# Studio Exercise Time Response & Frequency Response 1 st -Order Dynamic System RC Low-Pass Filter

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1 Studio Exercise Time Response & Frequency Response 1 st -Order Dynamic System RC Low-Pass Filter i i in R out Assignment: Perform a Complete e in C e Dynamic System Investigation out of the RC Low-Pass Filter 1

2 Measurements, Calculations, Manufacturer's Specifications Model Parameter ID Which Parameters to Identify? What Tests to Perform? Physical System Physical Model Math Model Experimental Analysis Assumptions and Engineering Judgement Physical Laws Model Inadequate: Modify Equation Solution: Analytical and Numerical Solution Actual Dynamic Behavior Compare Predicted Dynamic Behavior Modify or Augment Make Design Decisions Model Adequate, Performance Inadequate Model Adequate, Performance Adequate Design Complete Dynamic System Investigation 2

3 Zero-Order Dynamic System Model 3

4 Validation of a Zero-Order Dynamic System Model 4

5 1 st -Order Dynamic System Model τ = time constant K = steady-state gain t = τ Slope at t = 0 q o Kq = τ is t e τ Kq is q o t = = 0 τ 5

6 How would you determine if an experimentallydetermined step response of a system could be represented by a first-order system step response? qo() t = Kqis 1 e o () q t Kq q Kq is () t is t o 1 = e τ Kq is = e () t τ t τ qo t t t log10 1 = log10 e = Kqis τ τ Straight-Line Plot: q ( t) o log10 1 vs. t Kqis Slope = /τ 6

7 This approach gives a more accurate value of τ since the best line through all the data points is used rather than just two points, as in the 63.2% method. Furthermore, if the data points fall nearly on a straight line, we are assured that the instrument is behaving as a first-order type. If the data deviate considerably from a straight line, we know the system is not truly first order and a τ value obtained by the 63.2% method would be quite misleading. An even stronger verification (or refutation) of first-order dynamic characteristics is available from frequencyresponse testing. If the system is truly first-order, the amplitude ratio follows the typical low- and highfrequency asymptotes (slope 0 and 20 db/decade) and the phase angle approaches -90 asymptotically. 7

8 If these characteristics are present, the numerical value of τ is found by determining ω (rad/sec) at the breakpoint and using τ = 1/ω break. Deviations from the above amplitude and/or phase characteristics indicate non-first-order behavior. 8

9 What is the relationship between the unit-step response and the unit-ramp response and between the unit-impulse response and the unit-step response? For a linear time-invariant system, the response to the derivative of an input signal can be obtained by differentiating the response of the system to the original signal. For a linear time-invariant system, the response to the integral of an input signal can be obtained by integrating the response of the system to the original signal and by determining the integration constants from the zero-output initial condition. 9

10 Unit-Step Input is the derivative of the Unit-Ramp Input. Unit-Impulse Input is the derivative of the Unit- Step Input. Once you know the unit-step response, take the derivative to get the unit-impulse response and integrate to get the unit-ramp response. 10

11 System Frequency Response 11

12 Bode Plotting of 1 st -Order Frequency Response db = 20 log 10 (amplitude ratio) decade = 10 to 1 frequency change octave = 2 to 1 frequency change 12

13 Analog Electronics: RC Low-Pass Filter Time Response & Frequency Response i in e in R C i out e out e RCs + 1 R e in out i = in Cs 1 i out eout 1 1 = = when iout = 0 e RCs + 1 τ s + 1 in 13

14 Time Response to Unit Step Input Amplitude R = 15 KΩ C = 0.01 µf Time (sec) x 10-4 Time Constant τ = RC 14

15 Time Constant τ Time it takes the step response to reach 63% of the steady-state value Rise Time T r = 2.2 τ Time it takes the step response to go from 10% to 90% of the steady-state value Delay Time T d = 0.69 τ Time it takes the step response to reach 50% of the steady-state value 15

16 Frequency Response R = 15 KΩ C = 0.01 µf Gain db Phase (degrees) Frequency (rad/sec) Bandwidth = 1/τ ( ) ( ) Frequency (ra e K K 0 K ωτ + ωτ + ωτ ωτ + out 1 ( iω ) = = = tan ωτ e in i 1 1 tan 1 16

17 Bandwidth The bandwidth is the frequency where the amplitude ratio drops by a factor of = -3dB of its gain at zero or low-frequency. For a 1 st -order system, the bandwidth is equal to 1/ τ. The larger (smaller) the bandwidth, the faster (slower) the step response. Bandwidth is a direct measure of system susceptibility to noise, as well as an indicator of the system speed of response. 17

18 MatLab / Simulink Diagram Frequency Response for 1061 Hz Sine Input τ = 1.5E-4 sec Sine Wave 1 tau.s+1 First-Order Plant output output input input Clock t time 18

19 Amplitude Ratio = = -3 db Phase Angle = -45 Input Response to Input 1061 Hz Sine Wave amplitude Output time (sec) x

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### H(s) = s. a 2. H eq (z) = z z. G(s) a 2. G(s) A B. s 2 s(s + a) 2 s(s a) G(s) 1 a 1 a. } = (z s 1)( z. e ) ) (z. (z 1)(z e at )(z e at )

.7 Quiz Solutions Problem : a H(s) = s a a) Calculate the zero order hold equivalent H eq (z). H eq (z) = z z G(s) Z{ } s G(s) a Z{ } = Z{ s s(s a ) } G(s) A B Z{ } = Z{ + } s s(s + a) s(s a) G(s) a a