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Topic: Help with a function for plotting zeros and poles
Replies: 2   Last Post: Mar 19, 2014 4:25 AM

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emammendes@gmail.com

Posts: 143
Registered: 8/22/05
Re: Help with a function for plotting zeros and poles
Posted: Mar 19, 2014 4:25 AM
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Many many thanks

Ed

On Mar 18, 2014, at 6:45 PM, Bob Hanlon <hanlonr357@gmail.com> wrote:

> Sorry about the oversight.
>
> Clear[zeroPole]
>
> zeroPole[tf_TransferFunctionModel] :=
> First /@
> Map[{Re[#], Im[#]} &,
> Flatten[Through@{TransferFunctionZeros, TransferFunctionPoles}@tf,=


> 1], {3}];
>
> tf1 = TransferFunctionModel[(3 (13/8 + s))/(2 (3/2 (13/8 + s) +
> s (1 + s) (2 + s) (5 + s))), s];
>
> tf2 = TransferFunctionModel[(199 +
> 344 s)/(16 (s (1 + s) (2 + s) (5 + s) + 1/16 (199 + 344 s))), =

s];
>
> N@zeroPole[tf1]
>
> {{{-1.625, 0.}}, {{-0.5, 0.}, {-0.5, 0.}, {-5.08114, 0.}, {-1.91886, =

0.}}}
>
> N@zeroPole[tf2]
>
> {{{-0.578488, 0.}}, {{-0.5, 0.}, {-5.97986,
> 0.}, {-0.760068, -1.89264}, {-0.760068, 1.89264}}}
>
>
> Bob Hanlon
>
>
>
> On Tue, Mar 18, 2014 at 5:20 PM, Eduardo M. A. M. Mendes =

<emammendes@gmail.com> wrote:
> Dear Bob
>
> Many thanks but there is a problem: in the output of the new zeroPole =

function there is no distinction between poles and zeros (Please see the =
output of the old zeroPole function).
>
> Again I have no idea how to get this right. Changes on the level of =

Flatten won't do.
>
> Cheers
>
> Ed
>
>
> On Mar 18, 2014, at 12:18 PM, Bob Hanlon <hanlonr357@gmail.com> wrote:
>

>> Clear[zeroPole]
>>
>> zeroPole[tf_TransferFunctionModel] :=
>> {Re[#], Im[#]} & /@
>> Flatten[
>> Through@
>> {TransferFunctionZeros, TransferFunctionPoles}@
>> tf];
>>
>> tf1 = TransferFunctionModel[
>> (3 (13/8 + s))/(2 (3/2 (13/8 + s) + s (1 + s) (2 + s) (5 + s))), =

s];
>>
>> tf2 = TransferFunctionModel[
>> (199 + 344 s)/(16 (s (1 + s) (2 + s) (5 + s) + 1/16 (199 + 344 =

s))), s];
>>
>> N@zeroPole[tf1]
>>
>> {{-1.625, 0.}, {-0.5, 0.}, {-0.5, 0.}, {-5.08114, 0.}, {-1.91886, =

0.}}
>>
>> N@zeroPole[tf2]
>>
>> {{-0.578488, 0.}, {-0.5, 0.}, {-5.97986,
>> 0.}, {-0.760068, -1.89264}, {-0.760068, 1.89264}}
>>
>>
>> Bob Hanlon
>>
>>
>>
>> On Sat, Mar 15, 2014 at 3:46 AM, Eduardo M. A. M. Mendes =

<emammendes@gmail.com> wrote:
>> Hello
>>
>> Sometime ago I found a couple of functions that are used for plotting =

the poles and zeros of a transfer function. Here they are:
>>
>> =

xyPoints[values_]:=Module[{xy},xy=Flatten[Replace[values,{Complex[x_,y=
_]:>{x,y},x_?NumericQ:>{x,0}},{3}],1];Cases[xy,{_?NumericQ,_?NumericQ},{2}=
]
>> ];
>>
>> =

zeroPole[tf_]:=Module[{zp,zp0},zp0=Through@{TransferFunctionZeros,Tran=
sferFunctionPoles}@tf;
>> zp=FixedPoint[ReplaceAll[#,{}->{-100}]&,zp0];
>> xyPoints/@zp];
>>
>> zeroPole is a modification of the actual plot function (I have only =

removed the plot command).
>>
>> Here are two examples of using the functions
>>
>> tf1=TransferFunctionModel[(3 (13/8+s))/(2 (3/2 (13/8+s)+s (1+s) =

(2+s) (5+s))),s]
>> tf2=TransferFunctionModel[(199+344 s)/(16 (s (1+s) (2+s) (5+s)+1/16 =
(199+344 s))),s]
>>
>> N@zeroPole[tf1]
>> {{{-1.625,0.}},{{-0.5,0.},{-0.5,0.},{-5.08114,0.},{-1.91886,0.}}}
>>
>> N@zeroPole[tf2]
>> {{{-0.578488,0.}},{{-0.5,0.},{-5.97986,0.},{-0.760068-1.89264 =

I,0.},{-0.760068+1.89264 I,0.}}}
>>
>> The functions does what I expected for the first example, but not for =

the second example (the real and imaginary parts of the complex poles =
are not dealt with).
>>
>> Can someone tell me what is wrong? And how to modify xyPoints =

(Although I understand what the functions does I am not sure what to =
do)?
>>
>> Many many thanks
>>
>> Ed
>>
>>
>>

>
>






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