Lecture 8 Root mean square

Transcription

1 Lecture 8: ECEN 14 Introduction to Analog and Digital Electronics Lecture 8 oot mean square Concept of a sinusoidal signal Examples Application to rectified signals obert. McLeod, University of Colorado 91

2 Lecture 8: ECEN 14 Introduction to Analog and Digital Electronics A useful number for describing time-varying waveforms Although peak or peak-to-peak amplitude might seem like the most obvious descriptions of a time-varying waveform, there are two reasons to want something different. 1. What we often care most about is the average power we can extract, not instantaneous voltage or current.. We would like a D to compare waveforms of different shape on this basis (average power). Consider a square wave voltage with a 5% duty cycle (on for only 5% of the period). What average power would it deliver to a resistor? average sd t 3 ( ) + ( ) 1 ( ) 4 peak 4 peak peak So we observe that this AC voltage would deliver the same power as a DC voltage of peak /. his is the voltage description of this waveform. obert. McLeod, University of Colorado 9 4

3 Lecture 8: ECEN 14 Introduction to Analog and Digital Electronics oot mean square voltage of a sinusoidal function Q: What DC voltage would deliver the same average power to a load as a sinusoidal AC voltage p sin(ωt)? ( t) 1. AC ( t) p sin( π t ) ( t) ( t) ( t) [ sin( π t )] p dt dt 1. ( t) DC Equiv Equiv Setting them equal Equiv Equiv For a sine wave! t t obert. McLeod, University of Colorado 93

4 Lecture 8: ECEN 14 Introduction to Analog and Digital Electronics In general, then stands for root mean square because to calculate it, you take the square oot of the Mean of the Square of a signal. Why? Mean ( t) [ ] [ ( t ) ] Mean his works the same way for current I Mean I ( t) I [ ] [ I ( t) ] Mean So you can use quantities as if they were DC: average I I I Neat! his is why wall voltage of 1 is actually peak obert. McLeod, University of Colorado 94

5 Lecture 8: ECEN 14 Introduction to Analog and Digital Electronics voltage example he multimeter reads voltage of 1. he oscilloscope reads he average power delivered to the load is ( ) [ W ] obert. McLeod, University of Colorado 95 1K

6 Lecture 8: ECEN 14 Introduction to Analog and Digital Electronics applied to rectifiers Half-wave DC DC ( t) dt ( π t ) dt sin π π ( t) dt [ sin( t )] dt π 4 Full-wave DC DC 1 1 π ( t) dt sin( π t ) dt 1 π ( t) dt [ sin( t )] dt π DC of full-wave rectifier is twice that of half-wave rectifier of full-wave rectified output is same as input, ignoring diode forward bias drop. obert. McLeod, University of Colorado 96

7 Lecture 8: ECEN 14 Introduction to Analog and Digital Electronics Quiz 8.1 Q: Why is the voltage of a rectified waveform the same as the voltage of the original, bipolar waveform (assuming the same peak voltage in each case)? A: he instantaneous power delivered by the bipolar waveform does not depend on the sign of the voltage, so the same power is delivered in the positive cycle as the negative cycle. Changing the sign of the voltage does not change the power, so the voltages are also the same. obert. McLeod, University of Colorado 97

8 Lecture 8: ECEN 14 Introduction to Analog and Digital Electronics Quiz 8. Q: Line voltage of 1 is connected directly to a 5 Kohm resistor. he average power dissipated and peak current are? A: (rms) /.88 W Irms rms/ 4 ma Ipeak Sqrt() Irms 33.9 ma obert. McLeod, University of Colorado 98

9 Lecture 8: ECEN 14 Introduction to Analog and Digital Electronics DL t Q: he plot shows one period of a triangle wave with a peak voltage of.5. What is the voltage of this waveform? A: Since the waveform is symmetric, the average of the first half of the period is the same as the second half, which simplifies the calculation. he equation for this time is t, so we need the average of the function t^ over the period t to.5. he integral of t^ is t^3/3, so the average (.5^3/3-)/.5 1/1. Finally, taking the square root, we get 1/( sqrt(3)) or about.3 obert. McLeod, University of Colorado 99