How can I plot the transfer function? HELP Please.
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Hello, I have a problem with plot the transfer function. I have to plot the precision model's transfer function. To understand my function and precision model, I'll insert a photo.
Here, I must use
numerator = [];
denominator = [];
sys = tf(numerator,denominator)
stepplot(sys)
But I cannot write the numerator and denominator. I have numerator = KH * ( TL*s + 1 ) * exp(-t*s) and denominator = ( TI*s + 1 ) * ( ( (p^2) / (WN^2) ) + ( ( (2*zetaN) / (WN) ) * p ) + (1). As I understand, I must write these numerator and denominator as coefficients of s variable. But I couldn't write. Could you help me please? How can I plot the transfer function?
Also, I want to plot step function with 1 amplitude for compare it with transfer function. How can I do that? Is it just step(sys)?? Is this the same thing with stepplot?
If you could help me, I will be very glad. Thanks!
采纳的回答
Because you want to make plots, I'm going to assume that you have numerical values for each parameter in your model. For example, suppose we have:
Kh = 2; Tl = 1; Ti = 0.5; wn = 1; zetan = 0.7; tau = 0.2;
At this point, there are several options to form the transfer function F(s). Perhaps the easiest is to start with a product of low order terms:
>> F = Kh*tf([Tl 1],[Ti 1])*tf(1,[1/wn^2 2*zetan/wn 1])
F =
2 s + 2
-----------------------------
0.5 s^3 + 1.7 s^2 + 1.9 s + 1
Continuous-time transfer function.
Now the only part that is missing is the exp(-tau*s). That can be included by either setting the 'InputDelay' or 'OutputDelay' property of F or by mutiplying F by exp(-tau*s) explicilty. For example of the latter:
>> F = F*exp(-tau*tf('s'))
F =
2 s + 2
exp(-0.2*s) * -----------------------------
0.5 s^3 + 1.7 s^2 + 1.9 s + 1
Continuous-time transfer function.
If you'd rather see F expressed in terms of the variable p, you can do:
>> F.Variable='p'
F =
2 p + 2
exp(-0.2*p) * -----------------------------
0.5 p^3 + 1.7 p^2 + 1.9 p + 1
Continuous-time transfer function.
which is just for display purposes and doesn't change any mathematical properties of F. It's not clear what is meant by "plot the ... transfer function." Once F is defined, all sorts of different plots can be developed, most typically:
doc bode
doc step
doc impulse
更多回答（1 个）
Image Analyst
2021-4-13
You're using both p and s to describe the same p in the formula. Change s to p and I think it should work:
numerator = KH * ( TL*p + 1 ) * exp(-t*p)
term1 = TI*p + 1
term2 = p^2 / WN^2
term3 = 2 * zetaN * p / WN
denominator = term1 * (term2 + term3 + 1)
FH = numerator / denominator
If you'd rather call the independent variable s instead of p, then just change all the p to s.
numerator = KH * ( TL*s + 1 ) * exp(-t*s)
term1 = TI*s + 1
term2 = s^2 / WN^2
term3 = 2 * zetaN * s / WN
denominator = term1 * (term2 + term3 + 1)
FH = numerator / denominator
but don't mix s and p - that could be a disaster if they are separate variables and have different values.
3 个评论
hgrlk
2021-4-13
Image Analyst
2021-4-13
Sorry, I don't know anything about that function. You'll just have to read the documentation or hope someone else knows about it.
hgrlk
2021-4-14
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