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describing function analysis of mechanical systems with nonlinear ...

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0<br />

ω=30<br />

-2000 ω=40<br />

ω=50<br />

-4000<br />

ω=60<br />

linear<br />

F=50<br />

F=40<br />

F=30<br />

F=20<br />

10<br />

5<br />

Im (-1/N)<br />

-6000<br />

-8000<br />

-10000<br />

ω=70<br />

ω=80<br />

ω=90<br />

ω=100<br />

F=10<br />

-12000<br />

0 2000 4000 6000 8000 10000 12000 14000<br />

Re (-1/N)<br />

x(t)<br />

0<br />

-5<br />

-10<br />

65 70 75 80 85 90 95<br />

time (s)<br />

Fig. 3. Nyquist plot <strong>of</strong> −1/N(F,ω) for the system<br />

subjected to <strong>nonlinear</strong> friction and for an input<br />

force F = {10, 20, 30, 40, 50} N.<br />

Fig. 4 illustrates the variation <strong>of</strong> the Nyquist plots <strong>of</strong><br />

−1/N(F,ω) for the cases <strong>of</strong> the linear system and<br />

<strong>nonlinear</strong> friction. The log-log plots <strong>of</strong> Re{−1/N} and<br />

Im{−1/N} vs. the exciting frequency ω, for different<br />

values <strong>of</strong> the input force F = {10, 20, 30, 40, 50} N,<br />

reveal that we get results closer to the linear case the<br />

higher the excitation force F, particularly for the real<br />

component.<br />

|Re (-1/N)|<br />

10 4<br />

10 2<br />

10 0<br />

F=10 F=50<br />

Fig. 5. Time response <strong>of</strong> the output position x(t) <strong>of</strong><br />

the system <strong>with</strong> <strong>nonlinear</strong> friction <strong>with</strong> ω = 0.5<br />

rad/s, for F = 10 N (solid line) and F = 50 N<br />

(dashed line, scaled down by a factor <strong>of</strong> 10).<br />

Log(|F(x(t))|)<br />

10<br />

5<br />

0<br />

-5<br />

-10<br />

10 2<br />

10 1<br />

10 0<br />

ω (rad/s)<br />

10 -1<br />

10 -2<br />

0<br />

F=10 N<br />

1 2<br />

a)<br />

F=50 N<br />

3 4 5<br />

6 7<br />

8 9 10<br />

harmonic index k<br />

10 6 ω (rad/s)<br />

|Im (-1/N)|<br />

10 -2<br />

linear<br />

10 -4<br />

10 -2 10 -1 10 0 10 1 10 2<br />

10 4<br />

10 2<br />

F=10<br />

F=50<br />

10 0<br />

linear<br />

10 -2<br />

10 -4<br />

10 -2 10 -1 10 0 10 1 10 2<br />

10 6 ω (rad/s)<br />

Log(|F(x(t))|)<br />

10<br />

5<br />

0<br />

-5<br />

10 2<br />

10 1<br />

10 0<br />

ω (rad/s)<br />

10 -1<br />

10 -2<br />

0 1 2 3 4 5 6 7 8 9 10<br />

harmonic index k<br />

b)<br />

Fig. 6. Fourier transform <strong>of</strong> the output position x(t),<br />

F{x(t)}, over 20 cycles, vs. (ω, k), the exciting<br />

frequency ω and the harmonic frequency index k<br />

for: a) An input force F = 10 N and b) An input<br />

force F = 50 N.<br />

Fig. 4. Log-log plots <strong>of</strong> Re{−1/N} and Im{−1/N} vs.<br />

the exciting frequency ω, for F = {10, 20, 30, 40,<br />

50} N.<br />

Fig. 5 compares the time response for an input force<br />

<strong>of</strong> F = 10 N and F = 50 N. We conclude that for low<br />

forces the <strong>nonlinear</strong> effects <strong>of</strong> the static and Coulomb<br />

frictions are more significant, which is in accordance<br />

<strong>with</strong> the signal harmonic content F{x(t)} depicted in<br />

Fig.6.

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