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## AC WAVEFORMS

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**i**+1A 1 2 3 4 0 T(s) -1A AC WAVEFORMS • Alternating Current (AC) is current that periodically reverses direction • If the switch changes position every second: AC Fundamentals**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 1s 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**AC WAVEFORMS**i(t) 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**y**Positive angle x Negative angle REVIEW OF TRIG FUNCTIONS • Sine is a function of 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**90°**90° PROJECTION OF ROTATING RADIUS • 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 right-angled 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**r**b θ a USEFUL TRIG RELATIONSHIPS • Sin(θ180°) = -sin θ • Cos(θ180°) = -cos θ • Sin(θ180°) = -sin(θ) • Cos(θ180°) = -cos(θ) • Tan90° = +∞ • Tan(-90°) = -∞ • 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**T = 0.4s**T = 0.4s WAVEFORM PARAMETERS:PERIOD AND FREQUENCY i 0.3 0.4 0.6 t 0 0.1 0.2 0.5 • 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 periodT • 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**WAVEFORM PARAMETERS:RADIANS & ANGULAR FREQUENCY**y +1 y = sin = sin ωt 3π/2 π 0 π/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 2nπ 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**v(t)**+3V Peak value =3V Peak-to-peak value = 6V t 0 -3V WAVEFORM PARAMETERS:PEAK & INSTANTANEOUS VALUES • 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 • Vpand 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) = 3sin100t at t = 2ms? AC Fundamentals**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 (100t + 30°) V means v(t) is shifted left by 30° • Example: Find the instantaneous value at t = 0.25s of i(t) = 0.5 sin (8105t + 50°) A • Answer: 0.439 A AC Fundamentals**LAG AND LEAD**• When two waveforms have different phase angles, the one shifted farthest to the left is said to lead the other • v1(t) = 6 sin(ωt + 50°) leads v2(t) = 0.1 sin(ωt + 20°) because v1 is shifted left by 50°, while v2 is shifted left by 20° • v1 has a phase shift 30° greater than that of v2, i.e. v1leadsv2 by 30°, or v2lagsv1 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**80°**30° 50° 80° LAG AND LEAD EXAMPLE ωt • 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 • i1 = 75sin(ωt - 18°)A; i2 = 3sin(ωt - 31°)A • i1 shifted right by 18°, i2 shifted right by 31°. i1 lies 31-18=13° to left of i2. i1 leads i2 by 13° AC Fundamentals**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) A1 +15V A2 0.2 0.6 0.8 1.2 0 t(s) -3V T AC Fundamentals**i(t)**0.4A 0 0.1 0.2 0.3 0.4 0.5 t (s) 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 = 2Vp/ω V-s or 2Ip/ω A-s • What is the average of the following waveform AC Fundamentals**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 – Vpvolts • Maximum v(t) = Vdc + Vp volts • Find the average value of v(t), and derive mathematical expression as a function of time +12V Vdc 0 2 10 t (s) -2V AC Fundamentals**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) = 4.2sin(5000t + 45°)A The ac voltage supplied at a terminal has a frequency of 60Hz and effective value of 120V rms. Derive the mathematical expression for the voltage at the terminal, assuming zero phase angle EFFECTIVE (RMS) VALUES (1) AC Fundamentals**v(t)**+20V 2 4 7 9 12 0 1 5 6 10 11 t (s) -12V 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 AC Fundamentals**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**AC VOLTAGE AND CURRENT IN RESISTANCE: EXAMPLES**• The current in a 2.2 kΩ resistor is: • i(t) = 5sin(2π 100t + 45°) mA • 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 = 0.4ms? • The ac voltage across a 150Ω resistor is: • 39sin(2π 103t) V. At what value of t does the current through the resistor equal -0.26A? AC Fundamentals**XC**XC = 1/2πfC 0 f 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 = 1/2πfC Ohms • XCis 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 Capacitor is an open circuit when dc voltage is connected across it i.e. f = 0 AC Fundamentals**CAPACITORS AND AC (2)**• Example: The ac voltage across a 0.5F capacitor is: • v(t) = 16sin(2 103t) 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 20F capacitor is i(t) = 3sin(800t) 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° • vC(t) = Vpsin(ωt + )V • iC(t) = Ip sin(ωt + + 90°)A • Example: The voltage across a 0.01F capacitor is vC(t) = 240sin(1.25104t -30°)V. Write the mathematical expression for the current through it. AC Fundamentals**INDUCTORS AND AC**• Inductance resists or impedes the flow of ac current • Property called inductive reactanceXLwhere: • XL = ωL = 2πfL ohms; waveform: • 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° • iL(t) = Ipsin(ωt + ); vL(t) = Vpsin(ωt + + 90°) • Example: The current through an 80mH is 0.1sin(400t-25°)A. Write the mathematical expression for the voltage XL (Ω) XL f AC Fundamentals**AVERAGE POWER**• The power at any instant in time, instantaneous power: • p(t) = v(t)i(t) = [i(t)]2R = [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 12sin(377t)V across it. Use all the equations on the left. AC Fundamentals