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Chapter 15 EGR 272 – Circuit Theory II 1 Read: Ch. 15, Sect. 1-2; and Appendices D&E in Electric Circuits, 9th Edition by Nilsson Bode Plots We have determined the frequency response for several 1st and 2nd order circuits. Using the same methods for higher order circuits would become very difficult. A new method will be introduced here, called the Bode plot, which will allow us to form accurate “straight-line” approximations for the log-magnitude and phase responses quite easily for even high-order transfer functions. This technique will also show how various types of terms in a transfer function affect the logmagnitude and phase responses. Illustration - A Bode plot is used to make a good estimate of the actual response. 20log|H(jw)| Bode “straight-line” approximation Actual log-magnitude response (w on a log scale) w(rad/s) Chapter 15 Decibels EGR 272 – Circuit Theory II 2 V2 I2 P2 Note that the quantities , , and are unitless quantities. V1 I1 P1 However, when scaled logs of the quantities are taken, the unit of decibels (dB), is assigned. V2 20log10 log-magnitude (LM) of the voltage gain in dB V1 20log10 I2 log-magnitude (LM) of the current gain in dB I1 10log10 P2 log-magnitude (LM) of the power gain in dB P1 There are two types of Bode plots: • The Bode straight-line approximation to the log-magnitude (LM) plot, LM versus w (with w on a log scale) • The Bode straight-line approximation to the phase plot, (w) versus w (with w on a log scale) Chapter 15 EGR 272 – Circuit Theory II 3 Standard form for H(jw): Before drawing a Bode plot, it is necessary to find H(jw) and put it in “standard form.” Show the “standard form” for H(jw) below: H(s) K(s z1 )(s z2 ) (s z N ) (s p1 )(s p2 ) (s pM ) Chapter 15 EGR 272 – Circuit Theory II Example: Find H(jw) for H(s) shown below and put H(jw) in “standard form.” 10(s)(s + 200) H(s) (s 1000) 4 Chapter 15 EGR 272 – Circuit Theory II 5 The additive effect of terms in H(jw): The reason that Bode plot approximations are used with the log-magnitude is due to the fact that this makes individual terms in the LM additive. The phase is also additive. Show how the LM and phase of each term in 20log|H(jw)| is additive (or acts separately). Drawing Bode plots: To draw a Bode plot for any H(s), we need to: 1) Recognize the different types of terms that can occur in H(s) (or H(jw)) 2) Learn how to draw the log-magnitude and phase plots for each type of term. Chapter 15 EGR 272 – Circuit Theory II 6 5 types of terms in H(jw) 1) K (a constant) 2) w 1j w1 (a zero) 3) jw (a zero) or or 1 w 1 j w1 (a pole) 1/jw (a pole) 2 w w2 4) 1 j (a complex zero) 2 2 w0 w0 or 1 2 w w2 1 j 2 2 w0 w0 5) Any of the terms raised to a positive integer power. For example, w 1 j w1 2 (a double zero) Each term is now examined in detail. (a complex pole) Chapter 15 EGR 272 – Circuit Theory II 7 1. Constant term in H(jw) If H(jw) = K = K/0 Then LM = 20log(K) and (w) = 0 , so the LM and phase responses are: (w) LM (dB) 20log(K) 0o 0 1 10 100 w Summary: A constant in H(jw): • Adds a constant value to the LM graph (shifts the entire graph up or down) • Has no effect on the phase w Chapter 15 EGR 272 – Circuit Theory II 8 2. A) 1 + jw/w1 (a zero): The straight-line approximations are: If H(jw) 1 j w w1 2 w w 1 tan -1 w1 w1 2 w -1 w Then LM 20log 1 and (w) tan w1 w1 To determine the LM and phase responses, consider 3 ranges for w: 1) w << w1 2) w >> w1 3) w = w1 Chapter 15 EGR 272 – Circuit Theory II 9 The Bode approximations (LM and phase) for 1 + jw/w1 are shown below. (w) LM 90o 20dB slope = + 6dB/oct 45o = +20dB/dec 0dB w w1 10w1 0o 0.1w1 w1 slope = +45 deg/dec (for 2 decades) w 10w1 Discuss the amount of error between the actual responses and the Bode approximations. Summary: A 1 + jw/w1 (zero) term in H(jw): • Causes an upward break at w = w1 in the LM plot. There is a 0dB effect before the break and a slope of +20dB/dec or +6dB/oct after the break. • Adds 90 to the phase plot over a 2 decade range beginning a decade before w1 and ending a decade after w1 . The slope is +45 deg/dec or +13.5 deg/oct. EGR 272 – Circuit Theory II Chapter 15 10 1 2B) w 1 j w1 If H(jw) 1 w 1 j w1 (a pole): The straight-line approximations are: 10 2 w -1 w 1 tan w1 w1 w tan -1 2 w1 w 1 w1 1 1 -1 w Then LM 20log and (w) -tan 2 w1 w 1 w1 To determine the LM and phase responses, consider 3 ranges for w: 1) w << w1 2) w >> w1 3) w = w1 Chapter 15 EGR 272 – Circuit Theory II 11 1 The Bode approximations (LM and phase) for LM 0dB w1 10w1 (w) 0.1w1 0o w slope = -20dB/dec o -45 = - 6dB/oct -20dB are shown below. w 1 j w1 w1 10w1 w slope = -45 deg/dec (for 2 decades) -90o Discuss the amount of error between the actual responses and the Bode approximations. Summary: A 1 + jw/w1 (zero) term in H(jw): • Causes an downward break at w = w1 in the LM plot. There is a 0dB effect before the break and a slope of -20dB/dec or -6dB/oct after the break. • Adds -90 to the phase plot over a 2 decade range beginning a decade before w1 and ending a decade after w1 . The slope is -45 deg/dec or -13.5 deg/oct. Chapter 15 EGR 272 – Circuit Theory II Example: Sketch the LM and phase plots for the following transfer function. H(s) 10(s 100) 12 Chapter 15 EGR 272 – Circuit Theory II Example: Sketch the LM and phase plots for the following transfer function. 100 H(s) s 2000 13 Chapter 15 EGR 272 – Circuit Theory II 14 LM: Working with slopes on Bode plots: • If the LM has a slope of 20 dB/dec, the change in LM between two frequencies can be calculated using: w final LM 20 log w initial Example: Determine the value of LM2 below. LM 40 dB -20 dB/dec LM2 45 725 w (rad/s) Chapter 15 EGR 272 – Circuit Theory II 15 Phase: Working with slopes on Bode plots: • If the phase has a slope of 45 deg/dec, the change in phase (in degrees) between two frequencies can be calculated using: w final Phase 45 log w initial Example: Determine the value of 2 below. (deg) 90 -45 deg/dec 2 150 5200 w (rad/s) Chapter 15 EGR 272 – Circuit Theory II 16 Example: Sketch the LM and phase plots for the following transfer function. 200(s 50) H(s) s 400 (Solution provided on the following slide) EGR 272 – Circuit Theory II Chapter 15 17 200(s 50) H(s) s 400 Example: Sketch the Bode LM and phase plots for: Solution: First find H(jw) and put H(jw) into "standard form" 200(jw 50) 25(1 + jw/50) H(jw) = (standard form) jw 400 1 + jw/400 Log-magnitude: The LM will start level since there are no jw terms and will begin at LM = 20log(25) = 28 dB The LM will break up at w = 50 due to the zero. The pole at w = 400 will cancel the effect of the zero causing the graph to level out. Since 50 to 400 is three octaves and the slope is 6 dB/oct, the graph increases by 18 dB from 28 to 46 dB. Phase: The phase plots begins at 0 degrees since there are no jw terms. The phase begins increasing with a slope of 45 /dec at w = 5 (a decade before the zero). The phase levels out at w = 40 (a decade before the pole) as the decreasing slope due to the pole cancels the effect of the zero. The phase begins decreasing at w = 500 (a decade after the zero) as the zero stops increasing. The phase levels out at w = 4000 (a decade after the pole) as the pole stops decreasing. Since a zero causes a net increase of 90 in phase and a pole causes a net decrease of 90 in phase, the graph will end at 0 . (w) LM 46 dB o 90 slope = +20dB/dec 28 dB 50 400 w o 0 slope = - 45 deg/dec 5 40 500 4000 w Chapter 15 EGR 272 – Circuit Theory II 18 Example: Sketch the LM and phase plots on the 4-cycle semi-log graph paper shown below for the following transfer function. (Pass out 2 sheets of graph paper.) 2,000,000( s 2,000) H(s) (s 8,000)(s 50,000) Log-magnitude (LM) plot: Chapter 15 EGR 272 – Circuit Theory II Example: (continued) 2,000,000( s 2,000) H(s) (s 8,000)(s 50,000) Phase plot: 19 Chapter 15 EGR 272 – Circuit Theory II 20 3A) jw term in H(jw) If H(jw) = jw = w/90 Then LM = 20log(w) and (w) = 90 Calculate 20log(w) for several values of w to show that the graph is a straight line for all frequency with a slope of 20dB/dec (or 6dB/oct). The LM and phase for H(jw) = jw are shown below. w 0.1 1 10 100 1,000 10,000 100,000 20log(w) (w) LM 40 dB slope = +20dB/dec = + 6dB/oct 20 dB 0 dB 1 10 100 w 90o 0o 1 10 100 • So a jw (zero) term in H(jw) adds an upward slope of +20dB/dec (or +6dB/oct) to the LM plot. • And a jw (zero) term in H(jw) adds a constant 90 to the phase plot. w Chapter 15 EGR 272 – Circuit Theory II 21 3B) 1/(jw) term in H(jw) If H(jw) = 1/(jw) = (1/w)/-90 Then LM = 20log(1/w) = -20log(w) and (w) = -90 A few calculations could easily show that the graph of 20log(1/w)is a straight line for all frequency with a slope of -20dB/dec (or -6dB/oct). The LM and phase for H(jw) = 1/(jw) are shown below. (w) LM 0 dB -20 dB 10 100 w o 10 100 0 slope = -20dB/dec = - 6dB/oct -90o -40 dB • So a 1/(jw) (pole) term in H(jw) adds a downward slope of -20dB/dec (or 6dB/oct) to the LM plot. • And a 1/(jw) (pole) term in H(jw) adds a constant -90 to the phase plot. w Chapter 15 EGR 272 – Circuit Theory II Example: Sketch the LM and phase plots for the following transfer function. 200(s) H(s) s 5000 22 Chapter 15 EGR 272 – Circuit Theory II Example: Sketch the LM and phase plots for the following transfer function. 500(s 500) H(s) (s)(s 8000) 23 Chapter 15 EGR 272 – Circuit Theory II Calculation of exact points to check Bode Plots Evaluating H(jw) at a particular value of w is helpful to check Bode Plots. An example is shown below. Example: 5000(s + 200) Suppose H(s) = (s + 4000) w 250 1 + j 200 so H(jw) = w 1 + j 4000 Evaluating this function at w = 4000 rad/s yields: 4000 250 1 + j 250 1 + j20 200 = H(jw) = = 3539.9542.1 4000 1 + j 1 + j 4000 so (4000) = 42.1 LM(4000) = 20log(3539.95) = 70.98 dB 24 Chapter 15 EGR 272 – Circuit Theory II Example: Evaluate the H(jw) on the last page at w = 500 rad/s and w = 8000 rad/s. Compare the values with the Bode plots. Do they appear to be correct? 25 Chapter 15 EGR 272 – Circuit Theory II 26 5. Poles and zeros raised to an integer power in H(s) In the last class it was demonstrated that terms in H(jw) are additive. Therefore, a double terms (such as a pole or zero that is squared) simply acts like two terms, a triple term acts like three terms, etc. Illustration: Show that (1 + jw/w1)N results in the following responses: LM plot: Has a 0dB contribution before its break frequency Will increase at a rate of 20NdB/dec after the break There will be an error of 3NdB at the break between the Bode straight-line approximation and the exact LM Phase plot: Has a 0 degree contribution until 1 decade before its break frequency Will increase at a rate of 45Ndeg/dec for two decades (from 0.1w1 to 10w1). The total final phase contribution will be 90N degrees. Chapter 15 EGR 272 – Circuit Theory II 27 So (1 + jw/w1)N results in the following responses: w LM 90No 20N dB 0 dB slope = +20NdB/dec = + 6NdB/oct w1 10w1 w 45No 0o 0.1w1 w1 slope = +45N deg/dec (for 2 decades) w 10w1 Chapter 15 EGR 272 – Circuit Theory II Example: Sketch the LM plot for the following transfer function. 2x1012s2 H(s) (s 800)2 (s 2,000)2 28 Chapter 15 EGR 272 – Circuit Theory II Generating LM and phase plots using Excel and MATLAB: Refer to the handout entitled “Frequency Response” which includes detailed examples of creating LM and phase plots using Excel and MATLAB. Generating LM and phase plots using PSPICE: Refer to the handout entitled “PSPICE Example: Frequency Response (LogMagnitude and Phase)” which includes a detailed example of creating LM and phase plots using PSPICE. 29 Chapter 15 EGR 272 – Circuit Theory II 30 Bode Plots - Recall that there are 5 types of terms in H(s). The first four have already been covered. 5. Complex terms in H(s) A second order term in H(s) with complex roots has the general form s2 + 2s + wo2 . After factoring out the constant has the wo2 , the corresponding term in H(jw) is: 2 w w2 1 j (a 2nd-order pole or zero with complex roots) 2 2 w0 w0 Key point: The complex term above is identical to a double real pole or a double real zero in the Bode straight-line approximation, but they differ in the exact curves. Chapter 15 EGR 272 – Circuit Theory II 31 Complex peaks and damping ratio Recall that (zeta) = damping ratio where wo 2 w w2 2 w w2 then H(jw) 1 j is equal to H(jw) 1 j 2 2 2 w0 w0 w0 w0 and at the break frequency (w w o ) this yields H(jw) j2 so LM 20log(2 ) which yields a peak whose size depends on the value of Exercise: Complete the table below showing the value of the complex peak (for a complex zero) 1* 0.5 0.2 0.1 0.05 20log(2) = Value of complex peak * = 1 is not complex. It corresponds to a double-zero. Chapter 15 EGR 272 – Circuit Theory II The following graphs from the text (Electric Circuits, 9th Edition, by Nilsson) below show the Bode straight-line approximation and the exact curve for different values of . 32 Chapter 15 EGR 272 – Circuit Theory II Example: Sketch the LM responses for H1(s), which has a double zero, and H2(s), which has a complex zero. 104 H1 (s) (s 100)2 104 H2 (s) 2 (s 20s 104 ) A) Find H1(jw) and sketch the LM response 33 Chapter 15 EGR 272 – Circuit Theory II Example: Sketch the LM responses for H1(s), which has a double zero, and H2(s), which has a complex zero. 104 H1 (s) (s 100)2 104 H2 (s) 2 (s 20s 104 ) B) Find H2(jw) and sketch the LM response 34 Chapter 15 EGR 272 – Circuit Theory II 35 Filter order Unfortunately, we can’t build ideal filters. However, the higher the order of a filter, the more closely it will approximate an ideal filter. The order of a filter is equal to the degree of the denominator of H(s). (Of course, H(s) must also have the correct form.) LM Ideal LPF slope = H(s) = K (s + w C ) K 4th-order LPF H(s) = (s + w C )4 slope = -80dB/dec 3rd-order LPF slope = -60dB/dec H(s) = 2nd-order LPF slope = -40dB/dec K (s + w C )3 H(s) = K (s + w C )2 K 1st-order LPF H(s) = (s + w C ) slope = -20dB/dec wC w Chapter 15 EGR 272 – Circuit Theory II Recognizing filter types from the transfer function Low-pass filter (LPF) - Discuss the form of LM, H(jw), and H(s). High-pass filter (HPF) - Discuss the form of LM, H(jw), and H(s). 36 Chapter 15 EGR 272 – Circuit Theory II Band-pass filter (BPF) - Discuss the form of LM, H(jw), and H(s). Band-stop filter (BSF) - Discuss the form of LM, H(jw), and H(s). Also define a “notch” filter. 37 Chapter 15 EGR 272 – Circuit Theory II Practical Filter Examples: Discuss a stereo tuner, 5-band equalizer, 60 Hz interference block. 38 Chapter 15 EGR 272 – Circuit Theory II 39 Active and passive filters Passive filter - a filter constructed using passive circuit elements (R’s, L’s, and C’s) Active filter - a filter constructed using active circuit elements (primarily op amps) Analysis of active filters Active filters are fairly easy to analyze. Recall that a resistive inverting amplifier circuit has the following relationship: R2 ACL = _ Vin R1 Vo + Vo R =- 2 Vin R1 Replacing the resistors with impedances yields a transfer function, H(s): Z2 Vin(s) Z1 _ + H(s) = Vo(s) Vo ( s) Z =- 2 Vin ( s) Z1 Chapter 15 EGR 272 – Circuit Theory II 40 Example: Find H(s) =Vo(s)/Vin(s) for the circuit below and sketch the LM response. 10mF 200 W _ Vin Vo 10mF + Chapter 15 EGR 272 – Circuit Theory II 41 Impedance and Frequency Scaling Filter tables are commonly available that list component values for various types of filters. However, the tables could not possibly list component values for all possible cutoff frequencies. Typically, filter tables will list cutoff frequencies of perhaps 1 Hz or 1 kHz. The user can then scale the cutoff frequencies to the desired values using frequency scaling. Similarly, component values may be too large or too small (for example, a filter might be shown using 1 ohm resistors or 1F capacitors) so the components need to be scaled using impedance scaling. Chapter 15 EGR 272 – Circuit Theory II 42 Impedance Scaling – a procedure used to change the impedance of a circuit without affecting its frequency response. Consider the series RLC circuit shown below. R Z(s) sL 1 sC Note that Z(s) R sL R' K Z R Show that to scale the impedance by a factor KZ : L' K Z L C' C KZ 1 sC Chapter 15 EGR 272 – Circuit Theory II 43 Frequency Scaling – a procedure used to scale all break frequencies in the frequency response of a circuit. The frequency response is scaled without affecting the circuit impedance (the impedance is the same at the original and new break frequencies). Consider the series RLC circuit shown below. + Vi(s) sL 1 sC + R _ Vo(s) _ As seen in an earlier class, H(s) is described below. H(s) Vo (s) Vi (s) s2 R s L 1 R s LC L Note that H(s) described above is the transfer function of a band-pass filter. The LM is shown on the following page. Chapter 15 EGR 272 – Circuit Theory II 44 LM of the original band-pass filter (center frequency = wo): |H(jw)| |H(jwo)| w wo LM of the frequency-scaled band-pass filter (center frequency = KFwo): |H(jw/KF)| |H(jwo)| w wo KFwo Chapter 15 EGR 272 – Circuit Theory II Show that to scale the frequency by a factor KF : Development : 45 R' R L L' KF 1 C sC C' -j KF Z(jw) R jwL wC w and now substituting in place of w as indicatedon thepreviousgraph yields : KF If Z(s) R sL Chapter 15 EGR 272 – Circuit Theory II 46 Example: Impedance and frequency scaling. A) Determine H(s) = Vo(s)/Vi(s) B) Sketch the LM response + Vi(s) _ + 1 1F Vo(s) _ Chapter 15 EGR 272 – Circuit Theory II Example: (continued) C) A cutoff frequency of 1 rad/s and component values of 1 ohm and 1 Farad are not very useful. Scale the frequency so that the cutoff frequency is 500 rad/s. Also use a 1 kW resistor instead of a 1 W resistor. 47 Chapter 15 EGR 272 – Circuit Theory II Example: (continued) D) Draw the new circuit. Determine the new H(s) and verify that the LM response has been shifted as planned. E) Calculate the circuit impedance at the break frequency for the new circuit and compare it to the circuit impedance at the break frequency for the original circuit. 48