# Tag Info

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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 ...

7

You can apply a so-called all-pass transformation to a discrete-time low-pass prototype filter in order to convert it to other standard filters (such as high-pass, band-pass, and band-stop). This is accomplished by transforming the complex variable $z$ in the transfer function of the prototype filter by a function $G(z)$ which satisfies $|G(e^{j\omega})|=1$, ...

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

Your two transfer functions are in parallel, i.e they simply add up. So your feedback transfer function is simply $G(z) = H_1(z)+H_2(z)$. You want to makes sure that the magnitude of $G(z)$ is smaller than one. and the overall closed loop transfer function is $$H(z) = \frac{Y(z)}{X(z)} = \frac{1}{1+H_1(z)+H_2(z)}$$

4

It's true that the algebraic expression for the $\mathcal{Z}$-transform is generally not sufficient for computing the corresponding time-domain sequence. The additional information we need is the region of convergence (ROC). The ROC together with the expression for the $\mathcal{Z}$-transform uniquely determines the corresponding time-domain sequence. For ...

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

$$H(s) = \frac{R C s}{R C s + 1}$$ If $R C s$ is much lower than 1 (i.e., $R C s \ll 1$, in Math), then you can make an approximation: $$H(s) \simeq \frac{R C s}{1}$$ Basically, the effect of the $R C s$ in the denominator becomes insignificant. So, just pick a radian frequency well below $1 / R C$ and compute the magnitude of $H(s)$ at that point -- mark ...

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

Your transfer function looks correct. Note that you can rewrite $H(s)$ as $$H(s)=-\frac{3s}{5s+3}\tag{1}$$ or, equivalently, $$5sY(s)+3Y(s)=-3sX(s)\tag{2}$$ Since multiplication by $s$ corresponds to differentiation in the time domain you obtain from $(2)$ $$5y'(t)+3y(t)=-3x'(t)\tag{3}$$

3

You have discovered pole-zero cancellation, which is one of the reasons that state-space systems description was invented. Another reason is because when you go from a differential equation to a transfer function, you lose information about the structure of a system that you can choose to preserve in a state-space description. In general, you should be ...

3

I don't really understand the thing about cosines (as in: how is that helpful?) – a DFT is really just a mapping from a complex vector with $N$ elements to a complex vector with $N$ elements; and your calculation seems to be wrong, and I'm not sure where, but doing two of the three elements of the DFT manually might actually be enough to clear things up. \...

3

[EDIT: note that the note you refer to compute the discrete-time Fourier transform, via a continuous argument in frequency. And not a DFT. You are apparently computing a 3-point DFT] What I usually call the size-2 binomial filter is $\beta_1=\frac{1}{2}[1\;1]$, the 2-point moving average, whose Fourier representation is well-known, or easy to compute. $$... 3 The given transfer function has a specific form. With the denominator polynomial A(z)=2+bz^{-2}, you can write the transfer function as$$H(z)=\frac{z^{-2}}{2}\frac{A\left(\frac{1}{z}\right)}{A(z)}\tag{1}$$Since for z=e^{j\omega} (i.e., on the unit circle, where we evaluate the frequency response), we have |A(z)|=|A(z^{-1})|, the magnitude of H(e^{... 3 For continuous-time systems, you obtain the frequency response by evaluating the transfer function H(s) on the imaginary axis s=j\omega (assuming stability). In discrete-time, you get the \mathcal{Z}-transform instead of the Laplace transform, and the imaginary axis is replaced by the unit circle. So the frequency response is obtained by evaluating the ... 3 However when I look at the closed loop transfer function, I would say that this system is unstable for 𝐺𝐻=−1. In this case the transfer function becomes infinity so a bounded input will result in a unbounded (=infinity) output. This depends on your definition of stability. GH = -1 is called marginally stable because depending on how you look at it, it ... 2 Nowadays the easiest thing would be to use librosa for this task. It has the mel_to_stft function which does exactly what you want. As others have mentioned, this reconstruction is lossy and only approximate solution can be found. In librosa it is done using the Non-negative Lease Squares algorithm. A thing to keep in your mind: if you have extracted the ... 2 Explicitly making the imaginary part of your FFT coefficients 0 will change your magnitude response. It is very clear if you consider this:$$ X_k = a_k + jb_k\\ |X_k| = \sqrt{a_k^2 + b_k^2} $$Now, you're explicitly setting new coefficients to be X'_k = a_k, and hence |X'_k| = |a_k|. Taking the IFFT of your new frequency response will obviously not be ... 2 You give a transfer function in z^{-1}, but for the purposes of making a state-space model, you need a transfer function in z (more on that later). So your H becomes$$H(z) = \frac{b_0 z + b_1}{z^2 + a_1 z + a_2}$$When you invert that it's improper, which shows that you need to know information one time step into the future. That's not possible (or, ... 2 Obviously, you might need to use the given frequency somehow to solve the problem. Guessing alone will usually not suffice. Since this is a homework type problem, and since you still need to learn a lot of basics, I'll give you a few hints to get you started. First, the frequency domain behavior can be obtained from a (stable) transfer function by ... 2 Another way to see how the forward Euler method approximates a continuous-time system is by considering the "ideal" mapping of the s-plane to the z-plane (why?):$$z=e^{sT}\tag{1}$$For frequencies that are much smaller than the sampling frequency (i.e., |s|T\ll 1) we can approximate e^{sT} by its first order Taylor series:$$z\approx 1+sT\tag{2}$$... 2 For \mu > 0, z^L \in \left ( \sqrt[L]{\mu} \right) e^{j 2\pi n/L} \,\forall\, n \in 1 \cdots L . I.e., the roots are evenly spaced on a circle \sqrt[L]{\mu} in radius, and there's an L of a lot of them. 2 Frequency Response of Unknown System from Freq Chirp and FFT's My understanding from further discussion with the OP that he wants to specifically use an approach of providing a swept sine wave stimulus and use the FFT of this input and system output response to derive the transfer function. This may be for a system identification problem where the swept ... 2 HINT: There's no need to compute the frequency response by evaluating the transfer functions for z=e^{j\omega}. Just compute the poles and zeros and see where they are. Estimate the effect of the poles and zeros on the frequency response by trying to see how they influence the behavior of the transfer function on the unit circle. 2 No, at least not the way it's written. U_1(\omega) contains two frequency peaks (10Hz and 0Hz) and U_2(\omega) only one at \omega _2. G(\omega) can change the relative height of the peaks but the overall but not create new ones. A_2 and U_2(\omega) can vary the height and position of the peak but not the number of peaks. 2 HINT: The frequency response has the form$$H(e^{j\omega})=a+be^{-j\omega}+ce^{-2j\omega}+be^{-3j\omega}+ae^{-4j\omega}\tag{1}$$which can be rewritten as$$H(e^{j\omega})=e^{-2j\omega}\big[ae^{2j\omega}+be^{j\omega}+c+be^{-j\omega}+ae^{-2j\omega}\big]\tag{2}$$Now note that the term in brackets is purely real-valued. I trust that you can take it from here. 2 There are a few things I can note about your question. As far as I have always learned, the nyquist stability criterion is taken over the openloop transfer function. if you take the closed loop transfer function, you should count the encirclements of 0 instead (if i recall correctly). The formal definition of stability, as expressed by the Lyapunov's ... 1 Assuming that$$H(z) = A\frac{\prod_k (1-c_kz^{-1})}{\prod_l (1-d_l z^{-1})}, \: \: R_H$$you can perform Partial Fraction Expansion (PFE) to quickly get your impulse response h[n] (what you probably call anti-transformation) using Z-transform properties and tables of Z-transform pairs. If your transfer function is not rational, such as$$H(z) = \mathrm{...

1

First of all - I myself am not a pro in Control theory, but a mathematician - so I write what I think might be what your prof means. Part where i am quite certain: Not to be polynomial means, that there exists NO polynomial which describes the transfer function. E.g. let $H$ be your transfer function, then there exists no $n\in\mathbb{N}$ and $p\in P^n[X]$ ...

1

See the image (found in this article). Inject a signal into the loop somewhere (I show it below as going between the controller and the plant). Then measure the input to the plant and the output from the plant. Note that the image is a bit confusing -- it assumes that the normal command to the loop is 0 or some constant. It would be better to actually ...

1

or the transfer function from the input 𝑢 to the output, how can I do this ? Is that not what your $G$ is? If you want to find the transfer function from just one element of $u$ to the output, then either delete the columns of $B$ that don't pertain to that element of $u$ and get your transfer function, or just look at the column of the transfer function ...

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