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

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

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.

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

Invertible SystemIf a system has a unique relationship between its input and output, the system is called the invertible system. In other words, a system is said to be an invertible system only if an inverse system exists which when cascaded with the original system produces an output equal to the input of the first system. The block diagram representation of an invertible system is shown in Figure-1.Mathematically, an invertible system is defined as,     𝑥(𝑡) = 𝑇−1[𝑦(𝑡)] = 𝑇−1{𝑇[𝑥(𝑡)]}   … for continuous time system    𝑥(𝑛) = 𝑇−1[𝑦(𝑛)] = 𝑇−1{𝑇[𝑥(𝑛)]}   … for discrete time systemNon-Invertible SystemA ... Read More

Energy SignalA signal is said to be an energy signal if and only if its total energy E is finite, i.e., 0 < 𝐸 < ∞. For an energy signal, the average power P = 0. The nonperiodic signals are the examples of energy signals.Power SignalA signal is said to be a power signal if its average power P is finite, i.e., 0 < 𝑃 < ∞. For a power signal, the total energy E = ∞. The periodic signals are the examples of power signals.Continuous Time CaseIn electric circuits, the signals may represent current or voltage. Consider a voltage ... Read More

Even SignalA signal is said to be an even signal if it is symmetrical about the vertical axis or time origin, i.e., 𝑥(𝑡) = 𝑥(−𝑡); for all 𝑡     … continuous time signal𝑥(𝑛) = 𝑥(−𝑛); for all 𝑛     … discrete time signalOdd SignalA signal is said to be an odd signal if it is anti-symmetrical about the vertical axis, i.e., 𝑥(−𝑡) = −𝑥(𝑡); for all 𝑡    … continuous time signal𝑥(−𝑛) = −𝑥(𝑛); for all 𝑛     … discrete time signalDetermination of Even and Odd Components of a SignalContinuous-time CaseEvery signal need not be either purely ... Read More

Bounded SignalA signal whose magnitude is a finite value is called the bounded signal. A sine wave is an example of bounded signal.BIBO Stable SystemA system is called BIBO stable (or bounded-input, bounded-output stable) system, if and only if every bounded input to the system produces a bounded output.BIBO Stability CriterionFor a system to be BIBO stable, the necessary condition is given by the expression, $$\mathrm{\int_{-\infty }^{\infty}\left | h(t) \right |dt < \infty \; \;}\;\;...(1)$$Where, h(t) is the impulse response of the system. The condition given in the expression (1) is called the BIBO stability criterion.ProofConsider an LTI (linear time-invariant) ... Read More

Linear SystemA system is said to be linear if it obeys the principle of homogeneity and principle of superposition.Principle of HomogeneityThe principle of homogeneity says that a system which generates an output y(t) for an input x(t) must produce an output ay(t) for an input ax(t).Superposition PrincipleAccording to the principle of superposition, a system which gives an output 𝑦1(𝑡) for an input 𝑥1(𝑡) and an output 𝑦2(𝑡) for an input 𝑥2(𝑡) must produce an output [𝑦1(𝑡) + 𝑦2(𝑡)] for an input [𝑥1(𝑡) + 𝑥2(𝑡)].Therefore, for a continuous-time linear system, [𝑎𝑦1(𝑡) + 𝑏𝑦2(𝑡)] = 𝑇[𝑎𝑥1(𝑡) + 𝑏𝑥2(𝑡)] = 𝑎𝑇[𝑥1(𝑡)] + 𝑏𝑇[𝑥2(𝑡)]Also, ... Read More

Linear Time-Invariant (LTI) SystemA system that possesses two basic properties namely linearity and timeinvariant is known as linear time-invariant system or LTI system.There are two major reasons behind the use of the LTI systems −The mathematical analysis becomes easier.Many physical processes through not absolutely LTI systems can be approximated with the properties of linearity and time-invariance.Continuous-Time LTI SystemThe LTI systems are always considered with respect to the impulse response. That means the input is the impulse signal and the output is the impulse response.Consider a continuous-time LTI system as shown in the block diagram of Figure-1.Here, the input to the ... Read More