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Re: two pole passive low pass filters using transfer functio
What exactly is wrong? As far as I can tell, you don't have a "canonical" two pole low-pass filter because your transfer function has a pole at zero, another pole and a zero.
Well, if you know the open loop poles and zeros(and the gain) then you can write the transfer function. The closed loop transfer function is Hc = Ho/(1+b*Ho), where Ho is the open loop transfer fcn, b is the feedback path transfer fcn.
And yes you can quantify peaking from closed-loop transfer...
From what i can see, the regulator described in the patent is different from this one, that is a linear regulator whereas this one is non-linear. You can consider it a kind-of adaptive algorithm, at high errors gives a greater command, this should improve response time, but it could be a problem...
To answer the first question you must obviously compute the impulse response, to do that, you replace:
\[x[n] = \delta [n]\] where \[ \delta [n]\] is the unit impulse(kroneker delta)
and you get:
\[y[n] = h[n] = \delta [n]-2\delta [n-1]+3\delta[n-2]\]
that's it!
For the second point, use Matlab
Not quite.
Think of your fet transistor like a variable resistor witch is controlled by Vgs. If Vgs is smaller than the threadshot voltage than this resistor has a high value, thus very little current flows and the device is turned off. If you increase Vgs over that threadshot voltage than this...
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