Lecture 9 Time Domain vs. Frequency Domain

Transcription

1 . Topics covered Lecture 9 Time Domain vs. Frequency Domain (a) AC power in the time domain (b) AC power in the frequency domain (c) Reactive power (d) Maximum power transfer in AC circuits (e) Frequency response-filters (f) An example-suppressing 60 cycle noise. AC power in the time domain (a) Consider the power delivered by a given source in an arbitrary AC RLC network. The goal is to calculate the power delivered using the voltages and currents in the real time domain. In the next section we show how this same result can be obtained in a simpler way in the frequency domain characterized by complex phasors. (b) The voltage and current delivered by the given source can be written as V S (t) = V i cos(ωt + φ V ) I S (t) = I i cos(ωt + φ I ) () (c) The instantaneous power delivered is just the product of V S I S. P S (t) =V S I S = V i I i cos(ωt + φ V )cos(ωt + φ I ) () (d) Use the trigonometric identity cosa cosb = [cos(a + B)+cos(A B)] (3) (e) The power can be rewritten as P S (t) = V i I i [cos(φ V φ I )+cos(ωt + φ V φ I )] (4)

2 (f) Note that the first term is independent of time while the second term oscillates at a frequency equal to ω. Often times the quantity of most interest is not the instantaneous power, but the average power delivered over one cycle of oscillation. The average power isdefined as <P S > = T T 0 = V i I i T P S (t)dt T 0 T =π/ω [cos(φ V φ I )+cos(ωt + φ V φ I )] dt (g) In terms of the expression for P S, the second term is oscillatory and averages to zero. The first term is a constant (with respect to time) and can be taken out of the integral. The result is <P S >= V i I i cos(φ V φ I ) (6) 3. AC power inthe frequency domain (a) Let us next see if we can obtain the same result in a simpler way using complex phasors. (b) Note, in complex notation the voltage and current, (before taking the real parts), are given in terms of the complex phasors V i and I i by (c) Now form the product V S = V i e jωt = V i e j(ωt+φ V ) I S = I i e jωt = I i e j(ωt+φ I) Re(V S I S) = Re [ V i e j(ωt+φ V ) I i e j(ωt+φ I) ] = Re [ V i I i e j(φ V φ I ) ] (5) (7) = V i I i cos(φ V φ I ) (8) (d) Therefore a simple way to calculate the average power is ( ) <P S >= Re V iii (9)

3 (e) Finally, to make the expression look more like DC power we introduce the root mean square (rms) amplitude by the following definition [ T / A rms = A (t)dt] (0) T o (f) If A(t) =A 0 cos(ωt + φ) then A rms = [ T T 0 A 0cos (ωt + φ)dt] / = ( A 0 ) / = A 0 () (g) For our case V rms = V i I rms = I i () (h) We now make an obvious generalization by defining the rms phasor V rms = V i = V i e jωφ V I rms = I i = I i e jωφ I (3) (i) In terms of the rms phasors the average power has a simple DClike form with no factors of appearing anywhere 4. Reactive power <P s >= Re(V rms I rms ) (4) (a) We have just seen that Re(V rms I rms )isequal to the power delivered by a source. One can then ask whether or not there is any significance to the quantity Im Im(V rms I rms ) (b) The answer is yes but it involves a little bit of thinking and a little bit of algebra (c) To find the answer consider an AC source connected across an arbitrary RLC circuit without sources. At any given frequency the RLC circuit can be reduced to its Thevenin equivalent. For DC the equivalent would just be a resistor. For AC the equivalent is a resistor in series with some net reactance, which could be either inductive or capacitive. The equivalent circuit has the form 3

4 + V in(t) - I in(t) R REACTANCE Figure 9. (d) The voltage and current (in the time domain) are given by V in (t) = V i cos(ωt + φ V ) I in (t) = I i cos(ωt + φ I ) (5) (e) Now, the instantaneous power dissipated in the resistor is just P R (t) =RIin = R I i cos (ωt+φ I )= R I i [+cos(ωt+φ I )] (6) (f) Since the reactive elements do not on average dissipate power, the average resistive power per cycle must be equal to the average power delivered by the source <P S > = V i I i cos (φ V φ I ) = <P R >= R I i (7) (g) Next, we focus on the instantaneous rather than average power delivered by the source and do a little algebra P S (t) = V i I i [cos(φ V φ I )+cos(ωt + φ V + φ I )] = V i I i {cos(φ V φ I )[+cos(ωt +φ I )] cos(φ V φ I ) cos(ωt +ω I )+cos(ωt + φ V + φ I )} = P R (t)+ V i I i [cos(ωt + φ V + φ I ) cos(φ V φ I ) cos(ωt + φ I )] (8) 4

5 (h) The residual term is the power associated with the reactive elements in the Thevenin equivalent impedance. With a little algebra, this term can be simplified as follows P X (t) = V i I i = V i I i [cos(ωt + φ V φ I ) cos(φ V φ I ) cos(ωt +φ I )] [cos(ωt +φ I + φ V φ I ) cos(φ V φ I ) cos(ωt +φ I )] = V i I i [cos(ωt +φ I ) cos(φ V φ I ) sin(ωt +φ I ) sin(φ V φ I ) cos(φ V φ I ) cos(ωt +φ I )] = V i I i sin(φ V φ I ) sin(ωt +φ I ) (9) (i) As expected, the reactive power averages to zero over one cycle. Even so the source must deliver power over one half the cycle although it receives it back on the second half of the cycle. The requirement of delivering the extra power during half the cycle imposes additional requirements (and often costs) on the power supply. (j) Furthermore, in power transmission applications, the additional reactive current increases the magnitude of the total current over that of the resistive component itself. This leads to additional transmission line losses, which only depend on the magnitude of the current. (k) Thus, in many applications it is important to know, (and sometimes minimize) the magnitude of the reactive power. From the analysis above we see that the magnitude of the reactive power is given by P X (t) max = V i I i sin(φ V φ I ) (0) (l) In terms of phasor notation ( Vi I ) i P X (t) max = Im = Im(V rms I rms ) () 5

6 (m) The conclusion is that the quantity Im (V rms I rms )isequal to the magnitude of the reactive power 5. Maximum power transfer in AC circuits (a) In this section we consider the question of how to choose a load impedance such that the power delivered to it is maximized, assuming the load is connected to a circuit that contains an AC source with a complex source impedance. (b) In particular, assume there is a general AC RLC circuit with sources driving an as yet unspecified load. By standard analysis, the driving circuit and sources can be reduced to a Thevenin equivalent circuit as shown below I i Z s + V i - Z L + V - 0 Figure 9. (c) We assume the source and load impedance can be written as Z S = R S + jx S Z L = R L + jx L () (d) Here R S and R L are the source and load resistances while X S and X L are the corresponding reactances. Note that X S and X L are functions of ω and can be either positive (inductive) or negative (capacitive). Our goal is to derive an expression for R L and X L in terms of R S and X S such that maximum power is transferred to the load. (e) The derivation proceeds as follows. In phasor notation, the current flowing in the circuit and voltage across the load are given by I i = V i Z L + Z S 6

7 V o = Z L Z L + Z S V i (3) (f) The power dissipated in the load (assuming all voltage and current amplitudes are rms values) can be written as ( )( ) <P L > = Re(V o Ii Vi Z L V i )Re Z L + Z ( S ) = V i Z L Re Z S + Z L Z L + Z S = V i R L (R S + R L ) +(X S + X L ) (g) We see that any non-zero value of X S + X L reduces the power delivered to the load. Therefore maximum power is delivered when the load reactance cancels the source reactance X L = X S (5) (h) The resulting expression is just the old DC expression involving only resistors. This expression has a maximum when R L = R S (6) (i) Thus, the conditions for maximum power transfer, and the corresponding value of maximum power are given by (4) 6. Frequency response - filters Z L = R S jx S = Z S <P L > = V i 4R S (7) (a) There are many practical examples that make use of AC RLC circuits. In this section we show how such circuits can be used to build frequency filters. (b) A frequency filter is a circuit that allows a specific range of frequencies to be passed (or rejected) from an AC source to a given load. 7

8 (c) Before proceeding with any analysis we can show intuitively how our familiar series RLC circuit can be used to create a wide variety of filters. (d) The intuition is obtained by examining the behavior of the circuit at the following frequencies: ω = 0,ω = and the resonant frequency ω = ω 0. Recall that at the resonant frequency ω 0 = (/LC) / corresponding to Z L = Z C. (e) Consider the series RLC circuit shown below + V - i Z Z Z R C L Figure 9.3 (f) Recall also the impedance of each element as a function of frequency Resistor Z R = R Frequency independent Inductor Z L = jωl Short at ω =0,Openatω = Capacitor Z C =/jωc Open at ω =0,Short at ω = (g) We are now ready to consider different types of filters. It is convenient to do so in terms of the transfer function H(ω) defined as the ratio of the output voltage to the input voltage: H(ω) =V o /V i. Also for simplicity we assume that the resistance is chosen such that R =(L/C) / =Z 0. (h) Returning to our RLC circuit, assume the output is taken as the voltage across the capacitor. Then H(ω) = Z C Z R + Z L + Z C (8) 8

9 (i) Note that at ω =0,H =;atω =,H =0;atω = ω 0, H = /. This is a low pass filter H / ω/ω 0 Figure 9.4 (j) Next assume the output voltage is taken across the inductor. Then H(ω) = Z L Z R + Z L + Z C (9) (k) Note that at ω =0,H =0;atω =, H =;atω = ω 0, H = /. This is a high pass filter H / ω/ω 0 Figure 9.5 (l) Now assume the output voltage is taken across the resistor. Then H(ω) = Z R Z R + Z L + Z C (30) (m) Note that at ω =0,H =0;atω =,H =0atω = ω 0, H =. This is a band pass filter 9

10 H ω/ω 0 Figure 9.6 (n) Finally, assume the output is taken across the inductor + capacitor. Then Z L + Z C H(ω) = (3) Z R + Z L + Z C (o) Note that at ω =0,H =;atω =,H =;atω = ω 0, H = 0. This is a notch filter H ω/ω 0 Figure An example - suppressing 60 cycle noise (a) There are many practical applications where it is desirable to suppress 60H noise induced in low voltage AC detectors situated in close proximity to large machinery (b) One industrial application is as follows. The performance of electronic equipment designed for use in aircraft must be pre-tested under a number of environmental conditions that might arise in actual operation. One such condition is aircraft vibration. 0

11 (c) To test against vibration, the test unit is mounted on a large vibration table, essentially a tens-of-kilowatt loudspeaker. The unit s performance is examined under a range of frequencies and amplitudes such as might be experienced on an aircraft. The vibration table is driven by a high power variable frequency amplifier, which is itself powered by 40V,60H, AC. See the diagram below. (d) A small electromagnetic transducer is mounted on the test unit to monitor the frequency and amplitude of vibration. Typically, the transducer output is in the milli-volt range. This signal is to be used as a feedback signal to the amplifier to in order to maintain a pre-selected amplitude and frequency. (e) A difficulty that sometimes arises is that the close proximity of the high power 60H currents can induce a 60H noise signal on the transducer output, particularly if the transducer leads are not properly shielded. This 60H pickup can be on the order of 0.V. (f) One approach to resolve the problem is to insert a 60H notch filter between the transducer output and the feedback amplifier. This notch filter should have a relatively narrow bandwidth so that the filter would not affect even a relatively close vibration frequency of. 60 =66H. This allows the unit to be tested over a wide range of frequencies (H to 30kH) with only a very narrow range eliminated by the notch filter. (g) The solution proceeds as follows. Illustrated below is a schematic diagram of the system FEEDBACK PREAMPLIFIER TRANSDUCER TEST UNIT VIBRATION TABLE 40V 60H VARIABLE FREQUENCY AMPLIFIER Figure 9.8

12 (h) The equivalent circuit of the transducer-filter-preamplifier, including typical parameter values is illustrated below TRANSDUCER FILTER PREAMP 500Ω L V in R s C + V - outr L 000Ω Figure 9.9 (i) Let us assume that the transducer output voltage consists of the desired signal at f =66H plus a noise signal at 60H: V in (t) =0 3 cos(π 66 t)+0 cos(π 60 t) (3) (j) Our goal is to select values for L and C such that the 60H signal is completely suppressed, but the 66H signal is reduced in amplitude by no more than %. (k) First, lets convince ourselves that the parallel LC circuit actually works as a notch filter. Note that at low frequencies the inductor acts like a short circuit and the entire transducer output appears across the terminals of the preamplifier impedance. Similarly, at high frequencies, the capacitor acts like a short circuit and again the whole transducer signal appears across the preamplifier. (l) Note that at the resonant frequency ω =(/LC) / series and parallel LC circuits behave as follows Z = jωl + jωc = jωc ( ω LC) =0 (33) Figure 9.0

13 Z = (jωl)( jωl ) jωl + jωc = Figure 9. jωc ω LC = (34) (m) The transfer function for the circuit can be easily found using standard voltage divider relations H(ω) = V o V i = Z N (ω) = (jωl)( jωl ) jωl + jωc R L + R S R L + R S + Z N = jωl ω LC R L R L + R S (35) (n) This expression can be simplified by introducing normalized variables as follows ω0 = LC ξ = ( ) L / C R S + R L Ω = ω/ω 0 (36) (o) After some simple algebra the transfer function becomes H(Ω) = 3 H(Ω) = 4 9 Ω Ω +jωξ ( Ω ) (37) ( Ω ) +4Ω ξ (p) We see that if the 60H noise is to be completely suppressed, then the resonant frequency of the LC circuit must be set to 60H; that is, Ω = at 60H. This leads to one constraint on the values of L and C Ω= LC = (π 60) = (38) 3

14 (q) The second condition on L and C arises from the requirement that at 66H(Ω =.), the signal to the preamplifier should be reduced by no more than % of its actual value. Note that if there were no filter, the signal to the preamplifier would just be /3 ofthe transducer voltage because of the resistor voltage division. With the filter we require the signal to be at least 0.99 /3 at66h. In terms of the transfer function, this implies H(Ω =.) = (0.99) 4 9 = 4 [ ( Ω ) ] = ( Ω ) +4Ω ξ ξ Ω=. (39) (r) This expression can easily be inverted to obtain ξ. We find a second constraint on L and C [ ( 0.99 ] / ) ξ = = L C = 4(R S + R L )ξ =. (40) (s) The constraints on L and C can easily be solved simultaneously yielding the desired values L = H C = F (4) (t) Note that one practical problem with this solution is that since the notch filter has a very narrow bandwidth, the L and C must be of high precision so that the resonant frequency falls exactly at 60H. 4