Bump --> using tfdata on Discrete Transfer Function
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Sheikh
am 26 Jan. 2014
Bearbeitet: Sheikh
am 28 Jan. 2014
1. I am using tfdata to get the numerator and denominator coefficients of discrete transfer function. Note that the discrete transfer function matches the underlying continuous transfer function in Bode Diagram. [Ts=time period = 40e-6]
sysdxx=c2d(sys_xx,Ts,'tustin'); %getting Discrete system from continuous system sys_xx
[numdxx,dendxx]=tfdata(sysdxx,'v'); %getting num & den using tfdata
2. This is the continuous time system.
sys_xx =
(-3.71e05 s^15 - 4.784e08 s^14 - 2.247e13 s^13 - 2.501e16 s^12 - 4.589e20 s^11 - 4.249e23 s^10 - 3.745e27 s^9 - 2.865e30 s^8 - 1.161e34 s^7 - 7.184e36 s^6 - 1.12e40 s^5 - 5.53e42 s^4 - 1.47e45 s^3 - 6.161e47 s^2 - 1.661e47 s - 2.484e43)
/( s^16 + 9.043e05 s^15 + 1.002e10 s^14 + 1.114e14 s^13 + 5.156e17 s^12 + 3.396e21 s^11 + 8.22e24 s^10 + 3.55e28 s^9 + 4.978e31 s^8 + 1.385e35 s^7 + 9.979e37 s^6 + 1.805e41 s^5 + 3.511e43 s^4 + 3.205e46 s^3 + 1.786e48 s^2 + 5.594e47 s + 1.782e44)
sys_xx = num/den;
num = [-3.71e05 -4.784e08 -2.247e13 -2.501e16 -4.589e20 -4.249e23 -3.745e27 -2.865e30 -1.161e34 -7.184e36 -1.12e40 -5.53e42 -1.47e45 -6.161e47 -1.661e47 -2.484e43]
den = [1 9.043e05 1.002e10 1.114e14 5.156e17 3.396e21 8.22e24 3.55e28 4.978e31 1.385e35 9.979e37 1.805e41 3.511e43 3.205e46 1.786e48 5.594e47 1.782e44]
3. Then if I construct a discrete transfer function (or filter) from this numerator & denominator coefficients it does not match the original discrete system or the continuous-time system in bode diagram.
sysdxx2 = tf(numdxx,dendxx,Ts,'variable','z'); %reconstructing discrete system
4. If the discrete system is converted back to continuous time system, even then it does not match.
sysdxx3 = d2c(sysdxx2,'tustin');
5. "sysdxx" does not match with "sys_xx".
6. "sysdxx2" does not match with "sys_xx".
7. "sysdxx3" does not match with "sys_xx".
Anyone have any idea of what might be the reason.
Thanks.
3 Kommentare
Azzi Abdelmalek
am 27 Jan. 2014
Please write your continuous system as numerator and denominator, without variable s
Akzeptierte Antwort
Azzi Abdelmalek
am 26 Jan. 2014
Bearbeitet: Azzi Abdelmalek
am 26 Jan. 2014
You are not using the same variables, (numdxx and numkdxx). Also, if you want to make a comparison, just use the same variable z, instead of z^(-1). And why do you think that a discrete model will match a corresponding continuous model?
Look at this example
Ts=1
sys_xx=tf(1,[2 3]);
sysdxx=c2d(sys_xx,Ts,'tustin')
[numdxx,dendxx]=tfdata(sysdxx,'v')
sysdxx2 = tf(numdxx,dendxx,Ts)
sysdxx and sysdxx2 are identical
13 Kommentare
Azzi Abdelmalek
am 27 Jan. 2014
There is another problem with the c2d function. The poles of your continuous model are all stables, but the real part of some of them are near 0 (model at the limit of instability). If you check the poles of the discrete model obtained by c2d, you will notice that some of them are unstable. This is due to numerical errors. It's obvious that trying to get the original continuous model will be difficult from this unstable discrete model.
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