3-Lesson_Notes_Lecture_27 - EE 204 Lecture 26 The Root Mean...

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EE 204 Lecture 26 The Root Mean Square (RMS) Value The RMS Value of a Periodic Function of Time: The RMS [R oot M ean S quare] value of the periodic function () f t is defined as: 2 2 [( ) ] () o o avg tT t rms ft d t f T + = = The RMS value is also called the effective value eff rms f f = Example 1: Find the average and effective (rms) values of the sawtooth waveform. Solution: By inspection 25 av Y = For the interval 02 , t << 25 ; yt = then 22 0 1 T rms Yy d t T = 2 2 0 1 625 834, 2 td t == and 28.9 rms Y =
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Example 2: Find the average and effective (rms) values of the waveform shown. y t [s] 0.03 0.02 0.01 00 . 0 4 10 Solution: For . 0 1 , t << 10; y = For 0.01 0.03, t 0; y = the period is 0.03s 0.01 0 1 10(0.01) 10 3.33 0.03 0.03 av Yd t == = and 0.01 22 0 1 10 33.3 0.03 rms t therefore 5.77 rms Y = The RMS Value of the General Sinusoidal Function: Let us find the RMS value of the general sinusoidal function () cos( ) ft A t ω θ = + 2 2 1c o s ( 2 2) [( ) ] [c o s ( ) ] [ ] 2 avg avg avg rms t A t A f ωθ ++ =+ = = [1 cos(2 2 )] [1] [cos(2 avg avg avg rms AA tt f = + + = 10 2 rms A f [because the average value of a general sinusoid equals zero] 2 rms A f = The R.M.S. value is commonly used in sinusoidal circuits. For instance, the sinusoidal voltage ( ) cos( ) v vt V t = +
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Has the R.M.S. value 2 rms V V = The value V represents the peak voltage And the value 2 rms V V = represents the R.M.S. voltage Similarly, the sinusoidal current () cos ( ) i it I t ω θ = + Has a peak current I and an R.M.S. current 2 rms I I = Household voltage varies sinusoidally in time. The commonly used 120 V household voltage is actually an R.M.S measure, it is not a measure of the peak voltage. This means that if the outlets at a certain household provide120 V (actually 120 ( ) Vrm s ), then: 120 ( ) s 120 2 169.71 170 VV =×= ± (peak voltage) Figure 1 We shall use the units s & Irm s when the R.M.S. measure is used. Example 1: The outlets at a certain household provide 110 ( ) s at 60 Hz. Write an expression for the voltage () vt supplied by those outlets. Assume the phase angle to be zero.
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Solution: ( ) cos( ) cos( 0) cos v vt V t V t V t ω θω =+ = 2 110 ( ) 2 155.56 rms VV V r m s V = × = 2 2 60 120 376.991 377 [ / ] f rad s ππ π == × = = ± ( ) 155.56cos377 tV = (supply voltage as a function of time) Power Expressions in terms of the R.M.S. Voltage and Current: The power expressions must be revised when R.M.S. values are used.
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3-Lesson_Notes_Lecture_27 - EE 204 Lecture 26 The Root Mean...

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