# Second Order Transfer Function : Imaginary component in $R$?

I have been looking at the following transfer function:

$$H(z) = \tfrac12 - \tfrac12 z^{-2}$$

Given the usual method for finding $$\theta$$ and $$R$$ in the complex plane, I calculate that $$\theta = \frac{\pi}{2}$$ and $$R = j\sqrt{\frac12}$$

This would initially make me think that the conjugate zero pair here is 90° from the real axis, however this imaginary $$R$$ component has thrown me completely - the equivalent frequency response also seems odd to me as it shows zero at DC and Nyquist with a gentle curve upwards in between.

What is going on here? I think that $$R$$ in the complex plane can't be imaginary, but don't understand how to arrive at the correct angle and magnitude for the conjugate poles.

• Without showing us how you arrived at these values (and maybe explaining what you actually mean by 'theta' and 'R') we can't say where you went wrong. – Matt L. Aug 11 '19 at 20:17

HINT:

If you're looking for the zeros of $$H(z)$$, try $$z_0=1$$ or $$z_0=-1$$. If that works, try to figure out why, and derive it yourself. Then, think about what the magnitude and the phase of those zeros might be.

EDIT (in reaction to your comment): If you have a second-order polynomial with real-valued coefficients, then two complex conjugated zeros are not the only option; you can also have two real-valued zeros.

• So I was originally assuming a solution of the form: $H(z) = (1-Zz^{-1} )(1-Z^* z^{-1})$; where $Z$/$Z^*$ is a pair of complex conjugate zeros $Z = Re^{j\theta}$, whereas if I assume the following form: $H(z) = (1-Zz^{-1} )(1+Z^* z^{-1} )$; It returns: $H(z) = 1-2Rjsin(\theta)z^{-1}-R^2z^{-1}$; which gives me two real zeros at -1 and +1? (Doing this without direct inspection of the transfer function) – DamienBradley Aug 11 '19 at 21:56
• can you show the steps between the last and next-to-last equations, Eddy? – robert bristow-johnson Aug 12 '19 at 4:09
• Made a mistake, should be $H(z) = 1+2Rz^{-1}jsin(\theta)-R^2z^{-2}$. Found like so: $H(z) = (1-Zz^{-1})(1+Z^*z^{-1}) = 1-Zz^{-1}+Z^*z^{-1}-ZZ^*z^{-2}$. Then substituting $Z = Re^{-j\theta}, Z^* = Re^{j\theta}$, $H(z) = 1-Re^{-j\theta}z^{-1}+Re^{j\theta}z^{-1}-R^2z^{-2}$.With final result from sub. $e^{-j\theta}$ & $e^{j\theta}$ with $cos(\theta)-jsin(\theta)$ and $cos(\theta)+jsin(\theta)$. So in this case the zeros are real, but then in the case where theta is not pi / zero, such as in the following transfer function: H(z) = 1+z^{-1}-z^{-2}. I am unsure how to proceed with 2Rjsin(\theta) – DamienBradley Aug 12 '19 at 14:24
• My thought at the moment is that $jsin(\theta)$ is a 90 degree phase shift, so the answer involves that concept – DamienBradley Aug 12 '19 at 14:45

The polar representation of a nonzero complex number in the form

$$z = R e^{j \theta}$$ requires that $$R > 0$$ ; a positive real number. Setting $$R = j\sqrt{\frac{1}{2}}$$ violates this.

$$H(z) = \frac{1}{2} - \frac{1}{2} z^{-2}$$

and you want to find its zeros.

Let's find the zeros of the following transfer function instead :

$$G(z) = 1 - z^{-2}$$

it's very easy to see ( from $$a^2 - b^2 = (a-b)(a+b)$$ ) that

$$G(z) = 1 - z^{-2} = (1 - z^{-1})( 1 + z^{-1})$$

Then the zeros of $$G(z)$$ are :

$$G(z) = 0 \implies z = \{ -1 , 1 \}$$

from which you would argue that

$$z = 1 \implies R = 1 ~~,~~ \theta = 0$$ and $$z = -1 \implies R = 1 ~~,~~ \theta = \pi$$

And the zeros of $$H(z)$$ will be the same, since $$H(z) = \frac{1}{2} G(z)$$.