Sampling requirements for SysId

Am 1/28/2013 12:15 PM, schrieb Tim Wescott:
> On Mon, 28 Jan 2013 11:46:02 -0800, Peter Mairhofer wrote:
> 
> [...]
>> My observation from simple MATLAB experiments is:
>>
>> 1) I can use any subset of the equations and get a solution. This is in
>> one sense obvious because it is LS. On the other hand, it contradicts my
>> intuition. Suppose my Nyquist rate for input x(t) and the unknown system
>> h(t) is 10kHz and h(t) operates (amongst others) also at 4 kHz. If I
>> sample the output y(t) at 5kHz, how should the reconstruction algorithm
>> know about the 4kHz range?
> 
> Oy.  Nyquist didn't say that.  In fact, your above statement makes almost 
> no sense.  http://www.wescottdesign.com/articles/Sampling/sampling.pdf.
> 
> You state elsewhere that you are modeling the system h(t) as something 
> that can be reasonably represented as a FIR filter.  In that case, the 
> assumption that h(t) can be reasonably represented as a FIR filter is 
> only valid if you do your measurements at the same sampling rate as your 
> FIR filter is going to operate.

Maybe I should state the setup more precisly:
- I assume an ideal (mathematical) setup (exact pointwise samples, ideal
low-pass filters, infinite long signals)
- I assume no noise for now
- I assume perfect linearity
- I assume that my input signal x(t) perfectly bandlimited white noise
between 0-5kHz and is zero >5kHz.

The filter h(t) will operate on x(t), modifying any frequency content
between 0 and 5kHz (it can't do anything more because it is a perfect
linear system).

Because of the assumptions described above, I can convert the whole
setup to a discrete-time setup: x[n] represents x(t), y[n] y(t) where
the signals are sampled at n*Ts (Ts=1/10kHz). h(t) can be converted to
an IIR filter via bilinear transform (operating on x[n] the same way as
on x(t) between 1-5kHz).

The only non-ideal assumption: The IIR equivalent of h(t) is
approximated as an FIR filter h[n] with Q taps.

Assuming that h[n] models h(t) good enough, N samples of x[n] and y[n]
should recover the coefficients of h[n]:

h = X^+ * y

where (.)^+ denotes the Moore-Penrose pseudo inverse, X is a matrix
containg the tap-inputs of x[n] and y[n] contains N samples.

Stated differently: When I feed x[n] to my IIR equivalent of h(t) and
h[n] via y=X*h, both should produce a similar y[n] (as similar as h[n]
approximates the IIR equivalent of h(t)).

Now to what I meant in my original question: Is it possible to use just
any second or forth sample in y[n] and still recover h[n] perfectly?
Note that "every second sample" corresponds to a subsampling of 2 in the
ideal setup.

The answer is yes (in theory) because it is LS (which is solveable as
long as there are Q lin. independent rows).

What I did next is to sum over M consecutive samples of y[n] and take
the sum. This "sum-and-dump" corresponds to subsampling of factor M with
pre-lowpass filtering (in the ideal setup). In order to hold the
equality in y=X*h, I need to apply the summation in X as well. This
approach still recovered the h[n] under perfect numerical conditions
although I destroyed much frequency content because of the summation.
However, because of the summation in x[n] over M samples, the columns
become more and more orthogonal and the system therefore numerically
instable.

Please regard this as a tough-experiment rather than a question
regarding practical SysId.

Peter

On 28.01.2013 09:09, Peter Mairhofer wrote:
> Hi,
>
> [...]
>
> When using a simple LS method to solve for the coefficients, the
> overdetermined system of equations suggests that it is possible to drop
> rows (as long as there are enough linearly independent) and thus
> (implicitely) subsample y(t).
>
> Thanks,
> Peter
>


Theory:

The polynomial degree should as low as possible but high enough for 
fitting an ODE.


See Example:

Reducing the data points and changing the polynomial degree:

http://home.arcor.de/janch/20130131-control/default.html

-- 
Regards JCH