Re: Help with a function for plotting zeros and poles

• To: mathgroup at smc.vnet.net
• Subject: [mg132454] Re: Help with a function for plotting zeros and poles
• From: "Eduardo M. A. M. Mendes" <emammendes at gmail.com>
• Date: Wed, 19 Mar 2014 04:24:49 -0400 (EDT)
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• References: <20140315074645.B9C5E6A3E@smc.vnet.net> <CAEtRDSeEcZc=1WEvQ9jqOw+QVVE7YghNnAKURz0KyrDW1jT1XQ@mail.gmail.com> <93991AEA-C71B-4789-94D0-16E548CD1B29@gmail.com> <CAEtRDSeYHesumE2+LhEGyN_gezUzpZsQKa2oUP292Qi2JfrK9Q@mail.gmail.com>

```Many many thanks

Ed

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

>
> 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 at 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 at 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 at 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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