# Tag Info

17

One thing that really helped me understand poles and zeros is to visualize them as amplitude surfaces. Several of these plots can be found in A Filter Primer. Some notes: It's probably easier to learn the analog S plane first, and after you understand it, then learn how the digital Z plane works. A zero is a point at which the gain of the transfer ...

12

I think there are actually 3 questions in your question: Q1: Can I derive the frequency response given the poles of a (linear time-invariant) system? Yes, you can, up to a constant. If $s_{\infty,i}$, $i=1,\ldots,N,$ are the poles of the transfer function, you can write the transfer function as $$H(s)=\frac{k}{(s-s_{\infty,1})(s-s_{\infty,2})\ldots (s-s_{\... 10 Note the difference between the zeros at 0.3 \pi and at 0.8 \pi. The first one is clearly a zero crossing, much like abs(x) at x=0. At \theta = 0.8 \pi, however, the curve is tangent to the horizontal axis, much like x^2 at x=0. So you have a doulbe zero here. So your zeros are: 2 zeros at z = e^{\pm j 0.3 \pi} 2 double zeros at z = e^{\... 10 If you consider poles of an integral transform domain to be important to the solution of differential equations: (as usual,) Euler did it first, 1753. One "importance" of poles is that they're part of a very useful representation for linear systems. They must've appeared when people started looking at functions as built from generating functions, so that'd ... 9 First it's important to realize that many authors use the terms zero-input response and natural response as synonyms. This convention is used in the corresponding wikipedia article, and for instance also in this book. Even Proakis and Manolakis are not entirely clear about it. In the book you quoted you can find the following sentence on page 97: [...] ... 9 Short Answer: All the poles of a causal (right-sided) and stable LTI system must be inside the unit circle whereas all the poles of an acausal (left-sided) and stable LTI system must be outside the unit circle. Explanation: First, consider the following right-sided (causal) exponential sequence h[n] = a^n u[n] whose \mathcal{Z}-Transform is:$$H(z) = \...

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 ...

8

Here let me show you a simple procedure very similar to pole zero placement which will be helpful for your notch filter design. First, lets analyse the frequency response of a single zero and let $$H(z) = 1 - b z^{-1}$$ be a first order system with a single zero at $z = b$ where $b$ is a complex constant with a radius $r$ and phase angle $\phi$ radians; ...

8

It's not sufficient to only consider causality, you also need to check whether the inverse system is stable, otherwise it can't be implemented. If $G(z)$ has zeros on the unit circle, it cannot be inverted. If $G(z)$ has no zeros on the unit circle, but if there are zeros outside the unit circle, then there is no causal and stable inverse, because the zeros ...

7

Take the equation b/(x-c) with b non-zero. The ratio goes to infinity as x approaches c. So c is the location of a pole (something tall and pointy in a graph). Take the equation (x-b)/c with c non-zero. The ratio goes to zero as x gets closer to b. So b is the location of something commonly called "a zero". You can not only do this with scalar x, but ...

7

To answer this you need to understand what is a pole and what is a zero of a transfer function. Let's look at a simple 2 poles 2 zeros filter (also called biquad filter) transfer function : $$H(z) = \frac{b_0+b_1 z^{-1}+b_2 z^{-2}}{1+a_1 z^-1 +a_2 z^{-2}}$$ This can be factored as : \begin{align} H(z) &= \frac{ b_0 \, (1-q_1 z^{-1})(1-q_2 z^{-1})}{(... 6 The "poles-inside-unit-circle" stability criterion only applies to causal systems. Your system is not causal because it uses one sample from the future owing to the z term. The general technique to check for stability involves looking at the regions of convergence (ROC) of H(z). If the ROC includes the unit circle, then the system is stable. See also ... 5 Let me show you a simple way to see this property. Assume x[k] is a causal sequence and letx[\infty]=\lim_{k\rightarrow\infty}x[k]$$be finite. Then the sequence x[k] can be written as$$x[k]=x[\infty]u[k]+y[k]\tag{1}$$where u[k] is the unit step sequence, and y[k] is a causal sequence that decays to zero as k\rightarrow\infty. Taking the \... 5 it depends on how you map the analog filter to the digital filter and how the s-plane poles get mapped to the z-plane poles (ya know, the "bilinear transform" vs. "impulse invariant" vs. whatever else is out there). probably, if it were up to me, for the purpose of defining Q, i would map the poles as you would map s to z with$$ z = e^{sT} $$... 5 Suppose some H(q)=\frac{A(q)}{B(q)} where q is some complex variable and A,B are functions of q. Whether in the s or z planes, to evaluate the magnitude of H(q) at some q, you evalaute and sum all distances from q to the locations of the zeroes (i.e. the magnitude of A) and similarly for B and create the fraction above for that specific q. ... 5 First to clear up the OP's misunderstanding: the Nyquist Stability Criteria involves clockwise encirclements of -1, not the origin, and this would be the polar plot for the open-loop gain specifically. I've included some details below for those that are more interested. First a review of the basic equation relating Open Loop gain and Closed Loop Gain for a ... 5 You're comparing the transforms of two different functions. You consider the Fourier transform of the function x_1(t)=\cos(\omega_0 t), but you took the Laplace transform of the function x_2(t)=\cos(\omega_0t)u(t), where u(t) is the unit step function:$$X_1(j\omega)=\int_{-\infty}^{\infty}x_1(t)e^{-j\omega t}dt\\ X_2(s)=\int_{0}^{\infty}x_2(t)e^{-st}...

5

Let $H(s)$ be a transfer function of the form $$H(s) = \frac{1}{s-p}$$ where $p$, which is a pole of $H(s)$, can be written as a complex number $a+jb$. Taking the inverse Laplace transform of $H(s)$ gives the corresponding impulse response $h(t)$ (that is, the output of your system when given $\delta(t)$ as input). Noting $\mathcal{L}^{-1}$ the inverse ...

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 ...

5

It's actually quite straightforward: positive powers of $s$ (or, in discrete-time, $z$), correspond to poles at infinity. Negative powers give you zeros at infinity. Let's look at some examples. In continuous time, an ideal differentiator has the transfer function $$H(s)=s\tag{1}$$ Clearly, $\lim_{s\to\infty}H(s)=\infty$, hence you have a pole at infinity ...

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

The poles of low pass and high pass Butterworth filters are indeed the same if both filters have the same cut-off frequency. The difference between the two lies in the numerator. A low pass filter has a constant in the numerator, whereas a high pass filter has a constant times $s^N$, where $N$ is the filter order. The poles of a normalized Butterworth low ...

4

You'd have to figure out the frequency response of the filter. Here are two methods. I prefer Method 2 because it's quick and dirty, and you don't really care about the exact gain values in the frequency response, just the general shape to figure out the type of the filter. Method 1: Brute Force/Computer Assisted import scipy.signal as sp import numpy as ...

4

In this answer I'll try to show you how to qualitatively evaluate a given pole-zero plot by just looking at it. Of course, this method has its limits, but for relatively simple pole-zero plots you can very quickly decide on the type of the corresponding filter. You should know that for stable filters, the frequency response equals the transfer function $H(z)... 4 If the number of finite zeros is not greater than the number of finite poles then the transfer function is proper, i.e., the degree of the numerator polynomial is not greater than the degree of the denominator polynomial. If the degree of the numerator polynomial were greater than the degree of the denominator polynomial, we would get at least one pole at ... 4 Note that the pole locations would be the same for all$4$classic types of frequency-selective filters (low pass, high pass, band pass, band stop). It's the location of the zeros that determines which filter type it is. Since we don't see any zeros, we can assume that they are at infinity, hence, the filter is a low pass filter. For a high pass filter you ... 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 Repeated poles simply means there are more than one pole at the same location. If a pole is not repeated then it is a Distinct Pole. Consider the simple case of a cascade of two integrators (in s) which is similar to the cascade of two accumulators (in z). The Laplace transform of an integrator Is$\frac{1}{s}$, which is one pole at the origin. Two ... 4 One thing to consider when implementing an IIR filter, whatever the order, is quantization and limit cycles. Let me show you with a quick example with your original filter$y[n] = a*x[n]+(1-a)*y[n-1]$Let a = 0.005 and say that we use 16-bit signed coefficients.$a_{fixedpoint} = a * 32768 = 164\$ Let's assume that the input and output are 16-bit ...

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