System Analysis in the Frequency Domain PDF

C H 9 A P T E R System Analysis in the Frequency Domain CHAPTER OUTLINE CHAPTER OBJECTIVES Frequency Response of First-Order Systems 526 Frequency Response of Higher-Order Systems 538 9.3 Frequency Response Applications 550 9.4 Filtering Properties of Dynamic Systems 562 9.5 Response to General Periodic Inputs 569 9.6 System Identification from Frequency Response 574 9.7 Frequency Response Analysis Using MATLAB 579 9.8 Chapter Review 581 Problems 582 When you have ﬁnished this chapter, you should be able to 9.1 9.2 1. Sketch frequency response plots for a given transfer function, and use the plots or the transfer function to determine the steady-state response to a sinusoidal input. 2. Compute the frequencies at which resonance occurs, and determine a system’s bandwidth. 3. Analyze vibration isolation systems and the effects of base motion and rotating unbalance. 4. Determine the steady-state response to a periodic input, given the Fourier series description of the input. 5. Estimate the form of a transfer function and its parameter values, given frequency response data. 6. Use MATLAB as an aid in the preceding tasks. he term frequency response refers to how a system responds to a periodic input, such as a sinusoid. An input f (t) is periodic with a period P if f (t + P) = f (t) for all values of time t, where P is a constant called the period. Periodic inputs are commonly found in many applications. The most common perhaps is ac voltage, which is sinusoidal. For the common ac frequency of 60 Hz, the period is P = 1/60 s. Rotating unbalanced machinery produces periodic forces on the supporting structures, internal combustion engines produce a periodic torque, and reciprocating pumps produce hydraulic and pneumatic pressures that are periodic. Frequency response analysis focuses on harmonic inputs, such as sines and cosines. A sine function has the form A sin ωt, where A is its amplitude and ω is its frequency in radians per unit time. Note that a cosine is simply a sine shifted by 90◦ or π/2 rad, as T 525 526 CHAPTER 9 System Analysis in the Frequency Domain cos ωt = sin(ωt + π/2). Not all periodic inputs are sinusoidal or cosinusoidal, but an important result of mathematics, called the Fourier series, enables us to represent a periodic function as the sum of a constant term plus a series of cosine and sine terms of different frequencies and amplitudes. Thus we will be able to apply the results of this chapter to any such periodic input. The transfer function is central to understanding and applying frequency response methods. The transfer function enables us to obtain a concise graphical description of a system’s frequency response. This graphical description is called a frequency response plot. We analyze the response of ﬁrst-order systems to sinusoidal inputs in Section 9.1. In Section 9.2 we generalize the method to higher-order systems. Section 9.3 discusses several phenomena and applications of frequency response, including beating, resonance, the effect of base motion and rotating unbalance, and instrument design. In Section 9.4 we introduce the important concept of bandwidth, which enables us to develop a concise quantitative description of a system’s frequency response, much like the time constant characterizes the step response. Section 9.5 shows how to analyze the response due to general periodic inputs that can be described with a Fourier series. Experiments involving frequency response can often be used to determine the form of the transfer function of a system and to estimate the numerical values of the parameters in the transfer function. Section 9.6 gives some examples of this process. MATLAB has several useful functions for obtaining and for analyzing frequency response. These are treated in Section 9.7. ■ 9.1 FREQUENCY RESPONSE OF FIRST-ORDER SYSTEMS In this chapter we will frequently need to obtain the magnitude and angle of complex numbers. A complex number N can be represented in rectangular form as N = x + j y, where x is the real part and y is the imaginary part. The number can be plotted in two dimensions with x as the abscissa (on the horizontal axis) and y as the ordinate (on the vertical axis). We can think of the number as a two-dimensional vector whose head is at the point (x, y) and whose tail is at the origin. The vector length is the magnitude M of the number and the vector angle φ is measured counterclockwise from the positive real axis. In this form, the magnitude and angle of the number can be calculated from |N | = x 2 + y 2 . The complex conjugate of N is x − y j. See Figure 9.1.1a. Another form is the complex exponential form: N = Me jφ = M(cos φ + j sin φ) Note that the complex conjugate of N is Me− jφ . 9.1.1 PRODUCTS AND RATIOS OF COMPLEX NUMBERS The complex exponential form can be used to show that the magnitude and angle of a number consisting of products and ratios of complex numbers N1 , N2 , . . . , can be calculated as follows. N= M1 e jφ1 M2 e jφ2 M1 M2 j (φ1 +φ2 −φ3 −φ4 ) N1 N2 = = e jφ jφ 3 4 N3 N4 M3 e M4 e M3 M4 9. 1 Imaginary y Frequency Response of First-Order Systems Imaginary 4 x 1 yj 3 1 4j M 538 22 2 x Real 2488 3 Real M 2y x 2 yj 22 2 5j (a) 25 (b) Thus N = |N |e jφ , where |N | = M1 M2 M3 M4 φ = φ1 + φ2 − φ3 − φ4 So the magnitudes combine as products and ratios, and the angles combine as sums and differences. For example, consider the number N= −2 − 5 j 3 + 4j The magnitude of N can be calculated by computing the ratio of the magnitudes of the numerator and denominator, as follows: √ −2 − 5 j (−2)2 + (−5)2 29 |−2 − 5 j| √ = = = |N | = 2 2 3 + 4j |3 + 4 j| 5 3 +4 The angle of N , denoted by N , is the difference between the angle of the numerator and the angle of the denominator. These angles are shown in Figure 9.1.1b, which is a vector representation of the complex numbers in the numerator and denominator. From this diagram we can see that the angles are given by 5 = 180◦ + 68◦ = 248◦ 2 4 (3 + 4 j) = tan−1 = 53◦ 3 (−2 − 5 j) = 180◦ + tan−1 Thus and √ N= 29 195◦ = 5 N = 248◦ − 53◦ = 195◦ √ 29 (cos 195◦ + j sin 195◦ ) = −1.04 + 0.28 j 5 527 Figure 9.1.1 Vector representation of complex numbers. 528 CHAPTER 9 System Analysis in the Frequency Domain 9.1.2 COMPLEX NUMBERS AS FUNCTIONS OF FREQUENCY In our applications in this chapter, complex numbers will be functions of a frequency ω, as for example, N= −2 − j5ω 3 + j4ω but the same methods apply for obtaining the magnitude and the angle of N, which will then be functions of ω. Thus, √ √ 22 + 5 2 ω 2 4 + 25ω2 =√ |N | = √ 32 + 4 2 ω 2 9 + 16ω2 φ = N = (−2 − j5ω) − (3 + j4ω) 5ω 4ω = 180◦ + tan−1 − tan−1 2 3 Once a value is given for ω, we can compute |N | and φ. 9.1.3 FREQUENCY RESPONSE PROPERTIES The methods of this chapter pertain to any stable, linear, time-invariant (LTI) system. The basic frequency response property of such systems is summarized in Table 9.1.1 and Figure 9.1.2. Any linear, time-invariant system, stable or not, has a transfer function, say T (s). If a sinusoidal input of frequency ω is applied to such a system, and if the system is stable, the transient response eventually disappears, leaving a steady-state response that is sinusoidal with the same frequency ω as the input, but with a different amplitude and shifted in time relative to the input. To prove this result, suppose the system transfer function T (s) is of order n, the output is x(t), and the input is f (t) = A sin ωt. Then X (s) = F(s)T (s) = s2 Aω T (s) + ω2 Table 9.1.1 Frequency response of a stable LTI system. The transfer function T (s) with s replaced by jω is called the frequency transfer function. Therefore, the frequency transfer function is a complex function of ω, and it has a magnitude and an angle, just as any complex number. If the system is stable, the magnitude M of the frequency transfer function T ( jω) is the ratio of the sinusoidal steady-state output amplitude over a sinusoidal input amplitude. The phase shift of the steady-state output relative to the input is the angle of T ( jω). Thus, denoting the input by A sin ωt and the steady-state output by B sin(ωt + φ), we have B ≡ M(ω) A φ(ω) = T ( jω) |T ( jω)| = B = AM(ω) where |T ( jω)| and T ( jω) denote the magnitude and angle of the complex number T ( jω). Because of equation (1), M is called the amplitude ratio or the magnitude ratio. These results are illustrated in Figure 9.1.2. (1) (2) (3) 9. 1 Frequency Response of First-Order Systems Im Sinusoidal input Stable, linear system A sin t T(s) Steady-state response B sin(t 1 ) T( j) B 5 T( j)u A u 5 \ T( j) M5 M Re which can be expressed as a partial-fraction expansion. s2 C2 C1 Aω + + ··· T (s) = 2 +ω s + jω s − jω The factors containing s + jω and s − jω correspond to the term s 2 + ω2 , while the remaining terms in the expansion correspond to the factors introduced by the denominator of T (s). If the system is stable, all these factors will be negative or have negative real parts. Thus the response will have the form x(t) = C1 e− jωt + C2 e jωt + m Di e−αi t sin(βi t + φi ) + i=1 n Di e−αi t i=m+1 where the roots of T (s) are si = −αi ± βi j for i = 1, . . . , m and si = −αi for i = 1 + m, . . . , n. The steady-state response is thus given by xss (t) = C1 e− jωt + C2 e jωt We can evaluate C1 and C2 as follows. Note that T ( jω) = |T ( jω)|e jφ , where φ = T ( jω). A Aω A C1 = T (s) 2 (s + jω) = − T (− jω) = − |T ( jω)|e− jφ 2 s +ω 2j 2j s=− jω A Aω A T ( jω) = |T ( jω)|e jφ C2 = T (s) 2 (s − jω) = 2 s +ω 2 j 2 j s= jω The steady-state response can thus be expressed as xss (t) = − A A |T ( jω)|e− jφ e− jωt + |T ( jω)|e jφ e jωt 2j 2j = |T ( jω)|A e j (ωt+φ) − e− j (ωt+φ) 2j or x ss (t) = |T ( jω)|A sin(ωt + φ) because e j (ωt+φ) − e− j (ωt+φ) = sin(ωt + φ) 2j Thus at steady state, the output amplitude is |T ( jω)|A and the phase shift is φ = T ( jω). A speciﬁc example will help to clarify these concepts. 529 Figure 9.1.2 Frequency response of a stable linear system. 530 CHAPTER 9 System Analysis in the Frequency Domain 9.1.4 FREQUENCY RESPONSE OF τ ẏ + y = f (t) Consider the linear model having a time constant τ . τ ẏ + y = f (t) (9.1.1) The transfer function is T (s) = Y (s) 1 = F(s) τs + 1 (9.1.2) If the input is sinusoidal, f (t) = A sin ωt, we can use the Laplace transform to show that the forced response is Aωτ 1 −t/τ y(t) = sin ωt e − cos ωt + (9.1.3) 1 + ω2 τ 2 ωτ For t ≥ 4τ , the transient response has essentially disappeared, and we can express the steady-state response as A A (sin ωt − ωτ cos ωt) = √ sin(ωt + φ) 2 2 1+ω τ 1 + ω2 τ 2 where we have used the identity yss (t) = (9.1.4) sin(ωt + φ) = cos φ sin ωt + sin φ cos ωt with φ = − tan−1 ωτ . If we substitute s = jω into the transfer function (9.1.2), we obtain T ( jω) = The magnitude is |T ( jω)| = 1 1 + jωτ 1 = √ 1 1 + jωτ 1 + ω2 τ 2 (9.1.5) and the angle is φ = T ( jω) = − (1 + jωτ ) = − tan−1 (ωτ ) (9.1.6) These results are summarized in Table 9.1.2. So, if f (t) = A sin ωt, the steady state response of τ ẏ + y = f (t) obtained from (9.1.5) and (9.1.6) is identical to (9.1.4), as it should be. The point is that the transfer function provides an easier way to obtain the steady state response compared to the Laplace transform solution method, if the forcing function is sinusoidal. Table 9.1.2 Frequency response of the model τ ẏ + y = f (t). M= 1 |Y | = √ |F| 1 + ω2 τ 2 (1) φ = − tan−1 (ωτ ) (2) f (t) = A sin ωt (3) B = AM (4) yss (t) = B sin(ωt + φ) (5) 9. 1 Frequency Response of First-Order Systems 531 Note that for a stable system, the free response disappears eventually, so the steady state response is independent of the initial conditions. Frequency Response of a Mass with Damping E X A M P L E 9.1.1 ■ Problem Consider a mass subjected to a sinusoidal applied force f (t). The mass is m = 0.2 kg and the damping constant is c = 1 N · s/m. If v is the speed of the mass, then the equation of motion is 0.2v̇ + v = f (t) where f (t) = sin ωt and ω is the oscillation frequency of the applied force. The initial speed is v(0) = 0. Find the total response for two cases: (a) ω = 15 rad/s and (b) ω = 60 rad/s. ■ Solution This equation is of the form of (9.1.1) with τ = 0.2 and y = v. From (9.1.3) the response is 0.2ω v(t) = 1 + (0.2)2 ω2 e −5t 1 sin ωt − cos ωt + 0.2ω The transient response contains the exponential e−5t , which is essentially zero for t > 4/5 s. The steady-state response from (9.1.4) with A = 1 and y = v, is 1 vss (t) = sin(ωt + φ) 1 + (0.2)2 ω2 where from (9.1.6), φ = − tan−1 (0.2ω). The √ steady-state response can be seen to oscillate at the input frequency ω with an amplitude of 1/ 1 + 0.04ω2 , and a phase shift of φ relative to the input. Since the phase shift is negative, a peak in the input is followed by a peak in the response a time |φ|/ω later. If φ is in radians and ω is in radians per second, the time shift |φ|/ω will be in seconds. Figure 9.1.3(a) shows the total response for the case where ω = 15. Part (b) shows the case where ω = 60. Note that the transient response has in both cases essentially disappeared by t = 4/5 as predicted. Note also that the amplitude of the response is much smaller at the higher frequency. This is because the inertia of the mass prevents it from reacting to a rapidly changing input. Although the time shift is also smaller at the higher frequency, this does not indicate a faster response, because the peak attained is much smaller. Circuit Response to a Step-Plus-Cosine Input ■ Problem The model of the voltage v2 across the capacitor in a series RC circuit having an input voltage v1 is dv2 RC + v2 = v1 (t) dt Suppose that RC = 0.02 s and that the applied voltage consists of a step function plus a cosine function: v1 (t) = 5u s (t) + 3 cos 80t. Obtain the circuit’s steady-state response. ■ Solution We can apply the principle of superposition here to separate the effects of the step input from those of the cosine input. Note that if only a step voltage of 5 were applied, the steady-state response to the step would be 5 (this can be shown by setting dv2 /dt = 0 in the model to see that v2 = v1 = 5 at steady state). Next ﬁnd the steady-state response due to the cosine function 3 cos 80t. To do this, we can use the same formulas developed for the amplitude and phase shift E X A M P L E 9.1.2 532 Figure 9.1.3 Response of the model 0.2v̇ + v = sin ωt: (a) ω = 15 and (b) ω = 60. CHAPTER 9 System Analysis in the Frequency Domain 1 0.8 f(t) = sin(15t) 0.08 0.6 0.4 v(t) 0.2 0 –0.2 –0.4 –0.6 –0.8 –1 0 0.25 0.5 0.75 t 1 1.25 1.5 (a) 1 0.8 f(t) = sin(60t) 0.6 0.02 0.4 0.2 v(t) 0 –0.2 –0.4 –0.6 –0.8 –1 0 0.1 0.2 0.3 0.4 t 0.5 0.6 0.7 0.8 (b) due to a sine function, but express the response in terms of a cosine function. From (9.1.5) and (9.1.6) with A = 3, 3 B = AM = 1 + (80)2 (0.02)2 = 1.59 φ = − tan−1 [(80)(0.02)] = − tan−1 (1.6) = −1.012 rad The steady-state response is v2 (t) = 5 + 1.59 cos(80t − 1.012). Thus at steady state, the voltage oscillates about the mean value 5 V with an amplitude of 1.59 and a frequency of 80 rad/s. 9. 1 Frequency Response of First-Order Systems 533 9.1.5 THE LOGARITHMIC PLOTS Inspecting (9.1.5) reveals that the steady-state amplitude of the output decreases as the frequency of the input increases. The larger τ is, the faster the output amplitude decreases with frequency. At a high frequency, the system’s “inertia” prevents it from closely following the input. The larger τ is, the more sluggish is the system response. This also produces an increasing phase lag as ω increases. Figure 9.1.4 shows the magnitude ratio and phase curves for two values of τ . In both cases, the magnitude ratio is close to 1 at low frequencies and approaches 0 as the frequency increases. The rate of decrease is greater for systems having larger time constants. The phase angle is close to 0 at low frequencies and approaches −90◦ as the frequency increases. Logarithmic scales are usually used to plot the frequency response curves. There are two reasons for using logarithmic scales. The curves of M and φ versus ω can be sketched more easily if logarithmic axes are used because they enable us to add or subtract the magnitude plots of simple transfer functions to sketch the plot for a transfer function composed of the product and ratio of simpler ones. Logarithmic scales also can display a wider variation in numerical values. When plotted using logarithmic scales, the frequency response plots are frequently called Bode plots, after H. W. Bode, who applied these techniques to the design of ampliﬁers. Keep in mind the following basic properties of logarithms: x = log x − log y log (x y) = log x + log y log y log x n = n log x When using logarithmic scales the amplitude ratio M is speciﬁed in decibel units, denoted dB. (The decibel is named for Alexander Graham Bell.) The relationship between a number M and its decibel equivalent m is m = 10 log M 2 = 20 log M dB (9.1.7) 1 Figure 9.1.4 Magnitude ratio and phase angle of the model τ ẏ + y = f (t) for τ = 2 and τ = 20. 0.8 =2 M 0.6 0.4 = 20 0.2 0 0 0.5 1 1.5 2 2.5 3 (rad/time) 3.5 4 4.5 5 1 1.5 2 2.5 3 (rad/time) 3.5 4 4.5 5 (degrees) 0 –20 =2 –40 –60 = 20 –80 –100 0 0.5 534 CHAPTER 9 System Analysis in the Frequency Domain where the logarithm is to the base 10. For example, the M = 10 corresponds to 20 dB; the M = 1 corresponds to 0 dB; numbers less than 1 have negative decibel values. It is common practice to plot m(ω) in decibels versus log ω. For easy reference, φ(ω) is also plotted versus log ω, and it is common practice to plot φ in degrees. 9.1.6 LOGARITHMIC PLOTS FOR τ ẏ + y = f (t) From equation (1) of Table 9.1.2, M=√ 1 (9.1.8) 1 + ω2 τ 2 So we have m(ω) = 20 log √ 1 1 + τ 2 ω2 = 20 log(1) − 10 log(1 + τ 2 ω2 ) = −10 log(1 + τ 2 ω2 ) (9.1.9) Figure 9.1.5 shows the logarithmic plots for the two systems whose plots with rectilinear axes were given in Figure 9.1.4. Note that the shape of the m versus log ω curve is very different from the shape of the M versus ω curve. The logarithmic plots can be confusing for beginners; take time to study them. Remember that m = 0 corresponds to a magnitude ratio of M = 1. Positive values of m correspond to M > 1, which means that the system ampliﬁes the input. Negative values of m correspond to M < 1, which means the system attenuates the input. When you need to obtain values of M from a plot of m versus log ω, use the relation M = 10 m/20 (9.1.10) For example, m = 50 corresponds to M = 316.2; √ m = −15 corresponds to M = 0.1778; and m = −3.01 corresponds to M = 1/ 2 = 0.7071. m (decibels) 0 =2 –10 –20 –30 = 20 –40 –50 10–2 0 (degrees) Figure 9.1.5 Semilog plots of log magnitude ratio and phase angle of the model τ ẏ + y = f (t) for τ = 2 and τ = 20. 10–1 (rad/time) 100 101 100 101 =2 –20 –40 –60 –80 –100 10–2 = 20 10–1 (rad/time) 9. 1 Frequency Response of First-Order Systems m (decibels) 5 Figure 9.1.6 Asymptotes and corner frequency ω = 1/τ of the model 1/(τ s + 1). 0 –5 –10 20 dB 3.01 dB –15 –20 –25 10–1 100 101 100 101 (degrees) 0 –45 –90 –1 10 To sketch the logarithmic plot of m versus log ω, we approximate m(ω) in three frequency ranges. For τ ω 1, 1 + τ 2 ω2 ≈ 1, and (9.1.9) gives m(ω) ≈ −10 log 1 = 0 For τ ω 535 (9.1.11) 1, 1 + τ 2 ω2 ≈ τ 2 ω2 , and (9.1.9) gives m(ω) ≈ −10 log τ 2 ω2 = −20 log τ ω = −20 log τ − 20 log ω (9.1.12) This gives a straight line versus log ω. This line is the high-frequency asymptote. Its slope is −20 dB/decade, where a decade is any 10 : 1 frequency range. At ω = 1/τ , (9.1.12) gives m(ω) = 0. This is useful for plotting purposes but does not represent the true value of m at that point because (9.1.12) was derived assuming τ ω 1. For ω = 1/τ , (9.1.9) gives m(ω) = −10 log 2 = −3.01. Thus, at ω = 1/τ , m(ω) is 3.01 dB below the low-frequency asymptote given by (9.1.11). The low-frequency and high-frequency asymptotes meet at ω = 1/τ , which is the breakpoint frequency. It is also called the corner frequency. The plot is shown in Figure 9.1.6 (upper plot). The phase angle is given by φ = − tan−1 (ωτ ), and the curve of φ versus ω is constructed as follows. For ωτ 1, this equation gives φ(ω) ≈ tan−1 (0) = 0◦ . For ωτ = 1, φ(ω) = − tan−1 (1) = −45◦ , and for ωτ 1, φ(ω) ≈ − tan−1 (∞) = −90◦ . Because the phase angle is negative, the output “lags” behind the input sine wave. Such a system is called a lag system. The φ(ω) curves are easily sketched using these facts and are shown in Figure 9.1.6 (lower plot). 9.1.7 A COMMON FORM HAVING NUMERATOR DYNAMICS An example of a ﬁrst order system having numerator dynamics is the transfer function T (s) = K τ1 s + 1 τ2 s + 1 (9.1.13) 536 Figure 9.1.7 Electrical and mechanical examples having the transfer function form K(τ1 s + 1)/(τ2 s + 1). CHAPTER 9 System Analysis in the Frequency Domain R1 1 2 y k1 C vs x R2 vo k2 c (a) (b) An example of this form is the transfer function of the lead compensator circuit analyzed in Chapter 6, and shown again in Figure 9.1.7a. Its transfer function is R1 R2 Cs + R2 Vo (s) = (9.1.14) T (s) = Vs (s) R1 R2 Cs + R1 + R2 which can be rearranged as (9.1.13) where R1 R2 C τ2 τ1 = R 1 C τ2 = K = (9.1.15) R1 + R2 τ1 A mechanical example of this form is shown in Figure 9.1.7b. The equation of motion is c ẋ + (k1 + k2 )x = c ẏ + k1 y With y as the input, the transfer function is cs + k1 X (s) = Y (s) cs + k1 + k2 which can be rearranged as (9.1.13), where c c τ2 τ2 = K = τ1 = k1 k1 + k2 τ1 T (s) = E X A M P L E 9.1.3 A Model with Numerator Dynamics ■ Problem Find the steady-state response of the following system: ẏ + 5y = 4ġ + 12g if the input is g(t) = 20 sin 4t. ■ Solution First obtain the transfer function. T (s) = Y (s) 4s + 12 s+3 = =4 G(s) s+5 s+5 Here ω = 4, so we substitute s = 4 j to obtain T ( jω) = 4 Then, M = |T ( jω)| = 4 3 + 4j 3 + jω =4 5 + jω 5 + 4j √ |3 + 4 j| 32 + 4 2 = 4√ = 3.123 |5 + 4 j| 52 + 4 2 (9.1.16) (9.1.17) 9. 1 Frequency Response of First-Order Systems The phase angle is found as follows: 3 + jω 4 φ = T ( jω) = 5 + jω ω ω = 0◦ + tan−1 − tan−1 3 5 = 4 + (3 + jω) − (5 + jω) Substitute ω = 4 to obtain φ = tan−1 4 4 − tan−1 = 0.253 rad 3 5 Thus the steady-state response is yss (t) = 20M sin(4t + φ) = 62.46 sin(4t + 0.253) 9.1.8 SKETCHING PLOTS USING ASYMPTOTES Substituting s = jω into the transfer function (9.1.13) gives T ( jω) = K τ1 ωj + 1 τ2 ωj + 1 (9.1.18) Thus, (τ1 ω)2 + 1 M(ω) = |K | (τ2 ω)2 + 1 m(ω) = 20 log |K | + 10 log [(τ1 ω)2 + 1] − 10 log [(τ2 ω)2 + 1] (9.1.19) Thus, the plot of m(ω) can be obtained by subtracting the plot of τ2 s + 1 from that of τ1 s + 1. The scale is then adjusted by 20 log |K |. The sketches in Figure 9.1.8 are for K = 1, so that 20 log K = 0. The term τ1 s + 1 causes the curve to break upward at ω = 1/τ1 . The term τ2 s + 1 causes the curve to break downward at ω = 1/τ2 . If 1/τ1 > 1/τ2 , the composite curve looks like Figure 9.1.8a. If 1/τ1 < 1/τ2 , the plot is given by Figure 9.1.8b. The plots of m(ω) were obtained by using only the asymptotes of the terms τ1 s + 1 and τ2 s + 1 without using the 3-dB corrections at the corner frequencies 1/τ1 , and 1/τ2 . This sketching technique enables the designer to understand the system’s general behavior quickly. From (9.1.18), the phase angle is φ(ω) = K + (τ1 ωj + 1) − (τ2 ωj + 1) = K + tan−1 (τ1 ω) − tan−1 (τ2 ω) (9.1.20) where K = 0 if K > 0. The plot can be found by combining the plots of φ(ω) for K , τ1 s + 1, and τ2 s + 1, using the low-frequency, corner frequency, and high-frequency values of 0◦ , 45◦ , and 90◦ for the terms of the form τ s + 1, supplemented by evaluations of (9.1.20) in the region between the corner frequencies. This sketching technique is more accurate when the corner frequencies are far apart. 537 538 CHAPTER 9 System Analysis in the Frequency Domain 1 , 2 Figure 9.1.8 The log magnitude plots for the transfer function form (τ1 s + 1)/(τ2 s + 1) for (a) τ1 < τ2 and (b) τ1 > τ2 . 0 Slope 5 220 dB/decade m (dB) 220 log 1 2 1 2 5 1y2 5 1y1 log (a) 1 . 2 20 log 1 2 1 2 Slope 5 20 dB/decade m (dB) 0 5 1y1 log 5 1y2 (b) 9.2 FREQUENCY RESPONSE OF HIGHER-ORDER SYSTEMS The results of Section 9.1 can be easily extended to a stable, invariant, linear system of any order. The general form of a transfer function is T (s) = K N1 (s)N2 (s) . . . D1 (s)D2 (s) . . . (9.2.1) 9.2 Frequency Response of Higher-Order Systems where K is a constant real number. In general, if a complex number T ( jω) consists of products and ratios of complex factors, such that T ( jω) = K N1 ( jω)N2 ( jω) . . . D1 ( jω)D2 ( jω) . . . (9.2.2) where K is constant and real, then from the properties of complex numbers |T ( jω)| = |K ||N1 ( jω)||N2 ( jω)| . . . |D1 ( jω)||D2 ( jω)| . . . (9.2.3) In decibel units, this implies that m(ω) = 20 log |T ( jω)| = 20 log |K | + 20 log |N1 ( jω)| + 20 log |N2 ( jω)| + · · · − 20 log |D1 ( jω)| − 20 log |D2 ( jω)| − · · · (9.2.4) That is, when expressed in logarithmic units, multiplicative factors in the numerator of the transfer function are summed, while those in the denominator are subtracted. We can use this principle graphically to add or subtract the contribution of each term in the transfer function to obtain the plot for the overall system transfer function. For the form (9.2.1), the phase angle is φ( jω) = T ( jω) = K + N1 ( jω) + N2 ( jω) + · · · − D1 ( jω) − D2 ( jω) − · · · (9.2.5) The phase angles of multiplicative factors in the numerator are summed, while those in the denominator are subtracted. This enables us to build the composite phase angle plot from the plots for each factor. 9.2.1 COMMON TRANSFER FUNCTION FACTORS Most transfer functions occur in the form given by (9.2.1). In addition, the factors Ni (s) and Di (s) usually take the forms shown in Table 9.2.1. We have already obtained the frequency response plots for form 3. The effect of a multiplicative constant K , which is form 1, is to shift the m curve up by 20 log |K |. If K > 0, the phase plot is unchanged because the angle of a positive number is 0◦ . If K < 0, the phase plot is shifted down by 180◦ because the angle of a negative number is −180◦ . Table 9.2.1 Common factors in the transfer function form: T (s) = K Factor Ni (s) or Di (s) 1. Constant, K 2. s n 3. τ s + 1 4. s + 2ζ ωn s + 2 ωn2 = s ωn 2 s + 2ζ + 1 ωn2 , ωn ζ <1 N1 (s)N2 (s) . . . . D 1 (s)D 2 (s) . . . 539 540 CHAPTER 9 System Analysis in the Frequency Domain 9.2.2 OVERDAMPED CASE The denominator of a second-order transfer function with two real roots can be written as the product of two ﬁrst-order factors like form 3. For example, T (s) = 1 0.05 1 = = 2s 2 + 14s + 20 2(s + 2)(s + 5) (0.5s + 1)(0.2s + 1) (9.2.6) where the time constants are τ1 = 0.5 and τ2 = 0.2. E X A M P L E 9.2.1 Steady State Response of an Underdamped System ■ Problem a. Obtain the expressions for m(ω) and φ(ω) for the following transfer function. T (s) = b. X (s) 0.05 = F(s) (0.5s + 1)(0.2s + 1) Obtain the steady state response if f (t) = 14 sin 3t. ■ Solution a. First obtain M. M(ω) = |T ( jω)| = 0.05 0.05 = 2 (0.5ωj + 1)(0.2ωj + 1) (0.5ω) + 1 (0.2ω)2 + 1 Then m(ω) = 20 log M = 20 log 0.05 − 20 log (0.5ω)2 + 1 − 20 log (1) (0.2ω)2 + 1 or m(ω) = −26.0206 − 10 log[(0.5ω)2 + 1] − 10 log[(0.2ω)2 + 1] (2) The phase angle is φ(ω) = 0.05 − (0.5ωj + 1) − (0.2ωj + 1) = 0 − tan−1 0.5ω 0.2ω − tan−1 1 1 or φ(ω) = − tan−1 0.5ω − tan−1 0.2ω b. Equations (2) and (3) can be plotted versus ω, but the easiest way is to obtain the plots with MATLAB. How this is done is the subject of Section 9.7. The plots are shown in Figure 9.2.1. Because the input is given as f (t) = 14 sin 3t, we substitute ω = 3 into (1) and (3) to obtain 0.05 M = = 0.0238 2 (1.5) + 1 (0.6)2 + 1 φ = − tan−1 1.5 − tan−1 0.6 = −1.5232 rad Thus the steady state response is x(t) = 14M sin(3t + φ) = 14(0.0238) sin(3t − 1.5232) = 0.3332 sin(3t − 1.5232) (3) 9.2 Frequency Response of Higher-Order Systems Figure 9.2.1 Frequency response plots for the transfer function 0.05/(0.5s + 1)(0.2s + 1). Magnitude (dB) –20 –40 –60 –80 –100 Phase (deg) –120 0 –45 –90 –135 –180 10–2 541 10–1 100 101 Frequency (rad/s) 102 103 Consider the second-order model m ẍ + c ẋ + kx = f (t) Its transfer function is X (s) 1 = (9.2.7) 2 F(s) ms + cs + k If the system is overdamped, both roots are real and distinct, and we can write T (s) as 1/k 1/k T (s) = = (9.2.8) (m/k)s 2 + (c/k)s + 1 (τ1 s + 1)(τ2 s + 1) where τ1 and τ2 are the time constants of the roots. Substitute s = jω into (9.2.8). 1/k T ( jω) = (τ1 jω + 1)(τ2 jω + 1) The magnitude ratio is |1/k| M(ω) = |T ( jω)| = |τ1 jω + 1||τ2 jω + 1| Thus 1 m(ω) = 20 log M(ω) = 20 log − 20 log |τ1 ωj + 1| k (9.2.9) − 20 log |τ2 ωj + 1| T (s) = The phase angle is 1 − (τ1 ωj + 1) − (τ2 ωj + 1) (9.2.10) k The magnitude ratio plot in decibels consists of a constant term, 20 log |1/k|, minus the sum of the plots for two ﬁrst-order lead terms. Assume that τ1 > τ2 . Then for 1/τ1 < ω < 1/τ2 , the slope is approximately −20 dB/decade. For ω > 1/τ2 , the φ(ω) = Figure 9.2.2 Semilog plots of log magnitude ratio and phase angle of the model 1/(τ1 s + 1)(τ2 s + 1). CHAPTER 9 System Analysis in the Frequency Domain 0 m (dB) 542 –20 –40 –60 –80 = 1/1 log = 1/2 (degrees) 0 –45 –90 –135 –180 = 1/1 log v = 1/2 contribution of the term (τ2 ωj + 1) becomes signiﬁcant. This causes the slope to decrease by an additional 20 dB/decade, to produce a net slope of −40 dB/decade for ω > 1/τ2 . The rest of the plot can be sketched as before. The result is shown in Figure 9.2.2 (upper plot) for k = 1. The phase angle plot shown in Figure 9.2.2 (lower plot) is produced in a similar manner by using (9.2.10). Note that if k > 0, (1/k) = 0◦ . 9.2.3 UNDERDAMPED CASE We now consider the underdamped case of a second-order system that has two complex roots. E X A M P L E 9.2.2 Response with Two Complex Roots ■ Problem The model of a certain system is 6ẍ + 12ẋ + 174x = 15 f (t) a. Obtain its steady-state response for f (t) = 5 sin 7t. b. Obtain the expressions for m(ω) and φ(ω). ■ Solution a. The system’s transfer function is T (s) = X (s) 15 = 2 F(s) 6s + 12s + 174 Substitute s = 7 j. T (7 j) = −6(7)2 15 15 = + 12(7 j) + 174 −120 + 84 j 9.2 Frequency Response of Higher-Order Systems 543 Then 15 M = |T (7 j)| = = 0.1024 (120)2 + (84)2 The phase angle is φ = T (7 j) = (15) − (−120 + 84 j) = 0◦ − (−120 + 84 j) Noting that (−120 + 84 j) is in the second quadrant, we obtain −84 −1 φ = − π + tan = −2.531 rad 120 Thus the steady-state response is xss (t) = 5M sin(7t + φ) = 0.512 sin(7t − 2.531) b. Replacing s with jω gives T ( jω) = −6ω2 15 + 12ωj + 174 Thus, 15 M(ω) = (174 − 6ω2 )2 + 144ω2 and m(ω) = 20 log 15 − 10 log 174 − 6ω2 φ(ω) = (15) − ⎧ ⎪ ⎨− tan−1 m (dB) + 144ω2 174 − 6ω2 + 12ωj 12ω 174 − 6ω2 = 12ω ⎪ ⎩tan−1 − 180◦ 2 6ω − 174 The plots are shown in Figure 9.2.3. –10 –20 –30 –40 –50 –60 –70 –80 100 2 if 174 − 6ω2 > 0 if 174 − 6ω2 < 0 Figure 9.2.3 Semilog plots of log magnitude ratio and phase angle of the model 15/(6s 2 + 12s + 174). 101 102 101 102 (degrees) 0 –50 –100 –150 –200 0 10 544 CHAPTER 9 System Analysis in the Frequency Domain If the transfer function 1 X (s) = 2 F(s) ms + cs + k (9.2.11) has complex conjugate roots, it can be expressed as form 4 in Table 9.2.1 as follows: T (s) = k X (s) 1 1 = = 2 2 F(s) (m/k)s + (c/k)s + 1 (s/ωn ) + 2ζ (s/ωn ) + 1 where we have deﬁned the natural frequency and damping as usual to be k ωn = m c ζ = √ 2 mk (9.2.12) (9.2.13) (9.2.14) As shown in Chapter 2, the roots can be expressed in terms of these parameters as s = −ζ ωn ± jωn 1 − ζ 2 (9.2.15) The roots are complex if ζ < 1. Thus T (s) becomes T (s) = ωn2 k X (s) = 2 F(s) s + 2ζ ωn s + ωn2 (9.2.16) It is important to note that the reason for multiplying by k is that the quantity k X (s) represents a force, as does F(s). Therefore, the ratio k X (s)/F(s) represents a dimensionless quantity that is a function of only two parameters, ζ and ωn , instead of the three parameters m, c, and k. This allows us to obtain a more general understanding of the frequency response. For most applications of interest here, the quadratic factor given by form 4 in Table 9.2.1 occurs in the denominator; therefore, we will develop the results assuming this will be the case. If a quadratic factor is found in the numerator, its values of m(ω) and φ(ω) are the negative of those to be derived next. Replacing s with jω and dividing top and bottom of (9.2.16) by ωn2 gives T ( jω) = ( jω/ωn )2 1 1 = 2 + (2ζ /ωn ) jω + 1 1 − (ω/ωn ) + (2ζ ω/ωn ) j (9.2.17) To simplify the following expressions, deﬁne the following frequency ratio: r= ω ωn (9.2.18) Thus the transfer function can be expressed as T (r ) = 1 − r2 1 + 2ζ r j (9.2.19) The magnitude ratio is M= |1 − r2 1 1 = 2 + 2ζ r j| (1 − r )2 + (2ζ r )2 (9.2.20) 9.2 The log magnitude ratio is Frequency Response of Higher-Order Systems m = 20 log 1 1 − r 2 + 2ζ r j = −20 log (1 − r 2 )2 + (2ζ r )2 = −10 log (1 − r 2 )2 + (2ζ r )2 (9.2.21) The asymptotic approximations are as follows. For r 1 (that is, for ω ωn ), m ≈ −20 log 1 = 0. For r 1 (that is, for ω ωn ), m ≈ −20 log r 4 + 4ζ 2r 2 √ ≈ −20 log r 4 = −40 log r Thus, at low frequencies where ω ωn , the curve is horizontal at m = 0, while for high 1, the curve has a slope of −40 dB/decade, just as in the frequencies ω ωn where r overdamped case (Figure 9.2.4a). The high-frequency and low-frequency asymptotes intersect at the corner frequency ω = ωn . The phase angle plot can be obtained in a similar manner (Figure 9.4.2b). From the additive property for angles, we see that for (9.2.19), φ = − (1 − r 2 + 2ζ r j) Thus 2ζ r φ = − tan 1 − r2 where φ is in the third or fourth quadrant. For r 1, φ ≈ 0◦ . For r 1, φ ≈ −180◦ . ◦ At ω = ωn , φ = −90 independently of ζ . The curve is skew-symmetric about the inﬂection point at φ = −90◦ for all values of ζ . −1 9.2.4 RESONANCE The complex roots case differs from the real roots case in the vicinity of the corner frequency. To see this, note that M given by (9.2.20) has a maximum value when the denominator has a minimum. Setting the derivative of the denominator with respect to r equal to zero shows that the maximum M occurs at r = 1 − 2ζ 2 , which corresponds to the frequency ω = ωn 1 − 2ζ 2 . This frequency is the resonant frequency ωr . The peak of M exists only when the term under the radical is positive; that is, when ζ ≤ 0.707. Thus, the resonant frequency is given by 0 ≤ ζ ≤ 0.707 (9.2.22) ωr = ωn 1 − 2ζ 2 The peak, or resonant, value of M, denoted by Mr , is found by substituting r = 1 − 2ζ 2 into the expression (9.2.20) for M. This gives Mr = 2ζ 1 1 − ζ2 0 ≤ ζ ≤ 0.707 (9.2.23) If ζ > 0.707, no peak exists, and the maximum value of M occurs at ω = 0 where M = 1. Note that as ζ → 0, ωr → ωn , and Mr → ∞. For an undamped system, the roots are purely imaginary, ζ = 0, and the resonant frequency is the natural frequency ωn . These formulas are summarized in Table 9.2.2. 545 546 CHAPTER 9 System Analysis in the Frequency Domain 30 Figure 9.2.4 Semilog plots of log magnitude ratio and phase angle of the model 2 2 ωn /(s 2 + 2ζ ωn s + ωn ). = 0.01 20 = 0.1 m (dB) 10 = 0.5 0 = 0.7 –10 =1 –20 –30 –40 10–1 100 /n 101 (a) 0 = 0.01 = 0.1 –45 (degrees) = 0.5 = 0.7 –90 =1 –135 –180 10–1 100 /n 101 (b) Resonance occurs when the input frequency is close to the resonant frequency of the system. If the damping is small, the output amplitude will continue to increase until either the linear model is no longer accurate or the system fails. When φ is near −90◦ , the velocity ẋ is in phase with the input. This causes the large amplitude. Circuit designers take advantage of resonance by designing ampliﬁcation circuits whose natural frequency is close to the frequency of a signal they want to amplify (such as the signal from a radio station). Designers of structural systems and suspensions try to avoid resonance because of the damage or discomfort that large motions can produce. 9.2 Frequency Response of Higher-Order Systems 547 Table 9.2.2 Frequency response of a second-order system. Transfer Function: ms 2 ωn2 k = 2 + cs + k s + 2ζ ωn s + ωn2 k m Natural Frequency: ωn = Damping ratio: c ζ = √ 2 mk Resonant frequency: ωr = ωn Resonant response: Mr = 2ζ 1 − 2ζ 2 1 0 ≤ ζ ≤ 0.707 0 ≤ ζ ≤ 0.707 1 − ζ2 −1 φr = −tan 1 − 2ζ 2 ζ 0 ≤ ζ ≤ 0.707 In Figure 9.2.4a, the correction to the asymptotic approximations in the vicinity of the corner frequency depends on the value of ζ . The peak value in decibels is (9.2.24) m r = 20 log Mr = −20 log 2ζ 1 − ζ 2 At the resonant frequency ωr , the phase angle is φ|r =√1−2ζ 2 = − tan−1 1 − 2ζ 2 ζ (9.2.25) When r = 1 (at ω = ωn ), m|r =1 = −20 log 2ζ (9.2.26) The transfer function of a mass-spring-damper model m ẍ + c ẋ + kx = f (t) with an applied force f (t) has the form X (s) 1 = (9.2.27) F(s) ms 2 + cs + k This can be put into the form required by Table 9.2.2 by multiplying the numerator and denominator by k to obtain 1 k X (s) = (9.2.28) F(s) k ms 2 + cs + k Thus we can use the results of Table 9.2.1 for (9.2.27), but we must divide the table formula for Mr by k. The formula for φr is unchanged. Determining Maximum Response ■ Problem The model of a certain mass-spring-damper system is 2ẍ + c ẋ + 18x = f (t) = 14 sin ωt Determine its resonant frequency ωr and its peak response xpeak at resonance if ζ = 0.4. ■ Solution The transfer function is X (s) 1 = 2 F(s) 2s + cs + 18 E X A M P L E 9.2.3 548 CHAPTER 9 System Analysis in the Frequency Domain To put this in the form required by Table 9.2.2, we multiply and divide by 18 to obtain 1 18 X (s) = 2 F(s) 18 2s + cs + 18 So to use the expression for Mr̄ from Table 9.2.2, we must divide the result by 18. This gives Mr = 1 1 18 2ζ 1 − ζ 2 With ζ = 0.4, Mr = 1 1 = 0.0758 18 2(0.4) 1 − (0.4)2 Thus the peak response is xpeak = 14Mr = 14(0.758) = 1.0608 √ Note that ωn = 18/2 = 3. Thus the resonant frequency is given by ωr = ωn 1 − 2ζ 2 = 3 1 − 2(0.4)2 = 2.4739 Thus if f (t) = 14 sin 2.4739t, the amplitude of the steady state response will be 1.0608. Note that we did not need the value of the damping coefﬁcient c. E X A M P L E 9.2.4 Limits on Damping ■ Problem The model of a certain mass-spring-damper system is 5ẍ + c ẋ + 80x = f (t) = 7 sin ωt How large must the damping constant c be so that the maximum steady-state amplitude of x is no greater than 2, for an arbitrary value of ω? ■ Solution The transfer function is 1 X (s) = 2 F(s) 5s + cs + 80 To put this in the form required by Table 9.2.2, we multiply and divide by 80 to obtain 1 80 X (s) = 2 F(s) 80 5s + cs + 80 So to use the expression for Mr from the table, we must divide the result by 80. This gives Mr = 1 1 80 2ζ 1 − ζ 2 Because the amplitude of the forcing function is 7, the maximum response amplitude, which occurs at ω = ωr , is 7Mr , and it can be no greater than 2. Therefore, 7Mr = 7 1 =2 80 2ζ 1 − ζ 2 Solve for ζ by squaring each side to obtain ζ4 − ζ2 + 7 =0 320 9.2 Frequency Response of Higher-Order Systems 549 This is a quadratic in ζ 2 , and the solutions are ζ 2 = 0.9776, 0.0224. Because ζ must be positive, we have two possible solutions: ζ = 0.9887, 0.15. The ﬁrst solution is not acceptable because our formula for Mr is valid only for ζ ≤ 0,707. So the solution is ζ = 0.15. We now ﬁnd c. From the deﬁnition of the damping ratio, c ζ = 0.15 = √ 2 5(80) Thus gives c = 6. Thus if c = 6, there is no value of the forcing frequency ω that will produce a response amplitude greater than 2. Once you understand how each of the forms shown in Table 9.2.1 contributes to the magnitude and phase plots, you can quickly determine the effects of each term in a more complicated transfer function. A Fourth-Order Model ■ Problem Determine the effect of the parameter τ of the magnitude plot of the following transfer function. Assume that time is measured in seconds. τs + 1 T (s) = 4 3 2 s + 40.8s + 8337s + 4.1184 × 104 + 5.184 × 105 Is it possible to choose a value for τ to increase the gain at low frequencies? ■ Solution First determine the roots of the denominator. They are s = −2.4 ± 7.632 j −18 ± 88.10 j For the ﬁrst root pair, 7.632 −1 ζ = cos tan = 0.3 2.4 ωn = 2.42 + 7.6322 = 8 ωn = 182 + 88.12 = 90 Similarly, for the second pair, 88.1 −1 = 0.2 ζ = cos tan 18 Thus, T (s) can be expressed in factored form as T (s) = (s 2 τs + 1 + 4.8s + 64)(s 2 + 36s + 8100) The model has two resonant frequencies corresponding to the two root pairs. These frequencies are √ ωr1 = ωn 1 − 2ζ 2 = 8 1 − 0.18 = 7.24 rad/s √ ωr2 = ωn 1 − 2ζ 2 = 90 1 − 0.08 = 86.3 rad/s Because ζ is small for each factor, we can expect a resonant peak at these frequencies. However, we cannot use the formula (9.2.24) to calculate m r at each peak because it applies only to a second-order system. Each quadratic term contributes −40 dB/decade to the composite slope at high frequencies. The numerator term contributes a slope of +20 dB/decade at frequencies above 1/τ . So the composite curve will have a slope of 20 − 2(40) = −60 dB/decade at high frequencies. E X A M P L E 9.2.5 550 CHAPTER 9 System Analysis in the Frequency Domain –80 Figure 9.2.5 Log magnitude ratio plot of a fourth-order model. –100 = 1.25 m (dB) –120 –140 –160 =0 = 0.0125 –180 –200 100 101 (rad/s) 102 103 The numerator term causes the m curve to break upward at ω = 1/τ . If 1/τ is less than ωr1 , the m curve will break upward before the −40 dB/decade slope of the ﬁrst quadratic term takes effect. Figure 9.2.5 shows the m plot for three cases, including the case where τ = 1.25, which corresponds to a corner frequency of ω = 1/τ = 0.8. As compared with the case having no numerator dynamics (τ = 0), the choice of τ = 1.25 can be seen to raise the magnitude. For example, at ω = 7.24 rad/s, the resonant peaks for the two cases are −109 dB (M = 3.55 × 10−6 ) for τ = 0, and −89.6 (M = 3.31 × 10−5 ) for τ = 1.25. The amplitude ratio is 3.31 × 10−5 /3.55 × 10−6 = 9.34 times larger for τ = 1.25. Using a value of 1/τ that is larger than the smallest resonant frequency will not increase the amplitude ratio because the −40 dB/decade slope from the ﬁrst quadratic term will take effect before the +20 dB/decade slope of the numerator term makes its contribution. An example of this is shown in Figure 9.2.5 for τ = 0.0125 s, which corresponds to a corner frequency of 1/τ = 80 rad/s. 9.3 FREQUENCY RESPONSE APPLICATIONS In this section we present more examples illustrating the concepts and applications of frequency response. 9.3.1 A NEUTRALLY STABLE CASE While all real systems will have some damping, it is instructive and useful to obtain some results for the undamped case, where c = ζ = 0. This is because the mathematical results for the undamped case are more easily derived and analyzed, and they give insight into the behavior of many real systems that have a small amount of damping. If the model is stable, the free response term disappears in time. The results derived in Section 9.2 therefore must be reexamined for the case where there is no damping because this is not a stable case (it is neutrally stable). In this case, the magnitude and phase angle of the transfer function do not give the entire steady-state response. The free response for the undamped equation m ẍ + kx = F sin ωt is found with the 9. 3 Frequency Response Applications Laplace transform as follows: (ms 2 + k)X free (s) − m ẋ(0) − msx(0) = 0 ẋ(0) + sx(0) m ẋ(0) + msx(0) X free (s) = = 2 ms + k s 2 + ωn2 Thus, xfree (t) = ẋ(0) sin ωn t + x(0) cos ωn t ωn The forced response is found as follows: Fω s 2 + ω2 Fω 1 X forced (s) = 2 2 m s + ωn (s 2 + ω2 ) (ms 2 + k)X forced (s) = Assuming that ω = ωn , the partial fraction expansion is ωn ω Fω ωn X forced (s) = − m ω2 − ωn2 s 2 + ωn2 ω s 2 + ω2 Thus the forced response is xforced (t) = Fω m ω2 − ωn2 sin ωn t − ωn sin ωt ω or, with r = ω/ωn , xforced (t) = − Fr F sin ω sin ωt t + n k(1 − r 2 ) k(1 − r 2 ) Thus the total response is F r F 1 ẋ(0) − sin ωn t + x(0) cos ωn t + sin ωt x(t) = 2 ωn k 1−r k 1 − r2 (9.3.1) There is no transient response here; the entire solution is the steady-state response. 9.3.2 BEATING We may examine the effects of the forcing function independently of the effects of the initial conditions by setting x(0) = ẋ(0) = 0 in (9.3.1). The result is the forced response: F 1 (9.3.2) (sin ωt − r sin ωn t) k 1 − r2 When the forcing frequency ω is substantially different than the natural frequency ωn , the forced response looks somewhat like Figure 9.3.1 and consists of a higherfrequency oscillation superimposed on a lower-frequency oscillation. If the forcing frequency ω is close to the natural frequency ωn , then r ≈ 1 and the forced response (9.3.2) can be expressed as follows: x(t) = x(t) = 1 F F 1 (sin ωt − sin ωn t) = (sin ωt − sin ωn t) 2 2 k 1−r m ωn − ω2 551 552 CHAPTER 9 System Analysis in the Frequency Domain 0.4 Figure 9.3.1 Response when the forcing frequency is substantially different from the natural frequency. 0.3 0.2 x(t ) 0.1 0 –0.1 –0.2 –0.3 –0.4 0 10 20 30 t 40 50 60 Using the identity 1 1 2 sin (A − B) cos (A + B) = sin A − sin B 2 2 we see that ω + ωn ω − ωn t cos t = sin ωt − sin ωn t 2 2 Thus the forced response is given by 2F 1 ω − ωn ω + ωn t cos t (9.3.3) sin x(t) = 2 2 m ωn − ω 2 2 This can be interpreted as a cosine with a frequency (ω + ωn )/2 and a time-varying amplitude of 1 ω − ωn 2F t sin 2 2 m ωn − ω 2 The amplitude varies sinusoidally with the frequency (ω − ωn )/2, which is lower than the frequency of the cosine. The response thus looks like Figure 9.3.2. This behavior, in which the amplitude increases and decreases periodically, is called beating. The beat period is the time between the occurrence of zeros in x(t) and thus is given by the half-period of the sine wave, which is 2π/|ω − ωn |. The vibration period is the period of the cosine wave, 4π/(ω + ωn ). The concept of beating is important for tuning musical instruments. When a piano tuner strikes a piano wire and a tuning fork emitting a standard frequency, he or she listens for the beats caused by the difference between the two frequencies and then tightens or loosens the piano wire until the beating disappears. 2 sin 9.3.3 RESPONSE AT RESONANCE Equation (9.2.20) shows that when ζ = 0 the amplitude of the steady-state response becomes inﬁnite when r = 1; that is, when the forcing frequency ω equals the natural frequency ωn . The phase shift φ is exactly −90 degrees at this frequency. 9. 3 Frequency Response Applications Figure 9.3.2 Beating response when the forcing frequency is close to the natural frequency. 3 2 x(t) 1 0 –1 –2 –3 0 553 5 10 15 20 25 t 30 35 40 45 50 To obtain the expression for x(t) at resonance, we use (9.3.2) to compute the limit of x(t) as r → 1 after replacing ω in x(t) using the relation ω = ωn r . We must use l’Hôpital’s rule to compute the limit. F 1 (sin ωn r t − r sin ωn t) r →1 k 1 − r 2 d(sin ωn r t − r sin ωn t)/dr F lim = k r →1 d(1 − r 2 )/dr ωn t cos ωn r t − sin ωn t F lim = k r →1 −2r Fωn sin ωn t = − t cos ωn t 2k ωn x(t) = lim (9.3.4) The plot is shown in Figure 9.3.3 for the case m = 4, c = 0, k = 36, and F = 10. The amplitude increases linearly with time. The behavior for lightly damped systems is similar except that the amplitude does not become inﬁnite. Figure 9.3.4 shows the damped case: m = 4, c = 4, k = 36, and F = 10. For large amplitudes, the linear model on which this analysis is based will no longer be accurate. In addition, all physical systems have some damping, so c will never be exactly zero, and the response amplitude will never be inﬁnite. The important point, however, is that the amplitude might be large enough to damage the system or cause some other undesirable result. At resonance the output amplitude will continue to increase until either the linear model is no longer accurate or the system fails. 9.3.4 RESONANCE AND TRANSIENT RESPONSE Although the analysis in Section 9.2 was a steady-state analysis and assumed that the forcing frequency ω is constant, resonance can still occur in transient processes if the input varies slowly enough to allow the steady-state response to begin to appear. For example, resonance can be a problem even if a machine’s operating speed is well above its 554 CHAPTER 9 System Analysis in the Frequency Domain 4 Figure 9.3.3 Response near resonance for an undamped system. 3 2 x(t ) 1 0 –1 –2 –3 –4 0 1 2 3 4 5 t 6 7 8 9 10 1 Figure 9.3.4 Response near resonance for a damped system. 0.8 0.6 0.4 x(t) 0.2 0 –0.2 –0.4 –0.6 –0.8 –1 0 1 2 3 4 5 t 6 7 8 9 10 resonant frequency, because the machine’s speed must pass through the resonant frequency at startup. If the machine’s speed does not pass through the resonant frequency quickly enough, high amplitude oscillations will result. Figure 9.3.5 shows the transient response of the model 4ẍ + 3ẋ + 100x = 295 sin[ω(t)t] where the frequency ω(t) increases linearly with time, as ω(t) = 0.7t. Thus the frequency passes through the resonant frequency of this model, which is ωr = 4.97, at t = 7.1 s. The plot shows the large oscillations that occur as the input frequency passes close to the resonant frequency. 9. 3 Frequency Response Applications Figure 9.3.5 Transient response when an increasing forcing frequency passes through the resonant frequency. 15 x(t ) 10 555 (t ) 5 0 –5 –10 –15 0 1 2 3 4 5 t (s) 6 7 8 9 10 9.3.5 BASE MOTION AND TRANSMISSIBILITY The motion of the mass shown in Figure 9.3.6 is produced by the motion y(t) of the base. This system is a model of many common displacement isolation systems. Assuming that the mass displacement x is measured from the rest position of the mass when y = 0, the weight mg is canceled by the static spring force. The force transmitted to the mass by the spring and damper is denoted f t and is given by f t = c( ẏ − ẋ) + k(y − x) c k Base or (9.3.6) The transfer function is cs + k X (s) = (9.3.7) 2 Y (s) ms + cs + k This transfer function is called the displacement transmissibility and can be used to analyze the effects of the base motion y(t) on x(t), the motion of the mass. From (9.3.5) (9.3.8) Substituting for X (s) from (9.3.7) gives cs + k ms 2 Y (s) Y (s) = (cs + k) Ft (s) = (cs + k) Y (s) − ms 2 + cs + k ms 2 + cs + k Thus, the second desired ratio is Ft (s) ms 2 = (cs + k) 2 Y (s) ms + cs + k Machine x y m ẍ = f t = c( ẏ − ẋ) + k(y − x) Ft (s) = (cs + k) [Y (s) − X (s)] m (9.3.5) This gives the following equation of motion: m ẍ + c ẋ + kx = c ẏ + ky Figure 9.3.6 Base excitation. (9.3.9) 556 CHAPTER 9 Figure 9.3.7 Single-mass suspension model. This is the ratio of the transmitted force to the base motion. It is customary to use instead the ratio Ft (s)/kY (s), which is a dimensionless quantity representing how the base displacement y affects the force transmitted to the mass. Thus, Body m System Analysis in the Frequency Domain Ft (s) cs + k ms 2 = kY (s) k ms 2 + cs + k x k c Suspension Road y Datum level E X A M P L E 9.3.1 (9.3.10) The ratio Ft (s)/kY (s) is called the force transmissibility. It can be used to compute the transmitted force f t (t) that results from a speciﬁed base motion y(t). An example of base excitation occurs when a car drives over a rough road. Figure 9.3.7 shows a quarter-car representation, where the stiffness k is the series combination of the tire and suspension stiffnesses. The equation of motion is given by (9.3.6). Although road surfaces are not truly sinusoidal in shape, we can nevertheless use a sinusoidal proﬁle to obtain an approximate evaluation of the performance of the suspension at various speeds. Vehicle Suspension Response ■ Problem Suppose the road proﬁle is given (in feet) by y(t) = 0.05 sin ωt, where the amplitude of variation of the road surface is 0.05 ft and the frequency ω depends on the vehicle’s speed and the road proﬁle’s period. Suppose the period of the road surface is 30 ft. Compute the steady-state motion amplitude and the force transmitted to the chassis, for a car traveling at a speed of 30 mi/hr. The car weighs 3200 lb. The effective stiffness, which is a series combination of the tire stiffness and the suspension stiffness, is k = 3000 lb/ft. The damping is c = 300 lb-sec/ft. ■ Solution For a period of 30 ft and a vehicle speed of 30 mi/hr, the frequency ω is ω= 5280 30 1 (2π)30 = 9.215 rad/sec 3600 For the car weighing 3200 lb, the quarter-car mass is m = 800/32.2 slug. From (9.3.7) with s = jω = 9.215 j, k 2 + (cω)2 |X | = = 1.405 |Y | (k − mω2 )2 + (cω)2 Thus the displacement amplitude is |X | = 0.05(1.405) = 0.07 ft. The transmitted force is obtained from (9.3.9). |Ft | 2 mω2 = k + (cω)2 = 2960 |Y | (k − mω2 )2 + (cω)2 Thus the magnitude of the transmitted force is |Ft | = 0.05(2960) = 148 lb. 9.3.6 INSTRUMENT DESIGN When no s term occurs in the numerator of a transfer function, its magnitude ratio is small at high frequencies. On the other hand, introducing numerator dynamics can produce a large magnitude ratio at high frequencies. This effect can be used to advantage in instrument design, for example. The instrument shown in Figure 9.3.8 illustrates this point. With proper selection of the natural frequency of the device, it can be used either as 9. 3 Frequency Response Applications Figure 9.3.8 An accelerometer. Case x k 2 v m y k 2 z Structure a vibrometer to measure the amplitude of a sinusoidal input displacement z = A z sin ωt, or an accelerometer to measure the amplitude of the acceleration z̈ = −A z ω2 sin ωt. When used to measure ground motion from an earthquake, for example, the instrument is commonly referred to as a seismograph. The model was obtained in Section 6.6 and is T (s) = 557 −ms 2 −s 2 /ωn2 Y (s) = = Z (s) ms 2 + cs + k s 2 /ωn2 + 2ζ s/ωn + 1 (9.3.11) where y(t) is the output displacement, which is measured from the voltage v(t). At steady state, y(t) = A y sin(ωt + φ). The numerator N (s) = −s 2 /ωn2 gives the following contribution to the log magnitude ratio: jω 2 ω 20 log |N ( jω)| = 20 log (9.3.12) = 40 log ωn ωn This term contributes 0 dB to the net curve at the corner frequency ω = ωn , and it increases the slope by 40 dB/decade over all frequencies. Thus, at low frequencies, the slope of m(ω) corresponding to (9.3.11) is 40 dB/decade, and at high frequencies, the slope is zero. The plot is sketched in Figure 9.3.9. For the device to act like a vibrometer, this plot shows that the device’s natural ωn , where ω is the oscillation frequency of frequency ωn must be selected so that ω the displacement to be measured. For ω ωn , |T ( jω)| ≈ 40 log ω ω − 40 log = 0 dB ωn ωn and thus A y ≈ A z as desired. This is because the mass m cannot respond to highfrequency input displacements. Its displacement x therefore remains approximately constant, and the motion z directly indicates the motion y. To design a vibrometer having speciﬁc characteristics, we must √ know the lower bound of the input displacement frequency ω. The frequency ωn = k/m is then made much smaller than this bound by selecting a large mass and a soft spring (small k). However, these choices are governed by constraints on the allowable deﬂection. For example, a very soft spring will have a large distance between the free length and the equilibrium positions. 558 CHAPTER 9 Figure 9.3.9 Frequency response of an accelerometer. System Analysis in the Frequency Domain 10 5 0 –5 Accelerometer region Vibrometer region m (dB) –10 –15 –20 –25 –30 –35 –40 0.1 1 /n 10 An accelerometer can be obtained by using the lower end of the frequency range; ω, or equivalently, for s near zero, (9.3.11) gives that is, selecting ωn T (s) ≈ − s2 Y (s) = 2 ωn Z (s) or Y (s) ≈ − 1 2 s Z (s) ωn2 The term s 2 Z (s) represents the transform of z̈ so the output of the accelerometer is Ay ≈ Figure 9.3.10 Two-mass suspension model. Body m1 x1 c1 m2 x2 k2 With ωn chosen large (using a small mass and a stiff spring), the input acceleration amplitude ω2 A z can be determined from A y . 9.3.7 PHYSICAL INTERPRETATION OF THE PHASE PLOT Suspension k1 1 ω2 |z̈| = Az ωn2 ωn2 Wheel Road y Datum level The phase angle plot can give useful information about the motion of the system relative to the input, or the motion of the constituent masses relative to each other. Consider the following two-mass suspension model analyzed in Example 4.5.9 and shown again in Figure 9.3.10. The equations of motion are m 1 ẍ1 = c1 (ẋ2 − ẋ1 ) + k1 (x2 − x1 ) m 2 ẍ2 = −c1 (ẋ2 − ẋ1 ) − k1 (x2 − x1 ) + k2 (y − x2 ) We will use the following numerical values: m 1 = 250 kg, m 2 = 40 kg, k1 = 1.5 × 104 N/m, k2 = 1.5 × 105 N/m, and c1 = 1917 N · s/m. The transfer functions for these 9. 3 Frequency Response Applications Figure 9.3.11 Frequency respon

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