AC Waveforms. f = 1/T T = 1/f

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1 AC Waveforms Sinusoidal Waveforms Time Domain Descriptions The basic unit of a sine wave is a cycle the distance in time between any two repetitive points on the waveform. Period T The period T is the time required for 1 cycle often expressed as msec or µsec. Frequency f The frequency is a measure of the number of cycles that occur in 1 second. It was measured in CPS (cycles/sec) but the modern unit is the Hertz (Hz) often expressed as khz or MHz. Examples f = 1/T T = 1/f What is the frequency of sine waves with the following periods T? 1. T = 1 msec 2. T = 1 µsec 3. T = 50 µsec What is the period of sine waves with the following frequency f? 1. f = 25 khz 2. f = 120 Hz 3. f = 200 khz 1

2 Amplitude Domain Descriptions Commonly both voltage and current AC sinusoidal waveforms can be described using one or more of three quantities. Peak Value V P The peak value is the voltage from the 0 volt point to either the positive or negative maximums. Peak-to-Peak Value V P-P The peak-to-peak value is the voltage from the positive to the negative maximum. V P-P = 2 x V P V P = V P-P /2 RMS (Root Mean Square) Value - V RMS The RMS value of a sine wave is a value that indicates the power producing capability of the waveform. For example a 100 V RMS sine wave can produce the same power in a load resistor as a 100 VD DC voltage. V RMS = V P / 2 = V P-P /2 2 V P = 2 x V RMS V P-P = 2 2 x V RMS 2 = A voltmeter (DVM) set to AC volts measures an RMS Value 2 The factor 2 is derived by Integral Calculus.

3 Instantaneous Value There is actually a 4 th way of describing AC sinusoidal waveforms but this is not used for their measurement but as an algebraic description for theoretical derivations using mathematics. The Instantaneous Value describes the value of the sine wave as a function of time and is thus always changing. v(t) = V Max sin(ωt) Where v(t) is a function where v varies with t, V MAX is the peak value and ω is the frequency of the waveform. Average Value of a Sine Wave - V AVG A voltmeter (DVM) set to DC volts measures Average Value. The average value of a +5 V DC voltage is simply 5 V. For an AC sine wave the story is a bit different. Average value is actually a measure of the total area under the curve for any AC waveform. In the diagram shown, if one full cycle of the sine wave is considered, the area above the curve can be considered positive and, the area below the curve can be considered negative. The total area consists of these 2 equal, but opposite in sign, areas the sum of which is zero. Thus the average value of a sine wave over one full cycle considered to be zero. V AVG of a sine wave over 1 full cycle = 0 In higher level courses you will learn what happens to the average value if less than a full cycle is considered. 3

4 Circuit Laws for Sinusoidal Circuits All circuit laws learned so far still apply to circuits with sinusoidal waveforms. This includes Ohm s Law, KVL, KCL, Voltage Divider etc. Careful attention must be paid to ensuring that units are consistent. Examples 1. A resistor has 12 V P voltage across it and 100 ma RMS flowing through it. What is the value of the resistor? R = V/I = 12 V P / 100 ma RMS = 8.49 V RMS /100 ma RMS = 84.9 Ω 2. A 1 kω resistor is connected to a 12 V P-P voltage. What will a DVM set to AC volts measure? A DVM set to AC volts measures 12 V P-P = 4.24 V RMS 3. A 1 kω resistor is connected to a 20 V P-P voltage. What are the peak and RMS values of the current in the resistor? I = V/R = 20 V P-P /1 kω = 20 ma P-P = 10 ma P = 7.07 ma RMS Power in Sinusoidal Circuits The power dissipated in a resistor that is connected to a sinusoidal voltage source is P D = V RMS x I RMS Example 1. What power is dissipated in a 100 Ω resistor if V is 60 V P-P? V = 60 V P-P = 21.2 V RMS I = V/R = 21.2 V RMS /100 Ω = 212 ma RMS PD = 21.2 V RMS x 212 ma RMS = 4.50 W 2. What DC voltage will produce the same power as in Question 1 in a 100 Ω resistor? V DC = 21.2 V 4

5 AC Waveforms with DC Levels or Offsets More complex waveforms can be represented as a combination of an AC waveform plus a DC voltage level. For example consider adding the DC level and sine wave signals shown. +1 V DC 2 V P-P sine wave 2 V P-P sine wave with a +1V DC Level or Offset This waveform consists of an AC voltage level plus a DC voltage. 5

6 Square Waves A Square wave is generally seen as a rectangular waveform that varies with time. Parameters There are a number of parameters that can be specified and measured. Frequency Domain Parameters Period - T Period T, as specified before, is the time for 1 cycle of the waveform. Frequency Frequency f represents the number of cycles that occur in time (usually 1 second). Frequency is related to period f = 1/T. Amplitude The peak amplitude indicates the size of the waveform. In the diagram above V P = +5 V. The Peak-to Peak value is not used in the diagram above. There is a RMS for a square wave but it is not commonly used. 6

7 Duty Cycle Square waves can also vary with respect to how much time that they are at the high level voltage during a cycle. Time High (T HIGH ) and Time Low (T LOW ) Describing Duty Cycle requires some new terminology for the waveform - Time High (T HIGH ) and Time Low (T LOW ). The period of the waveform T is now T = T HIGH + T LOW Frequency Frequency f represents the number of cycles that occur in time (usually 1 second). Frequency is related to period f = 1/T. Duty Cycle The Duty Cycle D.C. is described as D.C. = T HIGH / (T HIGH + T LOW ) x 100% 7

8 Average Value The average value of the waveform is controlled by the Duty Cycle. The average value V AVG for a square wave where T HIGH = T LOW is V AVG = (+5 V + 0)/2 = +2.5 V In general the average value V AVG for a square wave is V AVG = Duty Cycle x High level voltage Examples Calculate the period, frequency, duty cycle and average value for the following 0 to +5 V square waves. 1. T HIGH = 250 µsec T LOW = 250 µsec Period T = T HIGH + T LOW = 250 µsec µsec = 500 µsec Frequency f = 1/T = 1/500 µsec = 2000 Hz Duty Cycle = T HIGH / (T HIGH + T LOW ) x 100% = 250 µsec/(250 µsec µsec) x 100 % = 50% Average value V AVG = 0.50 x +5 V = 2.5 V 2. T HIGH = 200 µsec T LOW = 50 µsec Period T = T HIGH + T LOW = 200 µsec + 50 µsec = 250 µsec Frequency f = 1/T = 1/250 µsec = 4 khz Duty Cycle = T HIGH / (T HIGH + T LOW ) x 100% = 200 µsec/(200 µsec + 50 µsec) x 100 % = 80% Average value V AVG = 0.80 x +5 V = 4.0 V 3. T HIGH = 2.5 msec T LOW = 7.5 msec Period T = T HIGH + T LOW = 2.5 msec msec = 10 msec Frequency f = 1/T = 1/10 msec = 100 Hz Duty Cycle = T HIGH / (T HIGH + T LOW ) x 100% = 2.5 msec/(2.5 msec msec) x 100 % = 25% Average value V AVG = 0.25 x +5 V = 1.25 V The results are plotted 8

9 Average Voltage (V) DC Motor Speed Control Duty Cycle (%) The speed of a DC motor can be controlled by the DC voltage connected to it. For example you could connect a DC motor to a DC power supply and vary the speed by adjusting the voltage of the power supply. It is not practical to use a variable DC power supply to vary the speed of a DC motor connected to a computer port. Pulse Width Modulation In Pulse Width Modulation the speed of a DC motor for example is controlled by the average DC voltage of a variable duty cycle square wave produced by a computer at an output port the higher the duty cycle => the higher the average DC voltage => the faster the motor turns. The relationship has a linear shape but can become non-linear at higher speeds. Motor Speed (rpm) Duty Cycle (%) From the examples 1, 2,, and 3 above, which case will cause the motor to run the fastest and the slowest? 9

10 Square Wave - Most General Case The square wave looked at earlier is a specific case the most general case is shown. T HIGH T LOW + V 2 -V 1 Period The period of the waveform T is still T = T HIGH + T LOW Frequency Frequency f represents the number of cycles that occur in time (usually 1 second). Frequency is related to period f = 1/T. Peak Value V P There are two peak values for this waveform positive peak and a negative peak and in general they are not the same value. Peak-to-Peak Value V P-P The peak-to-peak value is the voltage from the positive to the negative maximum. Duty Cycle The Duty Cycle (D.C.) is still described as D.C. = T HIGH / (T HIGH + T LOW ) x 100% 10

11 Average Value The average value of the waveform is controlled by the Duty Cycle. The average value V AVG this generalized square wave is now modified In general the average value V AVG for a square wave is V AVG = (D. C. x High level voltage) + ((1 D. C.) x Low Level Voltage) Example A square has a positive peak voltage V 2 of +10 V and a negative peak voltage V 1 of -5 V. The duty cycle is 60%. What is the average value? V AVG = (.60 x 10 V) +((1 -.60) x -5 v) = 6 V + (.40 x 5 V) = 6 V 2 V = 4 V Application Note Written by David Lloyd Computer Engineering Program Humber College 11