Overview of Bode Plots Transfer function review Piecewise linear approximations Firstorder terms Secondorder terms (complex poles & zeros)


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1 Overview of Bode Plots Transfer function review Piecewise linear approximations Firstorder terms Secondorder terms (complex poles & zeros) J. McNames Portland State University ECE 222 Bode Plots Ver
2 Transfer Function Review x(t) H(s) y(t) Recall that if H(s) is known and x(t) =A cos(ωt + φ), then we can find the steadystate solution for y(t): y ss (t) =A H(jω) cos (ωt + φ + H(jω)) J. McNames Portland State University ECE 222 Bode Plots Ver
3 x(t) Bode Plots H(s) y(t) Bode plots are standard method of plotting the magnitude and phase of H(s) Both plots use a logarithmic scale for the xaxis Frequency is in units of radians/second (rad/s) The phase is plotted on a linear scale in degrees Magnitude is plotted on a linear scale in decibels H db (jω) 2 log 1 H(jω) J. McNames Portland State University ECE 222 Bode Plots Ver
4 Decibel Scales It is important to become adept at translating between amplitude, H(jω), and decibels, H db (jω). Amplitude ( H(jω) ) Decibels (2 log 1 H(jω) ) 1 2 log 1 1 = 1 2 log 1 1 = 1 2 log 1 1 = 1 2 log 1 1 =.1 2 log 1.1 =.1 2 log 1.1 =.1 2 log 1.1 = log 1 2 = log 1 2 = log 1 2 = J. McNames Portland State University ECE 222 Bode Plots Ver
5 Example 1: Bode Plots 2 nf 1kΩ 1 kω v s (t) R L + v o (t)  1. Find the transfer function of the circuit shown above. 2. Generate the bode plot. J. McNames Portland State University ECE 222 Bode Plots Ver
6 Example 1: Workspace J. McNames Portland State University ECE 222 Bode Plots Ver
7 Example 1: Bode Plot Active Lowpass RC Filter H(jω) (db) H(jω) (degrees) Frequency (rad/s) J. McNames Portland State University ECE 222 Bode Plots Ver
8 w = logspace(1,5,5); H = 5e3./(j*w + 5e3); subplot(2,1,1); h = semilogx(w,2*log1(abs(h))); set(h, LineWidth,1.4); ylabel( H(j\omega) (db) ); title( Active Lowpass RC Filter ); set(gca, Box, Off ); grid on; set(gca, YLim,[5 25]); Example 1: MATLAB Code subplot(2,1,2); h = semilogx(w,angle(h)*18/pi); set(h, LineWidth,1.4); ylabel( \angle H(j\omega) (degrees) ); set(gca, Box, Off ); grid on; set(gca, YLim,[85 185]); xlabel( Frequency (rad/s) ); J. McNames Portland State University ECE 222 Bode Plots Ver
9 Example 2: Bode Plots 2 μf 1kΩ 1 μf 1kΩ v s (t) + R L  1. Find the transfer function of the circuit shown above. 2. Generate the bode plot. J. McNames Portland State University ECE 222 Bode Plots Ver
10 Example 2: Workspace J. McNames Portland State University ECE 222 Bode Plots Ver
11 H(jω) (degrees) H(jω) (db) Example 2: Bode Plot Active Lead/Lag RC Filter (rad/s) Frequency J. McNames Portland State University ECE 222 Bode Plots Ver
12 w = logspace(1,5,5); H = 2*(j*w+5)./(j*w + 1); Example 2: MATLAB Code subplot(2,1,1); h = semilogx(w,2*log1(abs(h))); set(h, LineWidth,1.4); ylabel( H(j\omega) (db) ); title( Active Lead/Lag RC Filter ); set(gca, Box, Off ); grid on; set(gca, YLim,[2 8]); subplot(2,1,2); h = semilogx(w,angle(h)*18/pi); set(h, LineWidth,1.4); ylabel( \angle H(j\omega) (degrees) ); set(gca, Box, Off ); grid on; set(gca, YLim,[1816]); xlabel( Frequency (rad/s) ); J. McNames Portland State University ECE 222 Bode Plots Ver
13 Bode Plot Approximations Until recently (late 198 s) bode plots were drawn by hand There were many rulesofthumb, tables, and template plots to help Today engineers primarily use MATLAB, or the equivalent Why discuss the old method of plotting by hand? It is still important to understand how the poles, zeros, and gain influence the Bode plot These ideas are used for transfer function synthesis, analog circuit design, and control systems We will discuss simplified methods of generating Bode plots Based on asymptotic approximations J. McNames Portland State University ECE 222 Bode Plots Ver
14 Alternate Transfer Function Expressions There are many equivalent expressions for transfer functions. H(s) = N(s) D(s) = b ms m + b m 1 s m b 1 s + b a n s n + a n 1 s n a 1 s + a = b m s ±l (s z 1)(s z 2 )...(s z m ) a n (s p 1 )(s p 2 )...(s p n ) ( )( ) ( ) 1 s = ks ±l z 1 1 s z s z m ( )( ) ( ) 1 s p 1 1 s p s p n This last expression is called standard form The first step in making bode plots is to convert H(s) to standard form J. McNames Portland State University ECE 222 Bode Plots Ver
15 Magnitude Components Consider the expression for the transfer function magnitude: H db (jω) = 2log 1 H(jω) = 2log 1 k s±l (1 s z 1 )...(1 s (1 s p 1 )...(1 s p n ) z m ) s=jω jω 1 ±l z = 2log 1 k jω jω z m 1 jω p jω p n = 2log 1 k ±l2 log 1 ω +2 log 1 1 jω z log 1 1 jω z m 2 log 1 1 jω p 1 2 log 1 1 jω p n J. McNames Portland State University ECE 222 Bode Plots Ver
16 Magnitude Components Comments H db (ω) = 2log 1 k ±l2 log 1 ω +2 log 1 1 jω z log 1 1 jω z m 2 log 1 1 jω p 1 2 log 1 1 jω p n Thus, H db (ω) can be written as a sum of simple functions This is similar like using basis functions {δ(t),u(t),& r(t)} to write an expression for a piecewise linear signal We will use this approach to generate our piecewise linear approximations of the bode plot Note that there are four types of components in this expression Constant Linear term Zeros Poles J. McNames Portland State University ECE 222 Bode Plots Ver
17 Magnitude Components: Constant H(jω) (db) ω (rad/sec) 4 The constant term, 2 log 1 k, is a straight line on the Bode plot. J. McNames Portland State University ECE 222 Bode Plots Ver
18 Magnitude Components: Linear Term H(jω) (db) ω (rad/sec) 4 The linear term, ±l 2 log 1 ω, is a line on the magnitude plot with a slope equal to ±l 2 db per decade. The xaxis intercept occurs at ω =1rad/s. Plot the bode magnitude plots for H(s) =s, 1 s,s2, 1 s 2. J. McNames Portland State University ECE 222 Bode Plots Ver
19 Magnitude Components: Real Zeros Consider two limiting conditions for a term containing a zero, 2 log 1 1 jω z First condition: ω z lim 2 log ω z 1 1 jω z = Thus, if ω z 1, then2 log 1 Second condition: ω z 1 jω z ω lim z 2 log 1 1 jω z. =2log1 jω z =2log 1 ω 2 log 1 z ω Thus, if z 1, then this term is linear (on a log scale) with a slope of 2 db per decade and an xaxis intercept at ω = z. J. McNames Portland State University ECE 222 Bode Plots Ver
20 Magnitude Components: Real Zeros Continued H(jω) (db) ω (rad/sec) 4 Our piecewise approximation joins these two linear asymptotic approximations at ω = z. Plot the piecewise approximation of the term 2 log 1 1 jω z. J. McNames Portland State University ECE 222 Bode Plots Ver
21 Magnitude Components: Real Zeros Continued 2 5 Bode Magnitude Real Zero: z = ±1 rad/s 4 3 Mag (db) Frequency (rad/sec) The approximation is least accurate at ω = z. The true magnitude is 3 db higher than the approximation at this corner frequency. J. McNames Portland State University ECE 222 Bode Plots Ver
22 Magnitude Components: Real Poles H(jω) (db) ω (rad/sec) 4 Consider two limiting conditions for a term containing a pole, 2 log 1 1 jω p This is just the negative of the expression for a zero The piecewise approximation is the mirror image of that for a zero J. McNames Portland State University ECE 222 Bode Plots Ver
23 Magnitude Components: Real Poles Continued 1 Bode Magnitude Real Pole: p = 1 rad/s 1 Mag (db) Frequency (rad/sec) The approximation is least accurate at ω = p. The true magnitude is 3 db less than the approximation at this corner frequency. J. McNames Portland State University ECE 222 Bode Plots Ver
24 Complex Poles & Zeros Complex poles and zeros require special attention Will discuss later You will not be expected to plot approximations with complex poles or zeros on exams There are essentially 3 steps to generating piecewise linear approximations of bode plots 1. Convert to standard form 2. Plot the components 3. Graphically add the components together J. McNames Portland State University ECE 222 Bode Plots Ver
25 Example 3: Magnitude Components Draw the piecewise approximation of the bode magnitude plot for H(s) = (s + 1)(s + 1)2 1s 2 (s + 1) J. McNames Portland State University ECE 222 Bode Plots Ver
26 Example 3: Workspace H(jω) 6 (db) ω (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
27 Example 3: Solution Bode Magnitude Example Mag (db) Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
28 Example 4: Magnitude Components Draw the piecewise approximation of the bode magnitude plot for H(s) = 1 11 s(s + 1) (s + 1)(s + 1)(s +1, ) 2 J. McNames Portland State University ECE 222 Bode Plots Ver
29 Example 4: Workspace H(jω) 6 (db) ω (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
30 Example 4: Solution Mag (db) Bode Magnitude Example Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
31 Phase Components H(jω) = k (jω)l (1 jω z 1 )...(1 jω z m ) (1 jω p 1 )...(1 jω p n ) Each of these terms can be expressed in polar form: a + jb = Ae jθ. Note that (jω) l = ω l j l = ω l (e j π 2 ) l = ω l e j π 2 l H(jω) = k e jη kπ (ωl e jl π 2 )N 1 e jθ 1...N m e jθ m D 1 e jφ 1...Dn e jφ n = k ω l N 1...N m D 1...D n exp ( j(η k π + l π 2 + θ θ n φ 1 φ n ) ) where η k = { k 1 k< J. McNames Portland State University ECE 222 Bode Plots Ver
32 Phase Components: Comments H(jω) = η k π + l π 2 + θ θ n φ 1 φ n ( = η k π + l π jω ) ( jω ) z 1 z m ( 1 jω ) ( 1 jω ) p 1 p n Thus the phase of H(jω) is also a linear sum of the phases due to each component We will consider each of the four components in turn Constant Linear term Zeros Poles J. McNames Portland State University ECE 222 Bode Plots Ver
33 H(jω) 18 9 Phase Components: Constant (deg) 9 ω (rad/sec) 18 The complex angle of the constant term, k, is either if k> or 18 if k<. J. McNames Portland State University ECE 222 Bode Plots Ver
34 H(jω) 18 9 Phase Components: Linear Term (deg) 9 ω (rad/sec) 18 The linear term, (jω) l = j l = l 9, is a constant multiple of 9 Plot the bode phase plots for H(s) =s, 1 s,s2, 1 s 2. J. McNames Portland State University ECE 222 Bode Plots Ver
35 Phase Components: Real Zeros Consider three limiting conditions for a term containing a zero, 1 jω z First condition: ω z lim ( 1 jω ) ω z z = ). ω Thus, if z 1, then ( 1 jω z Second condition: ω = z ( ) 1 jω ω= z z = (1 jη z )= η z 45 where η z = sign(z) Third condition: ω z ω lim z ( 1 jω z Thus, if ω z 1, then ( 1 jω z ) = ( jω z ) ηz 9. ) = ηz 9 J. McNames Portland State University ECE 222 Bode Plots Ver
36 Phase Components: Real Zeros Continued H(jω) 18 9 (deg) 9 ω (rad/sec) 18 Our piecewise approximation joins these three linear asymptotic approximations at ω =1 1 z and ω =1 z. Plot the piecewise approximation of the term ( 1 jω ) z. Assume that z is in the lefthand plane (i.e. Re{z} < ) J. McNames Portland State University ECE 222 Bode Plots Ver
37 Phase Components: Real Zeros in Left Plane Phase (deg) Frequency (rad/sec) The approximation is least accurate at ω =.1 z and ω =1 z. J. McNames Portland State University ECE 222 Bode Plots Ver
38 Phase Components: Real Zeros Continued 2 H(jω) 18 9 (deg) 9 ω (rad/sec) 18 If the zero is in the right half plane (i.e. Re{z} > ), then the phase approaches 9 asymptotically. Plot the piecewise approximation of the term ( 1 jω ) z. Assume that z is in the right half plane. J. McNames Portland State University ECE 222 Bode Plots Ver
39 Phase Components: Real Zeros in Right Plane Phase (deg) Frequency (rad/sec) The approximation is least accurate at ω =.1 z and ω =1 z. J. McNames Portland State University ECE 222 Bode Plots Ver
40 Phase Components: Real Poles Real poles in the left half plane have the same phase as real zeros in the right half plane. We will only discuss poles in the left half plane because only these systems are stable. First condition: ω p lim ω p ( 1 jω p ) = Second condition: ω = p ( 1 jω p = (1 sign(p) j) = (1 + j) = 45 ) ω= p Third condition: ω p ( lim 1 jω ω p p ) = ( jω p ) = j = 9 J. McNames Portland State University ECE 222 Bode Plots Ver
41 Phase Components: Real Poles H(jω) 18 9 (deg) 9 ω (rad/sec) 18 Plot the piecewise approximation of the term that p is in the lefthand plane (i.e. Re{p} < ) ( 1 jω p ). Assume J. McNames Portland State University ECE 222 Bode Plots Ver
42 Phase Components: Real Poles in Left Plane Phase (deg) Frequency (rad/sec) The approximation is least accurate at ω =.1 p and ω =1 p. J. McNames Portland State University ECE 222 Bode Plots Ver
43 Example 5: Phase Components H(jω) (deg) (rad/sec) Draw the piecewise approximation of the bode phase plot for H(s) = (s + 1)(s + 1)2 1s 2 (s + 1) J. McNames Portland State University ECE 222 Bode Plots Ver
44 Example 5: Solution 5 Phase (deg) Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
45 Example 6: Phase Components H(jω) (deg) (rad/sec) Draw the piecewise approximation of the bode phase plot for H(s) = 1 11 s(s + 1) (s + 1)(s + 1)(s +1, ) 2 J. McNames Portland State University ECE 222 Bode Plots Ver
46 Example 6: Solution 1 5 Phase (deg) Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
47 Example 7: Magnitude H(jω) 6 (db) ω (rad/sec) Plot the magnitude of H(s) =1 s +1 s + 1 J. McNames Portland State University ECE 222 Bode Plots Ver
48 Example 7: Phase H(jω) (deg) (rad/sec) Plot the phase of H(s) =1 s +1 s + 1 J. McNames Portland State University ECE 222 Bode Plots Ver
49 Example 7: Solution 4 Bode Example 1 Mag (db) Phase (deg) Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
50 Example 8: Magnitude H(jω) 6 (db) ω (rad/sec) Plot the magnitude of H(s) = 1 s 1 s + 1 J. McNames Portland State University ECE 222 Bode Plots Ver
51 Example 8: Phase H(jω) (deg) (rad/sec) Plot the phase of H(s) = 1 s 1 s + 1 J. McNames Portland State University ECE 222 Bode Plots Ver
52 Example 8: Solution 4 Bode Example 2 Mag (db) Phase (deg) Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
53 1 mh 1 mf Example 9: Circuit Example + v s (t) 11 Ω v o (t)  Draw the straightline approximations of the transfer function for the circuit shown above. Hint: Recall from one of the Transfer Functions Examples R L H(s) = s s 2 + R L s + 1 LC J. McNames Portland State University ECE 222 Bode Plots Ver
54 Example 9: Magnitude H(jω) 6 (db) ω (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
55 Example 9: Phase H(jω) (deg) (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
56 Example 9: Solution 2 Bode Example 3 Mag (db) Phase (deg) Frequency 1 2 (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
57 Complex Poles Complex poles can be expressed in the following form: C(s) = ω 2 n s 2 +2ζω n s + ω 2 n = 1 1+2ζ s ω n + ( ) 2 s ω n ω n is called the undamped natural frequency ζ (zeta) is called the damping ratio The poles are p 1,2 =( ζ ± ζ 2 1) ω n If ζ 1, the poles are real If <ζ<1, the poles are complex If ζ =, the poles are imaginary: p 1,2 = ±jω n If ζ<, the poles are in the right half plane (Re{p} > ) and the system is unstable J. McNames Portland State University ECE 222 Bode Plots Ver
58 Complex Poles Continued The transfer function C(s) can also be expressed in the following form C(s) = 1 1+2ζ s ω n + ( ) 2 = s ω n 1 1+ s Qω n + ( ) 2 s ω n where Q 1 2ζ The meaning of Q, thequality factor, will become clear in the following slides. 1 C(jω)= 1 ( ) 2 ω ω n + jω Qω n J. McNames Portland State University ECE 222 Bode Plots Ver
59 Complex Poles Magnitude 2 log 1 C(jω) = 2log 1 1 ( ) 2 ω + jω ω n Qω n 1 ( ) 2 = 2 log 1 1 ω2 ωn 2 + ( ω Qω n ) 2 For ω ω n, 2 log 1 C(jω) 2 log 1 1 =db For ω ω n, 2 log 1 C(jω) 2 log 1 ω 2 ω 2 n = 4 log 1 ω ω n db At these extremes, the behavior is identical to two real poles. J. McNames Portland State University ECE 222 Bode Plots Ver
60 For ω = ω n, Complex Poles Magnitude Continued ( ) 2 2 log 1 C(jω) = 2 log 1 1 ω2 ωn 2 + ( ω Qω n 2 log 1 C(jω n ) = 2 log 1 1 Q =2log 1 Q = Q db ) 2 J. McNames Portland State University ECE 222 Bode Plots Ver
61 Mag (db) Complex Poles Magnitude Complex Poles Q =.1 Q =.5 Q =.77 Q = 1 Q = 2 Q = 1 Q = Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
62 For ω ω n, For ω ω n, For ω = ω n, Complex Poles Phase 1 C(jω) = ( ) 2 ω 1 ω n + jω Qω n C(jω) 1 ω Qω n C(jω) 1 = = Qω n ω = 1= 18 C(jω) 1 Qj = 1 j = j = 9 At the extremes, the behavior is identical to two real poles. At other values of ω near ω n, the behavior is more complicated. J. McNames Portland State University ECE 222 Bode Plots Ver
63 Phase (deg) Complex Poles Phase Complex Poles Q =.1 Q =.5 Q =.77 Q = 1 Q = 2 Q = 1 Q = Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
64 Complex Zeros Left half plane: Inverted magnitude of complex poles Inverted phase of complex poles Right half plane: Inverted magnitude of complex poles Same phase of complex poles This is the same relationship real zeros had to real poles J. McNames Portland State University ECE 222 Bode Plots Ver
65 Complex Zeros Magnitude 8 Complex Zeros 6 4 Mag (db) Q =.1 Q =.5 Q =.77 Q = 1 Q = 2 Q = 1 Q = Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
66 Complex Zeros Phase 18 Complex Zeros Phase (deg) Q =.1 Q =.5 Q =.77 Q = 1 Q = 2 Q = 1 Q = Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
67 Complex Poles Maximum What is the frequency at which C(jω) is maximized? C(jω) = C(jω) = 1 ( ) ( ) 2 1+ jω Qω n + jω ω n 1 ( ) 2 ( 1 ω2 + ω 2 n ) 2 ω Qω n For high values of Q, the maximum of C(jω) > 1 This is called peaking The largest Q before the onset of peaking is Q = This curve is said to be maximally flat This is also called a Butterworth response In this case, C(jω n ) = 3 db and ω n is the cutoff frequency J. McNames Portland State University ECE 222 Bode Plots Ver
68 Complex Poles Maximum Continued If Q>.77, the maximum magnitude and frequency are as follows: ω r = ω n 1 1 Q 2Q 2 C(jω r ) = 1 1 4Q 2 ω r is called the resonant frequency or the damped natural frequency As Q, ω r ω n For sufficiently large Q (say Q>5) ω r ω n C(jω r ) Q Peaked responses are useful in the synthesis of highorder filters Complex zeros (in the left half plane) have the inverted magnitude and phase of complex poles J. McNames Portland State University ECE 222 Bode Plots Ver
69 5 mh R Complex Poles Example + v s (t) 2 nf v o (t)  Generate the bode plot for the circuit shown above. H(s) = 1 LC s 2 + R L s + 1 LC = ω 2 n s 2 +2ζω n s + ω 2 n = ω 2 n s 2 + ω n Q s + ω 2 n J. McNames Portland State University ECE 222 Bode Plots Ver
70 Complex Poles Example Continued 1 1 ω n = =1k rad/s LC ζ = R R LC = CL = R.1 2L 2 1 L L Q = LC R = 1 C R = 5 R R =5Ω ζ =.5 Q = 1 Very Light Damping R =5Ω ζ =.5 Q =1 Light Damping R = 77 Ω ζ =1.41 Q =.77 Strong Damping R =1kΩ ζ =1 Q =.5 Critical Damping R =5kΩ ζ =5 Q =.1 Over Damping J. McNames Portland State University ECE 222 Bode Plots Ver
71 Complex Poles Example Continued 2 4 Resonance Example Mag (db) R = 5 Ω R = 5 Ω 8 R = 77 Ω R = 1 kω 1 R = 5 kω R = 5 kω Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
72 Complex Poles Example Continued 3 Resonance Example 2 4 Phase (deg) R = 5 Ω R = 5 Ω R = 77 Ω R = 1 kω R = 5 kω R = 5 kω Frequency (rad/sec) J. McNames Portland State University ECE 222 Bode Plots Ver
73 Complex Poles Example Continued 4 Output (V) Resonance Example R = 5 Ω R = 5 Ω R = 77 Ω R = 1 kω R = 5 kω R = 5 kω Time (sec) J. McNames Portland State University ECE 222 Bode Plots Ver
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