Decomposition minimal phase & all pass

• Engineering
Hidd
Thread moved from the technical forums to the schoolwork forums (this includes for non-homework revision studying)
Homework Statement:
I have the following transfer function, and I would like to decompose it to a minimal phase G1 & all pass G2 transfer functions:

G(s) = (1-s) / (2 + 10s)

G(s) = G1 * G2
Relevant Equations:
All-pass ==> magnitude of G2(jw) =1
Minimum phase ==> Re{Zeros,poles}<0
Hey everybody!

I have put G1 = (1-s)/(2-10s) & G2 = (2-10s)/ (2 +10s)
but than I read that all poles and zeroes should be inside the unit circle, and I don't know how to move the Zero S_01 = 1 to the unit circle

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Gold Member
Is this a homework problem? Do you know what the all-pass G2 transfer function is?

Hidd
it's when the magnitude of G2(jw) =1

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DaveE
Gold Member
OK, but G1 isn't minimum phase since it has right-half plane poles and/or zeros.
Still, I doubt that G2 = (1 + s)/(1 - s) is allowed since it's not causal (RHP pole). Still it is technically correct, |G2|=1. Otherwise, I don't see how to get rid of the RHP zero in G without a RHP pole in G2.* Note that non-causal digital filters are used sometimes.

But it's been decades since I did this sort of filter design problem. It's not the sort of thing you'll ever see much in practice, IMO.

I think the unit circle comment relates to discrete time systems where z outside the unit circle is equivalent to s in the RHP.

edit: * Oops! confused the first time

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Hidd
That means that it's impossible to derive the all-pass & min-phase functions from G ?!

$$G(s) = \frac{(1-s)}{(2 + 10s)} = \frac{(1-s)}{(1+s)} ⋅ \frac{(1+s)}{(2 + 10s)}$$