How to check if two transfer functions are the same?

Question

Bill Tubbs 2020-8-12

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编辑： Bill Tubbs 2020-11-7

采纳的回答： Walter Roberson

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I want to verify that two transfer functions are equal.

For example:

Kc = 0.4262; Ti=1.35;
C1 = pidstd(Kc,Ti);
tf(C1)
s = tf('s');
Gc1 = Kc*(1+1/(Ti*s))
assert(tf(C1) == Gc1)

Raises:

Undefined operator '==' for input arguments of type 'tf'.

Also note:

Gc1
tf(C1)
Gc1 - tf(C1)

Gc1 =

0.5754 s + 0.4262

-----------------

1.35 s

Continuous-time transfer function.

ans =

0.4262 s + 0.3157

-----------------

s

Continuous-time transfer function.

ans =

0

Static gain.

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Answer 1

Walter Roberson 2020-8-12

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编辑：Walter Roberson 2020-8-12

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isequal(tfdata(tf(C1) - Gc1), {[0]})

This will probably fail if there are delays in the tf.

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Bill Tubbs 2020-8-12

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Understood, but why does the RHS have to be a cell? Or, why doesn't this work:

isequal(tfdata(tf(C1) - Gc1), 0)

Walter Roberson 2020-8-12

tf are deliberately designed to commonly be used in arrays, Number of Inputs by Number of Outputs of them. So tfdata is designed to return the numerators of each of the tf array elements in distinguishable form, with the elements commonly not being the same length as each other. Rather than build a multidimensional array of coefficients with the first or last dimension being padded out to the longest numerator, and each element padded with leading zeros to match the longest numerator, the tfdata function simply creates a cell array of the numerators, like arrayfun of tfdata with 'uniform, 0. In the case of a scalar tf, that gives you a scalar cell array. tfdata does not then special case that scalar cell to extract the contents.

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Answer 2

Bill Tubbs 2020-8-12

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编辑：Bill Tubbs 2020-8-12

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There is an answer on Stackoverflow:

However, it does not work directly in my case:

isequal(tf(C1), Gc1)

ans =

logical

0

Need to factor the numerator and denominator so they are the same and then it works:

factor = Gc1.denominator{1}(1)
Gc1_factored = tf(Gc1.num{1}/factor,Gc1.den{1}/factor)
isequal(Gc1_factored,tf(C1))

factor =

1.3500

Gc1_factored =

0.4262 s + 0.3157

-----------------

s

Continuous-time transfer function.

ans =

logical

1

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Paul 2020-8-13

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Disclaimer: Only talking about SISO systems that don't have delays, though I think most if not all of this will exetend to systems with input or output delays.

If you're going to use Walter's approach, then you may need to account for some tolerance, as is always a concern when checking for equality. Here's an example where the difference between the two tf's should be zero, but they are not due to rounding. Depending on how you come up with your two tf's that you're comparing, this may or may not be of interest to you.

>> sys1=tf(rss(10));
>> sys2=sys1;
>> sys2.num{1}=sys2.num{1}*pi;
>> sys2.den{1}=sys2.den{1}*pi;
>> isequal(tfdata(sys1 - sys2), {0})
ans =
  logical
   0

Of course the difference is small.

>> [~,~,k]=zpkdata(sys2-sys1)
k =
   1.8094e-14

Making your decision on equality based on k is probably the better way to go if you're deriving two tf's that should be equal in principle but aren't equal numerically.

As for normalizing the leading coefficient of the denominator, either do it the manual way as you did above or convert your tf to one of the other forms and convert back:

>> sys1=tf(3,3*[1 1]);
>> tf(zpk(sys1))
ans =
 
    1
  -----
  s + 1
 
Continuous-time transfer function.

If you go down this route, you might as well just leave it in the zpk form, unless you really need a tf for some reason. Keep in mind that the double conversion could yield slightly different coefficients than the manual approach due to rounding effects in the conversion algorithms.

See doc minreal if by "simpiify a tf" you mean get rid of poles and zeros that (should) cancel. Hey, I just discovered that minreal also simplifies the gain.

>> minreal(sys1)
ans =
 
    1
  -----
  s + 1
 
Continuous-time transfer function.

I haven't checked the inner workings of minreal so don't know if that's a preferred approach to get that leading coefficient to one.

Bill Tubbs 2020-8-13

Thanks Paul. I did not know about zpk or minreal and they will for sure be useful for these type of tf manipulations.

Bill Tubbs 2020-11-6

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I think it's worth highlighting that if using zpk during the equality check, it's important to also convert back to tf before the subtraction as Paul says in comment above. Otherwise, you can get some numerical errors creeping in.

E.g.

>> zpk(tf(1,[2 1])) - zpk(tf(1,[2 1]))
ans =
 
  1.1102e-16
  ----------
  (s+0.5)^2
 
Continuous-time zero/pole/gain model.

whereas

>> tf(zpk(tf(1,[2 1]))) - tf(zpk(tf(1,[2 1])))
ans =
 
  0
 
Static gain.

Something to do with how subtraction is handled for zpk objects compared to tf objects perhaps.

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Answer 3

Bill Tubbs 2020-11-6

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编辑：Bill Tubbs 2020-11-6

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Based on comments from Paul above I offer these functions as a solution:

function c = is_equal_tf(G1,G2)
    c = almost_zero_static_gain_tf(G1 - G2);
end
function c = almost_zero_static_gain_tf(G)
    [~,~,k] = zpkdata(G);
    c = abs(k) < 1e-10;
end

Obviously, doesn't work with delays.

It passes the following tests:

assert(is_equal_tf(tf('s'),tf('s')))
assert(is_equal_tf(tf(1,1),tf(-1,-1)))
assert(is_equal_tf(tf(2,[2 1]),tf(1,[1 0.5])))
assert(~is_equal_tf(tf(2.1,[2 1]),tf(1,[1 0.5])))
assert(is_equal_tf(tf(2,conv([2 1],[5 1])), tf(2,[10 7 1])))
assert(~is_equal_tf(tf(2,conv([2 1],[4 1])), tf(2,[10 7 1])))
assert(is_equal_tf(tf(1,[2 1]) - tf(1,[5 1]), tf([3 0],[10 7 1])))
assert(is_equal_tf(pidstd(0.42,1.35), 0.42*(1+1/(1.35*tf('s')))))
G1=tf(rss(5)); G2=G1;
assert(is_equal_tf(G1, G2))
G2.num{1} = G2.num{1}*pi;
G2.den{1} = G2.den{1}*pi;
assert(is_equal_tf(G1, G2))

Any exceptions or other cases I should test?

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Paul 2020-11-7

Well, based on my own experience you never know what a student will submit ;).

I'm not aware of any built-in way to display the transfer function in the form of (as+1)/(bs+1). You could probably write your own, but probably not worth it.

I suppose another alternative that doesn't rely on comparing transfer functions algebraically would be to come up with some criteria on the difference between the impulse responses for the two cases.

Bill Tubbs 2020-11-7

编辑：Bill Tubbs 2020-11-7

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The only way I know is to do it symbolically. There's a nice pair of functions, tf2sym and sym2tf in this package by Oskar Vivero Osornio to convert to/from symbolic functions. Then you can get it in a form where they are comparable by eye, as you suggest:

>> tf2sym(G1)
 
ans =
 
(34*s + 1)/(220*s^2 + 54*s + 2)
 
>> factor(tf2sym(G1))
 
ans =
 
[ 1/2, 34*s + 1, 1/(5*s + 1), 1/(22*s + 1)]

The thing I don't like with this approach is the need to continually switch between tf and symbolic, always redefining s in the same script.

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