AC WAVEFORMS Alternating Current AC is current that

AC WAVEFORMS • • Alternating Current (AC) is current that periodically reverses direction If the switch changes position every second: i +1 A 0 1 2 3 4 T(s) -1 A AC Fundamentals 1

AC WAVEFORMS • • The plot, or graph, of a current (or voltage) versus time is called a waveform The magnitude is the size of current or voltage (y-axis) Waveforms where the current changes magnitude, but not direction (all the values remain positive or negative) are referred to as pulsating DC Such waveforms can also be regarded as the superposition (addition) of an AC waveform and a DC level An AC voltage is one that periodically reverses polarity The voltage across the 10Ω resistor reverses polarity as 1 s intervals An AC voltage source produces an EMF whose polarity reverses at periodic intervals The AC waveform used the most in circuit theory is the sinusoidal waveform or sine wave AC Fundamentals 2

i(t) AC WAVEFORMS AC increasing and decreasing linearly with time, triangular waveform t Pulsating DC. Current does not reverse direction, doesn’t go negative t Sawtooth waveform t Pulsating DC. Not AC. AC superimposed on DC level t Max current in positive direction Sinusoidal AC t Min current in negative direction AC Fundamentals 3

REVIEW OF TRIG FUNCTIONS • Sine is a function of angle y Positive angle x Negative angle • • • Positive angles are measured in the anticlockwise direction from the positive x-axis. Negative angles are measured in a clockwise direction. In AC circuit theory, angles greater than 180° are expressed as negative equivalent angles. Eg. 225° = -135° (225 - 360) Angles more negative than -180° are expressed in equivalent positive angles Top two quadrants expressed in positive angles Bottom two quadrants expressed in negative angles Angle on negative x-axis is 180° 200°=200°-360°= -160°; -250°= -250°+360°= 110° 400°=400°-360° = 40°; 800°=800°-2(360°) = 80° Sin(-θ) = -sin(θ) AC Fundamentals 4

PROJECTION OF ROTATING RADIUS 90° • • • 90° As the radius of the circle rotates anticlockwise, the angle it generates between itself and the positive x-axis varies from 0 to 360° At any instant, the radius is the hypotenuse of a rightangled triangle, containing the angle θ Sin θ = 0 when θ = 0 and θ = 180° Sin (θ+90°) = cos θ Sin θ = cos (θ - 90°) Cos(- θ) = cos θ AC Fundamentals 5

USEFUL TRIG RELATIONSHIPS • • • Sin(θ 180°) = -sin θ Cos(θ 180°) = -cos θ Sin(θ 180°) = -sin(θ ) Cos(θ 180°) = -cos(θ ) Tan 90° = +∞ Tan(-90°) = -∞ θ Tan θ = b/a Sin θ = b/r; b = r sin θ Cos θ = a/r; a = r cos θ Sin-1(b/r) = θ or arcsin(b/r) = θ Cos-1(a/r) = θ or arcos(a/r) = θ Tan-1(b/a) = θ or arctan(b/a) = θ AC Fundamentals r b a 6

WAVEFORM PARAMETERS: PERIOD AND FREQUENCY i T = 0. 4 s 0. 3 0 0. 1 0. 4 0. 2 0. 6 t 0. 5 T = 0. 4 s • • An AC waveform can be considered to exist for all time Yet we need a reference time, t =0, where plot begins Helps us to express waveform mathematically Periodic AC waveform repeats at regular intervals The time required to complete a cycle is called the period T Period can be measured between any two corresponding points on successive cycles Frequency is 1/T (Hertz – Hz) 1 Hz = 1 cycle/second AC Fundamentals 7

WAVEFORM PARAMETERS: RADIANS & ANGULAR FREQUENCY y +1 y = sin ωt π 0 π/2 3π/2 2π = ωt rad -1 • • • Radian is the SI unit of angle: π radians = 180° 45° = 45(π/180) = π/4 rad = 0. 7854 rad 1 rad = 1(180/π) = 57. 296° Above is a sine function plot versus angle in radians 1 cycle repeats at every 2 nπ radians intervals Angular velocity (ω) is the amount of angle the plot sweeps through in a given amount of time ω = θ/t rad/s (rad s-1) angular frequency = 2πf θ = ωt rad Sine wave can be expressed as a function of time Sin θ = sin ωt = sin (2πf)t AC Fundamentals 8

WAVEFORM PARAMETERS: PEAK & INSTANTANEOUS VALUES v(t) +3 V Peak-to-peak value = 6 V Peak value =3 V t 0 -3 V • • Max value reached by AC waveform - peak value (pk) Peak-to-peak (p-p) value is the difference between positive peak and negative peak values (3 - -3 = 6) The peak value is also called amplitude Any sin function can be expressed as Vp and Ip are the peak values Lowercase i and v used for AC quantities Uppercase I and V used for DC quantities Instantaneous value of AC waveform is the value at specific instant of time: i(t) = 3 sin 100 t at t = 2 ms? AC Fundamentals 9

PHASE RELATIONS • Adding angle to angle θ in the sine function: sin(θ ) causes to sine waveform to shift left (+ ) or right (- ) • • ωt is in radians, but is expressed in degrees v(t) = 5 sin (100 t + 30°) V means v(t) is shifted left by 30° Example: Find the instantaneous value at t = 0. 25 s of i(t) = 0. 5 sin (8 105 t + 50°) A Answer: 0. 439 A • • AC Fundamentals 10

LAG AND LEAD • When two waveforms have different phase angles, the one shifted farthest to the left is said to lead the other • v 1(t) = 6 sin(ωt + 50°) leads v 2(t) = 0. 1 sin(ωt + 20°) because v 1 is shifted left by 50°, while v 2 is shifted left by 20° • v 1 has a phase shift 30° greater than that of v 2, i. e. v 1 leads v 2 by 30°, or v 2 lags v 1 by 30° • Lead-lag terminology derived from observation of relative positions of the waveforms when plotted versus time • The waveform with greater positive phase reaches its peak first (earliest in time), i. e. it leads the other AC Fundamentals 11

LAG AND LEAD EXAMPLE 80° ωt 30° 50° 80° • • v(t) is shifted left by 30°, i(t) is shifted right by 50° v(t) lies to the left by 80°, thus v(t) leads i(t) by 80° Equivalently i(t) lags v(t) by 80° Notice how v(t) reaches its peak value 80° before i(t) Phase comparisons of this type can only be made if the two waveforms have the same frequency ω i 1 = 75 sin(ωt - 18°)A; i 2 = 3 sin(ωt - 31°)A i 1 shifted right by 18°, i 2 shifted right by 31°. i 1 lies 3118=13° to left of i 2. i 1 leads i 2 by 13° AC Fundamentals 12

AVERAGE VALUES (1) • • Average value of waveform is the average of all its values over a period of time Computing the average over time involves adding all the values that occur in a specific time interval and dividing that sum by time This is done by computing the area of waveform over a period of time Area above time axis is a positive area Area below time axis is a negative area Algebraic signs must be taken into account when computing total (net) area Time interval is the period T Find the average value of the following AC waveform v(t) A 1 +15 V A 2 0 -3 V 0. 6 0. 8 1. 2 t(s) T AC Fundamentals 13

AVERAGE VALUES (2) • • • Sine waves are symmetrical about time axis, thus average is zero: positive area cancels negative area Must use calculus to compute average, and we take into consideration the area of a half cycle of a sinusoidal waveform, called a pulse Using calculus it can be shown that: Area = 2 Vp/ω V-s or 2 Ip/ω A-s What is the average of the following waveform i(t) 0. 4 A 0 0. 1 0. 2 0. 3 AC Fundamentals 0. 4 0. 5 t (s) 14

AVERAGE VALUES (3) • • • Average value of waveform also called dc value Knowledge of average value important in the design of high power devices such as power supplies and power amplifiers Waveforms in circuits regarded as sine waves with dc offset DC level (offset) is simply added to ac waveform Equation for ac voltage with dc component Vdc is v(t) = Vdc + Vp sin(ωt + ) Vdc can either be positive or negative Minimum v(t) = Vdc – Vp volts Maximum v(t) = Vdc + Vp volts Find the average value of v(t), and derive mathematical expression as a function of time +12 V Vdc 0 2 10 t ( s) -2 V AC Fundamentals 15

EFFECTIVE (RMS) VALUES (1) • • • Average values not useful for comparing sinusoidal ac waveforms because average value is zero Use measure called effective or root mean square (rms) value Measure shows how effective the waveform produces heat in a resistance RMS measure eliminates consideration of waveform polarity The DC voltage that causes the same heating in resistance R as an ac voltage is the effective value (rms value) of the ac voltage (likewise for current) • • Find the effective value of the ac current i(t) = 4. 2 sin(5000 t + 45°)A The ac voltage supplied at a terminal has a frequency of 60 Hz and effective value of 120 V rms. Derive the mathematical expression for the voltage at the terminal, assuming zero phase angle AC Fundamentals 16

EFFECTIVE (RMS) VALUES (2) • • • Relationship between peak and rms values given in previous equations valid only for sinusoids It is possible to compute effective value of some nonsinusoidal waveforms using root mean square method Square the waveform, find its average, then square root the answer [average(waveform)2] Find the effective value of the voltage of this waveform v(t) +20 V 2 0 1 4 7 5 9 6 12 10 11 t (s) -12 V AC Fundamentals 17

AC VOLTAGE AND CURRENT IN RESISTANCE • Ohm’s Law can be applied to an ac circuit containing a resistance to determine ac current in a resistance when ac voltage is applied across it • At every instant in time, the current in the resistor is the voltage at that instant divided by the resistance, i. e. instantaneous current is instantaneous voltage divided by the resistance • For a sinusoidal voltage • Ip = Vp/R • The voltage across and the current through the resistor have the same phase angle, they are in phase AC Fundamentals 18

AC VOLTAGE AND CURRENT IN RESISTANCE: EXAMPLES • • The current in a 2. 2 kΩ resistor is: i(t) = 5 sin(2π 100 t + 45°) m. A Write the mathematical expression for the voltage across the capacitor What is the effective value of the resistor voltage? What is the instantaneous value of the resistor at t = 0. 4 ms? The ac voltage across a 150Ω resistor is: 39 sin(2π 103 t) V. At what value of t does the current through the resistor equal -0. 26 A? AC Fundamentals 19

CAPACITORS AND AC (1) • • AC current through a capacitor depends not only on the voltage across it, but also on the frequency of that voltage The property of a capacitor that causes it to resist the flow of ac current through it is called capacitive reactance, denoted by XC. Its units are also Ohms XC = 1/ωC = 1/2πf. C Ohms XC is inversely proportional to frequency The greater the frequency, the smaller the reactance and thus the greater the current through the capacitor The lower the frequency, the greater the reactance Graph of reactance XC v frequency f XC XC = 1/2πf. C 0 Capacitor is an open circuit when dc voltage is connected across it i. e. f = 0 f AC Fundamentals 20

CAPACITORS AND AC (2) • • • Example: The ac voltage across a 0. 5 F capacitor is: v(t) = 16 sin(2 103 t) V What is the capacitive reactance of the capacitor? What is the peak value of the current through it? Example: The ac current through a 20 F capacitor is i(t) = 3 sin(800 t) A. What is the peak voltage across the capacitor? • The peak current through a capacitor does not occur at the same instant of time that the voltage across it reaches its peak value The current through a capacitor leads the voltage across it by 90° v. C(t) = Vpsin(ωt + )V i. C(t) = Ip sin(ωt + + 90°)A Example: The voltage across a 0. 01 F capacitor is v. C(t) = 240 sin(1. 25 104 t -30°)V. Write the mathematical expression for the current through it. • • AC Fundamentals 21

INDUCTORS AND AC • • • Inductance resists or impedes the flow of ac current Property called inductive reactance XL where: XL = ωL = 2πf. L ohms; waveform: XL (Ω) XL f • • XL directly proportional to frequency XL decreases with frequency, approaching 0Ω as frequency approaches zero (dc) Inductor is a short circuit when a dc voltage is connected across it Ohm’s Law: Ip = Vp/XL amperes Voltage across inductor leads current through it by 90° i. L(t) = Ipsin(ωt + ); v. L(t) = Vpsin(ωt + + 90°) Example: The current through an 80 m. H is 0. 1 sin(400 t 25°)A. Write the mathematical expression for the voltage AC Fundamentals 22

AVERAGE POWER • • The power at any instant in time, instantaneous power: p(t) = v(t)i(t) = [i(t)]2 R = [v(t)]2/R More useful measure is average power, which is the average of the instantaneous power over a period of time Example: Find the average power dissipated through a 50Ω with a voltage of 12 sin(377 t)V across it. Use all the equations on the left. AC Fundamentals 23