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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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