I require frequency response from transfer function = 1/(s^2+(b/m)*s+k/m) where b= 0.0011, k= 87.88 , m = 0.99 in matlab and i need freq in khz

i require a code for the given graph this is the frequency response from transfer function 1/(s^2+(b/m)*s+k/m) where b= 0.0011, k= 87.88 , m = 0.99

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Hi @sia
As you can see from the step response, it is quite oscillatory, and it takes approximately 7,000 seconds to settle down. I am curious to know what specific application you have in mind for this linear second-order system in the real world?
s = tf('s');
b = 0.0011;
k = 87.88;
m = 0.99;
G = 1/(s^2 + (b/m)*s + k/m)
G = 1 ------------------------ s^2 + 0.001111 s + 88.77 Continuous-time transfer function.
step(G)
stepinfo(G)
ans = struct with fields:
RiseTime: 0.1338 TransientTime: 7.0323e+03 SettlingTime: 7.0323e+03 SettlingMin: 1.1174e-05 SettlingMax: 0.0225 Overshoot: 99.9659 Undershoot: 0 Peak: 0.0225 PeakTime: 0.3316

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Sam Chak
Sam Chak 2023년 7월 13일
편집: Sam Chak 2023년 7월 13일
Hi @sia
If you simply want to obtain the plot for the "transfer function 1/(s^2 + (0.0011/0.99)*s + 87.88/0.99)," WolframAlpha can generate it for you without the technical details, saving your time and everyone else's. However, to be honest, generating the Bode plot requires fewer characters in MATLAB code compared to WolframAlpha. In other words, MATLAB is faster.
Hi sia, there is not much need to do any detailed calculation when your starting point is what it is. This system has a small amount of damping so its resonant frequency is very close to
omega = sqrt(k/m)
omega = 9.4217
f = omega/(2*pi)
f = 1.4995
which is what the plot shows. To get a resonant frequency around 1.5 kHz, either k has to increase by 10^6 or m has to decrease by 10^6, or some combination of those has to come up with a factor of 1e6 in the ratio (k/m). [the damping constant b will need a change as well]. So you should go back to the original physical situation. Maybe m is in milligrams instead of kilograms. Lots of possibilities.

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sia
2023년 7월 13일

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2023년 7월 18일

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