Export transfer function model to workspace from Simulink without linmod
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I want to export the trasfer function of a simple simulink program to the workspace. I know it can be done simply by using the commands
[a,b,c,d]=linmod('model')
sys=ss(a,b,c,d);
tf(sys)
Is there any way to avoid the linmod command? The reason I need to do this is to present this method for people who are not familiar with state-sspace models.

1 件のコメント
Azzi Abdelmalek
2013 年 10 月 18 日
Lazaros commented
Isn't there anyone who can help me with this? Thanks again in advance!
回答 (1 件)
Azzi Abdelmalek
2013 年 10 月 14 日
編集済み: Azzi Abdelmalek
2013 年 10 月 14 日
Edit
Your model name is transfer, and you have two transfer function named Transfer Fcn and Transfer Fcn1
To get the parameters of Transfer Fcn block
num=get_param('transfer/Transfer Fcn','Numerator');
den=get_param('transfer/Transfer Fcn','Denominator');
model=tf(str2num(num),str2num(den))
8 件のコメント
Azzi Abdelmalek
2013 年 10 月 14 日
Look at Edit
Lazaros Moysis
2013 年 10 月 14 日
Azzi Abdelmalek
2013 年 10 月 14 日
num=get_param('transfer/Transfer Fcn','Numerator');
den=get_param('transfer/Transfer Fcn','Denominator');
num1=get_param('transfer/Transfer Fcn1','Numerator');
den1=get_param('transfer/Transfer Fcn1','Denominator');
model=tf(str2num(num),str2num(den))
model1=tf(str2num(num1),str2num(den1))
modelg=model*model1
Lazaros Moysis
2013 年 10 月 14 日
編集済み: Lazaros Moysis
2013 年 10 月 14 日
Azzi Abdelmalek
2013 年 10 月 18 日
When you read the help of linmod, nothing is referring to an equivalent function, which make me think there is not another way. I think also Its the way simulink stores your model (linear or non linerar), even when you use a transfer function, I think, simulink stores it as state space representation.
Lazaros Moysis
2013 年 10 月 18 日
haryana thomas
2018 年 8 月 21 日
you should type this command in the matlab command line.
plant = linearize('name of simulink model')
tf(plant)
Raja Singh R
2020 年 8 月 13 日
Thomas, It is showing empty transfer function
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