Case Study 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 4 km/hr and 1 km/hr. ME375 Frequency Response - 1
Case Study Pantograph Model F c (t) m z ( ) ( ) mz + b + b z + k + k z bz k z 1 1 1 1 1 1 mz + bz + kz bz kz F () c t 1 1 k b z 1 For m 1 3. kg, b 1 15 N/(m/s), k 1 96 N/m, m 11.5 kg, b 75 N/(m/s), and k 958 N/m: k 1 m 1 b 1 Z( s).87( s + 9.78s+ 834) 4 3 F () s s + 16.3s + 179s + 8155s+ 347,75 c ME375 Frequency Response - 3 Case Study Frequency Response -6 Magnitude (db) -7-8 -9-1 -11 18 Phase (deg) 135 9 45 1 1 1 1 ME375 Frequency Response - 4
Frequency Response Forced Response to Sinusoidal Inputs Frequency Response of LTI Systems Bode Plots ME375 Frequency Response - 5 Forced Response to Sinusoidal Inputs Ex: Let s s find the forced response of a stable first order system: y + 5y 1u to a sinusoidal input: ut () sin() t Forced response: Y() s G() s U() s [ ] where Gs ( ) and Us ( ) Lsin( t) Y( s) PFE: A1 A A3 Y() s + + Compare coefficients to find A 1, A and A 3 : ME375 Frequency Response - 6 3
Forced Response to Sinusoidal Inputs Ex: (cont.) Use ILT to find y(t) ) : [ ] 1 1 yt ( ) L Y( s) L + + Useful Formula: Asin( ω t) + Bcos( ω t) A + B sin( ω t + φ) Where φ atan( BA, ) ( A+ jb) Using this formula, the forced response can be represented by 5t yt ( ) e + sin( t+ φ) ME375 Frequency Response - 7 Forced Response of 1st Order System 1.5 1 Output Input is sin(t) Input.5 Response -.5-1 -1.5-4 6 8 1 1 14 Time (sec) ME375 Frequency Response - 8 4
Forced Response to Sinusoidal Inputs Ex: Given the same system as in the previous example, find the forced response to u(t) ) sin(1 t). Y() s G() s U() s [ ] where Gs ( ) and Us ( ) Lsin(1 t) Y( s) ME375 Frequency Response - 9 Forced Response of 1st Order Systems Input is sin(1t) 1 Output.8.6.4 Response. -. -.4 -.6 -.8-1 Input.5 1 1.5.5 3 3.5 4 4.5 5 Time (sec) ME375 Frequency Response - 1 5
Frequency Response Ex: Let s s revisit the same example where y + 5y 1u and the input is a general sinusoidal input: sin(ω t). 1 ω 1 ω Ys () Gs () Us () s+ 5 s + ω s+ 5 ( s jω)( s+ jω) A1 A A3 Ys () + + s+ 5 s jω s+ jω Instead of comparing coefficients, use the residue formula to find A i s: 1 ( 5) ( ) 5 ( 5) 1 ω A s+ Y s s+ s ( 5) s+ s + ω s 5 ω A ( s jω) Y( s) ( s jω) Gs ( ) s jω s + ω s jω ω A3 ( s+ jω) Y( s) ( s+ jω) G( s) s jω s + ω s jω ME375 Frequency Response - 11 Frequency Response Ex: (Cont.) 1ω A1 5 + ω 1 1 1 1 A G( jω ) j jω + 5 j j 1 1 1 1 A3 G( jω ) j jω + 5 j j The steady state response Y SS (s)) is: A A3 YSS () s + s jω s+ jω 1 jω t jω t y () t L Y ( s) A e + A e SS [ ] SS 3 y ( t) G( jω ) sin( ω t+ φ) where φ G( jω) SS ME375 Frequency Response - 1 6
Frequency Response Frequency Response ME375 Frequency Response - 13 In Class Exercise For the current example, y + 5y 1u Calculate the magnitude and phase shift of the steady state response when the system is excited by (i) sin(t) ) and (ii) sin(1t). Compare your result with the steady state response calculated in the previous examples. Note: 1 1 Gs () G( jω ) s+ 5 jω + 5 1 G( jω ) and φ G( jω) atan( ω,5) ω + 5 ME375 Frequency Response - 14 7
Frequency Response Frequency response is used to study the steady state output y SS (t)) of a stable system due to sinusoidal inputs at different frequencies. In general, given a stable system: ( n) ( n 1) ( m) ( m 1) any + an 1y + + a1y + ay bmu + bm 1u + + bu 1 + bu m m 1 bms + bm 1s + + bs 1 + b Ns () bm( s z1)( s z) ( s zm) Gs () n n 1 as n + an 1s + + as 1 + a Ds () an( s p1)( s p) ( s pn) If the input is a sinusoidal signal with frequency ω, i.e. ut () Asin( ω t) then the steady state output y SS (t)) is also a sinusoidal signal with the same frequency as the input signal but with different magnitude and phase: y () t G( jω ) A sin( ω t+ G( jω)) SS u where G(jω) ) is the complex number obtained by substitute jω for s in G(s) ), i.e. m m 1 bm( jω) + bm ( jω) + + b ( jω) + b G( jω ) G( s) s jω n n 1 a ( jω) + a ( jω) + + a ( jω) + a n u 1 1 n 1 1 ME375 Frequency Response - 15 Frequency Response Input u(t) U(s) LTI System G(s) Output y(t) Y(s) u π/ω y SS π/ω t ut () Asin( ω t) u y () t G( jω ) A sin( ω t+ G( jω)) SS u t 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 yss ( t) and the sinusoidal input ME375 Frequency Response - 16 8
Frequency Response G Input u(t) Output y(t) G ME375 Frequency Response - 17 In Class Exercise Ex: 1st Order System The motion of a piston in a cylinder can be modeled by a 1st order system with force as input and piston velocity as output: f(t) v The EOM is: Mv + Bv f() t (1) Let M.1 kg and B.5 N/(m/s), find the transfer function of the system: () Calculate the steady state output of the system when the input is Input f(t) Steady State Output v(t) sin(ω t) G(jω) sin(ω t + φ ) sin(t) sin(t + sin(1t) sin(1t + sin(t) sin(t + sin(3t) sin(3t + sin(4t) sin(4t + sin(5t) sin(5t + sin(6t) sin(6t + ME375 Frequency Response - 18 9
In Class Exercise (3) Plot the frequency response plot 1.8-1 1.6 - Magnitude ((m/s)/n) 1.4 1. 1.8.6 Phase (deg) -3-4 -5-6 -7.4-8. -9 1 3 4 5 6 7 1 3 4 5 6 7 ME375 Frequency Response - 19 Example - Vibration Absorber (I) Without vibration absorber: EOM: z 1 M z Bz K z f t 1 1+ 1 1+ 1 1 () K 1 M 1 f(t) B 1 Let M 1 1 kg, K 1 1 N/m, B 1 4 N/(m/s). Find the steady state response of the system for f(t) ) (a) sin(8.5t) ) (b) sin(1t) ) (c) sin(11.7t). TF (from f(t) ) to z 1 ): Input f(t) Steady State Output z 1 (t) sin(ω t) G(jω) sin(ω t + φ ) sin(8.5t) sin(8.5t + sin(1t) sin(1t + sin(11.7t) sin(11.7t + ME375 Frequency Response - 1
Example - Vibration Absorber (I).1 f(t) sin(8.5 t).5 z 1 (m) -.5 -.1.4 f(t) sin(1 t) z 1 (m). -. -.4.5 f(t) sin(11.7 t) z 1 (m) -.5 5 1 15 5 3 35 4 45 5 Time (sec) ME375 Frequency Response - 1 Example - Vibration Absorber (II) With vibration absorber: K K 1 M M 1 f(t) z B B 1 z 1 TF (from f(t) ) to z 1 ): EOM: M 1z1+ ( B1+ B) z 1+ ( K1+ K) z1 Bz Kz f( t) M z + Bz + Kz Bz Kz 1 1 Let M 1 1 kg, K 1 1 N/m, B 1 4 N/(m/s), M 1 kg, K 1 N/m, and B.1 N/(m/s). Find the steady state response of the system for f(t) ) (a) sin(8.5t) ) (b) sin(1t) ) (c) sin(11.7t). Input f(t) Steady State Output z 1 (t) sin(ω t) G(jω) sin(ω t + φ ) sin(8.5t) sin(8.5t + sin(1t) sin(1t + sin(11.7t) sin(11.7t + ME375 Frequency Response - 11
Example - Vibration Absorber (II).4 f(t) sin(8.5 t). z 1 (m) -. -.4.4 f(t) sin(1 t) z 1 (m). -. -.4. f(t) sin(11.7 t) z 1 (m).1 -.1 -. 5 1 15 5 3 35 4 45 5 Time (sec) ME375 Frequency Response - 3 Example - Vibration Absorber (II) Take a closer look at the poles of the transfer function: The characteristic equation 4 3 1s + 5.1s + 1.4s + 5s+ 1 Poles: p1,.1± 8.5 j p.155 ± 11.7 j 3,4 What part of the poles determines the rate of decay for the transient response? (Hint: when p σ ± jω the response is e σt e jω t ) ME375 Frequency Response - 4 1
Example - Vibration Absorbers Frequency Response Plot No absorber added Frequency Response Plot Absorber tuned at 1 rad/sec added.5.5 Magnitude (m/n)..15.1.5 Magnitude (m/n)..15.1.5 4 6 8 1 1 14 16 18 4 6 8 1 1 14 16 18 Phase (deg) -45-9 -135 Phase (deg) -45-9 -135-18 4 6 8 1 1 14 16 18-18 4 6 8 1 1 14 16 18 ME375 Frequency Response - 5 Example - Vibration Absorbers Bode Plot No absorber added Bode Plot Absorber tuned at 1 rad/sec added Phase (deg); Magnitude (db) -3-4 -5-6 -7-8 -9-1 -45 Phase (deg); Magnitude (db) -3-4 -5-6 -7-8 -9-1 -45-9 -9-135 -135-18 1 1 1 1-18 1 1 1 1 ME375 Frequency Response - 6 13
Bode Diagrams (Plots) Bode Diagrams (Plots) A unique way of plotting the frequency response function, G(jω), ), w.r.t. frequency w of systems. Consists of two plots: Magnitude Plot : plots the magnitude of G(jω) ) in decibels w.r.t. logarithmic frequency, i.e. G( jω) log 1 G( jω) vs log ω db 1 Phase Plot : plots the linear phase angle of G(jω) ) w.r.t. logarithmic frequency, i.e. G( jω ) vs log To plot Bode diagrams, one needs to calculate the magnitude and phase of the corresponding transfer function. Ex: s + 1 Gs ( ) s + 1s 1 ω ME375 Frequency Response - 7 Bode Diagrams Revisit the previous example: s+ 1 ( jω) + 1 Gs () G( jω) s + 1 s jω( jω + 1) 5 3 G( jω) G( jω) w.1..5 1 5 1 5 1 Gj ( ω) log 1 Gjω ( ) G( jω) Phase (deg); Magnitude (db) 1-1 -3-5 1 5-5 -1 1-1 1 1 1 1 ME375 Frequency Response - 8 14