1. Slide 1EE100 Summer 2008Bharathwaj Muthuswamy EE100Su08 Lecture #12 (July 23 rd 2008) Outline –MultiSim licenses on Friday –HW #1, Midterm #1 regrade deadline: TODAY (5:00 pm PST). –QUESTIONS? –Circuit analysis using complex impedances: Examples –Frequency Response and Bode plots Reading –Chapter 9 from your book (skip 9.10, 9.11 (duh)), Appendix E* (skip second-order resonance bode plots) –Chapter 1 from your reader (skip second-order resonance bode plots)

2. Slide 2EE100 Summer 2008Bharathwaj Muthuswamy Example: Single Loop Circuit 20k  + - 1F1F10V  0  VCVC + - f=60 Hz, V C =? How do we find V C ? First compute impedances for resistor and capacitor: Z R = R= 20k  = 20k   0  Z C = 1/j (2  f x 1  F) = 2.65k   -90 

3. Slide 3EE100 Summer 2008Bharathwaj Muthuswamy Circuit Analysis Using Complex Impedances Suitable for AC steady state. KVL Phasor Form KCL Use complex impedances for inductors and capacitors and follow same analysis as in chap 2. Phasor Form KVL

4. Slide 4EE100 Summer 2008Bharathwaj Muthuswamy Impedance Example 20k  0  + - 2.65k  -90  10V  0  VCVC + - Now use the voltage divider to find V C :

5. Slide 5EE100 Summer 2008Bharathwaj Muthuswamy What happens when  changes? 20k  + - 1F1F10V  0  VCVC + -  = 10 Find V C

6. Slide 6EE100 Summer 2008Bharathwaj Muthuswamy Steady-State AC Analysis Find v(t) for  =2  3000 1k  0.1  F 5mA  0  + - V 1k  -j530k  5mA  0  + - V

7. Slide 7EE100 Summer 2008Bharathwaj Muthuswamy Find the Equivalent Impedance 5mA  0  + - VZ eq

8. Slide 8EE100 Summer 2008Bharathwaj Muthuswamy Change the Frequency Find v(t) for  =2  455000 1k  0.1  F 5mA  0  + - V 1k  -j3.5  5mA  0  + - V

9. Slide 9EE100 Summer 2008Bharathwaj Muthuswamy Find an Equivalent Impedance 5mA  0  + - VZ eq

10. Slide 10EE100 Summer 2008Bharathwaj Muthuswamy Series Impedance Z eq = Z 1 + Z 2 + Z 3 Z eq Z1Z1 Z3Z3 Z2Z2 L2L2 L1L1 Z eq = j  (L 1 +L 2 ) For example: C1C1 C2C2

11. Slide 11EE100 Summer 2008Bharathwaj Muthuswamy Parallel Impedance 1/Z eq = 1/Z 1 + 1/Z 2 + 1/Z 3 Z3Z3 Z1Z1 Z2Z2 Z eq C1C1 C2C2 For example: L2L2 L1L1

12. Slide 12EE100 Summer 2008Bharathwaj Muthuswamy Steady-State AC Node-Voltage Analysis C I 0 sin(  t) I 1 cos(  t) R L Try using Thevinin equivalent circuit. What happens if the sources are at different frequencies? +-+- VCVC

13. Slide 13EE100 Summer 2008Bharathwaj Muthuswamy Resistor I-V relationship v R = i R R ………….V R = I R R where R is the resistance in ohms, V R = phasor voltage, I R = phasor current (boldface indicates complex quantity) Capacitor I-V relationship i C = Cdv C /dt...............Phasor current I C = phasor voltage V C / capacitive impedance Z C:  I C = V C /Z C where Z C = 1/j  C, j = (-1) 1/2 and boldface indicates complex quantity Inductor I-V relationship v L = Ldi L /dt...............Phasor voltage V L = phasor current I L / inductive impedance Z L  V L = I L Z L where Z L = j  L, j = (-1) 1/2 and boldface indicates complex quantity

14. Slide 14EE100 Summer 2008Bharathwaj Muthuswamy Summary of impedance relationships RCL

15. Slide 15EE100 Summer 2008Bharathwaj Muthuswamy Thevenin Equivalent f=60 Hz 20k  + - 1F1F 10V  0  VCVC + - Z R = R= 20k  = 20k   0  Z C = 1/j (2  f x 1  F) = 2.65k   -90  + - Z TH V TH

16. Slide 16EE100 Summer 2008Bharathwaj Muthuswamy Bode Plots (Appendix E, Chapter 1 in Reader) OUTLINE –dB scale –Frequency Response for Characterization –Asymptotic Frequency Behavior –Log magnitude vs log frequency plot –Phase vs log frequency plot –Transfer function example

17. Slide 17EE100 Summer 2008Bharathwaj Muthuswamy Bel and Decibel (dB) A bel (symbol B) is a unit of measure of ratios of power levels, i.e. relative power levels. –The name was coined in the early 20th century in honor of Alexander Graham Bell, a telecommunications pioneer. –The bel is a logarithmic measure. The number of bels for a given ratio of power levels is calculated by taking the logarithm, to the base 10, of the ratio. –one bel corresponds to a ratio of 10:1. –B = log 10 (P 1 /P 2 ) where P 1 and P 2 are power levels. The bel is too large for everyday use, so the decibel (dB), equal to 0.1B, is more commonly used. –1dB = 10 log 10 (P 1 /P 2 ) dB are used to measure –Electric power, Gain or loss of amplifiers, Insertion loss of filters.

18. Slide 18EE100 Summer 2008Bharathwaj Muthuswamy Logarithmic Measure for Power To express a power in terms of decibels, one starts by choosing a reference power, P reference, and writing Power P in decibels = 10 log 10 (P/P reference ) Exercise: –Express a power of 50 mW in decibels relative to 1 watt. –P (dB) =10 log 10 (50 x 10 -3 ) = - 13 dB Exercise: –Express a power of 50 mW in decibels relative to 1 mW. –P (dB) =10 log 10 (50) = 17 dB. dBm to express absolute values of power relative to a milliwatt. –dBm = 10 log 10 (power in milliwatts / 1 milliwatt) –100 mW = 20 dBm –10 mW = 10 dBm

19. Slide 19EE100 Summer 2008Bharathwaj Muthuswamy From the expression for power ratios in decibels, we can readily derive the corresponding expressions for voltage or current ratios. Suppose that the voltage V (or current I) appears across (or flows in) a resistor whose resistance is R. The corresponding power dissipated, P, is V 2 /R (or I 2 R). We can similarly relate the reference voltage or current to the reference power, as P reference = (V reference ) 2 /R or P reference = (I reference ) 2 R. Hence, Voltage, V in decibels = 20log 10 (V/V reference ) Current, I, in decibels = 20log 10 (I/I reference ) Logarithmic Measures for Voltage or Current

20. Slide 20EE100 Summer 2008Bharathwaj Muthuswamy Note that the voltage and current expressions are just like the power expression except that they have 20 as the multiplier instead of 10 because power is proportional to the square of the voltage or current. Exercise: How many decibels larger is the voltage of a 9-volt transistor battery than that of a 1.5-volt AA battery? Let V reference = 1.5. The ratio in decibels is 20 log 10 (9/1.5) = 20 log 10 (6) = 16 dB. Logarithmic Measures for Voltage or Current

21. Slide 21EE100 Summer 2008Bharathwaj Muthuswamy The gain produced by an amplifier or the loss of a filter is often specified in decibels. The input voltage (current, or power) is taken as the reference value of voltage (current, or power) in the decibel defining expression: Voltage gain in dB = 20 log 10 (V output /V input ) Current gain in dB = 20log 10 (I output /I input Power gain in dB = 10log 10 (P output /P input ) Example: The voltage gain of an amplifier whose input is 0.2 mV and whose output is 0.5 V is 20log 10 (0.5/0.2x10 -3 ) = 68 dB. Logarithmic Measures for Voltage or Current

22. Slide 22EE100 Summer 2008Bharathwaj Muthuswamy Bode Plot Plot of magnitude of transfer function vs. frequency –Both x and y scale are in log scale –Y scale in dB Log Frequency Scale –Decade  Ratio of higher to lower frequency = 10 –Octave  Ratio of higher to lower frequency = 2

23. Slide 23EE100 Summer 2008Bharathwaj Muthuswamy

24. Slide 24EE100 Summer 2008Bharathwaj Muthuswamy Frequency Response: Why? The shape of the frequency response of the complex ratio of phasors V OUT /V IN is a convenient means of classifying a circuit behavior and identifying key parameters. Frequency Low Pass Gain Break point Frequency High Pass Gain Break point FYI: These are log ratio vs log frequency plots

25. Slide 25EE100 Summer 2008Bharathwaj Muthuswamy

26. Slide 26EE100 Summer 2008Bharathwaj Muthuswamy