Question

A square wave has amplitude 0 V for the low voltage and 4 V for the high voltage. Calculate the average voltage by integrating over one period of the waveform. Hint: the equation for a square wave will be a piecewise function and it will be convenient to start the integration where the voltage changes; for example in this problem we could define that during the first half of the period the voltage is 0 V and for the second half of the period the voltage is 4 V

Answer 1

V_{average} =
`(1)/(2) V_o` , V_{average} = 2 V

Step-by-step explanation:

he average or effective voltage of a wave is the value of the wave in a period

V_average = ∫ V dt

in this case the given volage is a square wave that can be described by the function

V (t) =
`\left \{ {{V=V_o \ \ \ t< \tau /2} \atop {V=0 \ \ \ \ t> \tau /2 } } \right.`

to substitute in the equation let us separate the into two pairs

V_average =
`\int\limits^(1/2)_0 {V_o} \, dt + \int\limits^1_(1/2) {0} \, dt`

V_average =
`V_o \ \int\limits^(1/2)_0 {} \, dt`

V_{average} =
`(1)/(2) V_o`

we evaluate V₀ = 4 V

V_{average} = 4 / 2)

V_{average} = 2 V