EE292: Fundamentals of ECE

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1 EE292: Fundamentals of ECE Fall 2012 TTh 10:00-11:15 SEB 1242 Lecture

2 2 Outline Review RMS Values Complex Numbers Phasors Complex Impedance Circuit Analysis with Complex Impedance

3 3 Sinusoidal Currents and Voltages Sinusoidal voltage v t = V m cos ω 0 t + θ V m - peak value of voltage ω 0 - angular frequency in radians/sec θ phase angle in radians This is a periodic signal described by T the period in seconds ω 0 = 2π T f the frequency in Hz = 1/sec f = 1 T ω 0 = 2πf Note: Assuming that θ is in degrees, we have tmax = θ/360 T.

4 4 Sinusoidal Currents and Voltages For consistency/uniformity always express a sinusoid as a cosine Convert between sine and cosine In Degrees sin x = cos x 90 In Radians sin x = cos x π 2 v t = 10sin (200t + 30 ) v t = 10cos(200t ) v t = 10cos (200t 60 )

5 5 Root-Mean-Square Values Apply a sinusoidal source to a resistance Power absorbed p t = v2 (t) R Energy in a single period T E T = p t dt 0 Average power (power absorbed in a single period) P avg = E T T = 1 T P avg = 1 T T 0 T 0 v 2 (t)dt R p t dt 2 = 1 T T 0 v 2 (t) R dt

6 6 Root-Mean-Square Values 1 T T 0 v 2 (t)dt P avg = R Define rms voltage V rms = 1 T T 0 2 v 2 (t)dt Similarly define rms current I rms = 1 T T 0 i 2 (t)dt

7 7 RMS Value of a Sinusoid Given a sinusoidal source v t = V m cos ω 0 t + θ V rms = 1 T T 0 v 2 (t)dt The rms value is an effective value for the signal E.g. in homes we have 60Hz 115 V rms power V m = 2 V rms = 163 V

8 8 Example 5.1 Voltage v t = 100cos (100πt) V Applied to a 50Ω resistance Find the rms voltage and average power and plot power ω 0 = 100π f = ω = 50 Hz 2π T = 1 = 1 = 20msec f 50 p t = v2 (t) R = cos 2 100πt = 200cos 2 100πt W V rms = V m 2 = = V P avg = = 100 W

9 9 Complex Numbers in Rectangular Form A complex number in rectangular form z = x + jy x real part y imaginary part A vector in the complex plane with x horizontal coordinate y vertical coordinate Source: Wikipedia

10 10 Complex Arithmetic in Rectangular Form z 1 = 5 + j5, z 2 = 3 j4 Addition Add the real and complex parts separately z 1 + z 2 = 5 + j5 + 3 j4 z 1 + z 2 = 8 + j z 1 z 2 = 5 + j5 3 j4 z 1 z 2 = 2 + j9 Multiplication Use j 2 = 1 z 1 z 2 = 5 + j5 3 j4 = 15 j20 + j15 j 2 20 = 15 j = 35 j5 Division Multiply num/den by complex conjugate term to remove imaginary term in denominator z 1 = 5+j5 z 2 (3 j4)

11 11 Complex Arithmetic in Rectangular Form Division Multiply num/den by complex conjugate term to remove imaginary term in denominator

12 12 Complex Numbers in Polar Form Number represented by magnitude and phase Magnitude the length of the complex vector Phase the angle between the real axis and the vector Polar notation z = re jθ Phasor notation z = r θ Source: Wikipedia

13 13 Conversion Between Forms Rectangular to polar form r 2 = x 2 + y 2 tanθ = y x x (degrees) x (radians) sin (x) cos (x) tan (x) π π Polar to rectangular form x = rcosθ y = rsinθ Convert to polar form z = 4 j4 r = = 4 2 θ = arctan y x = 45 π 4 60 π π π NaN arctan 1 = π 4 z = 4 2e jπ/4 Source: Wikipedia

14 14 Arithmetic in Polar Form z 1 = r 1 e jθ 1 = r 1 θ 1 z 2 = r 2 e jθ 2 = r 2 θ 2 Addition Convert to rectangular form to add/subtract and convert back to polar form Multiplication Multiply magnitudes and add phase (exponent terms) z 1 z 2 = r 1 e jθ 1r 2 e jθ 2 = r 1 r 2 e j(θ 1+θ 2 ) = r 1 r 2 (θ 1 +θ 2 ) Division Divide magnitude and subtract phase z 1 z 2 = r 1e jθ1 r 2 e jθ 2 = r 1 r 2 e j(θ 1 θ 2 ) = r 1 r 2 (θ 1 θ 2 )

15 15 Euler s Formula Relationship between a complex exponential and a sinusoid e jθ = cosθ + jsinθ cosθ = 1 2 ejθ + e jθ sinθ = 1 2j ejθ e jθ

16 16 Phasors A representation of sinusoidal signals as vectors in the complex plane Simplifies sinusoidal steady-state analysis Given v 1 t = V 1 cos (ωt + θ 1 ) The phasor representation is V 1 = V 1 θ 1 For consistency, use only cosine for the phasor v 2 t = V 2 sin ωt + θ 2 = V 2 cos ωt + θ 2 90 = V 2 = V 2 (θ 2 90 )

17 17 Adding Sinusoids with Phasors 1. Get phasor representation of sinusoids 2. Convert phasors to rectangular form and add 3. Simplify result and convert into phasor form 4. Convert phasor into sinusoid Remember that ω should be the same for each sinusoid and the result will have the same frequency

18 18 Example 5.3 v 1 t = 20 cos ωt 45 V 1 = v 2 t = 10 sin ωt + 60 V 2 = = Calculate V s = V 1 + V 2

19 19 Phasor as a Rotating Vector v t = V m cos ωt + θ = Re{V m e j ωt+θ } V m e j ωt+θ is a complex vector that rotates counter clockwise at ω rad/s v(t) is the real part of the vector The projection onto the real axis of the rotating complex vector Source: Wikipedia

20 20 Phase Relationships Given v 1 t = 3 cos ωt + 40 v 2 t = 4 cos ωt 20 In phasor notation V 1 = 3 40 V 2 = 4 20 Phasor diagram Since phasors rotate counter clockwise V 1 leads V 2 by 60 V 2 lags V 1 by 60

21 21 Phase Relationships from Plots Given plots of a pair of periodic waveforms 1. Find the shortest time interval t p between positive peaks in a pair of waveforms 2. The angle between the peaks is the phase difference θ = t p T 360

22 22 Complex Impedance Previously we saw that resistance was a measure of the opposition to current flow Larger resistance less current allowed to flow Impedance is the extension of resistance to AC circuits Inductors oppose a change in current Capacitors oppose a change in voltage Capacitors and inductors have imaginary impedance called reactance

23 23 Resistance From Ohm s Law v t = Ri t Extend to an impedance form for AC signals V = ZI Converting to phasor notation for resistance V R = RI R R = V R I R Comparison with the impedance form results in Z R = R Since R is real, the impedance for a resistor is purely real

24 24 Inductance I/V relationship v L t = L di L (t) dt Assume current i L t = I m sin ωt + θ I L = I m θ π 2 Using the I/V relationship v L t = LωI m cos ωt + θ V L = ωli m θ Notice current lags voltage by 90 Using the generalized Ohm s Law Z L = V L = ωli m θ I L I m θ π = ωl π = jωl 2 2 Notice the impedance is completely imaginary

25 25 Capacitance I/V relationship i C t = C dv C (t) dt Assume voltage v c t = V m sin ωt + θ V C = V m θ π 2 Using the I/V relationship i C t = CωV m cos ωt + θ I C = ωcv m θ Notice current leads voltage by 90 Using the generalized Ohm s Law π 2 Z C = V C π = 1 ωc 2 jωc Notice the impedance is completely imaginary = V m θ = 1 I C ωcv m θ

26 26 Circuit Analysis with Impedance KVL and KCL remain the same Use phasor notation to setup equations E.g. v 1 t + v 2 t v 3 t = 0 V 1 + V 2 V 3 = 0

27 27 Steps for Sinusoidal Steady-State Analysis 1. Replace time descriptions of voltage and current sources with corresponding phasors. (All sources must have the same frequency) 2. Replace inductances by their complex impedances Z L = ωl π = jωl. Replace 2 capacitances by their complex impedances Z C = 1 π = 1. Resistances have ωc 2 jωc impedances equal to their resistances. 3. Analyze the circuit by using any of the techniques studied in Chapter 2, and perform the calculations with complex arithmetic

28 28 Example 5.5 Find voltage v c (t) in steady-state Find the phasor current through each element Construct the phasor diagram showing currents and v s

29 29 Convert Sources and Impedances Voltage source V s = ω = 1000 Inductance Z L = jωl = j = j100 Ω Capacitance Z C = j Ω = 1 = 106 = jωc j1000(10μ) j10 4

30 30 Find Voltage v c (t) Find voltage by voltage divider V C = V s Z eq Z eq +Z L Equivalent impedance Z eq = Z R Z C

31 31 Phasor Diagram Source current Phasor diagram Capacitor current Resistor current

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