Electrical Engineering - Alternating Current and Voltage - Discussion

Here,
Average = (11×0.20) + (3×0.80) = 2.2 + 2.4 = 4.6V.
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for 20% of the time - 8v, for 80% of the time - 3v.
Let's consider the T-time period,
Average = (20%(T) x 8 + 80%(T) x 3)/T.
Average voltage = 4v.

Is baseline means offset?

I think the answer should be B.

Baseline voltage basically is DC riding voltages or we can say reference level for waveform (Amplitude shifting of +3 volt above zero reference level in XY plane).

How can you assume 8v level is for 1 sec and 3v level is for 4sec?

Plase explain anyone.

Let us assume that ON time i.e. 8V level is for 1 sec and OFF time.
i.e.3V level is for 4 sec. Therefore we get a square wave whose lower value is 3V and upper value is 8V.

Therefore for a complete cycle of 5 sec, we can find the average value by simple math's formula- Avg Value = Area/time. So we have Vavg = (3 * 4 + 8 * 1)/5 = 4V.

Draw the waveform and find out the area under the curve (in this case, square wave) and divide it by cycle time. It will be easier for understanding.

The answer should be option B = 4.6v.

First of all Duty cycle is 20% it means 0.02.

For that,

V average = Baseline + Duty cycle * Amplitude.
So, Avg = 3v + (0.02 ) * (8)
= 4.6 v

Baseline means, Usually waves start from y axis at 0 if base varies the waves start from that value means the y of wave is 3.

Imagine an oscilloscope screen your signal has lowest voltage of 3 volt, that is your base line.

Now you have 8 volt, so your peak to peak amplitude is 5 volt.

You have 20% duty cycle, 20% of the time the signal is on, thus the voltage (amplitude) is 0.2 x 5 Volt = 1 volt.

You have base line 3 volt mean the amplitude of the signal is 3volt + 1 volt = 4Volt.