# Controllability and observability of a transfer function

I have the following transfer function:

sys = 1/(s+1)^10 (s+0.8)^10


There is no pole zero cancellation, but still when i convert this into state-space form using tf2ss command, the state space model is not minimal. Why it is so?

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Why do you conclude that the ss model is not minimal? The state space model should be order 10. –  copper.hat Jul 31 '12 at 16:00
You might want to consider sys-1 to get rid of the feed-through term. –  copper.hat Jul 31 '12 at 16:01
What is the order of your ss model? –  copper.hat Aug 1 '12 at 9:38
the order of ss model is 20, but it's not minimal. The rank of controllability gramian = 20 while the rank of observability gramian = 10. which is strange!! –  Mohsin Aug 1 '12 at 10:31
How are you computing the rank? What are the singular values of the observability grammian? It is a bit strange. –  copper.hat Aug 1 '12 at 16:31

is this what you are looking for?

expand((s+0.8)^10) = s^10 + 8*s^9 + (144*s^8)/5 + (1536*s^7)/25 + (10752*s^6)/125
+ (258048*s^5)/3125 + (172032*s^4)/3125 + (393216*s^3)/15625
+ (589824*s^2)/78125 + (524288*s)/390625 + 1048576/9765625

expand((s+10)^10) = s^10 + 100*s^9 + 4500*s^8 + 120000*s^7 + 2100000*s^6
+ 25200000*s^5 + 210000000*s^4 + 1200000000*s^3
+ 4500000000*s^2 + 10000000000*s + 10000000000

[A B C D]=tf2ss([1 8 144/5 1536/25 10752/125 258048/3125 172032/3125
393216/15625 589824/78125 524288/390625 1048576/9765625],
[1 100 4500 120000 2100000 25200000 210000000 1200000000
4500000000 10000000000 10000000000])

rank([C;C*A;C*A^2;C*A^3;C*A^4;C*A^5;C*A^6;C*A^7;C*A^8;C*A^9])

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Indeed!! and also the rank of observability matrix. The one you mentioned is controllablity matrix. The question is that why the rank of these matrices are less than the order of the system, which is 20 in this case? when there is no pole zero cancellation. –  Mohsin Jul 31 '12 at 14:56
The transfer function is not strictly proper (numerator coefficient is equal to denominator coefficient; D = 1) I don't know if this is the cause, but I read that a transfer function should be strictly proper to be able to transfer to state space format. see en.wikipedia.org/wiki/Realization_(systems) –  wesley Jul 31 '12 at 15:14
Surely the order of the system is 10? –  copper.hat Jul 31 '12 at 15:57
Sorry i mis-interpret your message, both (s+10)^10 and (s+0.8)^10 are in the denominator, hence the order is 20. Numerator is just 1. –  Mohsin Aug 1 '12 at 9:09

This is not an answer, you but might note that you can write the system as: $$\hat{h}(s) = 1 - \frac{2}{(s+1)} + \frac{9}{5 (s+1)^2} - \frac{24}{25 (s+1)^3} + \frac{42}{125 (s+1)^4} - \frac{252}{3125 (s+1)^5} + \frac{42}{3125 (s+1)^6} - \frac{24}{15625 (s+1)^7} + \frac{9}{78125 (s+1)^8} - \frac{2}{390625 (s+1)^9} + \frac{1}{9765625 (s+1)^{10}},$$ from which you can read off a minimal realization pretty quickly.

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both (s+10)^10 and (s+0.8)^10 are in the denominator, hence the order is 20. Numerator is just 1 –  Mohsin Aug 1 '12 at 9:20
I see, you should format your equation a little better! –  copper.hat Aug 1 '12 at 9:37