Giao trinh bai tap bt phan doi luu tu nhien

Bode Plots The Bode plot is a quick method to graphically evaluate the frequency dependence of a circuit, once we know its “transfer function” G(s) (or equivalently G(ω)) Using simple rules we can construct both a magnitude response and a phase response, describing the behavior of the circuit in the sinusoidal steady state Let’s start with an example Consider a simple low pass circuit: R Vin Vout G(s ) = C Vout(s) 1/RC k = = Vin(s) s+1/RC s-p k = 1/RC p = -1/RC The transfer function G(s) has a single “pole” p and a low frequency gain (when s is small) of k/-p We wish to plot both the magnitude and phase frequency response of the circuit magnitude phase G( s) = k ( s − p) ∠G(s) = −∠(s − p) To find the frequency response, let s=jω Then and (s − p) = ω + p2 ω ∠(s − p) = arctan  − p The Bode plot plots the log magnitude and phase angle of the transfer function vs frequency, which is also on a log scale Why log magnitude? Basically it is easier to add graphically than to multiply or divide The contribution from a single “pole” or “zero” is expressed as an asymptotic approximation, and then the overall response is found by simply summing the individual responses What the asymptotic representations of (s − p) and ∠(s − p) look like? Magnitude Response: for ω > p, (s − p) ≅ ω -1- DLD 01/24/01 EE 216 Spring 2001 handout # The contribution from a pole in dB is −20log( (s − p) ) = −20log( ω + p2 ) A pole at p=-1 has a magnitude response like the following p=1 -20 dB/decade -10 -20 -30 -2 10 10 -1 10 Frequency (rad/sec) 10 10 2 The contribution from a zero in dB is +20log( (s − z) ) = +20log( ω + z ) A zero at z=-1 has a magnitude response like the following z=1 40 +20 dB/decade 20 -2 10 10 -1 10 Frequency (rad/sec) 10 10 ω ∠(s − p) = arctan  − p For negative, real poles or zeros (the ones we will see this semester in class), Phase Response: ∠(s − p) ≅ for ω < p /10 o ≅ 90 for ω > 10p o = 45o for ω = p The phase contribution from a pole at p=-1 is shown below -2- DLD 01/24/01 EE 216 Spring 2001 handout # p=1 -30 -45 deg/decade -60 -90 10 -2 10 -1 10 Frequency (rad/sec) 10 10 The phase contribution from a zero at z=-1 is shown below z=1 -270 -300 +45 dB/decade -330 -360 -390 10 -2 10 -1 10 Frequency (rad/sec) Here's the punchline: for G(s) = G(s) dB = 10 10 k(s − z1)(s − z ) (s − zn ) , (s − p1)(s − p2 ) (s − pm ) ∑ (s − z) zeros dB − ∑ (s − p) dB + k dB poles ∠G(s) = ∑ ∠(s − z) − ∑ ∠(s − p) zeros poles Going back to our low pass circuit, -3- DLD 01/24/01 EE 216 Spring 2001 handout # 1/RC k = s +1/RC s - p let's let R=1K and C=10 µF, so that k = p = -100 We can construct the Bode plot as follows G(s) dB = k dB − (s + 100) dB G(s) = ∠G(s) = −∠(s + 100) 50 k=100 total response -20dB/dec p=100 -50 10 10 10 Frequency (rad/sec) 10 10 -30 -45 deg/dec -60 -90 10 10 10 Frequency (rad/sec) 10 10 An often useful alternative expression for the transfer function in terms of the dc gain is G(s) = A o (s / z1 − 1)(s / z − 1) (s / z n − 1) , (s / p1 − 1)(s / p2 − 1) (s / pm − 1) Ao = kz1z2 z n / p1 p2 pm = dc gain Now the Bode plot is assembled using -4- DLD 01/24/01 EE 216 Spring 2001 handout # G(s) dB = ∑ (s / z − 1) zeros ∠G(s) = dB − ∑ (s / p − 1) + A o dB poles ∑ ∠(s / z − 1) − ∑ ∠(s / p − 1) zeros poles This has the advantage that all of the finite poles and zeros 'break' from the dB axis The dc gain term can be incorporated either as a horizontal line representing the appropriate gain, or by simply re-labeling the gain axis to take into account the gain at dc In this case, the poles and zeros all 'break' from the AodB axis As an example, consider a two pole transfer function specified by G(s) = 100 , (s + 1)(s + 20) p1 = -1, p2 = -20, k = 100,A o = 5, Ao dB = 14 The Bode magnitude plot consists of a horizontal line at +14 dB, from which the contrubution from the two poles at s=1 and s=20 break When adding up the responses we consider the line at +14 dB to be zero Only after the composite curve is drawn we then think of the absolute gain being +14 dB at low frequencies The Bode plot looks like the following 50 Ao=14dB -20dB/dec p=1 p=20 -40dB/dec -50 -1 10 10 Frequency (rad/sec) 10 10 -45deg/dec -90 -90deg/dec -45deg/dec -180 10 -1 10 Frequency (rad/sec) -5- 10 10 DLD 01/24/01 EE 216 Spring 2001 handout # If there are poles or zeros at zero frequency, then the dc gain is (-∞ dB) and the first representation of G(s) is perhaps the most convenient To summarize the rules for generating Bode plots: Magnitude Plot 1) For each pole, draw a straight line with slope –20 dB/decade (-6 dB/octave) intersecting the dB axis at the pole frequency This is the contribution for that pole at frequencies greater than the pole frequency 2) For each zero, draw a straight line with slope +20 dB/decade (+6 dB/octave) intersecting the dB axis at the zero frequency This is the contribution for that zero at frequencies greater than the zero frequency 3) Sum the resulting partial plots to get the overall magnitude response 4) Set the vertical scale by choosing a convenient flat-band region and relabeling the dB axis to reflect the calculated gain in that region Phase Plot 1) For each pole, draw the angle response curve with slope –45o/decade over two decades and centered at the pole frequency p Below 0.1p the angle response is 0o, and above 10p the angle response is –90o 2) For each zero, draw the angle response curve with slope +45o/decade over two decades and centered at the zero frequency z Below 0.1z the angle response is 0o, and above 10z the angle response is +90o 3) Sum the individual responses to get the composite phase response -6- DLD 01/24/01 EE 216 Spring 2001 handout # Now consider a high pass circuit: s s + 1/ RC k = 1; k dB = 0; z = 0; log(z) = −∞; p = −1/ RC C G(s) = Vin Vout R How we handle the zero at ω=0? The zero location is at -∞ on the log(ω) axis Similarly, the dc gain AodB is -∞ For this case we plot the magnitude response of the zero as a line with slope +20 dB/decade with no break point The angle response is a constant +90o since for any finite frequency the zero will contribute +90o of phase We proceed as usual for the magnitude and angle plots, except for the absolute gain of the magnitude plot To fix the absolute gain we choose a convenient frequency, typically in a region where the gain is flat, and evaluate the transfer function at that point We can adjust the axis labels to include this offset value For the high pass circuit, G(s) = looks like this: s , let 1/RC = 100 as before The Bode plot s + 1/ RC 50 -50 10 10 10 10 10 10 Frequency (rad/sec) -270 -300 -330 -360 -390 10 10 Frequency (rad/sec) -7- DLD 01/24/01 EE 216 Spring 2001 handout # Here are a few more examples of circuits and their Bode plots R1 Vin Vout C s R2 ( + 1) (s + ) 1/CR2 R1 R2 CR2 + G(s) = = s (s + ) ( + 1) C(R1+ R2) 1/C(R1 + R2) R2 let R1=9k R2=1k C= 1µF then z=1000 p=100 20 -20 10 10 10 Frequency (rad/sec) 10 10 50 -50 10 10 10 Frequency (rad/sec) -8- 10 10 DLD 01/24/01 EE 216 Spring 2001 handout # C Vin R1 Vout R=R1+R2 C R2 G(s) = If R1=1k R2=9k C=1µF then p=100, z1=0, and z2=1000, and k=0.1 ks(s − z2 ) ( s − p) -20 -40 -60 10 10 3 10 10 Frequency (rad/sec) 10 10 -270 -360 10 10 10 10 Frequency (rad/sec) -9- 10 10 DLD 01/24/01 ... zero will contribute +90o of phase We proceed as usual for the magnitude and angle plots, except for the absolute gain of the magnitude plot To fix the absolute gain we choose a convenient frequency,... contribution from a zero in dB is +20log( (s − z) ) = +20log( ω + z ) A zero at z=-1 has a magnitude response like the following z=1 40 +20 dB/decade 20 -2 10 10 -1 10 Frequency (rad/sec) 10 10... specified by G(s) = 100 , (s + 1)(s + 20) p1 = -1, p2 = -20, k = 100,A o = 5, Ao dB = 14 The Bode magnitude plot consists of a horizontal line at +14 dB, from which the contrubution from the two poles