lec10_handout

# lec10_handout - Dynamic System Response • Input/Output...

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Unformatted text preview: Dynamic System Response • Input/Output Model of Linear Dynamic Systems • Time Response of Dynamic Systems – Solutions to Differential Equations • Transient and Steady State Response • Frequency Response of Dynamic Systems – Review of Complex Variables – Frequency Response Function – Gain and Phase Characteristics • System Integration ME365 Dynamic System Response 1 Time Response – 2nd Order Example • 2st order system with zero initial conditions: – m = 0.023 kg, k = 10 N/m, b = 0.126 N s, K = 42.9 ME365 Dynamic System Response 2 Time Response – 2nd Order Example Step Response Step Response 400 1.4 350 1.2 300 1 250 Amplitude 450 1.6 0.8 200 0.6 150 0.4 100 0.2 0 50 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0 0.2 0.4 0.6 0.8 Time (sec) 1 1.2 1.4 1.6 1.8 2 Time (sec) ME365 Dynamic System Response 3 Frequency Response • Periodic (Sinusoidal) Signals cos( ) cos( ) t Amplitude 1.8 sin( ) sin( ) Im Im cos( sin( ) ) 1 ( 2 1 ( 2 ME365 Dynamic System Response ) Re A t ) 4 Frequency Response • Review of Complex Numbers (Euler Formula) Imaginary σ is the Real part ω is the Imaginary part A is the magnitude (⎪z⎥ ) φ is the phase ( Arg[z] ) Real ME365 Dynamic System Response 5 Frequency Response • Complex Arithmetic ; Ex: Finish the following complex arithmetic: (1) (3+j4)(4+j3) (2) (3+j4)/(4+j3) ME365 Dynamic System Response (3) (3+j4)(3-j4) 6 Frequency Response • Complex Conjugates • Differentiation Let are a complex conjugate pair. Note: ∗ ME365 Dynamic System Response 7 Frequency Response Input x(t) (temperature) x 2π/ω Linear System G(jω ) (Thermocouple) y 4 t ⇒Output: sin Input: x(t) = 3 sin(ωt) Steady-state output: sin ME365 Dynamic System Response Output y(t) (voltage) cos 4 sin 2 t sin cos cos cos 2 ⋅ 3sin ( ) 8 Frequency Response Input x(t) (temperature) 2π/ω x Linear System G(jω ) (Thermocouple) Output y(t) (voltage) y t ⇒ Input: x(t) = A sin(ωt) Steady-state output: sin t cos Output: sin sin cos cos ⋅ Asin ( ME365 Dynamic System Response ) 9 Frequency Response Q: How can we describe G such that we can predict (describe) the frequency dependent effect on magnitude and phase for periodic input signals? Let the input signal be a complex periodic signal: Let: x(t) = Ae jωt ⇒ Find the steady state output y(t) !! Assume y(t) = G Ae jωt , ( G can be a complex number! ) ME365 Dynamic System Response 10 Frequency Response • Frequency Response Function (FRF) Given linear system G: ⋯ ⋯ The Frequency Response Function G(jω) is: – Characterizes the steady state response of a linear time-invariant system to periodic input x(t) = Asin(ωt) : – Is the ratio between the complex output y(t) = Y e jωt of a linear system to a complex periodic input x(t) = X e jωt , i.e. G(jω) = Y / X . ME365 Dynamic System Response 11 Frequency Response Input x(t) Output y(t) Linear System G(jω ) ∠ Input: x(t) = Asin(ωt) ME365 Dynamic System Response ⇒ Steady State Output: yss(t) 12 Frequency Response • Obtaining FRF G(jω) from Differential Equations Given ODE of the system: ⋯ ⋯ (1) Let input x(t) = X e jωt . (2) Let the steady state output y(t) = Y e jωt . (3) Substitute x(t) and y(t) into the ODE and carry out the differentiation: ⋯ ⋯ (4) Solve for G = Y / X . ME365 Dynamic System Response 13 Frequency Response Ex: Frequency Response Function of a 2nd Order System (1) Let a(t) = Ae jωt. (2) Let v(t) = Ve jωt. (3) Substitute a(t) and v(t) into the ODE ... ME365 Dynamic System Response 14 Frequency Response • 1st Order System • 2nd Order System Differential Equation Differential Equation 1 2 Frequency Response Function Frequency Response Function – Magnitude – Magnitude – Phase – Phase ME365 Dynamic System Response 15 Frequency Response Ex: A thermocouple can be modeled by a 1st order ODE 0.1 0.003 Find: (1) The frequency response function (2) The steady state response of the thermocouple to an input temp. of 4 cos 30 (3) The steady state response of the thermocouple to a input temp. of 25 ME365 Dynamic System Response 4 cos 30 16 Frequency Response Ex: A thermocouple has a frequency response: 0.003 0.1 1 Find the differential equation that describes the thermocouple. ME365 Dynamic System Response 17 Frequency Response Ex: An accelerometer can be characterized by the following ODE: 4 400 800 where V is the output voltage and A is the acceleration. Find the following: (1) The frequency response function of the accelerometer. (2) The steady state response due to each of the following inputs: ME365 Dynamic System Response 18 Frequency Response Ex: (Continued) Sum all six input signals Summation of Input Signals 10 100 Total Response 80 8 6 4 2 60 0 -2 0 -4 -6 0.05 0.1 0.15 0.2 Output (V) Acceleration Sum all six output responses 40 20 0 -20 0 0.05 0.1 0.15 0.2 -40 -60 -8 -10 -80 Time (sec) -100 Time (sec) Q: What went wrong? A: Nothing went wrong, this is what it is supposed to do!!! ME365 Dynamic System Response 19 Frequency Response • 1st Order System Differential Equation Frequency Response Function – Magnitude (Gain) – Phase ME365 Dynamic System Response 20 Frequency Response • 2nd Order System Differential Equation Frequency Response Function – Magnitude (Gain) – Phase ME365 Dynamic System Response 21 Frequency Response - Bode Plot • 1st Order System Frequency Response Function – Magnitude (Gain) – Phase Bode Plot • Plot dB Magnitude vs Log Frequency () ⎜ ⎟ ⇒ dB M agnitude = 20 log 10 ⎛ G j ω ⎞ ⎝ ⎠ • Plot Phase Angle vs Log Frequency ME365 Dynamic System Response 22 Gain and Phase Characteristics dB Magnitude • 1st Order System • →0 20logK 20logK - 20 20logK - 40 20logK - 60 • 0.1 →∞ 1.0 10 1.0 10 100 1000 Phase (deg) -20 -40 -60 -80 0.1 100 1000 Frequency ωτ ME365 Dynamic System Response 23 ...
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## This note was uploaded on 12/26/2011 for the course ME 365 taught by Professor Merkle during the Fall '07 term at Purdue.

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