Manish Kumar Saini has Published 1143 Articles

Fourier Transform of Two-Sided Real Exponential Functions

Manish Kumar Saini

Manish Kumar Saini

Updated on 09-Dec-2021 07:01:15

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Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Fourier Transform of Two-Sided Real Exponential FunctionLet a two-sided real exponential function as, $$\mathrm{x(t)=e^{-a|t|}}$$The two-sided or double-sided real exponential function is defined as, $$\mathrm{e^{-a|t|}=\begin{cases}e^{at} & for\:t ≤ 0\e^{-at} & for\:t ≥ 0 \end{cases} =e^{at}u(-t)+e^{-at}u(t) }$$Where, the ... Read More

Fourier Transform of the Sine and Cosine Functions

Manish Kumar Saini

Manish Kumar Saini

Updated on 09-Dec-2021 06:57:00

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Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as, $$\mathrm{x(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t }dt}$$Fourier Transform of Sine FunctionLet$$\mathrm{x(t)=sin\:\omega_{0} t}$$From Euler’s rule, we have, $$\mathrm{x(t)=sin\:\omega_{0} t=\left[\frac{ e^{j\omega_{0} t}- e^{-j\omega_{0} t}}{2j} \right]}$$Then, from the definition of Fourier transform, we have, $$\mathrm{F[sin\:\omega_{0} t]=X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt=\int_{−\infty}^{\infty}sin\:\omega_{0}\: t\: e^{-j\omega t}dt}$$$$\mathrm{ \Rightarrow\:X(\omega)=\int_{−\infty}^{\infty}\left[ \frac{e^{j\omega_{0} t}-e^{-j\omega_{0} t}}{2j}\right] ... Read More

Fourier Transform of Single-Sided Real Exponential Functions

Manish Kumar Saini

Manish Kumar Saini

Updated on 09-Dec-2021 06:53:42

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Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as, $$\mathrm{X(\omega)= \int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Fourier Transform of One-Sided Real Exponential FunctionA single-sided real exponential function is defined as, $$\mathrm{x(t)=e^{-a t}u(t)}$$Where, $u(t)$ is the unit step signal and is defined as, $$\mathrm{u(t)=\begin{cases}1 & for\:t≥ 0 \0 & for\:t < 0 ... Read More

Fourier Transform of Signum Function

Manish Kumar Saini

Manish Kumar Saini

Updated on 09-Dec-2021 06:50:10

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Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Fourier Transform of Signum FunctionThe signum function is represented by $sgn(t)$ and is defined as$$\mathrm{sgn(t)=\begin{cases}1 & for\:t>0\-1 & for\:tRead More

Fourier Transform of Rectangular Function

Manish Kumar Saini

Manish Kumar Saini

Updated on 08-Dec-2021 07:21:30

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Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}\:dt}$$Fourier Transform of Rectangular FunctionConsider a rectangular function as shown in Figure-1.It is defined as, $$\mathrm{rect\left(\frac{t}{τ}\right)=\prod\left(\frac{t}{τ}\right)=\begin{cases}1 & for\:|t|≤ \left(\frac{τ}{2}\right)\0 & otherwise\end{cases}}$$Given that$$\mathrm{x(t)=\prod\left(\frac{t}{τ}\right)}$$Hence, from the definition of Fourier transform, we have, $$\mathrm{F\left[\prod\left(\frac{t}{τ}\right) \right]=X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}\:dt=\int_{−\infty}^{\infty}\prod\left(\frac{t}{τ}\right)e^{-j\omega t}\:dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\int_{−(τ/2)}^{(τ/2)}1\cdot e^{-j\omega t}\:dt=\left[\frac{e^{-j\omega ... Read More

Fourier Transform of a Triangular Pulse

Manish Kumar Saini

Manish Kumar Saini

Updated on 08-Dec-2021 07:13:24

20K+ Views

Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)\:e^{-j\omega t}dt}$$Fourier Transform of a Triangular PulseA triangular signal is shown in Figure-1 −And it is defined as, $$\mathrm{\Delta \left(\frac{t}{τ}\right)=\begin{cases}\left( 1+\frac{2t}{τ}\right); & for\:\left(-\frac{τ}{2}\right) Read More

Expressions for the Trigonometric Fourier Series Coefficients

Manish Kumar Saini

Manish Kumar Saini

Updated on 08-Dec-2021 07:04:21

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The infinite series of sine and cosine terms of frequencies $0, \omega_{0}, 2\omega_{0}, 3\omega_{0}, ....k\omega_{0}$is known as trigonometric Fourier series and can written as, $$\mathrm{x(t)=a_{0}+\sum_{n=1}^{\infty}a_{n}\:cos\:n\omega_{0} t+b_{n}\:sin\:n\omega_{0} t… (1)}$$Here, the constant $a_{0}, a_{n}$ and $b_{n}$ are called trigonometric Fourier series coefficients.Evaluation of a0To evaluate the coefficient $a_{0}$, we shall integrate the ... Read More

Fourier Series Representation of Periodic Signals

Manish Kumar Saini

Manish Kumar Saini

Updated on 08-Dec-2021 06:55:39

8K+ Views

What is Fourier Series?In the domain of engineering, most of the phenomena are periodic in nature such as the alternating current and voltage. These periodic functions could be analysed by resolving into their constituent components by a process called the Fourier series.Therefore, the Fourier series can be defined as under ... Read More

Fourier Cosine Series – Explanation and Examples

Manish Kumar Saini

Manish Kumar Saini

Updated on 08-Dec-2021 06:53:17

6K+ Views

The cosine form of Fourier series is the alternate form of the trigonometric Fourier series. The cosine form Fourier series is also known as polar form Fourier series or harmonic form Fourier series.The trigonometric Fourier series of a function x(t) contains sine and cosine terms of the same frequency. That ... Read More

Signals and Systems – What is Quarter Wave Symmetry?

Manish Kumar Saini

Manish Kumar Saini

Updated on 07-Dec-2021 07:49:56

5K+ Views

Quarter Wave SymmetryA periodic function $x(t)$ which has either odd symmetry or even symmetry along with the half wave symmetry is said to have quarter wave symmetry.Mathematically, a periodic function $x(t)$ is said to have quarter wave symmetry, if it satisfies the following condition −$$\mathrm{x(t)=x(-t)\:or\:x(t)=-x(-t)\:and\:x(t)=-x\left (t ± \frac{T}{2}\right )}$$Some examples ... Read More

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