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^{\...


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 ...


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 ...


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 ...


2

For strongly selective filters that have well defined pass and stop bands and narrow transition bands and which, therefore, can approximate ideal brickwall filters to some desired degree, poles and zeros are generally distributed around and closer to the unit circle. Normally you would like to have all of them inside the unit circle but due to other reasons, ...


2

This is easier to see graphically, on an s-plane, knowing that the phase is from the frequency response of the system and the frequency response is determined when we restrict s to be the $j\omega$ axis. Consider the simple ratio of a single pole and a single zero expressed as $$H(s) = \frac{s-z_1}{s-p_1}$$ Given s as any point on the s-plane, and $z_1$ ...


2

If the problem description is correct, i.e., if $h[n]$ is zero for $n<0$ and $n>7$, and if we assume that $h[7]\neq 0$ (and why should we assume otherwise?), then the filter has $8$ taps and it is a $7^{th}$ order FIR filter with $7$ zeros. Since it is a linear phase filter and the number of taps is even, it must be either a type II filter (even number ...


2

There's a good reason why band pass filters usually have an even filter order. As you may know, real-valued filters have either real-valued poles and zeros, or their poles and zeros occur in complex-conjugate pairs. If the order is odd, there must always be at least one real-valued pole and zero. In the case of a band pass filter, a real-valued pole is not ...


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

The $s$-plane is the complex plane associated with the Laplace transform, i.e., with transfer functions of continuous-time systems, whereas the $z$-plane is the complex plane associated with the $\mathcal{Z}$-transform, i.e., with transfer functions of discrete-time systems. The matched $Z$-transform is one way of transforming continuous-time systems to ...


2

Yes repeated poles simply means there are more than one pole at the same location. 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 integrators in cascade would be $\frac{1}{s^2}$ so has ...


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

It's pretty simple if you know a few basic things: a single pole on the real axis corresponds to an exponential sequence. The elements of that sequence either have the same sign, or their signs alternate, depending on the angle of the pole ($0$ or $\pi$). This should allow you to pair figures $(c)$ and $(e)$ with figures $(1)$ and $(6)$. a complex conjugate ...


1

I guess it's for a homework. 1 - You want to attenuate the 50-Hz component. Zeroes placed at 50 Hz will cancel the 50-Hz component, thus you need to place your zeros at 50 Hz. 2 - You want your 300-Hz signal to be unaffected by the zeros, i.e. you want it to keep the same amplitude and the same phase. If you only added the complex pair of zeroes at 50 Hz, ...


1

Regardless of causality and stability, if you count poles and zeros at the origin and at infinity, the total number of poles always equals the total number of zeros. I'll show a few examples to make this obvious. First, take a polynomial $$P(z)=(z-z_0)(z-z_1)\ldots (z-z_M)\tag{1}$$ It has $M+1$ zeros and no (finite) poles. However, it must have $M+1$ ...


1

If a discrete-time LTI system is causal (that is that the impulse response doesn't respond before the driving impulse), then the number of zeros must not exceed the number of poles. Consider this general transfer function: $$\begin{align} H(z) &= A\frac{(z-q_1)(z-q_2)(z-q_3)...(z-q_M)}{(z-p_1)(z-p_2)(z-p_3)...(z-p_N)} \\ \\ &= A\frac{z^M + b_1z^{...


1

We don't ignore poles on the contour. As mentioned in a comment, poles are avoided by modifying the contour as shown in the figure below, where a contour appropriate for a pole at $s=0$ is shown. Fig. 1: Nyquist contour for a pole at $s=0$ (from "Modern Control Engineering" by K. Ogata). The contour moves around the pole along a semi-circle centered at the ...


1

Causality places the transfer-function poles at $z=0$, for a FIR filter. A FIR filter need not necessarily be causal, in which case some or all of its poles reside at $z=\infty$ (if not at $z=0$). In any of these cases, the poles play no role in shaping frequency response, since they remain equidistant from the unit circle. (A value of $k$ in the ...


1

No it does not (always posses symmetry). However, for real systems whose coefficients $a_k ~,~ b_k$ of system function, $$H(z) = \sum_{n} h[n] z^{-n} = \frac{\sum_{k=0}^{M}b_k z^{-k}}{\sum_{k=0}^{N} a_k z^{-k}} $$ are real, then accordingly its roots (which are the poles and zeros of $H(z)$) are either real or complex-conjugates. And because of this ...


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