Z Transform - Do i always need 2 poles for every “peak”?

I am quite new to digital signal processing (and also Z transforms). I am reading about frequency response modelling and have some questions

• do we always need a pair of poles for each "peak" in the frequency response ?
• why is this so ?
• what happens if i dont have a pair of pole for each peak ?
• Please clarify what you mean by “peak”; do you mean a frequency when the frequency response goes to infinity or is just at a maximum? What about having a peak at DC? – Dan Boschen Oct 6 '18 at 12:49
• Also is your thinking restricted to only real inputs (sinusoids) or are you able to understand yet what a complex exponential frequency is? (Since you said this was all new to you) – Dan Boschen Oct 6 '18 at 12:50
• @DanBoschen I meant when the frequency goes to infinity. Does having a peak at DC or not change anything ? I know Euler's formula.. is that called a complex exponential frequency ? – Kong Oct 6 '18 at 12:55

To answer your question specifically, no you do not need two poles. A single pole anywhere on the frequency axis will cause a real or complex frequency input to go to infinity (peak). For a digital system, the frequency axis is the unit circle. As Fat32 very well summarized, if your impulse response (which the poles and zeros are the z-transform or Laplace transform of for digital or an analog system specifically) is real, then complex conjugate poles are required. So in the case of DC (or fs/2 for a digital system where fs is the sampling rate) only one pole can exist in these cases even with a real system, otherwise for any other real system with a single frequency peak, 2 poles must exist.

Regarding confusion with complex conjugate poles for real systems with complex roots, the following graphic may be of help:

The figure on the left shows a complex number as a phasor that has real and imaginary components, as drawn on a complex plane with a real axis in the x direction (I) and an imaginary axis in the y direction (Q). This is not a plot showing poles and zeros but simply plotting the magnitude and phase of a complex number given by $$Ae^{j\phi}$$. (This also demonstrates why must have two real numbers to represent a complex number and why we often see "I" and "Q" in hardware implementations of radio transceivers and other processing systems for complex numbers. In this case the real numbers are A and $$\phi$$ but could also be the actual values for I and Q.) For a real system, the Q axis (imaginary axis) must always be zero, or stated another way the phase component must be either 0° or 180°.

The figure on the right demonstrates multiplying the phasor by its complex conjugate. The short answer is when multiplying two complex numbers, the coefficients are multiplied and the exponents add (so adding the phase with its negative results in the phase = 0 consistent with a real number!). This is readily seen by representing the complex number in exponential format (and not see easy to see when representing as $$I+jQ$$.)

This is also demonstrated easily mathematically showing that a complex conjugate is required for the imaginary components to cancel:

$$(I+jQ)(I-jQ) = (I^2 + Q^2)$$

And as shown by the Quadratic Formula, the factoring of any real 2nd order polynomial where $$4ac > b^2$$ must result in complex conjugate roots. (using the a,b, and c as typically used in the Quadratic Formula).

For poles and zeros we are multiplying complex numbers specifically as represented by the factoring of the polynomials in the transfer function. Of interest as well is the case when we are adding phasors such as representing $$cos(\phi)$$ using Euler's identity with two complex phasors. In the case of adding phasors we simply add the I and Q components. In order to bring the Q axis to zero, we require adding another number that has the exact negative of the imaginary component. The complex conjugate does this such as given by Euler's identity: ($$Acos(\phi)$$ represents a real number while $$Ae^{j\omega t}$$ represents a complex number):

$$Acos(\phi) = \frac{A}{2}e^{j\omega t} + \frac{A}{2}e^{-j\omega t}$$

• Thank you. But it looks like I lack the understanding of something very simple: real vs imaginary systems/transfer functions and what they mean... because now I don't understand why complex conjugate poles are required for an impulse response that is real. What happens if I don't have 2 poles for an impulse response that is real ? – Kong Oct 6 '18 at 14:50
• @kong that comes from the fact that a polynomial of real coefficients can only have real roots or complex roots in conjugate pairs. That polynomial is the characteristic equation polynomial of the LCCDE equation associated with the real system, which has real coefficients by definition. – Fat32 Oct 6 '18 at 15:17
• @kong You CAN have an impulse response that is real and not have 2 poles, as long as all the poles themselves are real. If any poles have imaginary components, then a complex conjugate of that pole must exist if the impulse response is real. – Dan Boschen Oct 10 '18 at 12:21

If your system has a real impulse response, then pole & zeros will be in complex-conjugate pairs; so a pole at $$\omega = \pi/3$$ implies another pole at $$\omega = -\pi/3$$ .

Further, by the definition of poles, as the frequency $$\omega$$ get closer to the pole frequency, the frequency response magnitude increases (so as to say peaks!). So for every peak in the (positive frequency axis) then you would have at least two poles at the same frequency (one for positive frequency and one for the negative frequency). Note that there could be more, but at least two is required for a real system.

However note that, having $$2N$$ poles does not necessarily mean that the frequency response curve has $$N$$ peaks (sharp rising transitions). Indeed counter-examples are easiy to find, such as an Butterworth or elliptic IIR lowpass filters having no peaks, compared to a sharp isolated rising transition.

Note that in those filters the poles are bundled together such that their indivual peaks have overlapping effects on the total frequency response (yielding a flat passband rather than isolated peaks).