What is MATLAB doing when adding multiple ZPK transfer functions together?

I am trying to add multiple zpk transfer functions by directly accessing the zeros, poles, and gains without the use of the "parallel" or "+" operations (for runtime reduction), and I was wondering what "+" does:

zero1 = [1, 2];

pole1 = [3, 4];

gain1 = 2;

tf1 = zpk(zero1, pole1, gain1)

% tf1 =

%

% 2 (s-1) (s-2)

% -------------

% (s-3) (s-4)

zero2 = [-1, -2];

pole2 = [-3, -4];

gain2 = -2;

tf2 = zpk(zero2, pole2, gain2)

% tf2 =

%

% -2 (s+1) (s+2)

% --------------

% (s+3) (s+4)

tf = tf1 + tf2

% tf =

%

% 16 s (s-2.345) (s+2.345)

% ------------------------

% (s-3) (s-4) (s+3) (s+4)

I googled how to add two ZPK transfer functions and it said that it's simply linearly adding them together but Matlab is definitely doing something different (as "tf" has a zero at +/- 2.345 when neither "tf1" nor "tf2" have those zeros)

Answer 1

[Aquatris]

Yes but you are doing a fractional algebra. To refresh your memory:

 a/b + c/d = (a*d + c*b)/(b*d)

Since numerator stays b*d, you maintain the same poles. However your denominator is changed significanty, meaning you do not necessarily maintain the same zeros after the addition.

In transfer function terms

(s-2)/(s-5) + (s-1)/(s-3) = [(s-2)*(s-3) + (s-1)*(s-5)] / [(s-5)(s-3)]

So your numerator is the same, poles of your tf1 and tf 2. However your denominator that defines your zeros is now

(s-2)*(s-3) + (s-1)*(s-5) = 2*s^2 - 11*s +11

So zeros of your tf1 and tf2 are no longer the zeros when you add them together