Understanding Transfer Function Notation in Circuit Engineering

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Just a quick question about some notation used in my book.

The proper form of the transfer function used in my book is as follows:

[tex]\bar H(j\omega) = \frac{K_0(j\omega)^{\pm N} (1+j\omega\tau_1)(1+2\zeta_3(j\omega\tau_3)+(j\omega\tau_3)^2)\cdot\cdot\cdot }{(1+j\omega \tau_a)(1+2\zeta_b(j \omega \tau_b)+(j \omega \tau_b)^2 )\cdot \cdot \cdot}[/tex]

I'm kinda just being picky here, but I would like to understand the convention that they used.

Why the jump from [itex]\tau_1[/itex] to [itex]\tau_3[/itex], the choice of starting with [itex]\zeta_3[/itex] in the numerator. Just curious if someone could shed some light on this.

Thank you
 
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Is this in circuits, controls, vibrations...? What is the topic?

By the looks of it, it's a formulation from an infinite series.
 
Woops I forgot about this post :blushing:

Sorry, I should have specified where this came from. This is from a basic circuit engineering course, specfically from the book: "Basic Engineering Circuit Analsysis" 8th Edition, Irwin/Nelms.

It looks like an expansion of some sort. The lecture notes have been very good, so I haven't been reading chapters like I typically do, just skimming them...it looks like I missed a page or two.

[tex]\bar H(s) = \frac{N(s)}{D(s)}=\frac{K_0(s-z_1)(s-z_2)\cdot \cdot \cdot(s-z_m)}{(s-p_1)(s-p_2)\cdot\cdot\cdot(s-p_n)}[/tex] (1)

where:
[tex]s= j\omega[/tex]
[tex]N(s) =[/tex] a polynomial of degree m
[tex]D(s) =[/tex] a polynomial of degree n

Also, it says that in general (1) can be expressed in the form that I gave in the OP. Hope that helps clear things up.