In an exam task, I am asked to determine the transfer function of the following direct-time system and decide whether it's stable.


I think this system is canonical and the amplifiers 'on top' correspond to $b_2$, $b_1$ and $b_0$ from left to right in the following formula for the transfer function. Also, the amplifiers 'on the bottom' correspond to $-a_2$ and $-a_1$ similarly from left to right:

$$ H(z) = \frac{b_0 + b_1 z^{-1} + b_2 z^{-2}}{1 + a_1 z^{-1} + a_2 z^{-2}} $$

Therefore the transfer function of this system is

$$ H(z) = \frac{z^{-1} + 0.4 z^{-2}}{1 + (-0.5) z^{-1} + 0.06 z^{-2}} = \frac{z + 0.4}{z^2 + (-0.5)z + 0.06} $$

So it has a single zero at $z = -0.4$ and its poles are $p_1 = 0.3$ and $p_2 = 0.2$

I know I can draw the pole-zero plot for this function that will look like this:

pole-zero plot

How do I deduce the stability of the system from here?

I have learned things before like given the eigenvalues $\lambda_i$ of the system's $\underline{\underline{A}}$ matrix, a discrete-time system is asymptotically stable if $\forall \lambda_i : |\lambda_i| < 1$. Similarly, in continuous time, the criterion is $\forall \lambda_i : \mathrm{Re}\{\lambda_i\} < 0$.

But now I have poles, not eigenvalues. I have only found that if there is a pole that is to the right of the imaginary axis on the pole-zero plot, the system is unstable. However, here, I think this system should be asymptotaically stable (therefore also BIBO stable), but it has a pole in such a place.

Why is the transfer function given this way for such systems?


1 Answer 1


Poles in the right half-plane only cause instability if the system is a continuos-time system, because in that case, a pole $s_{\infty}$ contributes a signal

$$k\cdot e^{s_{\infty}t}u(t)\tag{1}$$

to the output signal, with some constant $k$, and with $u(t)$ denoting the unit step function. For $(1)$ to decay we require $\text{Re}\{s_{\infty}\}<0$, i.e., for a (causal) continuous-time system to be stable, all its poles must lie in the left half-plane.

In discrete time, however, a pole $z_{\infty}$ contributes a sequence

$$k\cdot z_{\infty}^nu[n]\tag{2}$$

to the output signal. Consequently, in order for $(2)$ to decay, we require $|z_{\infty}|<1$, i.e., for a (causal and) stable discrete-time system, all poles must be inside the unit circle.

Concerning the equivalence of poles and eigenvalues of the system matrix have a look at this answer.

  • $\begingroup$ Thank you for the explanation. When you say 'stable', do you refer to asymptotic stability? $\endgroup$
    – bp99
    Dec 1, 2018 at 13:15
  • 1
    $\begingroup$ @bertalanp99: Yes, even though (to my knowledge) that term isn't really used so much when we talk about LTI systems with inputs (as opposed to non-linear autonomous systems). If the poles are inside the unit circle the system is always BIBO-stable. $\endgroup$
    – Matt L.
    Dec 1, 2018 at 14:08
  • $\begingroup$ I see. I am asking because I have been taught something that if we don't know the system's layout (I don't know the correct terminology, see the first image I added), we cannot be sure from simply knowing the poles from the transfer function that the system is asymptotically stable. $\endgroup$
    – bp99
    Dec 1, 2018 at 15:36
  • 1
    $\begingroup$ @bertalanp99: Yes, I think so. An LTI system is asymptotically stable if all its poles are in the left half-plane. $\endgroup$
    – Matt L.
    Dec 2, 2018 at 9:25
  • 1
    $\begingroup$ @bertalanp99 You can only be sure of this if $H$ is minimal. Namely if the state space representation of the system is uncontrollable or unobservable, then the transfer function of the system will have pole-zero cancellation. So if the cancelled pole is unstable then you might not see this pole in $H$ but the system is still internally unstable. $\endgroup$
    – fibonatic
    Dec 3, 2018 at 2:45

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