"Average" can mean many things. If what you're meaning is the root mean square, you can calculate the same as for a sine wave:
integrate the square of the waveshape from 0 to the end of the first period, divide by the length of the period, and take the square root of this value.
For a 50% square wave (it's at amplitude a for 50% of the cycle, and at 0 for the other 50%, so the integral will be over only 1/2 the period):
sqrt[1/P * integral (a^2) dx, from 0 to 1/2*P] = sqrt[a^2/P * (x from 0 to 1/2*P)]
sqrt[a^2 / P * (P/2) = a/sqrt(2)
For 33.3%, integrate over 1/3 of the period, so RMS = a / sqrt(3).
RMS stands for Root Mean Square. Power is calculated as V2/R where V is the voltage and R is the resistive component of a load, This is easy toi calculate for a DC voltage, but how to calculate it for a sinusoidal voltage? The answer is to take all the instantaneous voltages in the sine wave, square them, take the mean of the squares, then take the square root of the result. This is defined as the "heating effect voltage". For a sine wave, this is 0.707 of the peak voltage.
rms values refer to "root mean square" mathematical values of the sine wave of electricity. This is essentially an "average" value of the voltage being measured as voltage in any circuit varies constantly.
Peak values times 0.707 gives the RMS value. This question cannot be answered without knowing the waveform of the voltage. If it is continuous direct current rms value is 100V. If it is a sine wave (like house current), the rms value is 100/ (sq. root(2)) or 70.7V approx. For a regular series of square pulses, it is 100V times the mark/(mark+space) ratio. Any other waveform, you need to calculate the root mean square value of the function. ( Square, integrate and take square root over one complete period.) If all the above fail, measure the heating effect in a 40K resistor. It will be 0.25 Watt or less - calculate from the result.
Calculate the area.
RMS and peak voltage for a square waveform are the same. There is a small caveat, and that is that you'd have to have a "perfect" square wave with a rise time of zero. Let's have a look. If we have a perfect square wave, it has a positive peak and a negative peak (naturally). And if the transition from one peak to the other can be made in zero time, then the voltage of the waveform will always be at the positive or the negative peak. That means it will always be at its maximum, and the effective value (which is what RMS or root mean square is - it's the DC equivalent or the "area under the curve of the waveform") will be exactly what the peak value is. It's a slam dunk. If we have a (perfect) square wave of 100 volts peak, it will always be at positive or negative 100 volts. As RMS is the DC equivalent, or is the "heating value for a purely resistive load" on the voltage source, the voltage will always be 100 volts (either + or -), and the resistive load will always be driven by 100 volts. Piece of cake.
A 10 KVA 3-phase UPS will have an input and output current that depends on the specific voltage of the system. You can calculate the current by dividing the apparent power (in this case 10 KVA) by the square root of 3 multiplied by the voltage. For example, for a 208V system, the input and output current would be approximately 28.8 amps.
The regulator is the square silver box on the middle of the firewall.
RMS stands for Root Mean Square. Power is calculated as V2/R where V is the voltage and R is the resistive component of a load, This is easy toi calculate for a DC voltage, but how to calculate it for a sinusoidal voltage? The answer is to take all the instantaneous voltages in the sine wave, square them, take the mean of the squares, then take the square root of the result. This is defined as the "heating effect voltage". For a sine wave, this is 0.707 of the peak voltage.
In its simplest form the equation to calculate the wattage of an electrical appliance is: Watts = voltage x current. If the appliance is in a AC supply use the Route mean square voltage (the stated AC voltage).
Another name for average voltage is the RMS (Root Mean Square). This is a voltage derived from the peak to peak voltage multiplied by .707. If the peak to peak voltage is 170 volts then the average voltage (RMS) would be 170 x .707 = 120 volts.
Generally, the rectified voltage (DC Voltage) is less than the supply voltage (230 VAC or 110 VAC). Therefore, there is a need to step down the mains voltage to the required value before rectification. AC Voltage is denoted by the Root Mean Square (RMS) value which is equal to the peak voltage of the sine wave divided by 1.4 (square root of 2). Therefore, the out put of a full wave rectifier with a smoothening filter (say, a condenser) will be about 1.4 times the RMS value of AC Voltage. For eg. if we need 12 VDC output from the rectifier, the AC Voltage output of the transformer (which will be the input to the rectifier) should be 12/1.4 ie. 8.6 V. However, since a stabilized DC power supply will usually have some kind of a voltage stabilizer, the output of the rectifier can be higher. Therefore, the transformer output can be 12 volts RMS in this case. The DC output will be roughly 12 x 1.4 = 16.8 V and the stabilized DC voltage can be maintained at 12 V DC irrespective of small fluctuations in the AC mains voltage.
The root mean square (RMS) voltage is a measure of the effective voltage of an alternating current. It is calculated by taking the square root of the average of the squares of the voltage values over a given period of time. This value represents the equivalent direct current voltage that would produce the same amount of power in a resistive load.
The effective voltage of an electrical circuit is the measure of the average voltage over a complete cycle of alternating current. It is also known as the root mean square (RMS) voltage.
The output of the voltage-controlled oscillator (VCO) in a phase-locked loop (PLL) circuit is a periodic waveform, typically a sine or square wave, whose frequency is controlled by the input voltage. This voltage is derived from the phase comparator in the PLL, which adjusts the VCO frequency to match the frequency of a reference signal. As the PLL locks onto the reference signal, the VCO output frequency stabilizes, ensuring that it is phase-locked with the reference. This output can be used for various applications, including frequency synthesis and modulation.
Yes,schmitt trigger has upper and lower threshold voltage for the reason of noise protection while square wave generator doesn't have these properties.
a TDC sensor has a 12v feed from either the ecu or an ignition relay. a ground and a signal output to the ecu. this signal can be called a digital signal ie its on or its off. off state being 0v on usually being 5v althiugh this voltage can be different depending on make of sensor and vehicle. The voltage output would show on a scope as a square wave. a TDC sensor has a 12v feed from either the ecu or an ignition relay. a ground and a signal output to the ecu. this signal can be called a digital signal ie its on or its off. off state being 0v on usually being 5v althiugh this voltage can be different depending on make of sensor and vehicle. The voltage output would show on a scope as a square wave. a TDC sensor has a 12v feed from either the ecu or an ignition relay. a ground and a signal output to the ecu. this signal can be called a digital signal ie its on or its off. off state being 0v on usually being 5v althiugh this voltage can be different depending on make of sensor and vehicle. The voltage output would show on a scope as a square wave.
RMS (Root Mean Square) value for power is equivalent to the average power output. Therefore, if a device has a power rating of 1200W RMS, it will output an average power of 1200 watts.