Symbolics - equation form

Forum|Forum|8 years ago
April 10, 2018
5 replies
3837 views

Hello, I am trying to manipulate some equations into a desired form, specifically in the s domain, into a 'bode' form, and similarly in the z domain.

I am having no success, to the extent that I am struggling in the case of the latter to group terms by the z variable, let alone getting in terms of z^-1. Here's what I am working with at present:

As a first step I'd like to get numerator as a poly in z. A second step is getting into a format of z^-n (n integer), and finally the denominator as (1 - poly in z^-n) also:

Have spent some time trying to manually manipulate the equation to no avail. Any step wise instructions would be greatly appreciated.

Am using v15 (15.0 (M005 [MC15_M005_20101105]))

Example textbook forms of a discrete filter is one of either:

z form.png

Regards,

Jack

PS have tried to upload my working file but I am gettign the following error: The contents of the attachment doesn't match its file type

PPS have tried to include images of the desired form of eqn and am getting the following error: Message cannot exceed 20,000 characters.

PPPS: Zip File attached in below self-reply.

PPPPS: Images now correctly attached via the menu rather than ctrl-c / ctrl-v

1_Doc1.pdf

Other

J

jackclift

Author

1-Visitor

another attempt to upload file

1_bode_PI.zip

L

LucMeekes

23-Emerald IV

It is well known that it is very hard to control the output of Mathcad's symbolic processor. If collect and factor don't provide the results you want, what you can do is manipulate parts of your equation, if necessary by hand, and check you results by subtracting it from the original equation. Though even that sometimes fails.

Success!
Luc

J

jackclift

Author

1-Visitor

Thanks Luc.

The other thing I was think might work is define another equation in z with the right number of z terms and somehow getting the two equations to solve e.g. OringinalPoly(z) - NewPoly(z) = 0 solve for coeff. of NewPoly?

Of course I have no idea how to go about that other than the idea...

L

LucMeekes

23-Emerald IV

You have an 'equals' (bold 😃 in your expression, which means that the result of c(z)=2*k.x+...etc is either True (1) or False (0), also for the symbolic processor. It also causes it not to collect the z terms.

Better use the definition operator ':' (which shows as ':=') and do like this: