# relationship between poles and zeros of a transfer function and its bode plot

Discussion in 'General Electronics Chat' started by sp2821, Jun 18, 2015.

1. ### sp2821 Thread Starter New Member

Jan 8, 2014
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0
Hi all,

I am having a hard time grasping concepts of poles/zeros of a transfer function and magnitude bode plot and relationship between them.
From my understanding, pole and zero are any values of s that makes denominator and numerator = 0, respectively.
This to me sounds like at zero, H(s) will be 0 and 20*log(0) will be -inf.
Or at pole, H(s) will be +inf and 20*log(+inf) will be +inf.

However, for example, if H(s)=K(s+2), the magnitude bode plot is flat from s=0 to s=2 with DC gain of K and have slope of +20dB/dc from s=2 to s=+inf.
Why is this? Shouldn't it look like a notch filter?

Thanks,

2. ### ramancini8 Member

Jul 18, 2012
447
119
First a notch filter requires more than one pole/zero to get up and down movement of the transfer function.
H(s) = K(s + 2)/(s + 6)(s + 8) yields the upward line until (s + 6) flattens out the curve; then (s+8) causes the curve to head for -infinity. Your poor notch filter.

3. ### tindel Active Member

Sep 16, 2012
576
196
Why do you think it should be a notch filter? I'm not following you there. A transfer function with a single zero is a differentiator.

4. ### tindel Active Member

Sep 16, 2012
576
196

Actually - I think you mean H(s) = K*s/(s+2)^2 will give you a poor mans notch filter at 2 radians.

s will be a differentiator out to 'DC' (-inf at s=0) and the first pole will flatten out the curve and the second pole will cause the curve to go downwards to -inf when s=+inf

5. ### ramancini8 Member

Jul 18, 2012
447
119
Same-same theory just said different.

6. ### tindel Active Member

Sep 16, 2012
576
196
Err... yeah we're on the same page, but what you described was a low pass filter with gain in a limited band - technically it did differentiate, but just for a small part of the frequency spectrum creating a small bandpass filter. I described a traditional band-pass filter [not a notch filter... oops].

A traditional notch filter would take the form H(s)=K*(s+1)^2/((s+0.1)(s+10))

Oct 2, 2009
5,450
1,066
Not quite!

8. ### sp2821 Thread Starter New Member

Jan 8, 2014
12
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Well, my question is why? If H(s)=K(s+2), H(s)=0 only when s=-2. But when
Mike could you explain why?

At zero, H(s) has value of 0.
At pole, H(s) has value of +inf.

So if you take log of magnitude of H(s), shouldn't 20*log(mag(H(s)) have value of -inf at zero and +inf at pole?