MIT OpenCourseWare http://ocw.mit.edu 2.004 Dynamics and Control II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.H ( s )u ( t ) = A s i n (w t + q )y ( t ) = A | H ( jw) | s i n (w t + q + ÐH ( jw) )s sL i n e a r S y s t e m| H ( s ) |jws| H ( jw) | i s a " s l i c e " o f t h e| H ( s ) | s u r f a c e a l o n g t h ei m a g i n a r y a x i s o f t h e s - p l a n e .� � Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Spring Term 2008 Lecture 311 Reading: Nise: 10.1 • • Class Handout: Frequency Response and Bode Plots 1 Sinusoidal Frequency Response (continued) In Lecture 30 we saw that the steady-state response of a linear system with transfer function H(s) to a sinusoidal input u(t) = A sin(ωt + θ) is yss(t) = A H(jω ) sin (ωt + θ + � H(jω))| | where H(jω) = {H(jω)} + �2 {H(jω)}| | ��2 H(jω) = arctan ��{H(jω)}� �{H(jω)} We note that H(jω) = H(s)s=jω ,|that is H(jω) is H(s) evaluated along the imaginary axis of the s-plane. 1copyright cD.Rowell 2008 31–1� � � 1.1 The Frequency Response of Systems with Zeros If a system has a transfer function N(s)H(s) = D(s) the frequency response function is N(jω)H(jω) = . D(jω) For complex a and b, a/b = a / b and (a/b) = b, so that | | | | | | � � a − � N(jω)��2 {N(jω)} + �2 {N(jω)} (1)|H(jω)| =||D(jω)|| = ��2 {D(jω)} + �2 {D(jω)} H(jω) = N(jω) − D(jω ) = arctan ��{N(jω)}� − arctan ��{D(jω)}� (2) �{N(jω)} �{D(jω)} Example 1 Find and plot the frequency response of s + 5 H(s) = s + 10 The frequency response function is jω + 5 H(jω) = jω + 10 and N(jω)√ω2 + 25 |H(jω)| = ||D(jω)|| = √ω2 + 100 H(jω) = N(jω) − D(jω) = arctan �ω � − arctan � ω �� � � 5 10 The following MATLAB commands were used to plot the frequency response: w=0:.2:100; sys=zpk(-5,-10,1) y=freqresp(sys,w); plot(w,squeeze(abs(y))) plot(w,squeeze(angle(y))) which produced the following plots: 31–20 2 0 4 0 6 0 8 0 1 0 00 2 0 4 0 6 0 8 0 1 0 000 . 20 . 40 . 60 . 81 . 02 01 81 61 41 21 086420F r e q u e n c y R e s p o n s e P h a s e ( d e g )F r e q u e n c y R e s p o n s e M a g n i t u d ef r e q u e n c y ( r a d / s )f r e q u e n c y ( r a d / s )We note that • When ω = 0, |H(jω )| = 0.5 and � H(jω) = 0◦. • When ω → ∞, |H(jω)| → 1 and � H(jω) → 0◦. Example 2 Find the frequency response functions for (i) a differentiator, and (ii) an integra-tor. 31–3(i) A differentiator. The transfer function is H(s) = s, so that H(jω) = jω . Then π |H(jω)| = ω, � H(jω) = 2 (or +90◦) (ii) An integrator. The transfer function is H(s) = 1/s, so that H(jω) = 1/jω. Then 1 π |H(jω)| = ω, � H(jω) = − 2 (or -90◦) 1.2 The Frequency Response of a Second-Order System Consider the unity-gain second-order system: ω2 H(s) = n s2 + 2ζωns + ω2 n The frequency response is ω2 H(jω) = n −ω2 + j2ζωnω + ω2 n so that ω2 (3)|H(jω)| =�(ωn 2 − ω2)2 n + (2ζωnω)2 2ζωnω � H(jω) = − tan−1 ω2 (4) n − ω2 Equations (3) and (4) show the following: When ω = 0, H(jω) = 1 | | � H(jω) = 0 When H(jω) 0ω → ∞, | | → � H(jω) → −π (or -180◦) 1 When ω = ωn, H(jω) =| | 2ζ π � H(jω) = − 2 (or -90◦) The response of the system to frequencies close to the undamped natural frequency clearly depends on the damping ratio ζ. For a lightly damped system ζ < 0.5, H(jωn) > 1 and | |the system demonstrates amplification due to resonance. Differentiation of Eq. (4) shows the ωpeak, the frequency of the peak response is not ωn but is in fact ωp = ωn�1 − 2ζ2 31–40 1 2 3 401234560 1 2 3 4- 1 8 0- 1 5 0- 1 2 0- 9 0- 6 0- 3 00N o r m a l i z e d f r e q u e n c yN o r m a l i z e d f r e q u e n c yÐH ( jw)| H ( jw ) |z = 0 . 1z = 0 . 1z = 0 . 10 . 5125w / wnw / wn12210 . 550 . 20 . 50 . 20 . 2for ζ < 1/√2, and 1 |H(jωp)| =2ζ�1 − ζ2 . Frequency response plots for several values of ζ are shown below: Example 3 A tall slender structure, excited by wind forces, is modeled as a mass-spring-damper system. Find the frequency response of the displacement x of the building to a sinusoidal wind loading. 31–50 1 0 2 0 3 0 4 0 5 000 . 0 10 . 0 20 . 0 30 . 0 40 . 0 50 . 0 60 . 0 7- 1 8 0- 1 6 0- 1 4 0- 1 2 0- 1 0 0- 8 0- 6 0- 4 0- 2 00 f r e q u e n c y ( r a d / s )F r e q u e n c y R e s p o n s e M a g n i t u d eF r e q u e n c y R e s p o n s e P h a s e ( d e g )mKBFs( w i n d f o r c e )m1Fs( w i n d f o r c e )x B - e q u i v a l e n t d a m p i n gc o e f f i c i e n tK - e q u i v a l e n t l a t e r a ls t i f f n e s sxThe transfer function is X(s) 1 1 H(s) = = H(jω) = Fs(s) ms2 + Bs + K → (K − mω2) + jBω 1 |H(jω)| = �(K − mω2)2 + (Bω)2 Bω � H(jω) = − tan−1 K − mω2 With …
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