Circuit:
Analysis:
The left part of the circuit is simply the voltage adder, this time I added a sine wave of 1 Hz with a sine wave of 100 Hz. The goal is to filter away the 100 Hz signal.
The low pass filter I used is a simple RC filter. Denote the voltage of the opamp output to be $ V_s $
$ 1\frac{dV_{out}}{dt} = I = \frac{V_s - V_{out}}{1} $
Set $ V_{s} = e^{i \omega t} $ and $ V_{out} = H(\omega) e^{i\omega t} $, we get:
$ H(\omega)i\omega e^{i\omega t} = e^{i \omega t} - H(\omega) e^{i\omega t} $
$ H(\omega)i\omega = 1 - H(\omega) $
Note that $ \omega $ is in the denominator, meaning for large $ \omega $, the frequency response has a very small gain, so it is effectively removed. Also note that we have some phase shift because the frequency response is complex.
Simulation:
LTSpice:
The model can be downloaded here.
Analysis:
The left part of the circuit is simply the voltage adder, this time I added a sine wave of 1 Hz with a sine wave of 100 Hz. The goal is to filter away the 100 Hz signal.
The low pass filter I used is a simple RC filter. Denote the voltage of the opamp output to be $ V_s $
$ 1\frac{dV_{out}}{dt} = I = \frac{V_s - V_{out}}{1} $
Set $ V_{s} = e^{i \omega t} $ and $ V_{out} = H(\omega) e^{i\omega t} $, we get:
$ H(\omega)i\omega e^{i\omega t} = e^{i \omega t} - H(\omega) e^{i\omega t} $
$ H(\omega)i\omega = 1 - H(\omega) $
$ H(\omega) = \frac{1}{1 + i\omega} $
Note that $ \omega $ is in the denominator, meaning for large $ \omega $, the frequency response has a very small gain, so it is effectively removed. Also note that we have some phase shift because the frequency response is complex.
Simulation:
LTSpice:
The model can be downloaded here.
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