Properties of linear system transfer function

Jan 10, 2007 7 Replies

Hello,



Consider a linear transfer function with N-1 real and negative poles and N real and negative zeros factorized in the standard way:



(z-z_1)*(z-z_2)*...*(z-z_N) Q(z) = K ----------------------------- (z-p_1)*(z-p_2)*...*(z-z_{N-1}),



where lim z-> infinity => Q(z)->infinity.



Are there then any theorems regarding such a transfer function? I am looking for properties of such transfer functions.



Thanks,



Peter


Stable and proper (zero at inf. freq.)

I used to think of them as of real-life functions. Nothing more comes to my mind about generalized properties.

Thank you for looking into my problem.

To specify the matters a bit more, my problem is not at all related to filtering, but the mathematical function I am looking can be written in the form above. So I am hoping to apply some theorems derived in linear systems theory to my problem.

Also, the poles and zeros are all real and negative, i.e., Q(z) is a ratio of two strictly increasing positive polynomials.

Now, what I am after is some facts about the variation / oscillations of this function. Ideally, I wanted it to have no local maximums for z>0, but unfortunately this is not true. I am thinking something along the lines of that since all the poles and zeros are for z0. Also, only real z's are of interest.

Looking it as a transfer function, I then would be interested in results on the magnitude, since everything is positive, i.e. |Q| = Q.

Thanks,

Peter.

Do you mean Q(z)->0 ? And 'z' is Z transform parameter?

No, Q(z)->infinity. The reason is that the top polynomial is of order N, but the denominator polynomial is of order N-1. In the end, since both polynomials are positive, Q(z) goes to infinity.

Could be that yes, I am looking for properties of |Q(z)|=Q(z). I.e., are there theorems describing the behavior of |Q(z)|, when given as above?

Peter

Yes, of course (I saw N+1 instead of N-1). So they are improper in this case. About the rest I give up. I don't understand your question. If z is complex so how could it be that modulus (real value) equals complex value |Q(z)|=3DQ(z) for all z? If z is not complex then how can you even think of modulus? Or maybe it's an absolute value ||?

Absolute value is the only way to make sense of it. X = |X| is just another way to write X >= 0.

Jerry

Then Q is not a transfer function but just real rational function.

Join the Discussion

Have something to add? Share your thoughts — no account required.

Didn't find your answer?

Ask the community — no account required