrm{x\left ( t \right )\:=\:sinh\: \omega t\: u\left ( t \right )\:=\:\frac{e^{\omega t }\:-\:e^{-\omega t}}{2}u\left ( t \right )} $$
Now, from the definition of the Laplace transform, we have,
$$\mathrm{X\left ( s \right )\:=\:L\left [ sinh\: \omega t\: u\left ( t \right ) \right ]\:=\:L\left [ \frac{e^{\omega t }\:-\:e^{-\omega t}}{2}u\left ( t \right ) \right ]}$$
$$\mathrm{\Rightarrow\: L\left [ sinh\: \omega t\: u\left ( t \right ) \right ]\:=\:\frac{1}{2}\left\{L\left [ e^{\omega t}u\left ( t \right ) \right ]\:-\:L\left [ e^{-\omega t}u\left ( t \right ) \right ] \right\}}$$
$$\mathrm{\Rightarrow\: L\left [ sinh\: \omega t\: u\left ( t \right ) \right ]\:=\:\frac{1}{2}\left [ \left ( \frac{1}{s\:-\:\omega } \right ) \right ]\:-\:\left [ \left ( \frac{1}{s\:+\:\omega} \right ) \right ] \:=\:\frac{1}{2}\left [ \frac{s\:+\:omega \:-\:s\:+\:\omega}{s^{2}\:-\:\omega^{2}} \right ]}$$
$$\mathrm{\therefore\: L\left [ sinh\: \omega t\: u\left ( t \right ) \right ]\:=\:\left (\frac{\omega }{s^{2}\:-\:\omega ^{2}} \right )}$$
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,
$$\mathrm{sinh\: \omega t\: u\left ( t \right )\overset{LT}{\leftrightarrow}\left ( \frac{\omega }{s^{2}\:-\:\omega ^{2}} \right )\:\: and\:\: ROC\:\rightarrow\: Re\left ( s \right )\:\gt\: 0}$$
Laplace Transform of Hyperbolic Cosine Function
Let,
$$\mathrm{x\left ( t \right )\:=\:cosh \: \omega t\: u\left ( t \right )\:=\:\frac{e^{\omega t }\:+\:e^{-\omega t}}{2} u\left ( t \right )}$$
Now, from the definition of the Laplace transform, we have,
$$\mathrm{X\left ( s \right )\:=\:L\left [ cosh\: \omega t\: u\left ( t \right ) \right ]\:=\:L\left [ \frac{e^{\:\omega t } \:+\: e^{-\:\omega t}}{2}u\left ( t \right ) \right ]}$$
$$\mathrm{\Rightarrow\: L\left [ cosh\: \omega t\: u\left ( t \right ) \right ]\:=\:\frac{1}{2}\left\{L\left [ e^{\omega t}u\left ( t \right ) \right ]\:+\:L\left [ e^{-\omega t}u\left ( t \right ) \right ] \right\}}$$
$$\mathrm{\Rightarrow\: L\left [ cosh\: \omega t\: u\left ( t \right ) \right ]\:=\:\frac{1}{2}\left [ \left ( \frac{1}{s\:-\:\omega } \right ) \:+\: \left ( \frac{1}{s\:+\:\omega} \right ) \right ] \:=\:\frac{1}{2}\left [ \frac{s\:+\: \omega \:+\:s\:-\:\omega}{s^{2}\:-\:\omega^{2}} \right ]}$$
$$\mathrm{\therefore\: L\left [cosh\: \omega t\: u\left ( t \right ) \right ]\:=\:\left (\frac{s}{s^{2}\:-\:\omega^{2}} \right )}$$
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,
$$\mathrm{cosh\: \omega t\: u\left ( t \right )\overset{LT}{\leftrightarrow}\left ( \frac{s }{s^{2}\:-\:\omega^{2}} \right )\:\: and\:\: ROC\:\rightarrow\: Re\left (s \right )\:\gt\: 0}$$