# Tag Info

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 ...

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

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 ...

4

Your assumption why IIR filters can have steeper transitions from passbands to stopbands compared to FIR filters of the same order is correct: IIR filter have poles away from the origin of the complex plane, and poles inside the unit circle close to zeros on the unit circle cause the corresponding frequency response to change rapidly with frequency. The FIR ...

3

Is this because of poles? Yes. A steep drop or rise in a filter's frequency-domain response can only be achieved with a filter that has a long memory. In a FIR filter, this long memory can only come from, well, being long. In IIR filter, this long memory can come from having poles that are close to the unit circle.

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 ...

3

What you have are not the poles and zeros, but simply the filter coefficients, i.e., the coefficients of the numerator and denominator polynomials. The poles are the roots of the denominator polynomial, and the zeros are the roots of the numerator polynomial. In Matlab they can be found by using the roots command: p = roots(a); z = roots(b); Note that in ...

3

This happens frequently if your poles are reasonably close to the unit circle. Consider the following example %% TF2ZP is problematic fs = 44100; % 6th order lowpass, fc = 50Hz, sampled at 44.1kHz [z,p,k] = cheby2(6,80,50*2/fs); % to transfer function [b,a] = zp2tf(z,p,k); % back to zpk [z1,p1,k1] = tf2zp(b,a); display([p p1]); Displaying the poles side ...

3

Here is one viewpoint in evaluating system stability. [1] System z-plane pole(s) lie outside the unit circle: System impulse response increases, with time, toward ±infinity. System frequency response does not exist. System is unstable. [2] System z-plane pole(s) lie on the unit circle: System impulse response remains non-zero and finite for all time. ...

3

The one and only condition for BIBO stability of a 1D discrete-time system, in the z-domain, is that its transfer functions's ROC (region of convergence) should include the unit circle : $|z| =1$. Therefore, it's a necessary and sufficient condition for BIBO stability of a 1D SISO system. There are no other conditions (to my knowledge). EDIT [based on ...

2

For $\mu > 0$, $z^L \in \left ( \sqrt[L]{\mu} \right) e^{j 2\pi n/L} \,\forall\, n \in 1 \cdots L$. I.e., the roots are evenly spaced on a circle $\sqrt[L]{\mu}$ in radius, and there's an $L$ of a lot of them.

2

Distinct poles do not need to share an x or y coordinate. They are classified as distinct as long as they do not share the same x AND y space. So any two poles are either distinct or repeated. The effect repeated poles have on the impulse response of a filter is a little complicated, but the short answer is that it does change. This brief article describes ...

2

You're right, the ROC of the Laplace transform of a two-sided signal is a strip in the complex plane. In your case, the imaginary axis is inside the ROC, and the ROC is limited by the poles in the right and left half-planes. If the ROC were the right half-plane, the signal would be right-sided, which is clearly not the case.

2

The system with impulse response given by $h[n] = \cos(\pi\sqrt{n})u[n]$ is BIBO-unstable because the sum $\sum_{n=-\infty}^\infty |h[n]]$ diverges instead of being convergent as is needed for BIBO-stability. Note that for all positive integers $k$, $h[k^2]=\cos(\pi k)$ has value $\pm 1$ and so $$\sum_{n=-\infty}^\infty |h[n]] = \sum_{n=0}^\infty |\cos((\pi\... 2 In particular, is the fixed point design more challenging for smaller a or larger a ? Smaller is worse: the closer a gets to 0, the closer the pol movess to the unit circle and the more time domain ringing you have. What are the main design considerations to think about ? A first order low-pass IIR filter is relatively benign. The sum absolute sum ... 2 Assuming that you understand the left-hand side of Eq. (8.47), for understanding the right-hand side you need to know that -1=e^{j\pi}, and that e^{j2k\pi}=1. So in order to obtain all 2N roots of (-1)^{\frac{1}{2N}} you rewrite -1 as$$-1=e^{j\pi}e^{j2\pi k}\tag{1}$$from which you get$$(-1)^{\frac{1}{2N}}=e^{j\frac{\pi}{2N}}e^{j\frac{2\pi k}{...

1

this is my original posting about implementing a fixed-point DC blocking first-order HPF using fraction saving. Lyons and Yates later did an IEEE Sig Proc article about DC blocking filters where this was one of the topics. // let's say sizeof(short) = 2 (16 bits) and sizeof(long) = 4 (32 bits) short x[], y[]; long acc, A, prev_x, prev_y; double pole; ...

1

It used to be that multiplication was a lot more expensive than addition or bitwise operations. For integer (and fixed point) implementation, this can be taken exploited like this: When $a=1/2$ an extremely efficient implementation can be made with a single add and a single shift. $$y[n] = (x[n] + y[n-1]) >> 1$$ Similarly, fractions with a power ...

1

because even though it has more poles than zeros WRONG. The Z-transform transfer function will always have equal number of poles and zeros. Your poles are at $z = -5/4$ and $z = 1/4$. Zeros are at $z = -1/2$ and $z = \infty$. Since ROC will always be concentric circle region without including poles, there are 3 possible ROC for the given transfer function. $... 1 These are all frequency selective filters. Note the zeros on the unit circle, which correspond to zeros of the frequency response. The filter's passband is determined by range of the angles of the poles, and the stopband is determined by the range of the angles of the zeros. As an example, let's consider PZ-diagram$\#1$: since the stopband is around DC (i.... 1 The LTI system defined by the impulse response $$h[n] = \cos(\pi \sqrt{n} ) u[n]$$ is unstable, as the absolute sum of the impulse response does not converge and diverges to infinity instead, i.e.; $$\sum_{n=-\infty}^{\infty} |h[n]| = \sum_{n=0}^{\infty} |\cos(\pi \sqrt{n}) u[n]| \longrightarrow \infty$$ 1 You are correct. If the region of convergence of a right-sided signal (like the one you have) does not include infinity, then the signal is not causal. 1 Marginally stable means that an otherwise stable system has one or more simple poles on the unit circle (in discrete time), or on the imaginary axis (in continuous time). The consequence of that is that transients don't decay, but they also don't grow without bounds. Marginally stable systems are unstable in the bounded-input bounded-output (BIBO) sense. 1 This is a very complicated problems and I don't think there exists a one-size-fits-all solution. You can try Matlab's$invfreqz()$and see if it works for your purposes https://www.mathworks.com/help/signal/ref/invfreqz.html In general this is a error minimization problem but the actual data and the way you set the your error function and the search ... 1 $$H(s) = {K*\frac{s-z}{s-p}}\\ \\$$ when |z| < |p|, you have a lead compensator. It adds phase between a certain band, and can help you improve your phase margin. It can also increase your bandwidth. You typically use it to improve your transient as you mentionned. You can think of it as a PD controller cascaded with a low-pass filter. The gain K is ... 1 You are missing a couple of zeros. First of all, you must also include the reciprocal of the one located at$z=-2$, that is one at$z=-0.5$You should also have one more zero coming from the fact that types 3 and 4 are anti-symmetric. This zero must be located at$z=1$, and is responsible for the minus sign in the anti-mirror image polynomial equation:$H(...

1

We have a zero at $z=0$ and poles at $z=1+i$ and $z=1-i$ Therefore the Transfer function is: $$H(z) = k \dfrac{z}{[z-(1+i)][z-(1-i)]}$$ $$\Rightarrow H(z) = k\dfrac{z}{(z-1)^2+1}=\dfrac{kz}{z^2-2z+2}$$ $$\Rightarrow \dfrac{Y(z)}{X(z)} = \dfrac{kz}{z^2-2z+2}$$ $$\therefore Y(z)[z^2-2z+2] = kzX(z)$$ This gives: $$z^2Y(z)-2zY(z)+2Y(z) = kzX(z)$$ Taking ...

1

A transfer function is defined as the ratio of the transform of output to the transform of input where all initial conditions are zero.

1

Without using equations. You look at a circuit that involves R and C ( as a simple exmple ). There is an applied INPUT VOLTAGE SOURCE. There is an INITIAL VOLTAGE applied to the capacitor. You are going to SOLVE for a current or a voltage using 2 DRAWINGS. And keep track of both computed values to make the final answer. [ 1st drawing ] You draw the circuit ...

Only top voted, non community-wiki answers of a minimum length are eligible