- Trending Categories
- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- MS Excel
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP
- Physics
- Chemistry
- Biology
- Mathematics
- English
- Economics
- Psychology
- Social Studies
- Fashion Studies
- Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Found 1024 Articles for Digital Electronics
7K+ Views
Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ is defined as, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Inverse Fourier TransformThe inverse Fourier transform of a continuous-time function is defined as, $$\mathrm{x(t)=\frac{1}{2\pi}\int_{−\infty}^{\infty}X(\omega)e^{j\omega t}d\omega}$$Properties of Fourier TransformThe continuous-time Fourier transform (CTFT) has a number of important properties. These properties are useful for driving Fourier transform pairs and also for deducing general frequency domain relationships. These properties also help to find the effect of various time domain operations on the frequency domain. Some of the important properties of continuous time Fourier transform are given in the table as −Property of CTFTTime Domain x(t)Frequency Domain X(ω)Linearity Property$ax_{1}(t)+bx_{2}(t)$$aX_{1}(\omega)+bX_{2}(\omega)$Time Shifting ... Read More
2K+ Views
Fourier SeriesIf $x(t)$ is a periodic function with period $T$, then the continuous-time exponential Fourier series of the function is defined as, $$\mathrm{x(t)=\sum_{n=−\infty}^{\infty}C_{n}\:e^{jn\omega_{0} t}\:\:… (1)}$$Where, $C_{n}$ is the exponential Fourier series coefficient, which is given by, $$\mathrm{C_{n}=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x(t)e^{-jn\omega_{0} t}dt\:\:… (2)}$$Modulation or Multiplication PropertyLet $x_{1}(t)$ and $x_{2}(t)$ two periodic signals with time period $T$ and with Fourier series coefficient $C_{n}$ and $D_{n}$. If$$\mathrm{x_{1}(t)\overset{FS}{\leftrightarrow}C_{n}}$$$$\mathrm{x_{2}(t)\overset{FS}{\leftrightarrow}D_{n}}$$Then, the modulation or multiplication property of continuous time Fourier series states, that$$\mathrm{x_{1}(t)\cdot x_{2}(t)\overset{FS}{\leftrightarrow}\sum_{k=−\infty}^{\infty}C_{k}\:D_{n-k}}$$Proof From the definition of continuous time Fourier series, we get, $$\mathrm{FS[x_{1}(t)\cdot x_{2}(t)]=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}[x_{1}(t)\cdot x_{2}(t)]e^{-jn\omega_{0} t}dt}$$$$\mathrm{\Rightarrow\:FS[x_{1}(t)\cdot x_{2}(t)]=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x_{1}(t)\left (\sum_{k=−\infty}^{\infty} C_{k} e^{jk\omega_{0} t}\right )e^{-jn\omega_{0} t}dt}$$$$\mathrm{\Rightarrow\:FS[x_{1}(t)\cdot x_{2}(t)]=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x_{1}(t)\left (\sum_{k=−\infty}^{\infty} C_{k} e^{-j(n-k)\omega_{0} t}\right )e^{-jn\omega_{0} ... Read More
10K+ Views
Fourier TransformThe Fourier transform of a continuous-time function $x(t)$ can be defined as, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Modulation Property of Fourier TransformStatement – The modulation property of continuous-time Fourier transform states that if a continuous-time function $x(t)$ is multiplied by $cos \:\omega_{0} t$, then its frequency spectrum gets translated up and down in frequency by $\omega_{0}$. Therefore, if$$\mathrm{x(t)\overset{FT}{\leftrightarrow}X(\omega)}$$Then, according to the modulation property of CTFT, $$\mathrm{x(t)\:cos\:\omega_{0}t\overset{FT}{\leftrightarrow}\frac{1}{2}[X(\omega-\omega_{0})+X(\omega+\omega_{0})]}$$ProofUsing Euler’s formula, we get, $$\mathrm{cos\:\omega_{0}t=\left [\frac{e^{j\omega_{0} t}+e^{-j\omega_{0} t}}{2} \right ]}$$Therefore, $$\mathrm{x(t)\:cos\:\omega_{0}t=x(t)\left [ \frac{e^{j\omega_{0} t}+e^{-j\omega_{0} t}}{2}\right ]}$$Now, from the definition of Fourier transform, we have, $$\mathrm{F[x(t)]=X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega_{0} t} \:dt}$$$$\mathrm{\Rightarrow\:F[x(t)\:cos\:\omega_{0} t]=\int_{−\infty}^{\infty}x(t)\:cos\:\omega_{0} t\:e^{-j\omega_{0} t}dt}$$$$\mathrm{\Rightarrow\:F[x(t)\:cos\:\omega_{0} t]=\int_{−\infty}^{\infty}x(t)\left [ \frac{e^{j\omega_{0} t}+e^{-j\omega_{0} t}}{2}\right ]e^{-j\omega t}dt}$$$$\mathrm{\Rightarrow\:F[x(t)\:cos\:\omega_{0} ... Read More
13K+ Views
Fourier TransformFor a continuous-time function $x(t)$, the Fourier transform can be defined as, $$\mathrm{X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j\omega t}dt}$$Linearity Property of Fourier TransformStatement − The linearity property of Fourier transform states that the Fourier transform of a weighted sum of two signals is equal to the weighted sum of their individual Fourier transforms.Therefore, if$$\mathrm{x_{1}(t)\overset{FT}{\leftrightarrow}X_{1}(\omega)\:\:and\:\:x_{2}\overset{FT}{\leftrightarrow}X_{2}(\omega)}$$Then, according to the linearity property of Fourier transform, $$\mathrm{ax_{1}(t)+bx_{2}(t)\overset{FT}{\leftrightarrow}aX_{1}(\omega)+bX_{2}(\omega)}$$Where, a and b are constants.ProofFrom the definition of Fourier transform, we have, $$\mathrm{F[x(t)]=X(\omega)=\int_{−\infty}^{\infty}x(t)e^{-j \omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=F[ax_{1}(t)+bx_{2}(t)]=\int_{−\infty}^{\infty}[ax_{1}(t)+bx_{2}(t)]e^{-j \omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\int_{−\infty}^{\infty}ax_{1}(t)e^{-j \omega t} dt+\int_{−\infty}^{\infty}bx_{2}(t)e^{-j \omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=a\int_{−\infty}^{\infty}x_{1}(t)e^{-j \omega t} dt+b\int_{−\infty}^{\infty}x_{2}(t)e^{-j \omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=aX_{1}(\omega)+bX_{2}(\omega)}$$$$\mathrm{\therefore\:F[ax_{1}(t)+bx_{2}(t)]=aX_{1}(\omega)+bX_{2}(\omega)}$$Or, it can also be written as, $$\mathrm{ax_{1}(t)+bx_{2}(t)\overset{FT}{\leftrightarrow}aX_{1}(\omega)+bX_{2}(\omega)}$$Frequency Shifting Property of Fourier TransformStatement – Frequency ... Read More
702 Views
Fourier SeriesIf $x(t)$ is a periodic function with period $T$, then the continuous-time exponential Fourier series of the function is defined as, $$\mathrm{x(t)=\displaystyle\sum\limits_{n=−\infty}^\infty\:C_{n}\:e^{jn\omega_{0}t}\:\:\:… (1)}$$Where, $C_{n}$ is the exponential Fourier series coefficient, which is given by, $$\mathrm{C_{n}=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}X(t)e^{-jn\omega_{0}t}\:dt\:\:… (2)}$$Linearity Property of Continuous Time Fourier SeriesConsider two periodic signals $x_{1}(t)$ and $x_{2}(t)$ which are periodic with time period T and with Fourier series coefficients $C_{n}$ and $D_{n}$ respectively. If$$\mathrm{x_{1}(t)\overset{FS}{\leftrightarrow}C_{n}}$$$$\mathrm{x_{2}(t)\overset{FS}{\leftrightarrow}D_{n}}$$Then, the linearity property of continuous-time Fourier series states that$$\mathrm{Ax_{1}(t)+Bx_{2}(t)\overset{FS}{\leftrightarrow}AC_{n}+BD_{n}}$$Proof By the definition of Fourier series of a periodic function, we get, $$\mathrm{FS[Ax_{1}(t)+Bx_{2}(t)]=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}[Ax_{1}(t)+Bx_{2}(t)]e^{-jn\omega_{0}t}\:dt}$$$$\mathrm{\Rightarrow\:FS[Ax_{1}(t)+Bx_{2}(t)]=A\left ( \frac{1}{T}\int_{t_{0}}^{t_{0}+T}\:x_{1}(t)\:e^{-jn\omega_{0}t}\:dt\right )+B\left ( \frac{1}{T}\int_{t_{0}}^{t_{0}+T}\:x_{2}(t)\:e^{-jn\omega_{0}t}\:dt \right )\:\:… (3)}$$On comparing equations (2) & (3), ... Read More
11K+ Views
What is GIBBS Phenomenon?The GIBBS phenomenon was discovered by Henry Wilbraham in 1848 and then rediscovered by J. Willard Gibbs in 1899.For a periodic signal with discontinuities, if the signal is reconstructed by adding the Fourier series, then overshoots appear around the edges. These overshoots decay outwards in a damped oscillatory manner away from the edges. This is known as GIBBS phenomenon and is shown in the figure below.The amount of the overshoots at the discontinuities is proportional to the height of discontinuity and according to Gibbs, it is found to be around 9% of the height of discontinuity irrespective ... Read More
5K+ Views
Fourier TransformThe Fourier transform of a continuous-time function can be defined as, $$\mathrm{X(\omega)=\int_{−\infty }^{\infty}\:X(t)e^{-j\omega t}\:dt}$$Differentiation in Frequency Domain Property of Fourier TransformStatement − The frequency derivative property of Fourier transform states that the multiplication of a function X(t) by in time domain is equivalent to the differentiation of its Fourier transform in frequency domain. Therefore, if$$\mathrm{X(t)\overset{FT}{\leftrightarrow}X(\omega)}$$Then, according to frequency derivative property, $$\mathrm{t\cdot x(t)\overset{FT}{\leftrightarrow}j\frac{d}{d\omega}X(\omega)}$$ProofFrom the definition of Fourier transform, we have, $$\mathrm{X(\omega)=\int_{−\infty }^{\infty}x(t)e^{-j\omega t}\:dt}$$Differentiating the above equation on both sides with respect to ω, we get, $$\mathrm{\frac{d}{d\omega}X(\omega)=\frac{d}{d\omega}\left [ \int_{−\infty }^{\infty}x(t)e^{-j\omega t}\:dt \right ]}$$$$\mathrm{\Rightarrow\:\frac{d}{d\omega}X(\omega)=\int_{−\infty }^{\infty} x(t)\frac{d}{d\omega}\left [e^{-j\omega t} \right ]dt}$$$$\mathrm{\Rightarrow\:\frac{d}{d\omega}X(\omega)=\int_{−\infty }^{\infty} x(t)(-jt)e^{-j\omega ... Read More
39K+ Views
Fourier TransformFor a continuous-time function $x(t)$, the Fourier transform is defined as, $$\mathrm{X(\omega)=\int_{−\infty }^{\infty}x(t)e^{−j\omega t}\:dt}$$Fourier Transform of Unit Step FunctionThe unit step function is defined as, $$\mathrm{u(t)=\begin{cases}1 & for\:t≥ 0\0 & for\:t< 0\end{cases}}$$Because the unit step function is not absolutely integrable, thus its Fourier transform cannot be found directly.In order to find the Fourier transform of the unit step function, express the unit step function in terms of signum function as$$\mathrm{u(t)=\frac{1}{2}+\frac{1}{2}sgn(t)=\frac{1}{2}[1+sgn(t)]}$$Given that$$\mathrm{x(t)=u(t)=\frac{1}{2}[1+sgn(t)]}$$Now, from the definition of the Fourier transform, we have, $$\mathrm{F[u(t)]=X(\omega)=\int_{−\infty }^{\infty}x(t)e^{-j\omega t} dt=\int_{−\infty }^{\infty} u(t)e^{-j\omega t} dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\int_{−\infty }^{\infty} \frac{1}{2}[1+sgn(t)]e^{-j\omega t}dt}$$$$\mathrm{\Rightarrow\:X(\omega)=\frac{1}{2}\left [ \int_{−\infty }^{\infty} 1 \cdot e^{-j\omega t} dt ... Read More
2K+ Views
We can create a Cucumber project template using Maven. This can be done by following the below steps −Step1− Click on the File menu in Eclipse. Then select the option New. Next click on Other.Step2− Click on Maven Project from the Maven folder. Then click on Next.Step3− Proceed with the further steps.Step4− Select maven-archetype-quickstart template. Then click on Next.Step5− Add GroupId as Automation, Artifact Id as Cucumber, and proceed.Step6− A project should get created with a Cucumber-type project structure. The Cucumber-related scripts should be written within the src/test/java folder.
1K+ Views
The mean square error (MSE) is defined as mean or average of the square of the difference between actual and estimated values.Mathematically, the mean square error is, $$\mathrm{\varepsilon =\frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}\left [ x(t) -\sum_{r=1}^{n}C_{r}g_{r}(t)\right ]^{2}dt}$$$$\mathrm{\varepsilon =\frac{1}{t_{2}-t_{1}}\left [ \int_{t_{1}}^{t_{2}}x^{2}(t)dt+\sum_{r=1}^{n}C_{r}^{2}\int_{t_{1}}^{t_{2}}g_{r}^{2}(t)dt-2\sum_{r=1}^{n}C_{r}\int_{t_{1}}^{t_{2}}x(t)g_{r}(t)dt\right ]\; ...(1)}$$$$\mathrm{\therefore C_{r}=\frac{\int_{t_{1}}^{t_{2}}x(t)g_{r}(t)dt}{\int_{t_{1}}^{t_{2}}g_{r}^{2}(t)dt}=\frac{1}{K_{r}}\int_{t_{1}}^{t_{2}}x(t)g_{r}(t)dt\; \; ...(2)}$$$$\mathrm{\therefore \int_{t_{1}}^{t_{2}}x(t)g_{r}(t)dt=C_{r}\int_{t_{1}}^{t_{2}}g_{r}^{2}(t)dt=C_{r}K_{r}\; \; ...(3)}$$Using equations (1) and (3), we have, $$\mathrm{\varepsilon =\frac{1}{t_{2}-t_{1}}\left [\int_{t_{1}}^{t_{2}} x^{2}(t)dt +\sum_{r=1}^{n}C^{2}_{r}K_{r}-2\sum_{r=1}^{n}C^{2}_{r}K_{r}\right ]}$$$$\mathrm{\Rightarrow \varepsilon =\frac{1}{t_{2}-t_{1}}\left [\int_{t_{1}}^{t_{2}} x^{2}(t)dt -\sum_{r=1}^{n}C^{2}_{r}K_{r}\right ]\; \; ...(4)}$$$$\mathrm{\Rightarrow \varepsilon =\frac{1}{t_{2}-t_{1}}\left [ \int_{t_{1}}^{t_{2}}x^{2}(t)dt-(C_{1}^{2}K_{1}+C_{2}^{2}K_{2}+\cdot \cdot \cdot +C_{n}^{2}K_{n}) \right ]\; \; \cdot \cdot \cdot (5)}$$Therefore, the mean square error can be evaluated using eqn. (5).Numerical ExampleA rectangular function is defined as, $$\mathrm{x(t)=\left\{\begin{matrix} 1\; \; for\, 0< t< ... Read More