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c16 - 18.03 Class 16 Frequency response[1 Frequency...

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18.03 Class 16, March 15, 2006 Frequency response [1] Frequency response: without damping First recall the Harmonic Oscillator: x" + omega_n^2 x = 0 : The spring constant is k = omega_n^2 . Solutions are arbitrary sinusoids with circular frequency omega_n , the "natural frequency" of the system. Drive it sinusoidally: x" + omega_n^2 x = omega_n^2 A cos(omega t) I am driving the system through the spring, with a plunger moving sinusoidally with amplitude 1 . cos(omega t) is the "physical signal," as opposed to the force, or "complete signal" omega_n^2 cos(omega t) . We regard the plunger position as the system input. We solved this by luckily trying x_p = B cos(omega t) and solving for B . Let's do it using ERF: z" + omega_n^2 z = omega_n^2 A e^{i omega t} z_p = A (omega_n^2 / (omega_n^2 - omega ^2)) e^{i omega t} No damping ===> denominator is real and so x_p = A (omega_n^2 / (omega_n^2 - omega^2)) cos(omega t) This is ok unless omega = omega_n , in which case the system is in resonance with the signal. [2] Bode Plots.

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c16 - 18.03 Class 16 Frequency response[1 Frequency...

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