Frequency Response of Linear Time Invariant Systems

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1 ME 328, Spring 203, Prof. Rajamani, University of Minnesota Frequency Response of Linear Time Invariant Systems Complex Numbers: Recall that every complex number has a magnitude and a phase. Example: z a + bj, j a is called the real part of z, a Re(z) b is called the imaginary part of z, b Im(z) Magnitude of z: z (a 2 + b 2 ) Re(z) 2 + Im(z) 2 Phase of z: z tan ( b a ) tan ( Im(z) Re(z) ) Bothe the magnitude and phase of a complex number are real. What is the steady state response of any LTI system for a sinusoidal input of frequency ω? Assume that the system is stable : All its poles have negative real parts. For example: X(s) ms 2 + bs + k F(s)

2 ME 328, Spring 203, Prof. Rajamani, University of Minnesota If F(s) ω s 2 + ω 2 (sinusoid) Then X(s) ms 2 + bs + k. ω s 2 + ω 2 After partial fraction expansion, and inverse Laplace transforms, we will find: x(t) Ae ζω nt sinω d t + Be ζω nt cosω d t + Csinωt + Dcosωt Amplitude of steady state oscillation C 2 + D 2 Thus, steady state oscillation is at the same frequency as the input frequency. Example: Given Y(s) 3 s + 4 U(s) If the input u(t) sin2t, find the steady state response y(t). Solution: u(t) sin2t U(s) 2 s Y(s) 3 s s A s Bs + C s A(s 2 + 4) + (Bs + C)(s + 4) 6 s 4 A(( 4) 2 + 4) 6 A As 2 + Bs B A 3 0 4A + 4C 6

3 ME 328, Spring 203, Prof. Rajamani, University of Minnesota 4C 6 4A C Y(s) 3 0. s s s s y(t) 3 0 e 4t 3 0 cos2t sin2t 3 0 e 4t Transient response 0 as t 3 0 cos2t + 3 sin2t Steady state response 5 Amplitude of steady state response? Phase of steady state response? 3 0 cos2t sin2t 0 0 [ 3 45 cos2t sin2t] [sinα cos2t + cosα sin2t] sin(2t + α) , cosα 6 45, sinα 3 Amplitude Phase α tan ( 3 6 ) tan ( 2 ) 45 It turns out that the amplitude and phase are given by: y G(jω) ω2 Check: G(s) in this example is y G(jω) ω2

4 ME 328, Spring 203, Prof. Rajamani, University of Minnesota Let s jω, j G(s) 3 s + 4 G(jω) 3 jω jω + 4. jω 4 jω 4 3(4 jω) ω ω jω ω (jω 4) 3(jω 4) j 2 ω 2 42 ω 2 6 ω 2 rad s, G(jω) j j j G(jω) ω2 ( ) + ( ) G(jω) ω2 tan ( ) tan ( 0 ) tan ( 2 ) 5 Thus, the amplitude and phase of the response turned out to be equal to G(jω) ω2 and G(jω) ω2 respectively. Generalized Result: If Y(s) G(s)U(s), where G(s) is stable with all poles having ve real parts, and u(t) sinωt, then y(t) G(jω) sin (ωt + G(jω))

5 ME 328, Spring 203, Prof. Rajamani, University of Minnesota April, 203 Calculating magnitude and phase of a complex number z a + bj, z (a 2 + b 2 ), z tan ( b a ) z a + bj (a + bj)(c dj) (a + bj)(c dj) c + dj (c + dj)(c dj) c 2 d 2 j 2 ac adj + bcj bdj2 c 2 + d 2 (ac + bd) + j(bc bd) c 2 + d 2 z Hence c 2 + d 2 (ac + bd)2 + (bc bd) 2 c 2 + d 2 a 2 c 2 + b 2 d 2 + 2acbd + b 2 c 2 + a 2 d 2 2abcd c 2 + d 2 a2 (c 2 + d 2 ) + b 2 (a 2 + d 2 ) c2 + d 2 a 2 + d 2 c 2 + d 2 a2 + d 2 c 2 + d 2 z a + bj c + dj z a2 + d 2 Mag of num c 2 + d2 Mag of den Likewise z (a + bj)(c + dj) z a 2 + d 2 c 2 + d 2 (Mag of first factor)(mag of second factor) For phase calculations, z (a + bj)(c + dj) z tan ( b a ) + tan ( d c )

6 ME 328, Spring 203, Prof. Rajamani, University of Minnesota z a + bj c + dj z tan ( b a ) tan ( d c ) Thus, calculating the magnitude and phase of a transfer function is easy and can be calculated from the magnitude and phase of the individual factors. Sample example: Let Y(s) G(s)U(s), with G(s) 9s + 4 3(s 2 + 4s + 3) If u(t) 6cos2t, find the steady state response y ss (t). Solution: G(jω) G(jω) ω2 9jω + 4 3(j 2 ω 2 + 4jω + 3) G(jω) ω j + 4 8j + 4 3( j + 3) j G(jω) ω2 tan ( 8 4 ) tan ( 24 ) radians 3 y ss (t) (6) cos(2t ), or y ss (t) cos(2t )

7 ME 328, Spring 203, Prof. Rajamani, University of Minnesota Frequency Response Plots or Bode Plots G(jω) can be plotted as a function of ω. G(jω) can be plotted as a function of ω. Together these are called the frequency response plot of G(s). The G(jω) plot describes the amplitude of steady state oscillations over all frequencies. Example: Input Force F(t) Output: x(t) mx + bx + kx F(t).. 3 Find the frequency response of this system. Find the Laplace transform of 3, using the initial conditions as zero. G(jω) ms 2 X(s) + bsx(s) + kx(s) F(s) X(s) F(s) ms 2 + bs + k.. 4 X(s) F(s) G(s) ms 2 + bs + k m(jω) 2 + bjω + k mω 2 + bjω + k (k mω 2 ) + j(bω)

8 ME 328, Spring 203, Prof. Rajamani, University of Minnesota G(jω) (k mω 2 ) 2 + b 2 ω 2 If G(jω) is plotted as a function of ω, the plot is as follows: It turns out we just need to learn the G(jω) Vs. ω plot for 5 different transfer functions (all st and 2 nd order transfer functions). We would then be able to interpret and understand frequency response of all systems. Likewise, we just need to learn the G(jω) Vs. ω plot for the same 5 transfer functions. We can then interpret the phase plot of all systems. For the mass-spring-damper system under consideration,

9 ME 328, Spring 203, Prof. Rajamani, University of Minnesota G(jω) (k mω 2 ) + j(bω) Hence G(jω) tan bω ( k mω 2) G(jω) provides the phase difference between F(t) and x(t) at various input frequencies ω. Agenda: To learn the magnitude frequency response plots for the following transfer functions ). G(s) s 2). G(s) s 2). G(s) Ts+ 4). G(s) Ts + 5). G(s) ω n 2 s 2 +2ζω n s+ω n 2 We will use these 5 magnitude frequency response plots to analyze a # of engineering problems.

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