Lecture 30: Sinusoidal System Response

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) A A s in ( w t + f ) A s in ( f ) 0 -A t T = 2 p w 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) 1 c D.Rowell 2008 copyright � 30–1 The Euler Formulas: We will frequently need the Euler formulas ejωt = cos (ωt) + j sin (ωt) e−jωt = cos (ωt) − j sin (ωt) or conversely � 1 � jωt e + e−jωt 2 � 1 � jωt e − e−jωt sin (ωt) = 2j cos (ωt) = 1.2 The Steady-State Sinusoidal Response vm (t) F (t) vm (t) 0 t In p u t F (t) m a s s m fr ic tio n B tr a n s ie n t re s p o n s e s te a d y -s ta te re s p o n s e o t R e s p o n s e 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 an dn y dn−1 y dy dm u dm−1 u du + a + . . . + a + a y = b + b + . . . + b1 + b0 u n−1 1 0 m m−1 n n−1 m m−1 dt dt dt dt dt dt with a complex exponential input u(t) = ejωt . Assume a particular solution yp (t) to be of the same form as the input, that is yp (t) = Aejωt and since dk yp = 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 + . . . + (b1 jω) + b0 ejωt 30–2 or A= bm (jω)m + bm−1 (jω)m−1 + . . . + b1 (jω) + b0 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. u (t) = e L in e a r S y s te m jw t H (s ) y s s ( t) = H ( jw ) e jw t 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 � 1� jωt −jωt yss (t) = H(jω)e + H(jω)e 2 (2) • We break up H(jω) into its real and imaginary parts, H(jω) = � {H(jω)} + j� {H(jω)} H(jω) = � {H(jω)} − j� {H(jω)} and use the Euler formula to write ejωt = cos (ωt) + j sin (ωt) e−jωt = 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) 30–3 (3) • 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 � � �2 {H(jω)} + �2 {H(jω)} � � � {H(jω)} H(jω) = arctan � {H(jω)} |H(jω)| = 0 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. A m p litu d e u (t) y (t) a m p litu d e r a tio : |u (t)| |H ( jw ) | = |y (t)| |u (t)| |y (t)| t T im e 0 D t p e r io d T p h a s e s h ift: 30–4 Ð H ( jw ) = - 2 p D t T 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 vm (t) m a s s m F (t) fr ic tio n B has a transfer function vm (s) 1 = F (s) ms + B where m = 1 kg, and B = 2 Ns/m. Find the steady-state response if F (t) = 10 sin(5t). H(s) = 1 s+2 so that the frequency response function is 1 2 − jω H(jω) = H(s)|s=jω = = 2 jω + 2 ω +4 H(s) = Then 1 |H(jω)| = √ , 2 ω +4 With ω = 5 rad/s, � � ω� H(jω) = arctan − . 2 vss (t) = 10 |H(jω)| sin(5t + � H(jω) 10 = √ sin(5t − arctan 2.5) 29 = 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 |H(jω)| = √ , and 2 ω +4 These functions are plotted below: � 30–5 � ω� H(jω) = arctan − . 2 0 .5 F r e q u e n c y R e s p o n s e M a g n itu d e 0 .4 0 .3 0 .2 0 .1 fre q u e n c y (ra d /s ) F re q u e n c y R e s p o n s e P h a s e (d e g ) 0 0 0 2 4 6 8 1 0 -1 0 -2 0 -3 0 -4 0 -5 0 -6 0 -7 0 -8 0 -9 0 Note 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 frequencies. 30–6 Appendix: Evaluation of H(−jω). We start with H(jω) = so that H(−jω) = Note that bm (jω)m + bm−1 (jω)m−1 + . . . + b1 (jω) + b0 an (jω)n + an−1 (jω)n−1 + . . . + a1 (jω) + a0 bm (−jω)m + bm−1 (−jω)m−1 + . . . + b1 (−jω) + b0 an (−jω)n + an−1 (−jω)n−1 + . . . + a1 (−jω) + a0 (jω)k = (−1)k/2 ω k j(−1)(k−1)/2 ω k (−jω)k = (−1)k/2 ω k −j(−1)(k−1)/2 ω k k even k odd k even k odd Thus in both H(jω) and H(−jω) • The terms with even powers of ±jω in the numerator and denominator of H(jω) and H(−jω) generate real terms, while • the terms with odd powers of ±jω generate imaginary terms. With these substitutions, comparison of H(jω) and H(−jω) shows • The real terms (even powers of ±jω) are the same, while • The imaginary terms (odd powers of ±jω) have opposite signs leading to the conclusion H(−jω) = H(jω). 30–7