Found 1006 Articles for Electronics & Electrical

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

Synchronous Motor V CurvesThe graphs plotted between armature current (Ia) and field current (If) for different constant loads are known as the V curves of the synchronous motor.The power factor of a synchronous motor can be controlled by changing the field excitation, i.e., by variation of field current (If). Also, the armature current (Ia) changes with the change in the excitation or field current (If).Now, let us assume that the synchronous motor is operating at no-load. If the field current (If) is increased from a small value, the armature current (Ia) decreases until Ia becomes minimum. The power factor of ... Read More

Synchronous Motor TorqueThe mechanical power developed (Pm) by any synchronous motor can be expressed as −$$\mathrm{P_{m}=\frac{2πN_{S}\tau_{g}}{60}\:Watts\:\:\:\:\:\:...(1)}$$Also, the mechanical power developed (Pm) is, $$\mathrm{P_{m}=VI_{a}Cos(δ-φ)\:\:\:\:\:\:...(2)}$$Where, NS is the synchronous speed in RPM.τgis the gross-torque in N-m.Therefore, the gross torque of the synchronous motor is given by, $$\mathrm{\tau_{g}=\frac{60}{2π}\frac{P_{m}}{N_{S}}=9.55\times\frac{P_{m}}{N_{S}}N-m\:\:\:\:\:\:...(3)}$$And the shaft torque is given by, $$\mathrm{\tau_{sh}=9.55\times\frac{P_{o}}{N_{S}}N-m\:\:\:\:\:\:...(4)}$$Where, Po is the mechanical power output at the shaft of the motor.From Eqns.(3)&(4), it can be noted that the torque is directly proportional to the mechanical power since the speed of the motor is constant i.e. synchronous speed (NS).Types of Torque in a Synchronous MotorIn order to select a ... Read More

A 3-phase synchronous motor is a 3-phase synchronous machine which is operated as a motor i.e. converts electrical energy input into mechanical energy output.A synchronous motor has a unique feature that is it runs at a constant speed equal to the synchronous speed at all load provided that the load on the motor does not exceed the limiting value. If the load on the motor exceeds the limiting value, then the motor will come to rest and the average torque developed by the motor becomes zero. Because of this, a synchronous motor is not inherently self-starting.A synchronous motor is a ... Read More

Let us first take a look at the block diagram of a synchronous motor −Main Features of Synchronous MotorFollowing are the characteristic features of a synchronous motor −A synchronous motor either runs at synchronous speed or not at all, i.e., while running, it maintains a constant speed from no-load to full load.The speed of a synchronous motor is independent of load.Synchronous motors are not inherently self-starting. Some auxiliary means have to be provided for starting.A synchronous motor will stall if, while running, the load on the shaft is increased beyond the maximum limit that the machine can drive.A synchronous motor ... Read More

An over-excited synchronous motor running on no-load is called the synchronous condenser. It is also known as synchronous capacitor or synchronous compensator or synchronous phase modifier.A synchronous motor can deliver or absorb reactive power by changing the DC excitation of its field winding. It can be made to draw a leading current from the supply with over-excitation of its field winding and therefore, it supplies lagging reactive power (or absorbs leading reactive power).Under-excited Synchronous MotorWhen the synchronous motor is under-excited, then it draws a lagging current form the source and hence supplies leading reactive power (or absorbs lagging reactive power). ... Read More