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Lecture 5a

Lecture 5a. Review: Types of Circuit Excitation Why Sinusoidal Excitation? Phasors. Linear Time- Invariant Circuit. Linear Time- Invariant Circuit. Linear Time- Invariant Circuit. Linear Time- Invariant Circuit. Digital Pulse Source.

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Lecture 5a

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  1. Week 5a

  2. Lecture 5a Review: Types of Circuit Excitation Why Sinusoidal Excitation? Phasors Week 5a

  3. Linear Time- Invariant Circuit Linear Time- Invariant Circuit Linear Time- Invariant Circuit Linear Time- Invariant Circuit Digital Pulse Source Types of Circuit Excitation Steady-State Excitation OR Sinusoidal (Single- Frequency) Excitation Transient Excitation Week 5a

  4. Why is Sinusoidal Single-Frequency Excitation Important? • Some circuits are driven by a single-frequency • sinusoidal source. • Example: The electric power system at frequency of • 60+/-0.1 Hz in U. S. Voltage is a sinusoidal function of time because it is produced by huge rotating generators powered by mechanical energy source such as steam (produced by heat from natural gas, fuel oil, coal or nuclear fission) or by falling water from a dam (hydroelectric). Week 5a

  5. Bonneville Dam (Columbia River) Where Much of California’s Electric Power Comes From Week 5a

  6. Turbine-generator sets at Bonneville Dam Week 5a

  7. Where 3-Phase Electricity Comes From Generator driven by falling water has 3 separate coils Direct current in the rotor (rotating coil) produces a magnetic field that generates currents in stationary coils A, B and C Output voltages from the 3 coils (they leave the generating plant on 3 separate cables) Week 5a

  8. Why Sinusoidal Excitation? (continued) • Some circuits are driven by sinusoidal sources whose frequency changes slowly over time. • Example: Music reproduction system (different notes). • And, you can express any periodic electrical signal as a • sum of single-frequency sinusoids – so you can • analyze the response of the (linear, time-invariant) • circuit to each individual frequency component and • then sum the responses to get the total response. Week 5a

  9. Representing a Square Wave as a Sum of Sinusoids • Square wave with 1-second period. (b) Fundamental compo- • nent (dotted) with 1-second period, third-harmonic (solid black) • with1/3-second period, and their sum (blue). (c) Sum of first ten • components. (d) Spectrum with 20 terms. Week 5a

  10. Single-frequency sinusoidal-excitation AC circuit problems • The technique we’ll show works on circuitscomposed of linear • elements (R, C, L) that don’t change with time  “linear • time-invariant circuits”. • The circuit is driven with independentvoltage and/or current • sources whose voltages or currents vary at a single frequency, f, • measured in Hertz (abbreviated Hz) this is the number of cycles the • voltages or currents execute per second. We can represent the • source voltages or currents as functions of time as • v(t) = V0cos(wt) or i(t) = I0cos(wt), • where w = 2pf is the angular frequency in radians per second. • Example: In the U. S. the AC power frequency, f, is 60 Hz and the peak voltage V0 is 170 V, so w = 377 radians/s and • v(t) = 170cos(377t) V. More generally, we might have sources • v(t) = V0sin(wt) or i(t) = I0cos(wt = f), where f is a phase angle. Week 5a

  11. We could solve our circuit equations using such functions of time, but we’d have to do a lot of tedious trigonometric transformations. Instead we use a mathematical trick to eliminate time dependence from our equations! The trick is based on a fundamental fact about linear, time-invariant circuits excited with sinusoidal sources: the frequencies of all the voltages and currents in the circuit are identical. Week 5a

  12. RULE: “Sinusoid in”-- “Same-frequency sinusoid out” is true for linear time-invariant circuits. (The term “sinusoid” is intended to include both sine and cosine functions of time.) • Intuition: Think of sinusoidal excitation (vibration) of a linear mechanical system – every part vibrates at the same frequency, even though perhaps at different phases. Circuit of linear elements (R, L, C) Output: Excitation: vS(t) = VScos(wt + f) Iout(t) = I0cost(wt + a) Given Given ? ? Given SAME SAME w Week 5a

  13. PHASORSYou can solve AC circuit analysis problems that involveCircuits with linear elements (R, C, L)plusindependentand dependent voltage and/or currentsourcesoperating at asingle angular frequency w = 2pf (radians/s) such as v(t) = V0cos(wt) or i(t) = I0cos(wt). By using any of Ohm’s Law, KVL and KCL equations, doing superposition analysis, nodal analysis or mesh analysis, ANDUsing instead of the terms below on the left (general excitation), the terms below on the right (sinusoidal excitation): Week 5a

  14. Resistor I-V relationship General excitation Sinusoidal excitation vR = iRR VR = IRR where R is the resistance in ohms, VR = phasor voltage across the resistor, IR = phasor current through the resistor, and boldface indicates complex quantity. Capacitor I-V relationship General excitation Sinusoidal excitation iC = CdvC/dt IC = VC /ZC where IC= phasor current through the capacitor, VC = phasor voltage across the capacitor, the capacitive impedanceZCin ohms is ZC = 1/jwC , j = (-1)1/2, and boldface capital letters are complex quantities. (Note: EE’s use j for (-1)1/2 instead of i, since i might suggest current) Week 5a

  15. Inductor I-V relationship General excitation Sinusoidal excitation vL = LdiL/dt VL = IL ZL whereVL is the phasor voltage across the inductor, IL is the phasor current through the inductor, the inductive impedance in ohms ZL is ZL = jwL , j = (-1)1/2 and boldface capital letters are complex quantities. Week 5a

  16. Example 1 We’ll explain what phasor currents and voltages are shortly, but first let’s look at an example of using them: Here’s a circuit containing an AC voltage source with angular frequency w, and a capacitor C. We represent the voltage source and the current that flows (in boldface print) as phasors VS and I -- whatever they are! + V S I C - We can obtain a formal solution for the unknown current in this circuit by writing KVL: -VS + IZC = 0 We can solve symbolically for I: I = VS/ZC = jwCVS Week 5a

  17. Note that so far we haven’t had to include the variable of time in our equations -- no sin(wt), no cos(wt), etc. -- so our algebraic task has been almost trivial. This is the reason for introducing phasor voltages and currents, and impedances! In order to “reconstitute” our phasor currents and voltages to see what functions of time they represent, we use the rules below. Note that often (for example, when dealing with the gain of amplifiers or the frequency characteristics of filters), we may not even need to go back from the phasor domain to the time domain – just finding how the magnitudes of voltages and currents vary with frequency w may be the only information we want. Week 5a

  18. Rules for “reconstituting” phasors (returning to the • time domain) • Rule 1: Use the Euler relation for complex numbers: • ejf = cos(f) + jsin(f), where j = (-1)1/2 • Rule 2: To obtain the actual current or voltage i(t) or v(t) • as a function of time • 1. Multiply the phasor I or V by ejwt, and • 2. Take the real part of the product • For example, if I = 3 amps, a real quantity, then • i(t) = Re[Iejwt] = Re[3ejwt] = 3cos(wt) amps where Re • means “take the real part of” imaginary j f real 1 Week 5a

  19. Rule 3: If a phasor current or voltage I or V is not purely real but is complex, then multiply it by ejwt and take the real part of the product. For example, if V = V0ejf, then v(t) = Re[Vejwt] = Re[V0ejfejwt] = Re[V0ej(wt + f)] = V0cos(wt + f) Week 5a

  20. Finishing Example 1 Apply this approach to the capacitor circuit above, where the voltage source has the value vS(t) = 4 cos(wt) volts.The phasor voltage VS is then purely real: VS = 4. The phasor current is I = VS/ZC = jwCVS = (wC)VSejp/2, wherewe use the fact that j = (-1)1/2 = ejp/2; thus, the current in a capacitor leads the capacitor voltage by p/2 radians (90o).Note: Often (especially in this class) we may not care about the phase angle, and will focus just on the amplitude of the voltage or current that we obtain. This will be particularly true of filters and amplifiers. + vS(t) = 4 cos(wt) i(t) C - Week 5a

  21. The actual current that flows as a function of time, i(t), is obtained by substituting VS = 4 into the equation for I above, multiplying by ejwt, and taking the real part of the product. i(t) = Re[j (wC) x 4ejwt] = Re[4(wC)ej(wt + p/2)] i(t) = 4(wC)cos(wt + p/2) amperes Note: We obtained the current as a function of time (the current waveform) without ever having to work with trigonometric identities! Week 5a

  22. Analysis of an RC Filter Consider the circuit shown below. We want to use phasors and complex impedances to find how the ratio |Vout/Vin| varies as the frequency of the input sinusoidal source changes. This circuit is a filter; how does it treat the low frequencies and the high frequencies? + + R V C V out in I - - Assume the input voltage is vin(t) = Vincos(wt) and represent it by the phasor Vin. A phasor current I flows clockwise in the circuit. Week 5a

  23. Write KVL: -Vin + IR +IZC = 0 = -Vin + I(R + ZC) The phasor current is thus I = Vin/(R + ZC) The phasor output voltage is Vout = IZC. Thus Vout = Vin[ZC /(R + ZC)] If we are only interested in the dependence upon frequency of the magnitude of (Vout / Vin) we can write | Vout / Vin | = |ZC/(R + ZC)| = 1/|1 + R/ ZC | Substituting for ZC, we have 1 + R/ ZC = 1 + jwRC, whose magnitude is Thus, Week 5a

  24. Explore the Result If wRC << 1 (low frequency) then | Vout / Vin | = 1 If wRC >> 1 (high frequency) then | Vout / Vin | ~ 1/wRC If we plot | Vout / Vin | vs. wRC we obtain roughly the plot below, which was plotted on a log-log plot: The plot shows that this is a low-pass filter. Its cutoff frequency is at the frequency w for which wRC = 1. Week 5a

  25. Notice that we’ve obtained a lot of information about how this particular RC circuit performs just by looking at the magnitude of the ratio of phasor output voltage to phasor input voltage (i.e., we haven’t had to study the phase angles associted with those phasor voltages. (In more detailed studies the phase angles can be important – but not in this course.) Does this behavior make sense from what we know about capacitors? YES! At low frequency a capacitor is like an open circuit – so the output voltage would equal the input voltage At high frequency a capacitor is like a short circuit – so the output voltage would be very small. Week 5a

  26. Here is a useful web site to explore: http://www.phys.unsw.edu.au/~jw/AC.html You’ll find some demonstrations dealing with phasors and impedances there. And Appendix A in Hambley is a review of complex numbers. Week 5a

  27. Why Does the Phasor Approach Work? • Phasors are discussed at length in your text (Hambley 3rd Ed., pp. 195-201) with an interpretation that sinusoids can be visualized as the real axis projection of vectors rotating in the complex plane, as in Fig. 5.4. This is the most basic connection between sinusoids and phasors. • We present phasors as a convenient tool for analysis of linear time-invariant circuits with a sinusoidal excitation. The basic reason for using them is that they eliminate the time dependence in such circuits, greatly simplifying the analysis. • Your text discusses complex impedances in Sec. 5.3, and circuit analysis with phasors and complex impedances in Sec. 5.4. Skim over this LIGHTLY. Week 5a

  28. Motivations for Including Phasors in EECS 40 • It enables us to include a lab where you measure the behavior of RC filters as a function of frequency, and use LabVIEW to automate that measurement. • It enables us to (probably) include a nice operational amplifier lab project near the end of the course to make an “active” filter (the RC filter is passive). • It enables you to find out what impedances are and use them as real EEs do. • The subject was also supposedly included (in a way) in EECS 20 which some of you may have taken. Week 5a

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