On MIT's open course a simple exercise with two questions is given. On the first part, they question about the properties of the following discrete system:

$$ y[n]=x[n]+0.5y[n−1]−2y[n−2] $$

The answer states that none of the following properties can be determined, as no initial conditions were given: stability, causality, linearity, time (in)variance.

I would be inclined to say that the system is at LTI, as the coefficients are constant and the output is simply defined by a recursive relationship to itself and the input. Both poles are inside the unit circle and the system should be stable if it is causal, as the ROC includes the unit circle.

  1. My first question is therefore: could you come with a(n) example(s) of initial conditions (or any argument for that matter) so that the system is: variant in time OR unstable OR not causal OR non-linear? The only conditions that I can think of that could make it behave strangely would be extreme ones such as +/- infinity. I assumed this was not the intent of the initial question.

Next, the following transfer function is given (unrelated to question above):

$$ H(e^{j\omega }) = \frac{1 - 2e^{-j\omega }}{(1 - \frac{3}{4}e^{-j\omega })(1 - 3e^{-j\omega })} $$

Here it is also asked about its properties. On the final answer they state that, given that a transfer function is given, one can imply that the system is linear and time invariant. So far so good. Next, however, it states that the system is both stable and not causal, with the following argument:

...Further, from the existence of a frequency response you may conclude that the frequency response converges on the unit circle, i.e., that the system is stable. Given that the system is stable the region of convergence is also implied resulting in a two-sided sequence, meaning that the impulse response is not causal.

Hence my second question:

  1. How was this concluded from this transfer function, given that no ROC constraint is there?

Thanks in advance.

  • $\begingroup$ You should ask 2 separate questions since they are not related. $\endgroup$
    – Ben
    Commented Apr 24, 2020 at 13:19
  • $\begingroup$ @Ben Both questions are related to discrete system properties and are part of the same courseware context that I pointed in the link. I think it doesn't make sense to split them. $\endgroup$
    – edwillys
    Commented Apr 24, 2020 at 15:22
  • $\begingroup$ @MaxFrost your ROC can be less than z =3 which can include the unit circle, and this is the case here. So system is very much stable, the response is two sided. RoC z= 3/4 to 3 $\endgroup$ Commented Apr 24, 2020 at 16:43
  • $\begingroup$ @Dspguysam Apologies for overlooking that possibility. Yes, you're right. I'll try not to commit such errors in future. $\endgroup$
    – MaxFrost
    Commented Apr 24, 2020 at 17:24
  • 1
    $\begingroup$ @MaxFrost not an issue friend, we ought to make mistakes to learn, I remember better when I correct a mistake $\endgroup$ Commented Apr 24, 2020 at 17:31

2 Answers 2


Regarding your first question, the poles of the system are not inside the unit circle (both poles have a radius of $\sqrt{2}$), so the system can't be causal and stable. If the initial conditions are non-zero, then the system is neither linear nor time-invariant. The explanation is simple: the output signal consists of a term determined by the input signal (and, obviously, the system properties), and another term that is independent of the input signal. That latter term is determined by the initial conditions. Consequently, that part of the output remains unchanged if we scale the input or shift it in time. However, a linear and time-invariant system's output must scale and shift with the input signal. So we can conclude that - according to the definitions used in the book - the system cannot be linear and time-varying if there are non-zero initial conditions.

Concerning the second example, stability is implied by definition since the frequency response (Fourier transform of the impulse response) exists as a rational function. The $\mathcal{Z}$-transform of the impulse response (i.e, the system's transfer function) is simply obtained by replacing $e^{j\omega}$ by $z$. From the transfer function it can be seen that the system has poles at $z=\frac34$ and at $z=3$. Clearly, the ROC must be the annulus between these two poles (since it must include the unit circle due to stability). That shape of the ROC implies that the inverse transform (i.e., the impulse response) is a two-sided, hence non-causal, sequence.

  • $\begingroup$ Ok for answer on question 2: The sole existence of H(z) guarantees the system to be stable, otherwise no such function. As of question 1: my calculations give a magnitude of sqrt(2)/2, which would imply stability, if the system is causal. I guess the question can be then asked as: how would it be possible for this system to be non-causal? $\endgroup$
    – edwillys
    Commented Apr 25, 2020 at 9:02
  • $\begingroup$ @edwillys: The difference equation can describe a non-causal system just as easily as it can describe a causal system. Just rearrange: $y[n-2]=\ldots$, and you see that the output at time $n-2$ only depends on future values of $y[n]$ and $x[n]$. $\endgroup$
    – Matt L.
    Commented Apr 25, 2020 at 9:05
  • $\begingroup$ @edwillys: Concerning Q2: it's not the existence of $H(z)$ that guarantees stability, but the existence of $H(z)$ for $|z|=1$, i.e., the existence of the Fourier transform. $\endgroup$
    – Matt L.
    Commented Apr 25, 2020 at 9:07

To your first question, a linear system is causal if we have initial conditions of rest that is zero. This is a sufficient and necceassy condition. This property doesn't require time invariance

Stability is a consequence of absolute summability of system response. It doesn't depend on the initial condition (of course have to be finite to make any sense).

To your second question the transfer function has a pole at z=3 and it is given that the system is stable. That means the ROC has to include the unit circle. That means that the ROC is to the left of the pole at z = 3, since ROC is towards the left of outermost pole hence system is not causal

  • $\begingroup$ My first question was related to an example/argument in which that system would be non-causal OR vary in time OR unstable OR non-linear. Preferably one example for each. Regarding the second question, how did you infer that the system is stable? $\endgroup$
    – edwillys
    Commented Apr 24, 2020 at 15:26
  • $\begingroup$ You will find many examples of the scenarios you mention in the book signal and systems by Allen oppenheim, he is from MIT as well. Regarding your second question, the given transfer function is already given in terms of the response at unit circle and is rational, which means it exist at unit circle, implying stability. $\endgroup$ Commented Apr 24, 2020 at 16:24
  • $\begingroup$ Ok for the argument regarding question 2. Rephrasing it a bit: the sole existence of a rational transfer function guarantees indirectly that the system is stable, otherwise there wouldn't be such a function. $\endgroup$
    – edwillys
    Commented Apr 25, 2020 at 8:52
  • $\begingroup$ Yes, the existence of a rational transfer function when $z =e^{j\omega}$ indicates stability, for other values of z or annulus it might or might not exist as a rational function $\endgroup$ Commented Apr 25, 2020 at 8:57

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