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Unformatted text preview: ME375 Handouts Case Study School of Mechanical Engineering Purdue University ME375 Frequency Response - 1 Case Study SUPPORT POWER WIRE DROPPERS Electric train derives power through a pantograph, which contacts the power wire, which is suspended from a catenary. During high-speed runs between New Haven, CT and New York City, the train experiences intermittent power loss at 42 km/hr and 100 km/hr. School of Mechanical Engineering Purdue University ME375 Frequency Response - 2 1 ME375 Handouts Case Study – Pantograph Model m1z1 + ( b1 + b2 ) z1 + ( k1 + k2 ) z1 − b2z2 − k2z2 = 0 Fc(t) m2z2 + b2z2 + k2z2 − b2 z1 − k2 z1 = −Fc (t) z2 m2 k2 b2 z1 m1 k1 b1 For m1 = 23.0 kg, b1 = 150 N/(m/s), k1 = 9600 N/m, m2 = 11.5 kg, b2 = 75 N/(m/s), and k2 = 9580 N/m: Z 2 ( s) −0.087( s 2 + 9.78s + 834) = 4 Fc ( s ) s + 16.3s 3 + 1709 s 2 + 8155s + 347,705 School of Mechanical Engineering Purdue University ME375 Frequency Response - 3 Case Study – Frequency Response -60 Magnitude (dB) -70 -80 -90 -100 Phase (deg) -110 180 135 90 45 0 0 10 1 10 Frequency (rad/sec) School of Mechanical Engineering Purdue University 10 2 ME375 Frequency Response - 4 2 ME375 Handouts Frequency Response • • • Forced Response to Sinusoidal Inputs Frequency Response of LTI Systems Bode Plots School of Mechanical Engineering Purdue University ME375 Frequency Response - 5 Forced Response to Sinusoidal Inputs Ex: Let’s find the forced response of a stable first order system: Let’ y + 5 y = 10u to a sinusoidal input: – Forced response: u (t ) = sin(2t ) Y (s) = G (s) ⋅U (s) where G ( s ) = and ∴ – PFE: Y (s) = Y ( s) = A1 U ( s ) = L [sin(2t ) ] = + A2 ⋅ + A3 ⋅ – Compare coefficients to find A1, A2 and A3 : School of Mechanical Engineering Purdue University ME375 Frequency Response - 6 3 ME375 Handouts 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 ⋅ e −5 t + y (t ) = ⋅ sin(2t + φ ) School of Mechanical Engineering Purdue University ME375 Frequency Response - 7 Forced Response of 1st Order System Input is sin(2t) 2 Output 1.5 Input 1 0.5 Response 0 -0.5 -1 -1.5 -2 0 2 4 6 8 Time (sec) School of Mechanical Engineering Purdue University 10 12 14 ME375 Frequency Response - 8 4 ME375 Handouts Forced Response to Sinusoidal Inputs Ex: Given the same system as in the previous example, find the forced response forced to u(t) = sin(10 t). Y ( s) = G ( s) ⋅U ( s) where G ( s ) = and ∴ U ( s ) = L [sin(10t ) ] = Y ( s) = School of Mechanical Engineering Purdue University ME375 Frequency Response - 9 Forced Response of 1st Order Systems Input is sin(10t) Output 1 0.8 0.6 Response 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 Input -1 0 0.5 1 1.5 2 2.5 Time (sec) 3 School of Mechanical Engineering Purdue University 3.5 4 4.5 5 ME375 Frequency Response - 10 5 ME375 Handouts Frequency Response Ex: Let’s revisit the same example where Let’ y + 5y =10u and the input is a general sinusoidal input: sin(ω t). 10 ω 10 ω ⋅ = ⋅ s + 5 s2 +ω2 s + 5 (s − jω)(s + jω) A A A Y (s) = 1 + 2 + 3 s + 5 s − jω s + jω Y (s) = G(s) ⋅U(s) = – Instead of comparing coefficients, use the residue formula to find Ai’s: 10 ω A = (s + 5)Y(s) s=−5 = (s + 5) = 1 (s + 5) s2 +ω2 s=−5 A2 = (s − jω)Y(s) s= jω = (s − jω)G(s) ω = s +ω2 s= jω A3 = (s + jω)Y(s) s=− jω = (s + jω)G(s) 2 ω s2 +ω2 s=− jω School of Mechanical Engineering Purdue University = ME375 Frequency Response - 11 Frequency Response Ex: (Cont.) 10ω 52 + ω 2 1 10 1 1 A2 = ⋅ = ⋅ G ( jω ) = ⋅ 2 j jω + 5 2 j 2j 10 −1 −1 −1 A3 = ⋅ = ⋅ G (− jω ) = ⋅ 2 j − jω + 5 2 j 2j A1 = The steady state response YSS(s) is: YSS ( s ) = A3 A2 + s − jω s + jω ⇒ ySS (t ) = L −1 [YSS ( s ) ] = A2 ⋅ e jω ⋅t + A3 ⋅ e− jω ⋅t ⇒ ySS (t ) = G ( jω ) ⋅ sin(ω ⋅ t + φ ) where φ = ∠G ( jω ) School of Mechanical Engineering Purdue University ME375 Frequency Response - 12 6 ME375 Handouts Frequency Response • Frequency Response School of Mechanical Engineering Purdue University ME375 Frequency Response - 13 In Class Exercise For the current example, y + 5 y = 10u Calculate the magnitude and phase shift of the steady state response when the response system is excited by (i) sin(2t) and (ii) sin(10t). Compare your result with the sin(2t sin(10t steady state response calculated in the previous examples. Note: 10 10 G(s) = s+5 G ( jω ) = ⇒ G ( jω ) = 10 ω 2 + 52 jω + 5 and φ = ∠G ( jω ) = −atan2(ω ,5) School of Mechanical Engineering Purdue University ME375 Frequency Response - 14 7 ME375 Handouts Frequency Response •Frequency response is used to study the steady state output ySS(t) of a stable system due to sinusoidal inputs at different frequencies. In general, given a stable system: an y ( n ) + an−1 y ( n −1) + + a1 y + a0 y = bm u ( m ) + bm−1u ( m−1) + + b1u + b0 u m −1 bm s + bm−1s + + b1s+ b0 N ( s ) bm ( s − z1 )( s − z2 ) ( s − zm ) = = an s n + an −1s n −1 + + a1s + a0 D( s ) an ( s − p1 )( s − p2 ) ( s − pn ) If the input is a sinusoidal signal with frequency ω , i.e. G(s) ≡ m u (t ) = Au 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: ySS (t ) = G ( jω ) ⋅ Au sin(ω ⋅ t + ∠G ( jω )) where G(jω) is the complex number obtained by substitute jω for s in G(s) , i.e. G ( jω ) = G ( s ) s = jω ≡ bm ( jω ) m + bm −1 ( jω ) m −1 + + b1 ( jω )+ b0 an ( jω ) n + an −1 ( jω ) n −1 + + a1 ( jω ) + a0 School of Mechanical Engineering Purdue University ME375 Frequency Response - 15 Frequency Response Input u(t) Output y(t) U(s) u LTI System G(s) Y(s) ySS 2π/ω t u (t ) = Au sin(ω ⋅ t ) 2π/ω t ⇒ ySS (t ) = G ( jω ) ⋅ Au sin(ω ⋅ t + ∠G ( jω )) − A different perspective of 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 ⎩ School of Mechanical Engineering Purdue University ME375 Frequency Response - 16 8 ME375 Handouts Frequency Response G Input u(t) Output y(t) G School of Mechanical Engineering Purdue University ME375 Frequency Response - 17 In Class Exercise Ex: 1st Order System (2) Calculate the steady state output of the system when the input is The motion of a piston in a cylinder can be modeled by a 1st order system with force Input f(t) Steady State Output v(t) as input and piston velocity as output: sin(ω t) ⎢G(jω)⎥ sin(ω t + φ ) sin(0t + sin(0t) f(t) sin(10t) sin(10t + sin(20t) sin(20t + The EOM is: Mv + Bv = f (t ) sin(30t) sin(30t + (1) Let M = 0.1 kg and B = 0.5 N/(m/s), find the transfer function of the system: sin(40t) sin(40t + sin(50t) sin(50t + sin(60t) sin(60t + v School of Mechanical Engineering Purdue University ME375 Frequency Response - 18 9 ME375 Handouts In Class Exercise (3) Plot the frequency response plot -20 1.4 -30 Phase (deg) 0 -10 1.6 Magnitude ((m/s)/N) 2 1.8 1.2 1 0.8 -40 -50 -60 0.6 -70 0.4 -80 0.2 -90 0 0 10 20 30 40 50 Frequency (rad/sec) 60 70 0 School of Mechanical Engineering Purdue University 10 20 30 40 50 Frequency (rad/sec) 60 70 ME375 Frequency Response - 19 Example - Vibration Absorber (I) Without vibration absorber: EOM: M 1 z1 + B1 z1 + K1 z1 = f (t ) z1 M1 K1 f(t) B1 TF (from f(t) to z1): Let M1 = 10 kg, K1 = 1000 N/m, B1 = 4 N/(m/s). Find the steady state response of the system for f(t) = (a) sin(8.5t) (b) sin(10t) (c) sin(11.7t). sin(8.5t sin(10t sin(11.7t Input f(t) Steady State Output z1(t) sin(ω t) ⎢G(jω)⎥ × sin(ω t + φ) sin(8.5t + sin(8.5t) sin(10t) sin(11.7t) School of Mechanical Engineering Purdue University sin(10t + sin(11.7t + ME375 Frequency Response - 20 10 ME375 Handouts Example - Vibration Absorber (I) f(t) = sin(8.5 t) 0.01 z1 (m) 0.005 0 -0.005 -0.01 f(t) = sin(10 t) z1 (m) 0.04 0.02 0 -0.02 -0.04 f(t) = sin(11.7 t) z1 (m) 0.005 0 -0.005 0 5 10 15 20 25 30 35 40 45 50 Time (sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 21 Example - Vibration Absorber (II) With vibration absorber: z2 M 2 z2 + B2 z2 + K 2 z2 − B2 z1 − K 2 z1 = 0 M2 K2 B2 z1 M1 K1 f(t) EOM: M 1 z1 + ( B1 + B2 ) z1 + ( K1 + K 2 ) z1 − B2 z2 − K 2 z2 = f (t ) B1 TF (from f(t) to z1): Let M1 = 10 kg, K1 = 1000 N/m, B1 = 4 N/(m/s), M2 = 1 kg, K2 = 100 N/m, and B2 = 0.1 N/(m/s). Find the steady state response of the system for f(t) = (a) sin(8.5t) (b) sin(10t) (c) sin(11.7t). sin(8.5t sin(10t sin(11.7t Input f(t) Steady State Output z1(t) sin(ω t) ⎢G(jω)⎥ × sin(ω t + φ) sin(8.5t + sin(8.5t) sin(10t) sin(11.7t) School of Mechanical Engineering Purdue University sin(10t + sin(11.7t + ME375 Frequency Response - 22 11 ME375 Handouts Example - Vibration Absorber (II) f(t) = sin(8.5 t) 0.04 z1 (m) 0.02 0 -0.02 -0.04 f(t) = sin(10 t) z1 (m) 0.004 0.002 0 -0.002 -0.004 f(t) = sin(11.7 t) 0.02 z1 (m) 0.01 0 -0.01 -0.02 0 5 10 15 20 25 30 35 40 45 50 Time (sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 23 Example - Vibration Absorber (II) Take a closer look at the poles of the transfer function: The characteristic equation 10 s 4 + 5.1s 3 + 2100.4 s 2 + 500 s + 100000 = 0 ⇒ Poles: p1,2 = −0.1 ± 8.5 j p3,4 = −0.155 ± 11.7 j What part of the poles determines the rate of decay for the transient response? transient σt e jωt → (Hint: when p = σ ± jω → the response is e School of Mechanical Engineering Purdue University ) ME375 Frequency Response - 24 12 ME375 Handouts Example - Vibration Absorbers Frequency Response Plot Absorber tuned at 10 rad/sec added Frequency Response Plot No absorber added 0.025 Magnitude (m/N) Magnitude (m/N) 0.025 0.02 0.015 0.01 0.005 0 0 2 4 6 8 10 12 14 16 18 0.02 0.015 0.01 0.005 0 20 0 4 6 8 10 12 14 Frequency (rad/sec) 16 18 20 2 4 6 8 10 12 14 Frequency (rad/sec) 16 18 20 0 -45 Phase (deg) Phase (deg) 0 -45 2 0 Frequency (rad/sec) -90 -135 -180 -90 -135 -180 0 2 4 6 8 10 12 14 Frequency (rad/sec) 16 18 20 School of Mechanical Engineering Purdue University ME375 Frequency Response - 25 Example - Vibration Absorbers Bode Plot No absorber added Bode Plot Absorber tuned at 10 rad/sec added -30 -40 -40 -50 -50 -60 Phase (deg); Magnitude (dB) Phase (deg); Magnitude (dB) -30 -70 -80 -90 -100 0 -45 -90 -70 -80 -90 -100 0 -45 -90 -135 -135 -180 10 -60 0 10 1 Frequency (rad/sec) 10 2 -180 10 0 School of Mechanical Engineering Purdue University 1 10 Frequency (rad/sec) 10 2 ME375 Frequency Response - 26 13 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. 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 ) vs log 10ω – Phase Plot : plots the linear phase angle of G(jω) w.r.t. logarithmic frequency, i.e. ∠G ( jω ) vs log 10ω 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 - 27 Bode Diagrams Revisit the previous example: 50 ( jω ) + 1 s +1 ⇒ G ( jω ) = G(s) = 2 s + 10 s jω ( jω + 10) 30 G ( jω ) = 10 0 ∠G ( jω ) = ω 0.1 0.2 0.5 1 2 5 10 20 50 100 G( jω ) 20log10 G( jω) ∠G( jω ) Phase (deg); Magnitude (dB) -10 -30 -50 100 50 0 -50 -100 -1 10 School of Mechanical Engineering Purdue University 10 0 10 Frequency (rad/sec) 1 10 2 ME375 Frequency Response - 28 14 ME375 Handouts Bode Diagrams Recall that if G( s) = Then bm s m + bm −1s m −1 + + b1s + b0 bm ( s − z1 )( s − z2 ) ( s − zm ) = an s n + an −1s n −1 + + a1s + a0 an ( s − p1 )( s − p2 ) ( s − pn ) G ( jω ) = = bm ( jω − z1 )( jω − z2 ) ( jω − z m ) an ( jω − p1 )( jω − p2 ) ( jω − pn ) bm 1 1 ⋅ ⋅ an ( jω − p1 ) ( jω − p2 ) 20 log10 ( G ( jω ) ) = 20 log10 ⋅ 1 ⋅ ( jω − z1 ) ⋅ ( jω − z2 ) ( jω − pn ) F b I + 20 log F 1 I + +20 log F 1 I GH ( jω − p ) JK GH a JK GH ( jω − p ) JK c ( jω − z ) h+ +20 log c ( jω − z ) h m 10 10 n +20 log10 and ∠ G ( jω ) = ∠ ⋅ ( jω − z m ) 1 1 10 n 1 bm ( jω − z1 )( jω − z 2 ) ( jω − z m ) an ( j ω − p1 )( jω − p2 ) ( jω − pn ) = ∠ ( j ω − z1 ) + ∠ ( jω − z 2 ) + +∠ ( jω − z m ) − ∠ ( jω − p1 ) − ∠ ( jω − p2 ) −∠ ( j ω − pn ) School of Mechanical Engineering Purdue University ME375 Frequency Response - 29 Example Ex: Find the magnitude and the phase of the following transfer function: function: G( s ) = 3s 3 + 12 s 2 + 9s = 2 s + 22 s 2 + 76 s + 80 3 ( School of Mechanical Engineering Purdue University )( ) ME375 Frequency Response - 30 15 ME375 Handouts 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 R G ( jω ) | S | ∠ G ( jω ) T p1 p1 τ > 0 , = -20 , Phase (deg); Magnitude (dB) G p1 ( s ) = τ >0 1 τ 2ω 2 + 1 = − atan2 (τω ,1) = − tan − 1 τω a f -40 0 -45 Q: By just looking at the Bode diagram, can you determine the time constant and the steady state gain of the system ? -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 - 31 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 10 0 10 1 10 2 10 Frequency (rad/sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 32 16 ME375 Handouts Bode Diagram Building Blocks 40 • 1st Order Real Zeros Transfer Function: τ >0 Frequency Response: G z1 ( jω ) = τ jω + 1 , R G | S∠ G | T 20 τ >0 z1 ( jω ) = τ ω +1 z1 ( jω ) = atan2 ( τω ,1) = tan − 1 τω 2 2 a f Phase (deg); Magnitude (dB) G z1 ( s ) = τ s + 1 , 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 - 33 Example • 1st Order Real Zeros 20 15 10 Phase (deg); Magnitude (dB) Transfer Function: G ( s ) = 0 .7 s + 0 .7 Plot the straight line approximation of G(s)’s Bode diagram: 5 0 90 45 0 -1 10 School of Mechanical Engineering Purdue University 0 10 1 10 Frequency (rad/sec) 2 10 ME375 Frequency Response - 34 17 ME375 Handouts Example • Lead Compensator 30 20 10 Phase (deg); Magnitude (dB) Transfer Function: 35s + 35 G (s) = s+5 Plot the straight line approximation of G(s)’s Bode diagram: 0 -10 90 0 -90 -1 0 10 1 10 2 10 10 Frequency (rad/sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 35 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 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 10ωb , a straight line from 0 deg to −90 deg (passing −45 deg at ωb). For frequency 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 10ωb , a straight line from 0 deg to 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 system, break frequency, and DC (steady-state) gain. This process should also be reversible, i.e. given a frequency, (steadygain. transfer function, be able to plot a straight line approximated Bode diagram. School of Mechanical Engineering Purdue University ME375 Frequency Response - 36 18 ME375 Handouts Bode Diagram Building Blocks • Integrator (Pole at origin) 20 Transfer Function: 1 G p0 (s) = s Frequency Response: 0 Phase (deg); Magnitude (dB) -20 1 G p 0 ( jω ) = jω R G ( jω ) | S | ∠ G ( jω ) T = p0 p0 1 ω = − 90 ° -40 -60 0 -45 -90 -135 0.1 1 10 100 1000 Frequency (rad/sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 37 Bode Diagram Building Blocks • Differentiator (Zero at origin) 60 Transfer Function: 40 Gz0 (s) = s G z 0 ( jω ) = jω R G | S∠ G | T z0 ( jω ) =ω z0 ( jω ) = 90 ° Phase (deg); Magnitude (dB) 20 Frequency Response: 0 -20 135 90 45 0 0.1 1 10 100 1000 Frequency (rad/sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 38 19 ME375 Handouts Example • Combination of Systems Transfer Function: 30 G (s) = 20 35s + 35 s(s + 5) 10 0 Phase (deg); Magnitude (dB) Plot the straight line approximation of G(s)’s Bode diagram: -10 90 0 -90 -1 0 10 1 10 2 10 10 Frequency (rad/sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 39 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 - 40 20 ME375 Handouts 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 j G p 2 ( jω ) = 2ζω ωn Phase (deg); Magnitude (dB) Frequency Response: G p 2 ( jω ) = 1 ⎛ ω2 ⎞ ⎟ + ⎜1 − ⎜ 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 0.01 ωn 0.1 ωn ωn 10 ωn 100 ωn Frequency (rad/sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 41 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 - 42 1 ME375 Handouts 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: ω2 ⎞ 2ζ ω ⎛ ⎟ 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 0.1 ωn School of Mechanical Engineering Purdue University ωn Frequency (rad/sec) 10 ωn 100 ωn ME375 Frequency Response - 43 Bode Diagrams of Poles and Zeros 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 20 0 -20 -40 p z p z 10 p 10 p z 180 Phase (deg) Let 1 G p (s) = Gz ( s ) 40 Magnitude (dB) 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. 0 z -180 0.1ωn ωn 10ωn Frequency (rad/sec) School of Mechanical Engineering Purdue University ME375 Frequency Response - 44 22 ME375 Handouts 2nd Order Bode Diagram Summary • 2nd Order Complex Zeros • 2nd Order Complex Poles 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) = ⇒ 0 dB from DC to ωn and a straight line with 40 dB/decade slope after ωn. – Phase Plot Approximation 1 2ζ 1 − ζ 0 deg from DC to ( 1 5 ) ω n . Between ( 1 5 )ζ ω n 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. ζ 2 – Phase Plot Approximation 0 deg from DC to ( 1 5 ) ω n . Between ( 1 5 ) ω n 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. ζ 1≥ ζ > 0 , – Mag. Plot Approximation n G p 2 ( jω r ) MAX = 2 – Break Frequency ωb = ωn [ rad/s] 0 dB from DC to ωn and a straight line with −40 dB/decade slope after ωn. Peak value occurs at: ω = ω 1 − 2ζ 2 r s 2 + 2ζωn s + ωn 2 ωn Gz 2 ( s ) = 2 ζ School of Mechanical Engineering Purdue University ME375 Frequency Response - 45 2nd Order System Frequency Response A Closer Look: G p 2 ( s) = ωn 2 2 s 2 + 2ζω n s + ω n , 1≥ ζ ≥ 0 Frequency Response Function: G p 2 ( jω ) = Phase: Magnitude: G p 2 ( jω ) = ωn 2 1 = 2 ( jω ) + 2ζω n ( jω ) + ω n ω2 ⎞ 2 ζω ⎛ j + ⎜1 − 2 ⎟ ωn ⎝ ωn ⎠ 2 2 ⎛ 2 ζω ⎞ ⎛ ω ⎞ ⎜ ω ⎟ + ⎜1 − 2 ⎟ ⎝ n ⎠ ⎝ ωn ⎠ 2 ( ) ζ ω ∠G p 2 ( jω ) = −∠ ⎡ 1 − ω 2 + j 2ωnω ⎤ ⎢ ⎥ n ⎣ ⎦ 1 2 2 ( ) ζ ω2 = − atan2 ⎡ 2ωnω , 1 − ω 2 ⎤ ⎢ ⎥ n ⎣ ⎦ The maximum value of |G(jω)| occurs at the Peak (Resonant) Frequency ωr : |G 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 - 46 23 ME375 Handouts 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 - 47 2nd Order System Frequency Response A Few Observations: • Three different characteristic frequencies: – Natural Frequency (ωn) – 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 underthe 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 - 48 24 ME375 Handouts 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 - 49 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 - 50 25 ...
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This note was uploaded on 08/28/2010 for the course ME 375 taught by Professor Meckle during the Spring '10 term at Purdue University-West Lafayette.

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