Here are two slides from Fundamentals of Power Electronics. When talking about approximation roots on n-th order polynomial $P(s)=1+a_1s+...+a_ns^n$ with the form of $P(s)=(1+\tau_1s)(1+\tau_2s)...(1+\tau_3s)$, it uses some math that I cannot figure out on myself.

Firstly, I'd like to show the result when every root separate with each other.


And here is the result if consider two close roots exists.


In the last step, I do not know why approximation is accurate leads to the last inequality. Could you give me some hint or explanation? Thanks.

Here is the original slide file.

  • $\begingroup$ in $P(x)=(1+t_1s)(1+t_2s)$... What is $t_i$? $\endgroup$ – NoChance Sep 8 at 4:27
  • 1
    $\begingroup$ @NoChance I'd like to describe $\tau_i$ as placeholders, or maybe unknwons, writing polynomial in the form of $P(x)=...(1+\tau_is)...$ then control theory results can be easily applied. When roots are separated, the result is $\tau_i{\approx}a_i/a_{i-1}$. $\endgroup$ – Page David Sep 8 at 4:47
  • $\begingroup$ There must be a "root" for this formula somewhere....I am interested! I guess a good start for me would be to come up with an example. $\endgroup$ – NoChance Sep 8 at 5:25

I tried to do the special case first.
If the first equality is not satisfied, the slide say use this:

$P(s) ≈ (1+a_1s+a_2s^2)(1+{a_3 \over a_2}s)\cdots(1+{a_{n} \over a_{n-1}}s)$

Product of first two terms must approximate its underlying cubic:

$(1+a_1s+a_2s^2)(1+{a_3 \over a_2}s) ≈ 1 + a_1s+ a_2s^2 + a_3s^3$

$s^1$ coefficient: $|a_1| ≫ |{a_3 \over a_2}|$
$s^2$ coefficient: $|a_2| ≫ |{a_1 a_3 \over a_2}| → |{a_2^2 \over a_3}| ≫ |a_1| $

Use the same idea for the kth term, consider only the cubic:

$(1+ {a_{k-1} \over a_{k-2}}s)(1 + {a_k \over a_{k-1}}s + {a_{k+1} \over a_{k-1}}s^2) ≈ 1 + {a_{k-1} \over a_{k-2}}s + {a_{k} \over a_{k-2}}s^2 + {a_{k+1} \over a_{k-2}}s^3$

$s^1$ coefficient: $\large |{a_{k-1} \over a_{k-2}}| ≫ |{a_{k} \over a_{k-1}}|$
$s^2$ coefficient: $\large |{a_{k} \over a_{k-2}}| ≫ |{a_{k+1} \over a_{k-1}}| → |{a_{k} \over a_{k-1}}| ≫ |{a_{k-2}\; a_{k+1} \over a_{k-1}^2}|$


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