ME 374, System Dynamics Analysis and Design Homework 9: Solution (June 9, 2008) by Jason Frye

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1 ME 374, System Dynamics Analysis and Design Homewk 9: Solution June 9, 8 by Jason Frye Problem a he frequency response function G and the impulse response function ht are Fourier transfm pairs herefe, G F{ht} hte jt dt It is reasonable to assume that ht will only be considered f t >, { e ht t/τ, t >, t < herefe, G G e t/τ e jt dt e j+t/τt dt j + /τ e j+t/τt e j+t/τ e j+t/τ j + /τ j + /τ j + /τ Ans b Now we are given the FRF from yt to xt whose magnitude is given by G G + j + + Ans F such as G log {G} db he magnitude behaves like a low-pass filter with cutoff frequency rad/s and rolls off at db/decade, as shown in Figure c he frequency response function given by cresponds to a first-der system since there is only one pole in the denominat As mentioned, this fms a low-pass filter with cutoff frequency rad/s herefe, the bandwidth is rad/s d he magnitude of the output spectrum of xt ie, the magnitude of X is shown in Figure a he frequency response function from the input yt to the output xt is G X Y herefe, the input can be determined from Y X G and Y X G G X

2 Figure : Magnitude of the frequency response function G he magnitude of / G is shown in Figure b Note that the magnitude of G and the magnitude of the output spectrum of xt are both constant db up to their cutoff frequency However, the cutoff frequency f the output spectrum of xt is much higher than that f G herefe, the magnitude of the input spectrum of yt will initially follow the magnitude of / G up to rad/s At that point, the magnitude of the input spectrum of yt is the combination of the magnitude of / G + db/decade and the magnitude of the output spectrum of xt -4 db/decade, a net roll off of - db/decade he magnitude of the input spectrum of yt is shown in Figure 3 a b Figure : a Magnitude of the output spectrum of xt; b Magnitude of / G Figure 3: Magnitude of the input spectrum of yt

3 Problem a F the given fce histy ft {, < t <, otherwise generated by the hammer, where is the duration of the hammer impact, the Fourier transfm F is F F{ft} fte jt dt e jt dt j e jt F j e j Ans b he amplitude of F is determined by first rewriting F as F j e j j cos j sin sin + j cos hen the amplitude is F sin + cos sin + cos cos + F cos Ans which is illustrated in Figure 4 Note that when π F, 4π,, nπ n,, 3,, nπ cos nπ/ Figure 4: Amplitude of F c Note that the 4-Hz resonance of the disk cresponds to 8π rad/s If 5 ms 5 s, then π π rad 8π rad/s 4 Hz 5 s 3

From b, Fπ/ If the output of the system is xt, then output spectrum of xt is X GF and X G F Since F4 Hz, X4 Hz as well herefe, Henry would not excite the 4-Hz resonance of the disk At 5 Hz, F5 Hz,

4 From b, Fπ/ If the output of the system is xt, then output spectrum of xt is X GF and X G F Since F4 Hz, X4 Hz as well herefe, Henry would not excite the 4-Hz resonance of the disk At 5 Hz, F5 Hz, and G5 Hz 5 herefe, Henry would excite the 5-Hz resonance of the disk d Now ms s At 4 Hz 8π rad/s: hen F4 Hz F8π cos8π 8π X4 Hz G4 Hz F4 Hz X4 Hz Ans At 5 Hz π rad/s: hen F5 Hz Fπ cosπ π 5π X5 Hz G5 Hz F5 Hz 5 5π X5 Hz 5 5π Ans Problem 3 a b Figure 5: a Model f vibration control; b Free-body diagram a F the system shown in Figure 5a, the equation of motion can be determined using the free-body diagram in Figure 5b Because xt and yt are taken as absolute displacements, the fces F c and F k are F c cẋ ẏ, F k kx y 4

5 hen, summing fces gives F ft Fc F k mẍ ft cẋ ẏ kx y mẍ mẍ + cẋ + kx cẏ + ky + ft Ans b When the system is passive ft, the equation of motion is mẍ + cẋ + kx cẏ + ky he frequency response function G from yt to xt is G X Y k + jc k m + jc Ans his can be written using the natural frequency n and damping coefficient ζ as G n + jζ n n + jζ n + jζ n n + jζ n, where hen, the amplitude of G is n k m, ζ c m n + 4ζ G n n + 4ζ n he amplitude near resonance can be determined by letting n, which gives + 4ζ G n Ans 4ζ he rate at which G rolls off f n is 4ζ G n 4 n Ans he amplitude of G is shown in Figure 6 c Now, with spring-fce cancellation applied ft ky, the equation of motion becomes mẍ + cẋ + kx cẏ he frequency response function G sp is then G sp jc k m + jc jζ n n + jζ n, Ans 5

6 Figure 6: Amplitude of G where again he amplitude of G sp is G sp n k m, ζ c m n ζ n n + 4ζ n ζ n n + 4ζ n he amplitude near resonance can be determined by letting n, which gives G sp n ζ Ans 4ζ he rate at which G sp rolls off f n is ζ G sp he amplitude of G sp is shown in Figure 7 n 4 n + 4ζ n Ans Figure 7: Amplitude of G sp 6

7 d Now, with damping-fce cancellation applied ft cẏ, the equation of motion becomes he frequency response function G damp is then mẍ + cẋ + kx ky G damp k k m + jc n n + jζ n, Ans where again he amplitude of G damp is n k m, ζ c m n G damp n n + 4ζ n n + 4ζ he amplitude near resonance can be determined by letting n, which gives he rate at which G damp rolls off f n is G damp n 4ζ ζ Ans n G damp 4 Ans n he amplitude of G damp is shown in Figure 8 Figure 8: Amplitude of G damp e During an earthquake, the goal is to minimize the displacement of the building relative to the ground when excited at resonance When applying spring-fce cancellation, G sp n herefe, X n Y n ie, the building moves with the ground When applying damping-fce cancellation, G damp n ζ > herefe, you would use spring-fce cancellation f F an isolation table, the goal is to minimize the absolute displacement of the table When applying spring-fce cancellation, G sp rolls off as f n When applying damping-fce cancellation, G damp rolls off as f n By ensuring that n, you would use damping-fce cancellation 7

8 g he Fourier transfm of is F F{ft} π π ft fte jt dt sin te jt dt π { sin t, < t < π, otherwise e j t e jt e jt dt j e j t e j+t dt j [ ] π j j ej t + j + e j+t π j e + e jπ e jπ + e jπ + e jπ + F e jπ Ans π j+ e e jπ e jπ he magnitude of F can be determined by first rewriting as [ ] F cos π j sin π [ ] cos π + j sin π hen, F cos π + sin π F cos π Ans Note that F when, 3,, as illustrated in Figure 9 Also, in the case when, lim F π When and if n, the spectrum of ft is in the rolloff ption of G, which is decreasing at a rate of As a result, the response will be attenuated ie, ft will present insignificant effects to the response xt 8

Figure 9: Magnitude of F a b Figure : a Suspension model in HDD; b Free-body diagram of recding head Problem 4 a From the model f the recding head suspension system illustrated in Figure a, note that

9 Figure 9: Magnitude of F a b Figure : a Suspension model in HDD; b Free-body diagram of recding head Problem 4 a From the model f the recding head suspension system illustrated in Figure a, note that yt is the relative displacement of the head to the disk surface hen, from the free-body diagram shown in Figure b, the equation of motion can be determined from Fy F k F k F c mẍ + ÿ k x + y k y cẏ mẍ + ÿ mÿ + cẏ + k + k y mẍ k x Ans b he frequency response function G can be written as G m k k + k m + jc Ans By defining quantities G can be written as G k m, k + k m, + jζ, ζ c m 9

10 hen, the magnitude of G is G Ans + 4ζ F ease of plotting, we can think of writing the magnitude as G G G, where G, G + 4ζ he plots of G and G are shown in Figure, and the plot of G is shown in Figure a b Figure : a Amplitude of G ; b Amplitude of G Figure : Amplitude of G he phase of G is { } G + jζ { } { + jζ } G arctan ζ arctan Ans

11 When < <, herefe, arctan ζ π π, arctan G π π π ie, displacement of the head is out of phase with the disk surface When < <, herefe, arctan ζ, arctan G ie, displacement of the head is in phase with the disk surface When >, herefe, G π π arctan ζ, arctan π ie, displacement of the head is out of phase with the disk surface he plot of the phase of G is shown in Figure 3 a b Figure 3: a Phase of G; b Alternative phase plot where G π c he recding head follows the disk surface is in phase with the disk surface when < < he width of this frequency range can be increased by setting, that is k /m k + k /m k k k k + k Alternatively, from Figure, we see that the magnitude of G is constant and G < f herefe, Y < X ie, the displacement between the recding head and the disk surface is small o increase the bandwidth using this approach, we would want to make d With the bump on the disk surface, the Fourier transfm of { h sinπt/, < t < xt, t > k m as large as possible

12 is X {xt} xte jt dt h sin h j πt e jt dt e jπt/ e jπt/ e jt dt e jπ/ t e jπ/+t dt h j h j jπ/ ejπ/ t + jπ/ + e jπ/+t h e jπ/ h e jπ/ π/ π/ + h e jπ e j h e jπ e j π/ π/ + h e j + h + e j + π/ π/ + X π h e j + Ans o find the magnitude, first rewrite X as X h π cos j sin + hen, h X π cos + + sin h X π cos + Note that X when 3π, 5π, 7π, etc he plot of the amplitude of X is shown in Figure 4 Figure 4: Amplitude of X

13 e Recall from c that G remains constant when Also, note that Y G X herefe, the minimum the disk can have without significantly exciting the head into large vibrations can be determined from 3π > 3π Ans 3

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