ECE 45 Average Power Review

Transcription

1 UC San Diego J. Connelly Complex Power ECE 45 Average Power Review When dealing with time-dependent voltage and currents, we have to consider a more general definition of power. We can calculate the instantaneous power at any point in time, but we often also care about the average power over an interval of time, which is independent of time but generally does depend on the interval. P inst (t) v(t)i(t) P avg t 2 t t2 t P inst (t)dt When dealing with sinusoidal voltages and currents, we usually care about the average power in a period, i.e. t 2 T +t, where T is the period of the sinusoid. In this case, the expression for average power simplifies to P avg 2 Re{VI } wherev and I are the phasor representations ofv(t) andi(t) respectively. Maximum Power Suppose we want to maximize the average power delivered to a particular component in a circuit by adjusting the impedance of the component. In order to do this, we represent the circuit as a Thevenin equivalent, and we use impedance matching to find the values of current and voltage which maximize the power. That is we set Z th wherez th is the Thevenin impedance of the circuit. Example (a) For anyω 0, find the outputv o (t) when v in (t) cos(ωt). (b) How does the output behave asω 0 and ω (c) When v in (t) 2cos 2 (t), find v o (t). (d) Suppose a load consisting of a resistor in series with a capacitor is placed across the output terminals. Whenv in (t) cos(2t), find the values of the resistor and the capacitor that maximize the average power deliver to the load. (e) What is the maximum average power delivered to the load in this case? R R 2 Vc Z Z 2 v in (t) C L v o (t) Vo Please report any typos/errors to j2connelly@uscd.edu

2 WhereR Ω, R 2 3Ω, C F, L H (a) By taking the phasor transform of the circuit with respect to ω and using a voltage divider, we can write the phasor of the output voltage as: By KCL at the nodev C : V C V C V C 2 + V C 2 + () V C V C + +Z ( R V C ( + ) ). (2) +2 + By setting equal and 2: is the phasor ofv in (t) cos(ωt), so and (2 + )(+ )+ jωl R +R 2 ω 2 R LC +jω(l+r R 2 C) jωl (R 2 +jωl)(+jωr C)+R jω 4 ω 2 +4jω jω (jω +2) 2 and so, solving for magnitude and phase of yields: Transforming back to the time-domain gives us: (b) We have ω 4+ω 2, π ( ω ) 2 2 tan. 2 v o (t) ω (ωt+ 4+ω cos π ( ω )) tan. 2 lim ω 0 ω 4+ω 2 0 and ω lim ω 4+ω 0 2 so the output is 0 at very high and very low frequencies This can be verified by looking at how the components in the circuit behave at high and low frequencies.

3 (c) Note that ( ) e 2cos 2 jt +e jt 2 (t) 2 2+ej2t +e j2t +cos(2t) 2 2 Then v in (t) v i, (t)+v i,2 (t), where v i, (t) and v i,2 (t) cos(2t). Since v i, (t) and v i,2 (t) are different frequencies we need to use super position to find v o (t). v i, (t) is a sinusoid of frequency ω 0, i.e. a DC signal, so pluggingω 0 into our expression forv o (t) yields v o, (t) 0 v i,2 (t) is a sinusoid of frequencyω 2 2, so pluggingω 2 into our expression forv o (t) yields 4 cos(2t) Thus by superpositionv o (t) v o, (t)+v o,2 (t) 4 cos(2t). (d) By the maximum power theorem, we need to find the Thevenin impedance of the circuit and match the load impedance. To find Z th, set V i 0 and solve for the effective impedance at ω 2. We have Z th (Z // +Z 2 )// Z // R /jωc R +jωc 2j + 6j +4 Z 2 +Z // 2j + Z th +j 3 2 The value ofoad that maximizes power to the load is oad Z th j3/2. We also haveoad R Load + jωc Load. ThusR L and C L /3F. (e) By using a voltage divider in the Thevenin equivalent circuit, the voltage across the load is: and the current through the load is Hence the average power is oad 2 j3 V Load V th V th oad +Z th 4 I Load V th V th oad +Z th 2 P avg 2 Re{V LoadI Load } 2 Re { Vth 2 (2 j3) 8 } V th 2 To find V th, we use our results from part c. that showed that when the input is v in cos(2t), the (open circuit) output is v o cos(2t)/4, so the phasor of the open circuit voltage is. V th /4 Thus P avg 28 8

4 Example 2 Find the voltagev r (t) in the circuit below, when (a) v in (t) /3 (b) v in (t) sin(t) (c) v in (t) +2sin(t). v in (t) L C R v r (t) where R Ω, C /2F,L 2H. Forω 0, assume thatv in (t) Acos(ωt+θ). Then by taking the phasor transformation of the circuit, we have V R Voltage divider: // V r. // + where 2, 2 jω, and 2jω. Since (t) Acos(ωt+θ), we have Ae jθ and V R Ae jθ + / + / (a) In this case, we haveω 0 and /3, so Ae jθ +2jω +(jω) 2 Aejθ (jω +) 2 V r /3 v r (t) /3. (b) In this case, we haveω andθ π/2, so e jπ/2 and V r e jπ/2 ( 2e jπ4 ) 2 2 v r(t) 2 cos(t). (c) Here we utilize the fact an RLC circuit is linear (i.e. super-position). Let v () in (t) /3 and v(2) in resistor when v (k) in (t) sin(t) and for each k,2, let v(k) r (t) be the voltage across the (t) is the input voltage. Then by parts (a) and (b), we havev () r (t) /3 and v (2) r (t) 2 cos(t). We havev in (t) 3v () in (t)+2v(2) in (t), so by linearity and time invariance, we have v r (t) 3v () r (t)+2v (2) r (t) cos(t).

5 Example 3 Recall the Norton Equivalent of an RLC circuit is a current source in parallel with a resistor and a capacitor or an inductor. Find the value ofc for which the Norton Equivalent is a current source in parallel with only a resistor (i.e. the Thevenin Impedance is purely real). What arei sc (t) and R th in this case? R v(t) C L i(t) v o (t) i sc (t) v o (t) R th where v(t) cos(4t+π/3), i(t) sin(4t+5π/6),r Ω, and L /4H. C?? V I Since v(t) and i(t) are both sinusoidal with frequency 4, we can take the phasor transform with respect to ω 4. Then by using a source transformation on the voltage source and, we have: I sc I +V/ and V/ I Z eff // // + + +j4c j. IfZ eff is real, thenc /4F. Finally, V/ +I // // i sc (t) i(t)+ v(t) R 2cos(4t+π/3) R th Ω. Alternatively We can solve fori sc by shorting the output terminals: V + I I sc We can solve for Z th by setting V 0 and I 0 and solving for the effective impedance across the output terminals: