八度控制软件包tf

Question

Using the octave/matlab control toolbox:

octave.exe:1> pkg load control

I define the same transfer function in two different ways:

octave.exe:2> a = tf('1/(s + 1)')

Transfer function 'a' from input 'u1' to output ...

 y1:  1/(s + 1)

Continuous-time model.
octave.exe:3> b = 1 / (tf('s') + 1)

Transfer function 'b' from input 'u1' to output ...

        1
 y1:  -----
      s + 1

Continuous-time model.

And then evaluate it at s = j :

octave.exe:4> a(1)
ans =  0 + 1i
octave.exe:5> b(1)
ans =  0.50000 - 0.50000i

Why are these different!?

Answer 1

I think the way you define a is incorrect. I am not sure why it doesn't error out when you run the command, but it's not how you should define a transfer function. if we consider the following:

>> a = tf(1,[1 1])

Transfer function 'a' from input 'u1' to output ...

        1
 y1:  -----
      s + 1

Continuous-time model.
>> a(1)
ans =  0.50000 - 0.50000i
>> b = 1/(tf('s')+1)

Transfer function 'b' from input 'u1' to output ...

        1
 y1:  -----
      s + 1

Continuous-time model.
>> b(1)
ans =  0.50000 - 0.50000i
>> c = tf('1/(s+1)')

Transfer function 'c' from input 'u1' to output ...

 y1:  1/(s+1)

Continuous-time model.
>> c(1)
ans =  0 + 1i
>> s = tf('s')

Transfer function 's' from input 'u1' to output ...

 y1:  s

Continuous-time model.
>> d = 1/(s+1)

Transfer function 'd' from input 'u1' to output ...

        1
 y1:  -----
      s + 1

Continuous-time model.
>> d(1)
ans =  0.50000 - 0.50000i

You'll notice how c in my example ( a in yours) is not displayed the same as all the other transfer function, it's all on one line. Maybe it's treating the input 1/(s+1) as a string? I really don't know.

Anyway, the point is that all 3 other ways of defining the transfer function are correct and equivalent, and all give the same and correct result.