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.A- A0T = tA s i n (f)A s i n (wt + f)2pw� Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Spring Term 2008 Lecture 301 Reading: Nise: 10.1 • 1 Sinusoidal Frequency Response 1.1 Definitions Consider a sinusoidal waveform f(t) = A sin (ωt + φ) where A is the amplitude (in appropriate units) ω is the angular frequency (rad/s) φ is the phase (rad) In addition we can define T the period T = 2π/ω (s) f the frequency, ( f = 1/T = ω/2π) (Hz) 1copyright cD.Rowell 2008 30–1F ( t )0tI n p u ttom a s smF ( t )f r i c t i o nBR e s p o n s es t e a d y - s t a t er e s p o n s et r a n s i e n tr e s p o n s ev ( t )mv ( t )mThe Euler Formulas: We will frequently need the Euler formulas ejωt = cos (ωt) + j sin (ωt) e−jωt = cos (ωt) − j sin (ωt) or conversely cos ( ωt) =1 �ejωt + e−jωt� 2 sin ( ωt) = 1 �ejωt − e−jωt� 2j 1.2 The Steady-State Sinusoidal Response Assume a system, such as shown above, is excited by a sinusoidal input. The total response will have two components a) a transient component, and a steady-state component y(t) = yh(t) + yp(t). We define the steady-state component as the particular solution yp(t). Let the system dif-ferential equation be dny dn−1y dy dmu dm−1u du an dtn + an−1 dtn−1 + . . . + a1 dt + a0y = bm dtm + bm−1 dtm−1 + . . . + b1 dt + b0u with a complex exponential input u(t) = ejωt. Assume a particular solution yp(t) to b e of the same form as the input, that is yp(t) = Aejωt and since dkyp = A(jω)k ejωt dtk substitution into the differential equation gives: �an(jω)n + an−1(jω)n−1 + . . . + a1(jω) + a0� Aejωt = �bm(jω)m + bm−1(jω)m−1 + . . . + (b1jω) + b0� ejωt 30–2H ( s )u ( t ) = ey ( t ) = H ( jw) ejwtjwts sL i n e a r S y s t e mor bm(jω)m + bm−1(jω)m−1 + . . . + b1(jω) + b0A = an(jω)n + an−1(jω)n−1 + . . . + a1(jω) + a0 Examination of this equation shows its similarity to the transfer function H(s), in fact A = H(s)s=jω = H(jω)|so that the steady-state response yss(t) is yss(t) = yp(t) = Aejωt = H(jω)ejωt, (1) or in other words, the steady-state response to a complex exponential input is defined by the transfer function evaluated at s = jω, or along the imaginary axis of the s-plane. Note that H(jω) is in general complex. We now extend this argument to a real sinusoidal input, for example u(t) = cos (ωt) = (ejωt + e−jωt)/2. The principle of superposition for linear systems allows us to express the response as the sum of the two responses to the complex exponentials: yss(t) = 1 �H(jω)ejωt + H(−jω)e−jωt� 2 We now proceed as follows: We show that H(− jω) = H(jω) where H(jω) denotes the complex conjugate (see the • Appendix), so that yss(t) = 1 �H(jω)ejωt + H(jω)e−jωt� (2)2 • We break up H(jω) into its real and imaginary parts, H(jω) = H(jω) = and use the Euler formula to write � {H(jω)} + j� {H(jω)} � {H(jω)} − j� {H(jω)} ejωt e−jωt = = cos (ωt) + j sin (ωt) cos (ωt) − j sin (ωt) • We combine the real and imaginary parts of Eq. (2) to conclude yss(t) = � {H(jω)} cos(ωt) − � {H(jω)} sin(ωt) (3) 30–30| u ( t ) || y ( t ) |Ttu ( t ) y ( t )T i m ep e r i o dDt0ÐH ( jw ) = - | H ( jw) | = a m p l i t u d e r a t i o :| y ( t ) || u ( t ) |p h a s e s h i f t :2pDtTA m p l i t u d e� We then use the trig. identity • a cos θ − b sin θ = √a2 + b2 cos(θ + φ) to write Eq. (3) as yss(t) = H(jω) cos (ωt + � H(jω)) (4)| | where H(jω) = ��2 {H(jω)} + �2 {H(jω)} |H(jω) | = arctan ��{ H(jω)}� �{ H(jω)} Equation (4) states the answer we seek. It shows that The steady-state sinusoidal response is a sinusoid of the same angular frequency • as the input, The response differs from the input by (i) a change in amplitude as defined by • |H(jω)|, and (ii) an added phase shift � H(jω). H(jω) is known as the frequency response function. H(jω) is the magnitude of the frequency | |response function, and � H(jω) is the phase. 30–4m a s smF ( t )f r i c t i o nBv ( t )m| | � Note that if H(jω) > 1 the sinusoidal input is amplified, while if H(jω) < 1 the input is | | | |attenuated by the system. Example 1 The mechanical system has a transfer function vm(s) 1 H(s) = = F (s) ms + B where m = 1 kg, and B = 2 Ns/m. Find the steady-state response if F (t) = 10 sin(5t). 1 H(s) = s + 2 so that the frequency response function is H(jω) = H(s)s=jω = 1 =2 − jω |jω + 2 ω2 + 4 Then 1 ω H(jω) = , H(jω) = arctan � � .| |√ω2 + 4 � − 2 With ω = 5 rad/s, vss(t) = 1010|H(jω)| sin(5t + � H(jω) = √29 sin(5t − arctan 2.5) = 1.857 sin(5t − 1.1903) Example 2 Plot the variation of H(jω) and H(jω) from ω = 0 to 10 rad/s. From above 1 ω H(jω) = , and H(jω) = arctan � � .| |√ω2 + 4 � − 2 These functions are plotted below: 30–5f 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 )0 2 468 1 00- 1 0- 2 0- 3 0- 4 0- 5 0- 6 0- 7 0- 8 0- 9 00 . 50 . 40 . 30 . 20 . 10Note that • As the input frequency ω increases, the response magnitude decreases. • At low frequencies the phase is a small negative number, but as the frequency increases the phase lag increases and apparently is tending toward −90◦ at high …
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