Electronics Engineering

If we need to add two signals in time domain, we perform convolution. A better way, is to convert the two signals from time domain to frequency domain. This can be achieved by FAST FOURIER TRANFORM. Once both the signals have been converted to frequency domain, they can simply be multiplied. Since Convolution in time domain is similar to multiplying in Frequency domain. Once both the signals have been multiplied, they can be converted back to time domain by Inverse Fourier Transform method. Thus achieving accurate results.

The secondary winding leakage inductance limits the current during a short. It seems that the current through the primary is limited by winding resistance and leakage resistance when the secondary is shorted.

In the common emitter configuration, gain is hFe or collector resistance divided by emitter resistance, whichever is less. Placing a capacitor across the emitter resistor effectively makes the emitter resistor less, for higher frequencies, so the gain is higher for higher frequencies. This creates a high pass filter, or a low cut filter, depending on what you want to call it.

The only thing that limits the number of inputs and output instructions is the size/amount of the PLC memory.

inductors are more expensive and complex.and they take up space. Their effect can be replicated by active circuits.

main difference is that clamper circuit contains a caoacitor while clipper not

clipper cut off the specific portion of wave

Yes, if the resistance remains constant. Power is voltage times current, and current is voltage divided by resistance, so power is voltage squared divided by resistance. In essence, the power increases as the square of the voltage.

The reason why resistor voltage decreases while a capacitor discharges is because the resistor acts like a source of electrical energy. As the capacitor discharges, it draws energy from the resistor, which causes the voltage across the resistor to decrease. This is because the capacitor is acting like a drain, and is taking energy out of the resistor, thus causing the voltage across the resistor to decrease.

The resistor and capacitor work together in order to create a discharge circuit. This is done by connecting the capacitor to the resistor, and then to a voltage source. The voltage source supplies the energy to the resistor, and then the resistor transfers this energy to the capacitor. As the capacitor discharges, it takes energy from the resistor, which causes the voltage across the resistor to decrease.

In order to understand this process better, it is important to understand the basics of Ohm's Law. Ohm's Law states that the voltage across a resistor is equal to the current through the resistor multiplied by the resistance. As the capacitor discharges, it takes energy from the resistor, which means that the current through the resistor decreases, and therefore the voltage across the resistor will also decrease.

it won't flow

There is always a transition in the middle of an interval

the differences between UJT and FET are :-

1. structural :- a. there is only one p-channel in the UJT where as two in JFET

b. the p-channel of UJT is more highly doped when compared to p-channel in JFET

2. functional :- UJT always works in forward biased condition (gate is forward biased) where as JFET always work in rverse bias condition (gate is reverse biased)

connections are to be made per the crkt

the main of this crkt is to make strong or to bost the signal

comming to the crkt input is given with ac supply and they are connected to resistor and capacitor when the ac suppply is given it contains slight amount of dc supply so to block the dc supply capacitor is used and the dc supply is given at the collector so that resistor present at c and ground gets biased then another capacitor used at c to ground for removing the dc supply then out put seen on cro across collector to ground we get the perfect amplified signal

the more we give the coupled amplifer the most amplfed sgnal we get

When a circuit has a gap in it, everythig stops working

because the electricity wont be able to flow around the whole circuit

since torque is proportional to flux and armature current . Flux for dc shunt is constant . torque is proportional to armature current only. And initially armature current is very low hence it cannot be started at load .

Waves of very low frequency are not used for data transmission in computer networks because they may attenuate and fail to transmit data. However, sometimes such low frequency waves may be used if repeaters are installed along the transmission route.

Frequency

That is called a parallel connection and will double the power if the batteries is the same size and the same voltage, the voltage will be the same as one battery, so if it is two 12 volt batteries the voltage will be 12 volt. It is dangerous to connect two batteries of different voltage in parallel like a12 volt and a 6 volt the 6 volt will then draw current from the 12 volt and it can overheat and even explode.

Any device that outputs a voltage higher than its input voltage. This device can be capacitive, inductive, or other.

a) to act as a switching device; a change in the bias voltage at the base- emitter junction can cause an increase in signal flowing through the transistor through the collector terminal

and this cause the output voltage at the collector terminal to change; eg to drop to a low voltage level, this is seen as the transistor device switching on to maximum conduction rate and a low level output at the collector.

b) to act an amplifer . whether the input signal is an ac type signal in which case its a signal power amplifer function; eg sound amplifer control circuits , or whether a dc input signal, in which case the input dc signal is amplfied to be reproduced as a bigger signal at the output collector terminal

If you're looking for a definition, it's: the voltage at which, a diode can be considered a "short circuit" or low-value resistor

It varies with each diode, but most have approximately 0.6 or 0.7 Volts across them when you get almost 1mA flowing FORWARD through them. For light emitting diodes (LEDs), it varies between diodes and is largely dependent on the colour of the light. Green ones typically have 1.3V @ 1mA, red = 1.8V @ 1mA, and higher for other colours. Infrared LEDs usually have 1.1V @ 1mA. Higher cutoff voltages occur at higher forward currents, meaning that at 1mA, Vf might be 1.8V for a certain diode, but at 10mA Vf is maybe 1.9V.

One important side note is that reverse current is still possible, but is so small it's usually negligible. Also, it's not recommended to force current backwards through a diode (exception: Zener diodes) because it usually requires a higher voltage to accomplish this.

The cutoff voltage of a diode is the maximum voltage that the diode can withstand in the revers biase above which the device will be destroyed.

The incremental resistance of a diode is the inverse of the slope of the V-I curve at the operating point.

A normally closed switch is a switch that is normally closed. When you operate it, it opens.

A capacitor opposes a change in voltage, but it will help to look at both the device and at a circuit up close to see what's going on.

Any capacitor is two "plates" separated by a dielectric or insulator. Connect a wire to each plate and you've got the device. In a direct current circuit, the voltage source will cause current flow in only one direction. A common battery is a good example. Let's look further.

When a capacitor is connected in a DC circuit and the circuit is energized, the voltage source will want to cause current to flow in only the one direction. In the initial moment when the power is switched on, electrons will flow in the circuit. Electrons will leave the negative terminal of the source and enter the positive terminal. The current flow will travel through the wire, and electrons will "pile up" on one of the plates of the capacitor. As electrons are "piling up" on one plate, their presence there will create an electric field across the dielectric to the other plate. This electric field will cause electrons on that other plate to leave. The capacitor is charging, and the voltage source will, for the first instant of time, think that things are "fine" and current will flow. But as the capacitor charges, current flow drops off, and it eventually stops when the voltage across the plates equals the source voltage.

In review, as the DC power is switched on in a circuit with a capacitor in it, current will flow "normally" for the first instant. But as the first electrons arrive on one plate and force them off the other plate, current in the circuit will begin dropping off. The voltage developing across the plates of the capacitor opposes the battery voltage. Eventually the capacitor is charged and all current flow has stopped. There is some math that says something slightly different, but for all practical purposes, the capacitor is considered fully charged in a very short period of time. This will depend on circuit resistance and the ability of the source to deliver current, of course. But that capacitor will, when charged, not "pass" any more current. The voltage across the plates is equal to (an opposing) the source voltage, and no more electrons can get onto the negative plate to force more off the positive plate.