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According to my understanding, continuous first order systems do not exhibit oscillatory behavior because a first order system has only one energy storage element and thus oscillation is not possible. I was simulating the below difference equation for a first order discrete system:

$$y[n] = \tfrac{1}{10} u[n] - \tfrac{7}{10} y[n-1]$$

where $u[n]$ is a unit step function. As can be seen, this equation describes a first order difference equation. However, when plotting $y[n]$, I can see that it oscillates at the start and settles next:

enter image description here

However, when I simulate the following difference equation then this oscillation disappears:

$$y[n] = \tfrac{1}{10} u[n] + \tfrac{7}{10} y[n-1]$$

enter image description here

I am not sure how first order discrete time systems are different from their continuous counterparts, and why would the sign change the stability of the system. Any explanation?

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    $\begingroup$ this is a good question about an edge case issue. putting a single pole $p$ on the negative real-axis of the $z$-plane so that $-1 < p < 0$ is all of stable and realizable. but it has no counterpart with an analog filter with poles on the $s$-plane. as best as i can understand, that single pole on the $z$-plane sorta corresponds to two poles on the $s$-plane at $\pm$ Nyquist. it's a first-order pole that looks like a second-order pair of poles. $\endgroup$ – robert bristow-johnson Feb 15 '20 at 2:07
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For continuous-time systems, a pole at location $s_0=\sigma_0+j\omega_0$ will create a time-domain contribution of the form

$$e^{s_0t}=e^{\sigma_0t}e^{j\omega_0t}\tag{1}$$

which is a damped oscillation if the pole is in the left half-plane (i.e., $\sigma_0<0$), and if the pole is not on the real axis (i.e., $\omega_0\neq 0)$. For $\omega_0=0$ there is no oscillation. This is the case for a real-valued first-order system, which can only have a real-valued pole.

For discrete-time systems, a pole at $z_0=re^{j\omega_0}$ ($r\ge 0$) causes a sequence of the form

$$z_0^n=r^ne^{j\omega_0n}\tag{2}$$

If the pole is inside the unit circle (i.e., $r<1$) that sequence will decay in amplitude. Note that the sequence will not oscillate only if $\omega_0=0$, i.e., if the pole is on the positive real axis. If the pole is real, but if it is negative, we have $\omega_0=\pi+2\pi k$, i.e., we have an oscillation with maximum frequency, corresponding to an alternating sequence.

In sum, for continuous-time systems if the pole is on the real axis, it will not cause oscillations. For discrete-time systems, there are no oscillations only if the pole is on the positive real axis. A pole on the negative real axis causes oscillations with maximum frequency. Consequently, a discrete-time real-valued first-order system can exhibit oscillations if the pole happens to be on the negative real axis.

For additional insight consider the exact mapping between the $s$-plane and the $z$-plane:

$$z_0=e^{s_0T}\tag{3}$$

where $T$ is the sampling interval. Inversion of $(3)$ gives

$$s_0=\frac{\ln(z_0)}{T}\tag{4}$$

A $z$-plane pole $z_0=-r$ ($r>0$) on the negative real axis maps to a complex valued pole in the $s$-plane:

$$s_0=\frac{\ln(-r)}{T}=\frac{\ln(r)}{T}\pm j\frac{\pi}{T}\tag{5}$$

The frequency $\omega_0=\pi/T$ corresponds to half the sampling frequency, i.e., the maximum frequency of the corresponding discrete-time system.

As a final note, oscillations have nothing to do with whether the system is stable or not (stability in the bounded-input bounded-output sense). For stability, it only matters if the contributions of the poles to the output signal decay or not.

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  • $\begingroup$ Thank you for the great explanation. Any idea why the oscillation dies out in the example above? $\endgroup$ – HaneenSu Feb 15 '20 at 15:32
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    $\begingroup$ @HaneenSu: It dies out because the pole radius is $r=0.7$, and $(0.7)^n$ goes to zero for $n\to\infty$. As long as the pole is inside the unit circle, the system is stable, i.e., all transients decay. $\endgroup$ – Matt L. Feb 15 '20 at 15:45
  • $\begingroup$ Would 2 purely complex poles be mapped as a single discrete pole, or would you still get 2 discrete poles, albeit repeated. $\endgroup$ – Ben Feb 15 '20 at 17:42
  • $\begingroup$ In case I wasn't clear. 2 poles at pi/T $\endgroup$ – Ben Feb 15 '20 at 17:52
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    $\begingroup$ @Ben: Yes, it doesn't have a real-valued first-order continuous-time equivalent. $\endgroup$ – Matt L. Feb 16 '20 at 15:17
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In the first case you have a pole at the location (-0.7). In the second case, your pole is at 0.7.

Having a pole at -0.7 means that the natural frequency of your system is fs/2, that's why you have an oscillation at fs/2. Since the pole is stable, i.e. inside the unit circle, the oscillation eventually dies down.

Edit : You happen to have a pole at fs/2, the maximum frequency, with low-damping, which is why it oscillates. Continuous systems don't have a maximum frequency. This discrete system does not have a continuous equivalent... At least, that's my intuition.

unit circle and natural frequency here

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  • $\begingroup$ why would a continuous system not exhibit this oscillation, which eventually dies out? $\endgroup$ – HaneenSu Feb 15 '20 at 0:56
  • $\begingroup$ the reason why this phenomenon would not happen in a continuous system because this is a discretization issue. $\endgroup$ – Ben Feb 15 '20 at 1:01
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A real continuous time system has more restrictions than a discrete time system.

The reason is that a first order continuous system of the form $\dot{x} = f(x)$ cannot pass to the 'other side' of an equilibrium. A discrete time system has no such limitations.

It has nothing to do with the poles of the system, which need not be linear.

The discrete time systems $x_{n+1} = {1 \over 2} x_n$ and $x_{n+1} = -{1 \over 2} x_n$ are both exponentially stable. The first system could be a sampled version of a first order continuous time system, the second could not.

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