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Unformatted text preview: ME575 Handouts Frequency Response • • • • Forced Response to Sinusoidal Inputs Frequency Response of LTI Systems Bode Plots Quantify TF Modeling Errors and Bounds ME575 Session 7 – Frequency Response School of Mechanical Engineering Purdue University Slide 1 Frequency Response ME575 Session 7 – Frequency Response School of Mechanical Engineering Purdue University Slide 2 1 ME575 Handouts Forced Response to Sinusoidal Inputs Ex: Let’s find the forced response of a stable first order system: Let’ & y + 3 y = 6u to a sinusoidal input: u (t ) = 5sin(4t ) Y (s) = G (s) ⋅U (s) – Forced response: where G ( s ) = and ∴ – PFE: Y (s) = Y ( s) = A1 U ( s ) = L [5sin(4t ) ] = + A2 ⋅ + A3 ⋅ – Compare coefficients to find A1, A2 and A3 : ME575 Session 7 – Frequency Response School of Mechanical Engineering Purdue University Slide 3 Forced Response to Sinusoidal Inputs Ex: (cont.) – Use ILT to find y(t) : ⎡ y (t ) = L −1 [Y ( s ) ] = L −1 ⎢ ⎣ ⋅ + ⋅ + ⋅ ⎤ ⎥ ⎦ = Useful Formula: A sin(ω ⋅ t ) + B cos(ω ⋅ t ) = A + B sin(ω ⋅ t + φ ) Where φ = atan2( B, A) = ∠( A + jB) 2 2 – Using this formula, the forced response can be represented by −3 y (t ) = 1 24 t + 4 ⋅ e3 ME575 Session 7 – Frequency Response ⋅ sin(4t + φ ) 1442443 School of Mechanical Engineering Purdue University Slide 4 2 ME575 Handouts Frequency Response Ex: Let’s revisit the same example where Let’ & y + 3y = 6u and the input is a general sinusoidal input: Asin(ω t). 6 Aω 6 Aω ⋅ = ⋅ s + 3 s2 +ω2 s + 3 (s − jω)(s + jω) B B B Y (s) = 1 + 2 + 3 s + 3 s − jω s + jω Y (s) = G(s) ⋅U(s) = – Instead of comparing coefficients, use the residue formula to find residue to 6 Aω Bi’s: B = (s + 3)Y(s) = (s + 3) = 1 s =−3 (s + 3) s2 +ω2 s =−3 B2 = (s − jω)Y(s) s = jω = (s − jω)G(s) Aω = s +ω2 s = jω B3 = (s + jω)Y(s) s =− jω = (s + jω)G(s) ME575 Session 7 – Frequency Response 2 Aω = s +ω2 s =− jω 2 School of Mechanical Engineering Purdue University Slide 5 Frequency Response Ex: (Cont.) 6 Aω 32 + ω 2 6 A A A B2 = ⋅ = ⋅ G ( jω ) = ⋅ 2 j jω + 3 2 j 2j −A 6 −A −A B3 = ⋅ = ⋅ G (− jω ) = ⋅ 2 j − jω + 3 2 j 2j B1 = The steady state response YSS(s) is: YSS ( s ) = B3 B2 + s − jω s + jω ⇒ ⇒ ME575 Session 7 – ySS (t ) = L −1 [YSS ( s )] = B2 ⋅ e jω ⋅t + B3 ⋅ e − jω ⋅t ySS (t ) = A G ( jω ) ⋅ sin(ω ⋅ t + φ ) Frequency Response where φ = ∠G ( jω ) School of Mechanical Engineering Purdue University Slide 6 3 ME575 Handouts Frequency Response Frequency response is used to study the steady state output ySS(t) of a stable is system due to sinusoidal inputs at different frequencies. In general, given a stable system: & & y ( n ) + an −1 y ( n −1) + L + a1 y + a0 y = bm u ( m ) + bm −1u ( m −1) + L + b1u + b0 u bm s m + bm −1s m −1 + L + b1s+ b0 N ( s ) bm ( s − z1 )( s − z2 )L ( s − zm ) = = ( s − p1 )( s − p2 )L ( s − pn ) D( s) s n + an −1s n −1 + L + a1s + a0 If the input is a sinusoidal signal with frequency ω , i.e. i.e. G(s) ≡ u (t ) = A sin(ω ⋅ t ) then the steady state output ySS(t) is also a sinusoidal signal with the same frequency as the input signal but with different magnitude and phase: phase: ySS (t ) = G ( jω ) ⋅ A sin(ω ⋅ t + ∠G ( jω )) where G(jω) is the complex number obtained by substitute jω for s in G(s) , i.e. for G ( jω ) = G ( s ) s = jω ≡ ME575 Session 7 – Frequency Response bm ( jω ) m + bm −1 ( jω ) m −1 + L + b1 ( jω )+ b0 ( jω ) n + an −1 ( jω ) n −1 + L + a1 ( jω ) + a0 School of Mechanical Engineering Purdue University Slide 7 Frequency Response Input u(t) Output y(t) U(s) u LTI System G(s) Y(s) ySS 2π/ω 2π/ω t u (t ) = A sin(ω ⋅ t ) t ⇒ ySS (t ) = G ( jω ) ⋅ A sin(ω ⋅ t + ∠G ( jω )) − A different perspective on the role of the transfer function: Amplitude of the steady state sinusoidal output ⎧ ⎪ G ( jω ) = Amplitude of the sinusoidal input ⎨ ⎪ ∠G ( jω ) = Phase difference (shift) between y (t ) and the sinusoidal input SS ⎩ ME575 Session 7 – Frequency Response School of Mechanical Engineering Purdue University Slide 8 4 ME575 Handouts Frequency Response G Input u(t) Output y(t) G ME575 Session 7 – Frequency Response School of Mechanical Engineering Purdue University Slide 9 Bode Diagrams (Plots) • Magnitude Plot : plots the magnitude of G(jω) in decibels w.r.t. logarithmic frequency, i.e. G ( jω ) dB = 20log10 G ( jω ) vs log10ω • Phase Plot : plots the linear phase angle of G(jω) w.r.t. logarithmic frequency, i.e. ∠G ( jω ) vs log10ω To plot Bode diagrams, calculate the magnitude and phase of the corresponding transfer function. ME575 Session 7 – Frequency Response School of Mechanical Engineering Purdue University Slide 10 5 ME575 Handouts Bode Diagrams Transfer Function: bm s m + bm −1 s m −1 + L + b1 s + b0 bm ( s − z1 )( s − z 2 ) L ( s − z m ) = ( s − p1 )( s − p2 ) L ( s − pn ) s n + an −1 s n −1 + L + a1 s + a0 G (s) = Frequency Response G ( jω ) = bm ( jω − z1 )( jω − z 2 ) L ( jω − z m ) ( jω − p1 )( jω − p2 ) L ( jω − pn ) = bm ⋅ 1 1 1 L⋅ ⋅ ⋅ ( jω − z1 ) ⋅ ( jω − z 2 ) L ⋅ ( jω − z m ) ( jω − p1 ) ( jω − p2 ) ( jω − pn ) Bode Magnitude: ⎛ 1 20 log10 ( G ( jω ) ) = 20 log10 ( bm ) + 20 log10 ⎜ ⎜ ( jω − p ) 1 ⎝ ⎛ ⎞ ⎞ 1 ⎟ + L + 20 log10 ⎜ ⎟ ⎜ ( jω − p ) ⎟ ⎟ n ⎠ ⎝ ⎠ + 20 log10 ( ( jω − z1 ) ) + L + 20 log10 ( ( jω − z1 ) ) Bode Phase: ∠ G ( jω ) = ∠ bm ( jω − z1 )( jω − z 2 ) L ( jω − z m ) ( jω − p1 )( jω − p 2 ) L ( jω − p n ) = ∠ bm + ∠ ( jω − z1 ) + L + ∠ ( jω − z m ) − ∠ ( jω − p1 ) − L − ∠ ( jω − p n ) ME575 Session 7 – Frequency Response School of Mechanical Engineering Purdue University Slide 11 Bode Diagram Building Blocks • 1st Order Real Poles 0 -3 Transfer Function: 1 , τ s +1 Frequency Response: 1 , G p1 ( jω ) = τ jω + 1 τ >0 -20 τ >0 1 ⎧ G p 1 ( jω ) = ⎪ τ 2ω 2 + 1 ⎨ ⎪ ⎩ ∠ G p 1 ( jω ) = − a tan2 (τω ,1) = − tan − 1 (τω ) Q: By just looking at the Bode diagram, can you determine the time constant and the steady state gain of the system ? Phase (deg); Magnitude (dB) G p1 ( s ) = -40 0 -45 -90 0.01/τ 0.1/τ 1/τ 10/τ 100/τ Frequency (rad/sec) ME575 Session 7 – Frequency Response School of Mechanical Engineering Purdue University Slide 12 6 ME575 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 Gz1 ( s ) = τ s + 1 , τ > 0 τ >0 – Break Frequency 1 ωb = [ rad/s] τ τ – Mag. Plot Approximation – 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 ωb . Between 1 10 0 dB from DC to ωb and a straight line with 20 dB/decade slope after ωb. ωb and 10ωb , a straight line from 0 deg to straight −90 deg (passing −45 deg at ωb). For frequency higher than 10ωb , straight line on −90 deg. – Phase Plot Approximation 1 0 deg from DC to 10 ωb . Between 1 10 ωb and 10ωb , a straight line from 0 deg to straight 90 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. ME575 Session 7 – Frequency Response School of Mechanical Engineering Purdue University Slide 13 Bode Diagram Building Blocks • 2nd Order Complex Poles 40 Transfer Function: 20 Frequency Response: -20 ωn2 G p2 (s) = 2 , 1≥ ζ ≥ 0 s + 2ζ ω n s + ω n 2 j G p 2 ( jω ) = 2ζ ω ωn Phase (deg); Magnitude (dB) G p 2 ( jω ) = 1 ⎛ ω2 ⎞ + ⎜1 − 2 ⎟ ωn ⎠ ⎝ 1 4ζ ω 2 ωn 2 2 ⎛ ω2 ⎞ + ⎜1 − 2 ⎟ ωn ⎠ ⎝ ζ ∠ G p 2 ( jω ) = − tan −1 ⎡ 2ωnω ⎢ ⎣ 0 2 -40 -60 0 -45 -90 (1 − )⎤⎥⎦ ω2 ωn 2 -135 -180 0.1ωn ωn 10ωn Frequency (rad/sec) ME575 Session 7 – Frequency Response School of Mechanical Engineering Purdue University Slide 14 7 ME575 Handouts 2nd Order Bode Diagram Summary • 2nd Order Complex Poles ωn 2 G p 2 ( s) = 2 , 1≥ ζ > 0 s + 2ζ ω n s + ω n 2 – Break Frequency ωb = ωn [ rad/s ] – Mag. Plot Approximation • 2nd Order Complex Zeros Gz 2 ( s ) = ωb = ωn [ rad/s] 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 ) ωn . Between ( 1 5 ) ωn and 5ζ ωn , a straight line from 0 deg to − straight 180 deg (passing −90 deg at ωn). For frequency higher than 5ζ ωn , straight line on −180 deg. Frequency Response 1≥ ζ > 0 – Mag. Plot Approximation – Phase Plot Approximation ME575 Session 7 – , – Break Frequency 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 ( jωr ) = MAX 2ζ 1 − ζ 2 ζ s 2 + 2ζωn s + ωn 2 ωn 2 ζ ζ 0 deg from DC to ( 1 5 ) ωn. Between ( 1 5 ) ωn ζ ω , a straight line from 0 deg to and 5 n straight 180 deg (passing 90 deg at ωn). For frequency higher than 5ζ ωn , straight line on 180 deg. School of Mechanical Engineering Purdue University ζ Slide 15 2nd Order System Frequency Response A Few Observations: • Three different characteristic frequencies: different – Natural Frequency (ωn) – Damped Natural Frequency (ωd): ωd = ωn 1 − ζ 2 – Resonant (Peak) Frequency (ωr): ωr = ωn 1 − 2ζ 2 ωr ≤ ωd ≤ ωn • When the damping ratio ζ > 0.707, there is no peak in the Bode 0.707 magnitude plot. DO NOT confuse this with the condition for overoverdamped and under-damped systems: when ζ < 1 the system is underunderthe underdamped (has overshoot) and when ζ > 1 the system is over-damped (no the overovershoot). • As ζ → 0 , ωr → ωn and |G(jω)|ΜΑΧ increases; also the phase transition from 0 deg to −180 deg becomes sharper. ME575 Session 7 – Frequency Response School of Mechanical Engineering Purdue University Slide 16 8 ME575 Handouts Quantify TF Modeling Errors • Nominal System Model Y ( s ) = Go ( s )U ( s ) • Actual (Calibration) System Model Y ( s) = G ( s)U ( s ) • Modeling Errors in terms of Transfer Functions Y ( s) = G ( s)U ( s ) = (Go ( s ) + Gε ( s ))U ( s ) =Go ( s )(1 + GΔ ( s ))U ( s ) • Bounds for Modeling Errors GΔ ( jω ) < ε (ω ) ME575 Session 7 – Frequency Response School of Mechanical Engineering Purdue University Slide 17 Quantify TF Modeling Errors - Examples • Missing High-Frequency Pole HighBode Plots of GΔ ( s) for α = 0.2, 1, 20 Go (s) = F(s), G(s) = 1 F(s) α s +1 GΔ (s) = ME575 Session 7 – Frequency Response School of Mechanical Engineering Purdue University −αs αs +1 Slide 18 9 ME575 Handouts Quantify TF Modeling Errors - Examples • Error in Time-Delay TimeBode Plots of GΔ ( s) for τ = 0.2τ 0 , 5τ 0 Go (s) = e−τos F(s), G(s) = e−τ s F(s) −(τ −τo ) s GΔ (s) = e τ τo (1− )τo s −1 = e −1 ⎛ ωτ ⎛ τ ⎞ ⎞ GΔ ( jω) = 2 sin ⎜ o ⎜1− ⎟ ⎟ ⎜2 ⎟ ⎝ τo ⎠ ⎠ ⎝ ⎡⎛ ⎛ τ ⎞⎞ ∠GΔ ( jω) = atan2 ⎢sin ⎜ωτo ⎜1− ⎟ ⎟ , ⎜ ⎟ ⎢⎝ ⎝ τo ⎠ ⎠ ⎣ ⎛ ⎛ τ ⎞ ⎞⎤ −1+ cos ⎜ωτo ⎜1− ⎟ ⎟⎥ ⎜ ⎟ ⎝ τo ⎠ ⎠⎥ ⎝ ⎦ ME575 Session 7 – Frequency Response School of Mechanical Engineering Purdue University Slide 19 Quantify TF Modeling Errors - Examples • Missing Resonance Effect Bode Plots of GΔ ( s) for ζ = 0.1, 0.2, 0.7 Go (s) = F(s), G(s) = 2 ωn F(s) 2 s2 + 2ζωn s + ωn GΔ (s) = ME575 Session 7 – Frequency Response School of Mechanical Engineering Purdue University −s(s + 2ζωn ) 2 s + 2ζωn s + ωn 2 Slide 20 10 ...
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