---+ DAN Bode Plots
Motivation
Since DAN is an integral control loop, its transfer function can be written down analytically. DAN is linear, so the Bode plot contains all the information on the system (barring rails and things I haven't thought of).
Derivation of DAN Transfer function
The time sampling is discrete. We can enforce some initial condition and write down the differential equation describing the DAN feedback path. Here
a[t] describes the nuller amplitude and
x[t] describes the current through the bolometer (to be nulled out).
%BEGINLATEX%
$$
a[n] - a[n-1] = K \left( x[n] - a[n-1] \right)
$$
%ENDLATEX%
Taking the Z-transform gives the transfer function
%BEGINLATEX%
$$
H(z) = \frac{A(z)}{X(z)} = \frac{K}{1 + (K-1) z^{-1}}
$$
%ENDLATEX%
Note that for 0 <
K < 2, the poles lie within the unit circle and we can write its Fourier equivalent
%BEGINLATEX%
$$
H(z) = \frac{K}{1 + (K-1) e^{-i \omega}}
$$
%ENDLATEX%
For K>2, there are poles outside the unit circle: the system is unstable to perturbations and no Fourier representation exists.
Bode Plots
Z-domain
Plotting the amplitude and phase of
H(z) gives the following plot
 |
| Fig. 1: Bode Plot of DAN. |
Note that transfer function is 1 at DC, and drops off at higher frequency. The phase starts going from 0 to -90 as expected for integral control, but never quite makes it to -90. Instead the phase change goes back to 0 and Nyquist. This is due to the discrete-time nature of the system and is not seen in the S-domain equivalent Bode plot.
--
TijmenDeHaan - 03 Aug 2010
This topic: ColdFeedback
> WebHome > DanBodePlots
Topic revision: r2 - 2011-06-15 - TijmenDeHaan
png 