Frequency Response

Frequency Response - ME375 Handouts Case Study School of...

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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 & & m1&&1 + ( b1 + b2 ) z1 + ( k1 + k2 ) z1 − b2z2 − k2z2 = 0 z Fc(t) & & m2&&2 + b2z2 + k2z2 − b2 z1 − k2 z1 = −Fc (t) z 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 Si • 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: & y + 5 y = 10u to a sinusoidal input: – Forced response: u (t ) = sin(2t ) Y (s) = G (s) ⋅U (s) where G ( s ) = and ∴ Y (s) = Y ( s) = A1 U ( s ) = L [sin(2t ) ] = – PFE: + 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 −5 y (t ) = 1 24 t + 4 ⋅ e3 ⋅ sin(2t + φ ) 1442443 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 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 & y + 5y =10u and the input is general sinusoidal input: sin( 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: residue to 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.) (Cont.) 10ω 52 + ω 2 1 10 1 1 A2 = ⋅ = ⋅ G ( jω ) = ⋅ 2 j jω + 5 2 j 2j −1 −1 −1 10 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 Calculate the magnitude and phase shift of the steady state response when the system system is excited by (i) sin(2t) and (ii) sin(10t). Compare your result with the 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 is due to sinusoidal inputs at different frequencies. In general, given a stable system: general given stable system: & & an 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 ) = = an s n + an −1s n −1 + L + a1s + a0 D( s ) an ( s − p1 )( s − p2 )L ( s − pn ) If the input is a sinusoidal signal with frequency ω , i.e. i.e. G(s) ≡ 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: diff 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. for G ( jω ) = G ( s ) s = jω ≡ bm ( jω ) m + bm −1 ( jω ) m −1 + L + b1 ( jω )+ b0 an ( jω ) n + an −1 ( jω ) n −1 + L + 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: 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 1st The motion of a piston in a cylinder can be modeled by 1st order system with force modeled by a 1st order system with force as as input and piston velocity as output: (2) Calculate the steady state output of the system when the input is Steady State Output v(t) ⎢G(jω)⎥ sin(ω t + φ ) sin(0t + sin(10t) f(t) Input f(t) sin(ω t) sin(0t) sin(10t + sin(20t) sin(20t + v The EOM is: & Mv + Bv = f (t ) sin(30t) sin(30t + (1) Let M = 0.1 kg and B = 0.5 N/(m/s), 0.1 0.5 find the transfer function of the system: sin(40t) sin(40t + sin(50t) sin(50t + sin(60t) sin(60t + 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&&1 + B1 z1 + K1 z1 = f (t ) z 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). 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) sin(11 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 M2 K2 B2 z1 M1 K1 f(t) B1 TF (from f(t) to z1): EOM: & & M 1&&1 + ( B1 + B2 ) z1 + ( K1 + K 2 ) z1 − B2 z2 − K 2 z2 = f (t ) z & & M 2 &&2 + B2 z2 + K 2 z2 − B2 z1 − K 2 z1 = 0 z 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 kg, 100 0.1 the steady state response of the system for f(t) = (a) sin(8.5t) (b) sin(10t) (c) 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 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) sin(11 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? (Hint: when p = σ ± jω → the response is eσt e jωt → the ) 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 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 Bode Plot No absorber added Bode Plot Absorber tuned at 10 rad/sec added -40 -50 -50 -60 -60 Phase (deg); Magnitude (dB) -30 -40 Phase (deg); Magnitude (dB) -30 -70 -80 -90 -100 0 -45 -90 -80 -90 -100 0 -45 -90 -135 -135 -180 10 -70 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 ...
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