transfer function forms of expression

he
2012 3 17
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Consider the following transfer function
Zero/pole/gain:
118614.6077 (s+6.35e005) (s+725)
-----------------------------------
(s+3.05e005) (s+1.119e004) (s+1600)
How to transforme above formula to the following formula?
10 (1+1/6.35e005s) (1+1/725s)
-----------------------------------
(1+1/3.05e005s) (1+1/1.119e004s) (1+1/1600s)
What about it?
124000 s^2 (s+6.353e005) (s+3.18e005) (s+725.8)
--------------------------------------------------------
s^2 (s+3.18e005) (s+1467) (s^2 + 3.165e005s + 3.897e010)

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Walter Roberson
Walter Roberson 2012-3-17

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The first two formula you show are not equivalent. In the second form, each of the s+X terms has been divided by s, giving 1 + X/s . That was applied on the top and the bottom both, so the division by s on the top cancels the division by s on the bottom (unless s is 0). This leaves us with all the terms being equivalent except for the constants. The constant 118614.6077 of the first cannot be reconciled with the constant 10 of the second.
For the third formula, I would speculate that you should be dividing by s^2 for each term. For example, (s+6.353e005) becoming (1/s+6.353e005/s^2). If you do that, then "s" will not appear with any positive power, which would be consistent with the pattern of your second expression.

更多回答（1 个）

he
he 2012-3-17

0 个投票

I do not want this form
Simple:
s=tf('s');
wz1=12.4*10^4
wz2=1.45*10^3
wz3=318*10^3
A1=(1+s/wz2)/((s/wz2)*(1+s/wz3))
A2= (0.0006897*s + 1)/(2.169e-0090*s^2 + 0.0006897*s)
bode (A1,A2)
hold on
These two formulas is the same
but
*first,Bode diagram is not the same
and
I want to take A2 transform to A1 forms of expression *

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