# Tag Info

8

I agree with Peter K.'s answer, but I would like to add one important point: the two statements in the question are only true for causal systems. The most general statement about stability for LTI systems described by rational transfer function is: An LTI system with a rational transfer function is stable if the region of convergence (ROC) of its transfer ...

7

short answer: all the poles of a causal (right-sided) and stable LTI system must be inside the unit circle whereas all the poles of an acausal (left-sided) and stable LTI system must be outside the unit circle. The explanation: Consider a causal LTI system with right-sided impulse response $h[n]$ whose transfer function can be expressed in terms of ...

7

The magnitude of that complex exponential is 1. Recall from complex algebra: any complex number can be expressed as $z = r e^{j \phi}$ where $|z|=r$ is its magnitude and $\arg z = \phi$ is the argument. Using this note that $$|e^{-j\Omega \lambda}| = 1$$ which is why it "disappeared".

6

The acronym BIBO, when used by university DSP professors, is merely a fancy (academic) term that means stable. So when when you encounter the term BIBO in the literature of DSP you can replace that term with the word "stable". As for your final question, the answer is yes. Think of this: You have an audio signal that you'd like to apply to the input of a ...

5

No it does not satisfy the condition. Simply take an example: $$n = 1 \implies y = x$$ Hence the output value at the present time $n=1$ depends on a future value of the input at time $n=2$. This violates the causality principle.

5

First to clear up the OP's misunderstanding: the Nyquist Stability Criteria involves clockwise encirclements of -1, not the origin, and this would be the polar plot for the open-loop gain specifically. I've included some details below for those that are more interested. First a review of the basic equation relating Open Loop gain and Closed Loop Gain for a ...

5

Why does a system with no poles have a finite support? If a system doesn't have finite poles, then its transfer function is of the form: $$H(z) = \frac{Y(z)}{X(z)} = a_Nz^N+a_{N-1}z^{N-1}+...+a_1z+a_0$$ So if you go back to the time domain: $$y[n]=a_Nx[n+N]+a_{N-1}x[n+N-1]+...+a_1x[n+1]+a_0x[n]$$ Note that if $x[n]=\delta[n]$, then $y[n]=h[n]$, the ...

4

The two are both true, but they are for different cases. Case 1 is true for continuous-time systems, and the transform is the Laplace transform and the variable is the derivative operator, $s$. Case 2 is true for discrete-time systems, and the transform is the $z$-transform and the variable is the delay operator, $z$.

4

For BIBO stability in the case of discrete time, there is a necessary and sufficient condition given by $\sum |h[n]| < \infty$ that is if the impulse response is absolute summable then the system is BIBO stable. Clearly $$h[n] = \delta[n] - \delta[n-1]$$ and it has a finite support, the impulse response is absolutely summable and therefore BIBO stable.

4

An improper system cannot be causal and stable. If the order of the numerator is greater than the order of the denominator, you'll always have at least one pole at infinity. Consequently, not all poles are in the left half-plane (or inside the unit circle in the case of discrete-time systems). The system in your example is clearly unstable: $$H(s)=\frac{s^... 4 What you are missing is that this is about a discrete-time system, because we're talking about poles and zeros in the complex z-plane and about poles inside or outside the unit circle. So there is no differential equation, but there is a difference equation:$$y[n]=\frac12y[n-1]+x[n]\tag{1}$$The corresponding impulse response is$$h[n]=\left(\frac12\...

4

They are independent of each other. Continuous systems: For stability, the ROC (region of convergence) must include the jw-axis of the s-plane. Causal systems have a ROC which is a right-sided plane, with $Re(s)>\alpha$. Here $\alpha$ is the real part of the "most to the right" pole. Due to this, for a continuous system to be causal and stable, all its ...

4

The IIR filter doesn't have to be unstable, but it has the potential of being so; unlike the FIR case which doesn't have even the potential. One reason for the (potential) unstability of an IIR (adaptive) filter is the numerical issues due to coefficient quantization. When the poles are closer to unit circle this will be critical. This is especially ...

4

Although what @Fat32 wrote is correct, I think the potential instability of IIR filters is not the main reason for the instability of an adaptive IIR filter. After all, we can calculate the poles in each iteration and put a hard constraint to avoid poles out of the unit circle. Even within the case of the FIR filters - which are unconditionally stable- we ...

3

one thing about a non-minimum phase system (with a rational transfer function), is that it can be thought of as the series concatenation (or cascade) of a minimum-phase system, having identical magnitude response as the given non-min-phase filter, with an all-pass filter. the APF will have a poles that cancels specific zeros of the min-phase system that are ...

3

Let us try with another hints: could you imagine a bounded input signal which could result in a non bounded output? general suggestion whenever analyzing a system: try a few "simple to compute" input signals, looking a the outputs, this could help you guessing some properties: an impulse, a ramp, a step, etc.

3

It is generally not true that the relation $Y(\omega)=X(\omega)H(\omega)$ is independent of the system's stability. For systems with a rational transfer function (i.e., systems that can be described by differential or difference equations), the frequency response, which is the Fourier transform of the impulse response, only exists for stable systems. For ...

3

The question has an answer on Math SE. The argument goes as follows: \begin{eqnarray} \int_{-\infty}^\infty \left| \frac{\sin t}{\pi t} \right|dt &=& \frac{1}{\pi}\sum_{n=-\infty}^\infty \int_{n\pi}^{(n+1)\pi} \left| \frac{\sin t}{t} \right| dt \\ &>& \frac{1}{\pi} \sum_{n=-\infty}^\infty \frac{1}{|n+1|\pi}\int_{n\pi}^{(n+1)\pi} |\sin t|...

3

Your idea of replacing the pole $p$ outside the unit circle (i.e., $|p|>1$) with a pole $1/p^*$ inside the unit circle is correct. However, as mentioned in this answer, you need to be careful with the scaling. In order to obtain the correct scaling, a factor $(z-p)$ in the denominator must be replaced by a factor $\pm |p|(z-1/p^*)$. In that way the ...

3

The I/O relation $$y[n] = \cos(x[n])$$ represents a stable system. Because by definition of $\cos(x)$, irrespective of the input $x[n]$, its output will always be bounded by $$-1 \leq \cos(x[n]) \leq 1$$ for all $x[n]$ for all $n$. Even if $x[n]$ is undefined ( such as would be the case with $x[n] = 1/n$ at $n=0$ as Marcus pointed out) we for sure know ...

3

You're conflating the discrete-time definition of a system with the continuous-time representation of a system. Your discrete-time $$Y(z)\cdot\big(z-\frac{1}{2}\big)=X(z)\cdot z$$ does not transform to: $$\dot{y}-\frac{1}{2}y=\dot{x}$$ but to: $$y[n+1] - \frac{1}{2} y[n] = x[n+1]$$ or $$y[n] = \frac{1}{2} y[n-1] + x[n]$$ which has an impulse ...

3

You're right that the Laplace transform is not more general than the Fourier transform. They are just different. There are several (theoretically) important functions for which the Laplace transform doesn't exist, but the Fourier transform does. A few examples are $x(t)=e^{j\omega_0t}$ $x(t)=\sin(\omega_0t+\phi)$ $x(t)= \textrm{sinc}(\omega_0t)$ $x(t)=\frac{... 3 Let's try this in Octave/Matlab: a = [1,-1.601089848140876,0.668366341586459]; b = [0.016819123361395717,0.033638246722791434,0.016819123361395717]; r = roots(a); abs(r) ans = 0.81754 0.81754 Since the roots are all inside the unit circle, your filter is stable. The magnitude response looks like it should for a Butterworth filter. However, as you'... 3 A system with simple distinct poles on the imaginary axis (and note that the origin is on the imaginary axis) and no poles in the right half-plane is called marginally stable. If you have poles with multiplicity greater than$1$on the imaginary axis, or if there are poles in the right half-plane, then the system is unstable. For discrete-time systems, the ... 2 In audio processing at least this is always important. If your output goes out of bounds from an unstable IIR or mechano-acoustically the output feeds back to input, you're hosed. Feedback will squeal in the ear, or the filter will stop doing its job. 2 The problem here is that for the given functions, the usual Nyquist contour (see figure below) results in Nyquist plots from which no decision can be made about stability. Nyquist contour ($M\to\infty$) (from Signals and Systems by A.V.Oppenheim) The reason is the fact that for all three functions the limit for$|s|\to\infty$equals zero. So the function ... 2 Since this is most likely homework, here is a hint. Write the integral you have displayed in the form$\int_{-\infty}^\infty x(\tau)h(t-\tau) d\tau$where you get to choose what the function$h(\cdot)$is to make it all work out. Then,$h(t)$is the impulse response of the LTI system, Do you know the criterion for BIBO stability of an LTI system in terms of ... 2 As correctly pointed out in a comment, this system is definitely not BIBO-stable. The output signal becomes arbitrarily large (in magnitude) when$x(t)$approaches zero. So a bounded input signal can result in an unbounded output signal, and, consequently, the system is unstable. 2 Simply because, when$C$is a real number,$|e^{jC}|=1$. Here,$C=\Omega\lambda$, apparently real numbers. Another interpretation is: if you change the phase of a signal (like a mere time-shift), it turns into a constant modulus in the Fourier domain, and provides some invariance, used for instance recently in data classification with Invariant Scattering ... 2 As you've pointed out, inversion leads to poles at locations of the zeros of the original transfer function and vice versa. Assuming that$G(z)$is causal and stable (i.e., it has all its poles inside the unit circle), we have to distinguish$3$cases:$G(z)\$ has at least some zeros outside the unit circle. This means its inverse has some poles outside the ...

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