Found 451 Articles for Electron

Addition and Subtraction of Continuous-Time SignalsThe sum of two continuous time signals 𝑥1(𝑡) and 𝑥2(𝑡) can be obtained by adding their values at every instant of time. Likewise, the difference of two continuous time signals 𝑥1(𝑡) and 𝑥2(𝑡) can be obtained by subtracting the values of one signal (say 𝑥2(𝑡)) from another signal (say 𝑥1(𝑡)) at every instant of time.Addition of SignalsLet two continuous time signals 𝑥1(𝑡) and 𝑥2(𝑡) as shown in Figure-1. The sum of these two signals 𝑥1(𝑡) + 𝑥2(𝑡) is also shown in Figure-1.ExplanationThe sum of these two signals can be obtained by considering different time intervals ... Read More

Causal SignalA continuous time signal 𝑥(𝑡) is called causal signal if the signal 𝑥(𝑡) = 0 for 𝑡 < 0. Therefore, a causal signal does not exist for negative time. The unit step signal u(t) is an example of causal signal as shown in Figure-1.Similarly, a discrete time sequence x(n) is called the causal sequence if the sequence x(n) = 0 for n < 0.Anti-Causal SignalA continuous-time signal x(t) is called the anti-causal signal if x(t) = 0 for t > 0. Hence, an anti-causal signal does not exist for positive time. The time reversed unit step signal u(-t) is ... Read More

Causal SystemA system whose output or response at any time instant (t) depends only on the present and past values of the input but not on the future values of the input is called the causal system. For a causal system, the output or response does not begin before the input signal is applied. This is why, a causal system is also called a non-anticipated system.The causal systems are real time systems and they can be physically realised. For a causal system, the impulse response of the system is zero for negative time (i.e., t < 0) because the impulse ... Read More

A sinusoidal function or sinusoidal signal is a function that describes a smooth periodic oscillation.Continuous-Time Sinusoidal SignalA sinusoidal signal which is defined for every instant of time is called continuous-time sinusoidal signal. The continuous time sinusoidal signal is given as follows −𝑥(𝑡) = 𝐴 sin(𝜔𝑡 + 𝜑) = 𝐴 sin(2𝜋𝑓𝑡 + 𝜑)Where, A is the amplitude of the signal. That is the peak deviation of the signal from zero.ω=2πf is the angular frequency in radians per seconds.f is the frequency of the signal in Hz.φ is the phase angle in radians.All the continuous-time sinusoidal signals are periodic signal. The time ... Read More

What is Time Shifting?Time shifting or Shifting of a signal in time means that the signal may be either delayed in the time axis or advanced in the time axis.Time Shifting of Continuous-Time SignalThe time shifting of a continuous time signal x(t) is represented as, 𝑦(𝑡) = 𝑥(𝑡 − 𝑡0)The time-shifting of a signal results in the time delay or time advancement. The above expression shows that the signal y(t) can be obtained by time shifting the signal x(t) by t0 units. If t0 is positive in the above expression, then the shift of the signal is to the right ... Read More

What is Time Reversal of a Signal?The time reversal of a signal is folding of the signal about the time origin (or t = 0). The time reversal or folding of a signal is also called as the reflection of the signal about the time origin (or t = 0). Time reversal of a signal is a useful operation on signals in convolution.Time Reversal of a Continuous-Time SignalThe time reversal of a continuous time signal x(t) is the rotation of the signal by 180° about the vertical axis. Mathematically, for the continuous time signal x(t), the time reversal is given ... Read More

Discrete Time SignalsThe signals which are defined only at discrete instants of time are known as discrete time signals. The discrete time signals are represented by x(n) where n is the independent variable in time domain.Representation of Discrete Time SignalsA discrete time signal may be represented by any one of the following four ways −Graphical RepresentationFunctional RepresentationTabular RepresentationSequence RepresentationGraphical Representation of Discrete Time SignalsConsider a discrete time signal x(n) with the values, x(−3) = −2, x(−2) = 3, x(−1) = 0, x(0) = −1, x(1) = 2, x(2) = 3, x(3) = 1This discrete time signal can be represented graphically ... 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 signalProperties of Even and Odd SignalsAddition and Subtraction Properties of Even and Odd SignalsThe addition or subtraction of two odd signals is also ... Read More

Consider a synchronous motor is operating at lagging power factor. The voltage equation of a synchronous motor is given by, $$\mathrm{V=E_{f}+I_{a}Z_{S}\:\:\:\:\:\:...(1)}$$Where, $$\mathrm{V=V\angle 0°\:and\:E_{f}=E_{f}\:\angle-δ}$$$$\mathrm{\therefore\:I_{a}=\frac{V-E_{f}}{Z_{S}}\:\:\:\:\:\:...(2)}$$$$\mathrm{\Longrightarrow\:I_{a}=\frac{V\angle 0°-E_{f}-δ}{Z_{S}\angleθ_{Z}}=\frac{V}{Z_{S}}\angle-θ_{Z}-\frac{E_{f}}{Z_{S}}\angle-(δ+θ_{Z})}$$$$\mathrm{\therefore\:I^{*}_{a}=\frac{V}{Z_{S}}\angleθ_{Z}-\frac{E_{f}}{Z_{S}}\angle(δ+θ_{Z})\:\:\:\:\:\:...(3)}$$Complex Power Output per Phase of a Synchronous MotorThe complex power output of a synchronous motor is given by, $$\mathrm{S_{o}=E_{f}I^{*}_{a}=P_{o}+jQ_{o}\:\:\:\:\:\:...(4)}$$$$\mathrm{\Longrightarrow\:S_{o}=E_{f}\:\angle-δ\left(\frac{V}{Z_{S}}\angleθ_{Z}-\frac{E_{f}}{Z_{S}}\angle(δ+θ_{Z})\right)}$$$$\mathrm{\Longrightarrow\:S_{o}=\left(\frac{VE_{f}}{Z_{S}}cos(θ_{Z}-δ)+j\frac{VE_{f}}{Z_{S}}sin(θ_{Z}-δ)\right)-\left(\frac{E^{2}_{f}}{Z_{S}}cosθ_{Z}+j\frac{E^{2}_{f}}{Z_{S}}sinθ_{Z}\right)}$$$$\mathrm{\therefore\:S_{o}=\left(\frac{VE_{f}}{Z_{S}}cos(θ_{Z}-δ)-\frac{E^{2}_{f}}{Z_{S}}cosθ_{Z}\right)+j\left(\frac{VE_{f}}{Z_{S}}sin(θ_{Z}-δ)-\frac{E^{2}_{f}}{Z_{S}}sinθ_{Z}\right)\:\:\:\:\:\:...(5)}$$Real Power Output per Phase of the Synchronous MotorBy equating the real part of equation(5), we get the real power output of the synchronous motor, i.e., $$\mathrm{P_{o}=\frac{VE_{f}}{Z_{S}}cos(θ_{Z}-δ)-\frac{E^{2}_{f}}{Z_{S}}cosθ_{Z}}$$$$\mathrm{\because\:cosθ_{Z}=\frac{R_{a}}{Z_{S}}}$$$$\mathrm{\therefore\:P_{o}=\frac{VE_{f}}{Z_{S}}cos(θ_{Z}-δ)-\frac{E^{2}_{f}}{Z^{2}_{S}}R_{a}\:\:\:\:\:\:...(6)}$$But, $$\mathrm{θ_{Z}=(90°-α_{Z});cos(θ_{Z}-δ)=cos(90°-δ+α_{Z})=sin(δ+α_{Z})}$$$$\mathrm{\therefore\:P_{o}=\frac{VE_{f}}{Z_{S}}sin(δ+α_{Z})-\frac{E^{2}_{f}}{Z^{2}_{S}}R_{a}\:\:\:\:\:\:...(7)}$$Reactive Power Output per Phase of the Synchronous MotorBy equating the imaginary part of Equation(5), we obtain the reactive power output of the synchronous motor, i.e., $$\mathrm{Q_{o}=\frac{VE_{f}}{Z_{S}}sin(θ_{Z}-δ)-\frac{E^{2}_{f}}{Z_{S}}sinθ_{Z}}$$$$\mathrm{\because\:sinθ_{Z}=\frac{X_{S}}{Z_{S}}}$$$$\mathrm{\therefore\:Q_{o}=\frac{VE_{f}}{Z_{S}}sin(θ_{Z}-δ)-\frac{E^{2}_{f}}{Z^{2}_{S}}R_{a}\:\:\:\:\:\:...(8)}$$But, $$\mathrm{θ_{Z}=(90°-α_{Z});sin(θ_{Z}-δ)=sin(90°-δ+α_{Z})=cos(δ+α_{Z})}$$$$\mathrm{\therefore\:Q_{o}=\frac{VE_{f}}{Z_{S}}cos(δ+α_{Z})-\frac{E^{2}_{f}}{Z^{2}_{S}}X_{S}\:\:\:\:\:\:...(9)}$$Also, for a synchronous ... Read More

The phasor diagram at lagging power factor and the equivalent circuit diagram of a cylindrical synchronous motor are shown in Figure-1 and Figure-2, respectively.The terminal voltage (V) is taken as reference phasor and the excitation voltage (Ef) lags the terminal voltage (V) by an angle δ so that$$\mathrm{V=V\angle0°\:and\:E_{f}=E_{f}\:\angle-δ}$$Applying KVL in the loop of equivalent circuit, we get, $$\mathrm{V=E_{f}+I_{a}Z_{S}\:\:\:\:\:\:...(1)}$$$$\mathrm{\therefore\:I_{a}=\frac{V-E_{f}}{Z_{S}}\:\:\:\:\:\:...(2)}$$$$\mathrm{\Longrightarrow\:I_{a}=\frac{V\angle0°-E_{f}\:\angle-δ}{Z_{S}\angleθ_{Z}}=\frac{V}{Z_{S}}\angle-θ_{Z}-\frac{E_{f}}{Z_{S}}\angle-(δ+θ_{Z})}$$$$\mathrm{\therefore\:I^*_{a}=\frac{V}{Z_{S}}\angleθ_{Z}-\frac{E_{f}}{Z_{S}}\angle(δ+θ_{Z})\:\:\:\:\:\:...(3)}$$Therefore, the expressions for various input powers to a synchronous motor are given as follows −Complex Power Input per Phase to the Synchronous MotorThe complex input power to a synchronous motor is given by, $$\mathrm{S_{i}=VI^*_{a}=P_{i}+jQ_{i}\:\:\:\:\:\:...(4)}$$From Eqns.(3) and (4), we get, $$\mathrm{S_{i}=\frac{V^{2}}{Z_{S}}\angleθ_{Z}-\frac{VE_{f}}{Z_{S}}\angle(δ+θ_{Z})}$$$$\mathrm{\Longrightarrow\:S_{i}=\left(\frac{V^{2}}{Z_{S}}cosθ_{Z}+j\frac{V^{2}}{Z_{S}}sinθ_{Z}\right)-\left(\frac{VE_{f}}{Z_{S}}cos(δ+θ_{Z})+j\frac{VE_{f}}{Z_{S}}sin(δ+θ_{Z})\right )}$$$$\mathrm{\therefore\:S_{i}=\left[\frac{V^{2}}{Z_{S}}cosθ_{Z}-\frac{VE_{f}}{Z_{S}}cos(δ+θ_{Z})\right]+j\left[\frac{V^{2}}{Z_{S}}sinθ_{Z}-\frac{VE_{f}}{Z_{S}}sin(δ+θ_{Z})\right]\:\:\:\:\:\:...(5)}$$Real Input ... Read More