Bode Plot - ME375 Handouts Bode Diagrams (Plots) • Bode...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ME375 Handouts Bode Diagrams (Plots) • Bode Diagrams (Plots) A unique way of plotting the frequency response function, G(jω), w.r.t. frequency ω of systems systems. Consists of two plots: – Magnitude Plot : plots the magnitude of G(jω) in decibels w.r.t. logarithmic frequency, i.e. G ( j ω ) dB = 20 log 10 G ( j ω ) v s log 10ω – Phase Plot : plots the linear phase angle of G(jω) w.r.t. logarithmic frequency, i.e. ∠ G ( jω ) v s log 1 0ω To plot Bode diagrams, one needs to calculate the magnitude and phase of the corresponding transfer function. Ex: s +1 G(s) = s 2 + 10 s School of Mechanical Engineering Purdue University ME375 Frequency Response - 1 Bode Diagrams Recall that if G( s) = Then bm s m + bm −1s m −1 +L+ b1s + b0 bm ( s − z1 )( s − z2 )L ( s − zm ) = an s n + an −1s n −1 +L+ a1s + a0 an ( s − p1 )( s − p2 )L ( s − pn ) G ( jω ) = = bm ( jω − z1 )( jω − z2 )L ( jω − z m ) an ( jω − p1 )( jω − p2 )L ( jω − pn ) bm 1 1 1 ⋅ ⋅ ⋅ ( jω − z1 ) ⋅ ( jω − z2 ) L⋅ ( jω − z m ) L⋅ an ( jω − p1 ) ( jω − p2 ) ( jω − pn ) 20 log10 ( G ( jω ) ) = 20 log10 F b I + 20 log F 1 I +L+20 log F 1 I GH ( jω − p ) JK GH a JK GH ( jω − p ) JK c ( jω − z ) h+L+20 log c ( jω − z ) h m 10 10 n +20 log10 and ∠ G ( jω ) = ∠ 1 1 10 n 1 bm ( jω − z1 )( jω − z 2 )L ( jω − z m ) an ( jω − p1 )( j ω − p2 )L ( jω − pn ) = ∠ ( jω − z1 ) + ∠ ( jω − z 2 ) +L +∠ ( jω − z m ) − ∠ ( jω − p1 ) − ∠ ( jω − p2 )L −∠ ( jω − pn ) School of Mechanical Engineering Purdue University ME375 Frequency Response - 2 1 ME375 Handouts Example Ex: Find the magnitude and the phase of the following transfer functions: G ( s) = ( jω ) + 1 s +1 ⇒ G ( jω ) = s 2 + 10 s jω ( jω + 10) G ( jω ) = ∠G ( jω ) = G( s ) = 3s 3 + 12 s 2 + 9s = 2 s 3 + 22 s 2 + 76s + 80 ( )( ) School of Mechanical Engineering Purdue University ME375 Frequency Response - 3 Bode Diagram Building Blocks • 1st Order Real Poles 0 -3 Transfer Function: τ s+1 Frequency Response: G p1 ( jω ) = 1 p1 -20 τ jω + 1 , = τ >0 1 R G ( jω ) | S | ∠ G ( jω ) T p1 τ>0 , τ 2ω 2 + 1 = − a tan2 (τω ,1) = − tan − 1 τ ω af Phase (deg); Magnitude (dB) 1 G p1 ( s ) = -40 0 -45 -90 0.01/τ 0.1/τ 1/τ 10/τ 100/τ Frequency (rad/sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 4 2 ME375 Handouts Example • 1st Order Real Poles Transfer Function: 20 50 G (s) = s+5 Plot the straight line approximation of G(s)’s Bode diagram: 15 10 Phase (deg); Magnitude (dB) 5 0 0 -45 -90 -1 0 10 1 10 2 10 10 Frequency (rad/sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 5 Bode Diagram Building Blocks 40 • 1st Order Real Zeros Transfer Function: RG | S∠ G | T z1 ( jω ) = τ 2ω 2 + 1 z1 ( jω ) = a tan2 ( τω ,1) = tan − 1 τω af 20 Phase (deg); Magnitude (dB) G z1 ( s ) = τ s + 1 , τ > 0 Frequency Response: G z1 ( jω ) = τ jω + 1 , τ > 0 3 0 90 45 0 0.01/τ 0.1/τ 1/τ 10/τ 100/τ Frequency (rad/sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 6 3 ME375 Handouts Example • 1st Order Real Zeros 20 Transfer Function: 15 G ( s ) = 0 .7 s + 0 .7 Plot the straight line approximation of G(s)’s Bode diagram: 10 Phase (deg); Magnitude (dB) 5 0 90 45 0 -1 0 10 10 School of Mechanical Engineering Purdue University 1 10 Frequency (rad/sec) 2 10 ME375 Frequency Response - 7 Example • Lead Compensator 30 20 10 0 Phase (deg); Magnitude (dB) Transfer Function: 35s + 35 G (s) = s+5 Plot the straight line approximation of G(s)’s Bode diagram: -10 90 0 -90 -1 10 0 10 1 10 2 10 Frequency (rad/sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 8 4 ME375 Handouts 1st Order Bode Diagram Summary • 1st Order Poles 1 , τ s +1 – Break Frequency 1 ωb = [ rad/s] G p1 ( s ) = • 1st Order Zeros Gz 1 ( s ) = τ s + 1 , τ > 0 τ >0 – Break Frequency Frequency 1 ωb = [ rad/s] τ τ – Mag. Plot Approximation 0 dB from DC to ωb and a straight line with −20 dB/decade slope after ωb. – Phase Plot Approximation 1 0 deg from DC to 10 ω b . Between 1 ω and 10 b straight 10ωb , a straight line from 0 deg to −90 deg (passing −45 deg at ωb). For frequency 45 higher than 10ωb , straight line on −90 deg. – Mag. Plot Approximation 0 dB from DC to ωb and a straight line with 20 dB/decade slope after ωb. – Phase Plot Approximation 1 0 deg from DC to 10 ω b . Between 1 ω 10 b and 10 and 10ωb , a straight line from 0 deg to 90 straight line from deg to 90 deg deg (passing 45 deg at ωb). For frequency higher than 10ωb , straight line on 90 deg. Note: By looking at a Bode diagram you should be able to determine the relative order of the system, its break frequency, and DC (steady-state) gain. This process should also be reversible, i.e. given a (steadytransfer function, be able to plot a straight line approximated Bode diagram. School of Mechanical Engineering Purdue University ME375 Frequency Response - 9 Bode Diagram Building Blocks • Integrator (Pole at origin) Transfer Function: 1 G p0 (s) = s Frequency Response: 0 p0 p0 = 1 ω = − 90 ° Phase (deg); Magnitude (dB) -20 1 G p 0 ( jω ) = jω R G ( jω ) | S | ∠ G ( jω ) T 20 -40 -60 0 -45 -90 -135 0.1 1 10 100 1000 Frequency (rad/sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 10 5 ME375 Handouts Bode Diagram Building Blocks • Differentiator (Zero at origin) 60 Transfer Function: 40 Gz0 (s) = s Phase (deg); Magnitude (dB) 20 Frequency Response: G z 0 ( jω ) = jω RG | S∠ G | T z0 ( jω ) =ω z0 ( jω ) = 90 ° 0 -20 135 90 45 0 0.1 1 10 100 1000 Frequency (rad/sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 11 Example • Combination of Systems Transfer Function: 30 35s + 35 G (s) = s(s + 5) 20 10 0 Phase (deg); Magnitude (dB) Plot the straight line approximation of G(s)’s Bode diagram: -10 90 0 -90 -1 10 0 1 10 10 2 10 Frequency (rad/sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 12 6 ME375 Handouts Example • Combination of Systems 40 Transfer Function: 2500 s(s2 + 55s + 250 ) 0 Plot the straight line approximation of G(s)’s Bode diagram: -40 Phase (deg); Magnitude (dB) G (s) = -80 -120 0 -90 -180 -270 -1 10 10 0 1 10 2 10 3 10 Frequency (rad/sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 13 Bode Diagram Building Blocks • 2nd Order Complex Poles 20 Transfer Function: 0 ω n2 Gp2 (s) = 2 s + 2ζω n s + ω n 2 , 1≥ ζ ≥ 0 -40 G p 2 ( jω ) = Phase (deg); Magnitude (dB) Frequency Response: G p 2 ( jω ) = 1 2ζω ⎛ ω2 ⎞ ⎟ j + ⎜1 − ω n ⎜ ω n2 ⎟ ⎠ ⎝ 1 4 ζ 2ω 2 ωn 2 -20 ⎛ ω2 ⎞ + ⎜1 − ⎜ ω 2⎟ ⎟ ⎝ n⎠ 2 -60 -80 0 -90 ⎡ζ ω2 ⎟⎤ ⎜ ∠ G p 2 ( jω ) = − tan −1 ⎢ 2ω ω ⎛ 1 − ω 2 ⎞ ⎥ n⎠ ⎣n ⎝ ⎦ -180 0.01ωn 0.1ωn ωn 10ωn 100ωn Frequency (rad/sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 14 7 ME375 Handouts Example • Second-Order System Second- 40 Transfer Function: 2500 s2 + 10s + 2500 Plot the straight line approximation of G(s)’s Bode diagram: 0 G (s) = Phase (deg); Magnitude (dB) -40 -80 -120 0 -90 -180 10 0 1 10 2 10 10 3 4 10 Frequency (rad/sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 15 Bode Diagram Building Blocks • 2nd Order Complex Zeros 80 Phase (deg); Magnitude (dB) Transfer Function: 2 s 2 + 2ζ ω n s + ω n Gz 2 ( s ) = , 1≥ ζ ≥ 0 2 ωn Frequency Response: 2ζ ω ⎛ ω2 ⎞ ⎟ G z 2 ( jω ) = j + ⎜1 − ⎜ ω n ⎝ ω n2 ⎟ ⎠ G z 2 ( jω ) = 1 G p 2 ( jω ) = 4 ζ 2ω 2 ωn 2 ⎛ ω2 ⎞ ⎟ + ⎜1 − ⎜ ω n2 ⎟ ⎝ ⎠ 2 60 40 20 0 -20 180 90 ∠ G z 2 ( j ω ) = −∠ G p 2 ( j ω ) ⎡ζ = tan −1 ⎢ 2ω ω ⎣n ⎛1 − ⎜ ⎝ ω2 ω n2 ⎞⎤ ⎟⎥ ⎠⎦ 0 0.01ωn 0.1ωn School of Mechanical Engineering Purdue University ωn Frequency (rad/sec) 10ωn 100ωn ME375 Frequency Response - 16 8 ME375 Handouts Bode Diagrams of Poles and Zeros Let 1 G p (s) = Gz ( s ) R G ( jω ) = 1 | G ( jω ) ⇒S | ∠ G ( jω ) = −∠ G ( jω ) T R20 log e G ( jω ) j = − 20 log c G ( jω ) h | ⇒S | ∠ G ( jω ) = −∠ G ( jω ) T 40 Magnitude (dB) Bode Bode Diagrams of stable complex zeros are the mirror images of the Bode diagrams of the identical stable complex poles w.r.t. the 0 dB line and the 0 deg line, respectively. 20 0 -20 -40 p z z 10 p p 10 z 180 Phase (deg) p 0 z -180 0.1ωn ωn 10ωn Frequency (rad/sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 17 2nd Order Bode Diagram Summary • 2nd Order Complex Poles • 2nd Order Complex Zeros ωn , 1≥ ζ > 0 2 s + 2 ζω n s + ω n – Break Frequency ω b = ω n [ rad/s] – Mag. Plot Approximation 2 G p 2 ( s) = 2 0 dB from DC to ωn and a straight line with −40 dB/decade slope after ωn. Peak value occurs at: ω r = ω n 1 − 2ζ 2 1 ⇒ G p 2 ( jω r ) MAX = 2ζ 1 − ζ 2 – Phase Plot Approximation 0 deg from DC to ( 1 5 )ζ ω n . Between ( 1 5 )ζ ω n straight and 5ζ ωn , a straight line from 0 deg to −180 deg (passing −90 deg at ωn). For frequency higher than 5ζ ωn , straight line on −180 deg. Gz 2 ( s ) = s 2 + 2ζωn s + ωn 2 ωn 2 , 1≥ ζ > 0 – Break Frequency ωb = ω n [ rad/s] – Mag. Plot Approximation 0 dB from DC to ωn and a straight line with 40 dB/decade slope after ωn. – Phase Plot Approximation 0 deg from DC to ( 1 5 )ζ ω n . Between ( 1 5 )ζ ω n deg from DC to Between straight and and 5ζ ωn , a straight line from 0 deg to 180 deg (passing 90 deg at ωn). For frequency higher than 5ζ ωn , straight line on 180 deg. School of Mechanical Engineering Purdue University ME375 Frequency Response - 18 9 ME375 Handouts 2nd Order System Frequency Response A Closer Look: G p 2 ( s) = ωn 2 2 s + 2 ζω n s + ω n 2 , 1≥ ζ ≥ 0 Frequency Response Function: G p 2 ( jω ) = Magnitude: G p 2 ( jω ) = ωn 2 1 = 2 ( jω ) 2 + 2ζω n ( jω ) + ω n ω2 ⎞ 2 ζω ⎛ + ⎜1 − 2 ⎟ j ωn ⎝ ωn ⎠ Phase: 2 ⎛ 2 ζω ⎞ ⎛ ω ⎞ ⎜ ω ⎟ + ⎜1 − 2 ⎟ ⎝ n ⎠ ⎝ ωn ⎠ 2 ( ) ζ ω2 ∠G p 2 ( jω ) = −∠ ⎡ 1 − ω 2 + j 2ωnω ⎤ ⎢ ⎥ n ⎣ ⎦ 2 ⎡ 2 ζω , 1 − ω 2 ⎤ = − atan2 ⎢ ωn ωn ⎥ ⎣ ⎦ 1 2 ( ) The maximum value of |G(jω)| occurs at the Peak (Resonant) Frequency ωr : Peak ω r = ω n 1 − 2ζ 2 and G p 2 ( jω r ) = School of Mechanical Engineering Purdue University 1 2ζ 1 − ζ 2 ME375 Frequency Response - 19 2nd Order System Frequency Response 40 20 Phase (deg); Magnitude (dB) 0 -20 -40 -60 0 -45 -90 -135 -180 0.1ωn ωn 10ωn Frequency (rad/sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 20 10 ME375 Handouts 2nd Order System Frequency Response A Few Observations: • Three different characteristic frequencies: different – Natural Frequency (ωn) Frequency – Damped Natural Frequency (ωd): ω d = ω n 1 − ζ 2 – Resonant (Peak) Frequency (ωr): ω r = ω n 1 − 2ζ 2 ωr ≤ωd ≤ωn • When the damping ratio ζ > 0.707, there is no peak in the Bode magnitude plot. 0.707 DO NOT confuse this with the condition for over-damped and under-damped overundersystems: when ζ < 1 the system is under-damped (has overshoot) and when ζ > 1 th (h the the system is over-damped (no overshoot). over• As ζ → 0 , ωr → ωn and ⏐G(jω)⏐ΜΑΧ increases; also the phase transition from 0 deg to −180 deg becomes sharper. School of Mechanical Engineering Purdue University ME375 Frequency Response - 21 Example • Combination of Systems Transfer Function: G (s) = 2000(s2 + s + 25) s( s + 2 0 0 )( s 2 + 1 0 s + 2 5 0 0 ) Plot the straight line approximation of G(s)’s Bode diagram: School of Mechanical Engineering Purdue University ME375 Frequency Response - 22 11 ME375 Handouts Example 40 Magnitude (dB) 20 0 -20 -40 -60 -80 180 Phase (deg) 90 0 -90 -180 -270 -1 10 10 0 1 10 Frequency (rad/sec) School of Mechanical Engineering Purdue University 10 2 10 3 ME375 Frequency Response - 23 12 ...
View Full Document

This note was uploaded on 12/22/2011 for the course ME 375 taught by Professor Meckle during the Fall '10 term at Purdue University-West Lafayette.

Ask a homework question - tutors are online