Lecture 46 Bode Plots of Transfer Functions:II A. Low Q Approximation for Two Poles w o

Transcription

1 Lecture 46 Bode Plots of Transfer Functions:II A. Low Q Approximation for Two Poles w o w L =Q - w o πf o w h =Qw o w L ~ RC w h w L f(l) w h f(c) B. Construction from T(s) Asymptotes the Actual T(s) Analytical Forms. Erickson sproblems 8. and 8.3. Problem 8.5 C. Algebra on a Graph Graphically Combining Asymptotes to Find Z(total). Series impedance s. Parallel Impedance s 3. Series / Parallel 4. Voltage Dividers: Division of Asymptotes D. Measuring T(s) and Impedance. Narrow Band Tracking Voltmeter. Methods for Measuring T(s) a. Overview and Simple X C Measurement b. Choice of proper injection point for T(s) Measurements is artful c. Measure Z out of a System d. Grounding Issues In Measurement R L

2 Lecture 46 Bode Plots of Transfer Functions:II A. Low Q Approximation for Two Pole T(s) Often our two pole transfer functions have widely separated poles in frequency space allowing some nice approximate solutions to G(s). The T(s) function, V (s) = V (s) + S Qw + ( S o w ) o, needs to be solved for the pole locations. From the pole locations we can sketch the Bode plot by inspection when the poles are far apart. When the poles are close a resonant response can be constructed from the asymptotic curves by employing a method called Q peaking.

3 3 We can solve for the two roots. The roots can be put into the Q form as follows in order to better see how the poles change as Q of the circuit changes. Can you see what occurs for low Q? 3

4 The high frequency pole of the two pole pair is. 4 The low frequency pole can be shown to be Qf 0. Both poles appear in the low Q approximation Bode plot as follows: Note above, by knowing that Q is low and the value of f 0 we can rapidly draw the Bode plot of T(s). When the exact value of Q is specified so are the two pole locations. F 0 is easily known from L and C values. 4

5 5 As Q reduces below / the roots separate from their original position,both located at w o, to widely separate and ultimately isolated. For Q < ½ we find: w low = Q wo F(Q) w high = wo Q w L = Qw o w o w H = w Q = R L /RC w H f(c) w L f(l) Notice as L 0 For very large C this pole heads out w L 0 to high w Future use of low Q approximation in analyzing two-pole transfer functions will occur. Specifically, the DCM mode of operation of the boost circuit in Chapter 0 Figure 0.7b, is reproduced below on page 6. Looking ahead we will find an equivalent circuit as shown below whose transfer function,v O / d we need to solve. The very low Q approximation we have been playing with will be employed. That is we will end up the discussion with a live example. It may be hard to see because the transfer function is from the input duty cycle to the output voltage. We are not yet familiar with this concept. o 5

6 6 vg il L + v - i r DCM boost switch network small-signal ac model + + jd gv gv jd r v C v i - - R Find G vd (s) = V o(s) d(s) V g=0 L shorts the input We will find the output voltage to duty cycle transfer function: o vd vd z G (s) = G (s- w ) + as+ a s with two poles and a RHP zero For now forget about the right-half plane zero. In standard form the denominator will be: + s Qw + ( s w ), w o = o o LC We can rapidly sketch out the pole locations using the low Q approximation method. w high = R L or above the f sw as L 6

7 w low = 0 for very high C RC This allows us to see circuit conditions for which we go from two poles to a single pole. G vd (s) has a single pole when L values are chosen very low, see Erickson Table 0.3 for other conditions. 7 G db 0 db f=qfo/f(q) ~Qfo fo -0 db/dec f=fof(q)/q ~fo/q -40 db/dec From T(s) plots we can estimate system stability, via the Nyquist criteria. Explain the Nyquist criteria. B. Construction from T(s) Asymptotes the Actual T(s) Analytical Forms. Given the three T(s) plots below Problems 8. / 8.3 of Erickson. Find analytic expressions for the low f asymptote in terms of G and the break frequencies. fo +0 db/dec G f -40 db/dec Q Q f G 7

8 8 f -0 db/dec f -40 db/dec Q f3 G Solutions to Erickson Problems 8. and 8.3 follow. Given the Bode plots on the left find analytical expressions for low f asymptotes. Also get G(s) in factored pole zero form. First find T(s) s 0 then T(s) low frequency asymptote. (a) fo G G(w) +0 db/dec G wo G s G(s) = G s + w 0 w s as s w (b) f -40 db/dec Q Double pole Q f G Double zero 8 w G w G(w) 0 constant + w Q s + w s G(s) = G + w as s 0 G w s w + Q s s w = G w w s 0 s w

9 9 (c) f -0 db/dec Pole f Pole G(w) G(s) = G -40 db/dec Q f3 + w 3 Q s + w s 3 + w + + s G Double zero 3 w s w3 G ww 0 constant G w s s = G w s 0 s w w w w 3. Erickson Problem 8.5 Given experimental data of Figure 8.56 find the proper asymptotes. After some noodling, Zero f = 80Hz pole f = 50Hz pole f 3 = 4.4KHz RHP zero f 4 = 50KHz (since A and A 3 w K(s + w )(s w 4) So F A(s) = (s + w )(s + w ) 3 From graph, K = 35.3 db 58.3 Right Half Plane zero pushes 9 gain up vs. f, Phase down

10 3. Given T(s) Plot, Find the analytical form in the standard format with poles and zeros. This is Erickson problem 8. for plot(c) found on page 35 of the text db/dec Double pole f Q -0 db/dec f Inverted zero G G(s) = G + w s + s Q w s + w C. Algebra on a Graph Graphically Adding Asymptotes to calculate Z(total). Combining Series R-C Elements Plot each Z(f) part separately first Zin 0 ohms uf 0

11 f crossover = πrc Algebra on a graph For Z(total) in series always Take the bigger of the two. Series R - L - C Plot each, X(f),individually. Then for a series connection always take the largest value for Z(total)=Z(f). K mh Zin 0.uF Plotted on the left is the case when the resistance, R, is larger than the value of R 0 = wl= /wc. What if R lies below, R 0, the crossing of wl and /wc? Then we take a nose dive from the reactive Z asymptotes to reach R and observe Q peaking in the impedance plot at ω 0. Clearly Z in at resonance is R f o + jwl cancels -j/wc. Both are equal in magnitude. R o = = wl, which occurs only at ω 0 wc The relative value or R compared to R 0 is then the key to determining whether or not we see Q peaking in the Z(f) plots. The case of Q peaking for the series R-L-C is seen on page, while the non-peaking plots are shown above.

12 For Z(total) take the largest except just at resonance when R is less than R o = wl = wc Note: Q = R o / R and is plotted symmetrically. What if R > R o? Return to top of page and you are done. If we combine series X(f) plots using the largest value for the total impedance, what do you think would be the rule for combining parallel X(f) plots?

13 3. Combining Parallel Impedance s Plot each X(f) separately and then combine for Z(total) 3 0ohms mh 0.uF Take the smaller of the three for Z(total) in parallel What if R > /w c = w L = R o 4. Combining Series / Parallel Combinations R 0k C 33nF R 5k C 3.3nF 3

14 First do wc R and for Z(total) in parallel Take the smaller of the two Next do w C R and for Z(total) in parallel Take the smaller of the two R to is one curve w C R to is the nd curve w C Add these two curves in series 4 For series combinations Z(total) = the higher of the two f is one curve f is nd curve Take the higher of the two Note in the middle a new frequency at the crossing f 3 = since its π R C R " w C 5. Voltage Dividers : Division of Asymptotes V o(s) z = Z V in (s) z + z + z zin z in the is series combo seen at V in Vin Z Vo In some other cases we may already know Z out (looking in from 4

15 V o ) so we want to express the transfer function differently: V o(s) Z Z out = V in (s) Z Z Z = Z + z Sometimes it s easier to do a Series Combo or a Parallel Combo Consider the two pole circuit model below where L effective = L/(D ) as occurs in the boost or buck-boost circuit model. That is, the duty cycle choice effects the effective inductance see by the small signal model. For small values of duty cycle, D, to achieve DC operation goals L appears bigger in the dynamic model than its circuit value. By equations alone it is sometimes difficult to see these duty cycle, D, design choice effects on ac models. vg(s) e(s)d(s) j(s)d(s). : M(D). + ve(s) - Zin He(s) Le C + v(s) - Le = L/D' R Zout = R C Le Z Z 5 Z Algebra on the graph however provides a visualization aid. From the secondary of the transformer to the output. V o(s) z out = = z = z V (s) z + z z z L/D' C R Zout e Now lets plot versus frequency Zout via algebra on a graph z wl (D ) in 5 Z Z Z

16 Z out is the smallest of three asymptotes because it represents all three elements in parallel. Z is just an increasing positive slope but it is offset by the duty cycle,d, choice. For D D just shift the wl/ (D ) asymptote as shown upwards toward the left. zout z ratio? ratio for w < w o is clearly unity but for w > w o what is this z (w < w ) w C w L = L C out o = z (w > w o) c w c } Sharp 40 db/decade slope 6 Z out and Z plotted together versus f Z out /Z plotted versus f Section C, I hope, illustrates the value of algebra on a graph to better visualize changes in transfer functions: * Effects of varying quiescent point via duty cycle, D,which in turn changes the dynamic model. * Effect of changing element values like C, L, or R on the transfer functions 6

17 D. Measuring Transfer Functions and Impedance s 7. Narrow Band Tracking Voltmeter Vin Mixer Narrow Band Filter fbase Useful tool for PWM Converters where signals are typically complex with various frequency components present. Vout switching spikes Slow varying output at signal frequency with fsw ripple We will need to make Z(s) and T(s) measurements versus f. t 7

18 The required test equipment for T(s) and Z(s) measurements includes: 8 The front panel of a network analyzer looks like: The Z(s) or T(s) data is in db and angles. We also have a probe signal V Z (f) available. 8

19 . Measuring T(s) Overview and Simple X C Measurement 9 Below we measure X(f) for a Capacitor. For a tantalum capacitor we find X C (f): C values and ESR values are both found. 9

20 a. Choice of proper injection point for T(s) Measurements is artful We use the signal frequency to inject an input into a system to measure the transfer function, T(s). To avoid undesired loading when we break a feedback loop, V z (f) is injected and V x /V y is measured versus f(applied) as shown below schematically. 0 vy vz vx into the unknown system we could use a feedback loop for signal injection V y is the return signal from the unknown system. The above measurement scheme is for a feedback loop analysis where we need to inject a signal to measure the loop gain. Plots of V o /V in amplitude versus f and V o /V in phase versus f can both be measured The DC bias shown should simulate actual operating conditions before we broke the loop for signal injection, so it will have no effect on the measurement V o (s)/v in (s) 0

21 c. Measure Z out Z o o(s) = V (s) i o(s) V x from i o, V y from V o o Z o(s) = V (s) i o(s) V y(s) = V (s) x The test set-up requires us to simulate the input impedance that normally drives the circuit to accurately measure Z OUT.

22 The DC bias simulates V o of prior stage into input. Injection of V z via Z s, composed of R s & /wc, creates a current into Z o. AC input to the device must be zero. Value of Z s = (R + ) jwc and amplitude of V z do not effect Z out measurement d. Grounding Issues. Statement of the problem i out above the injected into Z o at the top terminal in the figure above comes back towards - V z along the bottom return path. The current return can choose either the -V y path or the -V z path when it exits the G(s) at the bottom terminal. i out splits < ki o to - V z ground < ( - k)i o to - V y ground But the sum of both paths is I o. K depends on the relative impedance s of the two paths Z probe into - V y gnd Z rz into - V z gnd Z is actual impedance of the sample Z o measured is: Z o = Z + Z probe(into- V y) Z rc. This situation

23 sets a lower limit on the smallest Z you can accurately measure. Z >> Z probe Z rc Easy solution for low Z: Buffer V z with a transformer Finally, For HW#3 Due in week:. Answer any Questions asked throughout lectures Chapter 8 of Erickson do Problems 6 and Explain the measurement below in as much detail as possible. 3 3