I'm given poles at ${1+i}$ and ${1-i}$, and zero at $0$.

I have to find the difference equation and find out whether the system is stable.

Now I found in a similar question, which structure I tried to follow ( How to find system function, H(z) in the z-domain, given zero-pole plot of the system? ), that we can express $H(z)$ as $$H(z)=G_0z^{-1}\frac{(z-z_0)(z-z_1)\ldots}{(z-p_0)(z-p_1)\ldots} $$

It's clear to me that the numerator of H(z) is simply = z, but I can't wrap my head around the denominator.

I tried to go for: (z- (1-i))*(z- (1+i)) - my prof looked over my shoulder, nodded and told me to simply multiply it, adding "there will be no i in the polynomial". Now I really don't find any combination of tricks to achieve this, and I've been stuck for several hours right now.

I'd very much appreciate a hint!


So you're given two complex-conjugate poles $p=1+i$ and $p^*=1-i$. Your denominator polynomial is then given by


I trust that you can do the multiplication (for general $p$) and verify that the result is purely real-valued for any pair of complex-conjugate poles.


Thank you very much for your fast reply!

In the meantime, I came across a really embarassing knowledge gap: $$i^2 = -1$$ (yeah, I'm that blank).

That way, I can transform $$(z-1-i)*(z-1+i)$$ to $$z^2-z+zi-z+1-i-zi+i-i^2$$ and further to $$z^2-2z+2$$

Since transfer function $H(z)$:

$$ H(z) = \frac{Y(z)}{X(z)} $$

I conclude the difference equation should be

$$ y[n] = b_0 x[n] - a_1 y[n-1] - 2 * a_2 y[n-2] + 2 \tag{2} $$

This leads me to a result and even more to read about. I am sorry, but apparently I didn't understand how to solve for 'general' $$p$$ yet. Thank you very much again for your time and help!

  • 2
    $\begingroup$ also, in the electrical engineering subculture, we use "$i$" for current in the time domain, like "$i(t)$" or "$v(t)\,=\,R\,i(t)$" so then to reduce confusion (among EEs, because this increases confusion for everyone else), we use the letter "$j$" for the imaginary unit. or $$ j^2 = -1 $$ $\endgroup$ – robert bristow-johnson Jun 26 '18 at 23:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.