- #1
FrankJ777
- 140
- 6
Hi all. I'm trying to relearn S-plane analysis and filter transfer functions. What I'm having problems with is simple algebra i think, because it's been a while. Right now I've been reading http://www.ee.up.ac.za/main/_media/en/undergrad/subjects/eli220/polezero.pdf. What I'm confused about, among other things, is factoring the transfer function H(s).
In their example:
H(s) = [itex]\frac{2s + 1}{(s^2 + 5S + 6)}[/itex]
factored
H(s) = [itex]\frac{1}{2}[/itex] [itex]\frac{s+1/2}{(s+3)(s+2)}[/itex]
I don't understand how they get that. The way I see it, because:
2s+1 = 2(s+1/2)
and
[itex]s^{2}[/itex]+5s+6 = (s+3)(s+2)
so
H(s) = [itex]\frac{2s + 1}{(s^2 + 5S + 6)}[/itex] = [itex]\frac{2}{1}[/itex] [itex]\frac{s+1/2}{(s+3)(s+2)}[/itex]
Also It seems that:
[itex]\frac{2s + 1}{(s^2 + 5S + 6)}[/itex] ≠ [itex]\frac{1}{2}[/itex] [itex]\frac{s+1/2}{(s+3)(s+2)}[/itex]
So I guess what I don't understand is why it seems that instead of factoring out 2 from the numerator, we are dividing the numerator by 2 and multiplying the denominator by 2.
Can someone please set me straight. I know I did this a few years ago and it seemed to make more sense then, than it does now.
Thanks a lot.
In their example:
H(s) = [itex]\frac{2s + 1}{(s^2 + 5S + 6)}[/itex]
factored
H(s) = [itex]\frac{1}{2}[/itex] [itex]\frac{s+1/2}{(s+3)(s+2)}[/itex]
I don't understand how they get that. The way I see it, because:
2s+1 = 2(s+1/2)
and
[itex]s^{2}[/itex]+5s+6 = (s+3)(s+2)
so
H(s) = [itex]\frac{2s + 1}{(s^2 + 5S + 6)}[/itex] = [itex]\frac{2}{1}[/itex] [itex]\frac{s+1/2}{(s+3)(s+2)}[/itex]
Also It seems that:
[itex]\frac{2s + 1}{(s^2 + 5S + 6)}[/itex] ≠ [itex]\frac{1}{2}[/itex] [itex]\frac{s+1/2}{(s+3)(s+2)}[/itex]
So I guess what I don't understand is why it seems that instead of factoring out 2 from the numerator, we are dividing the numerator by 2 and multiplying the denominator by 2.
Can someone please set me straight. I know I did this a few years ago and it seemed to make more sense then, than it does now.
Thanks a lot.