Hi,

Perhaps I don't understand your question well, but it seems that you give the answer yourself, when you say that the zeros are introduced to sharpen the transition.

Basically, zeros are very useful in getting sharper cutoff for a filter.

I think you are making the mistake of confusing complex zeros with real zeros. In filter design, you add complex zeros or rather zeros on the imaginary axis to enhance cutoff. A "zero" of a filter is by definition, a frequency where the magnitude of the frequency response goes to "zero" It will increase only for a zero on the real axis, and the real axis does not correspond to real-world frequencies.

Hope this helps.
Vivek

Hi Sezi,

I think you too are confusing your opamp Bode plots with filter frequency response plots

A zero as I already mention is defined as a frequency where your frequency response magnitude goes to 0, while at a pole, it goes to Inf.

Now, all the real-world frequencies (and this is what we are interested in when making filters) are on the imaginary axis on the s-plane.

The zero that you are thinking of lies on the real axis of the s-plane, where the frequencies are actually not real-world frequencies, so they are imaginary

I guess it is quite a nightmare with real-world frequencies on the imaginary axis, and the real axis not really containing any real-world frequencies

Think carefully.

Vivek

Hi Sezi,

This is just a basic concept, but many people get confused about it. If you are familiar with basic Laplace transforms and bode plots, then this should be enough. Otherwise, try reading any book on signals or controls. There are several of these by Oppenheim, Kuo etc.

But no book will really discuss this matter. Just consider the following examples:

Imagine a linear system with N zeros and M poles

H(s) = (s-z1)(s-z2)....(s-zN)/(s-p1)(s-p2)...(s-pM)

Now, if the value of 's' were to be equal to any of the zi, then H(s)=0, and thus the zi values are called zeros of the system.

Conversely, for the case where 's' equals any of the pi values, the denominator is zero and H(s) tends to infinity.

Remember that in the discussion above, I allow 's' to assume some values.

If s = a + j*w, then all real-world frequencies can be found by setting a=0, and sweeping w.

If you now consider the 2 systems shown below:

Hreal(s) = (s-z1)/(s-p1), where z1 and p1 are real numbers, say z1=1, p1=-2

and Himag(s) =(s-z21)(s-z22)/(s-p2)(s-p22),
where z21= 0+j1, z22=0-j1 , p21=0-j2, p22=0+j2.

Now try to calculate H(jw) for both the systems by setting s=0+jw. You will get the answer to your questions.

The respective pole and zero magnitudes are same for both cases, but the location is different. Hreal(s) has real poles and zeros, while Himag(s) has imaginary poles and zeros. This makes all the difference.

Regards
Vivek