# Tag Info

9

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 $|G(e^{j\omega})|=1$, ...

6

Note that in this case you can see that the system is causal only from the given implementation. It's important to understand that you can't see it from the difference equation (if no initial conditions are given), and in general you can't see it from the transfer function either (if no region of convergence is given). The only case for which the expression ...

6

In your original code you defined the sampling interval to be $T_s=0.05$ (idd1 = iddata(ddout, ddin, 0.05);). Yet, according to the data file, the time step between sample points appears to be $T_s=0.003$. Changing $T_s$ from $0.5$ to $0.003$ should compress the step response by a factor of $0.05/0.003=16.67$, which compares favorably to the relation between ...

5

What you are looking for are what we, in the audio space, call comb filters. Comb filters may or may not have a feedback path, just like FIR and IIR filters. In fact, there is a generalized theory about designing comb filters based simply on designing any digital filter. In fact, you can think of every LTI digital filter as being a comb filter where the ...

4

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 $H(e^{... 4 Or another way, visually, with the plot of the magnitude response of your filter shown below you see that it goes to zero at$\pi/2$. So what you could do is use the relationship between the angular frequency$\omega$in [radians/sample] and the relative frequency$f$in [cycles/sample] $$\omega = 2\pi f = 2\pi \left(\frac FF_s\right)$$ Then compute$F_s$... 4 Purely by inspection of the block diagram the system is causal, because the output is the sum of the current input sample and stuff that's delayed -- there's no$z$blocks in there to predict the future, just$z^{-1}$block to react to the past. Also by your method of finding the transfer function, the system is causal -- with a$3^{rd}$order numerator and ... 4 This is unfortunately a difficult problem. A microphone doesn't have a single frequency response. It has a different one for each direction of incidence. Especially for second order microphones (dipoles, cardioids, etc.), the frequency response also depends heavily on the distance and radiation impedance of the source. So first you need to decide: what ... 4 The transfer function is$H(s) = \frac{16.94s + 579.5}{s^2 + 507.2s + 1224}$This transfer function has 2 poles, one slow pole at -2.4248 and a fast pole at -504.7752. The function has a slowish zero at -34.2. Good news, your poles and zero are all in the left-half plane. It is much easier to control a system with zeroes and poles in the left-half plane ... 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 That purple array is giving the impulse response over time. You can get it directly from the difference equation. Assume initial rest,$y[-1]=0$, then write out the impulse response for$n=0, 1, 2, .... If you do that you will get: \begin{align} h[0]&=1+0 \\ h[1] &= 0 + 0.5 \\ h[2] &= 0 + 0.25 \\ h[3] &= 0 + 0.125 \\ ... \end{align} 3 Basically the transfer function is given by (Continuous time case): $$H \left( j \omega \right) = K \frac{ \left( j \omega - {q}_{1} \right) \left( j \omega - {q}_{2} \right) }{ \left( j \omega - {p}_{1} \right) \left( j \omega - {p}_{2} \right) }$$ Now, given your frequency data sampled an N $points$ {\left\{ {\omega}_{i} \right\}}_{i = 1}^{N} $we ... 3 However when I look at the closed loop transfer function, I would say that this system is unstable for 𝐺𝐻=−1. In this case the transfer function becomes infinity so a bounded input will result in a unbounded (=infinity) output. This depends on your definition of stability.$GH = -1$is called marginally stable because depending on how you look at it, it ... 3 [EDIT: note that the note you refer to compute the discrete-time Fourier transform, via a continuous argument in frequency. And not a DFT. You are apparently computing a 3-point DFT] What I usually call the size-2 binomial filter is$\beta_1=\frac{1}{2}[1\;1], the 2-point moving average, whose Fourier representation is well-known, or easy to compute. $$... 3 I don't really understand the thing about cosines (as in: how is that helpful?) – a DFT is really just a mapping from a complex vector with N elements to a complex vector with N elements; and your calculation seems to be wrong, and I'm not sure where, but doing two of the three elements of the DFT manually might actually be enough to clear things up. \... 3 The general approach would be to find the zeros of the filter's transfer function. If you do things right it will turn out that there are two complex conjugate zeros on the unit circle at \pm j, i.e., at half the Nyquist frequency. Now you need to choose the sampling frequency such that 80 Hertz corresponds to half the Nyquist frequency. 3 If the OP is actually interested in selecting only one individual frequency from the even or odd harmonics, then a moving average filter (MAF) would be ideal since this can provide a null at every other frequency when the frequencies are harmonically related. The low-pass MAF with all the coefficients as 1 will pass DC at f=0 and provide nulls spaced by 1/... 2 HINT: There's no need to compute the frequency response by evaluating the transfer functions for z=e^{j\omega}. Just compute the poles and zeros and see where they are. Estimate the effect of the poles and zeros on the frequency response by trying to see how they influence the behavior of the transfer function on the unit circle. 2 Explicitly making the imaginary part of your FFT coefficients 0 will change your magnitude response. It is very clear if you consider this:$$ X_k = a_k + jb_k\\ |X_k| = \sqrt{a_k^2 + b_k^2} $$Now, you're explicitly setting new coefficients to be X'_k = a_k, and hence |X'_k| = |a_k|. Taking the IFFT of your new frequency response will obviously not be ... 2 No, at least not the way it's written. U_1(\omega) contains two frequency peaks (10Hz and 0Hz) and U_2(\omega) only one at \omega _2. G(\omega) can change the relative height of the peaks but the overall but not create new ones. A_2 and U_2(\omega) can vary the height and position of the peak but not the number of peaks. 2 HINT: The frequency response has the form$$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}$$Now note that the term in brackets is purely real-valued. I trust that you can take it from here. 2 There are a few things I can note about your question. As far as I have always learned, the nyquist stability criterion is taken over the openloop transfer function. if you take the closed loop transfer function, you should count the encirclements of 0 instead (if i recall correctly). The formal definition of stability, as expressed by the Lyapunov's ... 2 A transfer function describes an LTI system. As such, the given system can be described by a transfer function. However, if there are non-zero initial conditions, the system is no longer linear because there's a contribution in the output that does not depend on the input signal but only on the initial conditions. Consequently, the transfer function cannot ... 2 Your filter is a 5-tap delay line FIR filter. The FIR filter's coefficients are [2 6 1 5 1]. If x is your filter's input sequence and you execute the following MATLAB command: h = [2 6 1 5 1]; you can then implement the filter (to produce a y output sequence) using the following MATLAB command: y = filter(h, 1, x); You can see your filter's frequency ... 2$$\begin{align} Y(z) &= E(z) + \frac{1}{1-z^{-1}}(X(z) - z^{-1} Y(z)) \\ &= E(z) + \frac{1}{1-z^{-1}}X(z) - \frac{z^{-1}}{1-z^{-1}} Y(z) \\ \\ Y(z) + \frac{z^{-1}}{1-z^{-1}} Y(z) &= E(z) + \frac{1}{1-z^{-1}}X(z) \\ \left(1 + \frac{z^{-1}}{1-z^{-1}}\right) Y(z) &= E(z) + \frac{1}{1-z^{-1}}X(z) \\ \frac{1}{1-z^{-1}} Y(z) &= E(z) + \frac{1}... 2 The closed loop poles are the roots of the polynomial $$D(s)=s^2+2s+2+K\tag{1}$$ and, according to the root locus plot, they ares_{1,2}=-1\pm 2j$. Consequently, we get $$2+K=|1+2j|^2=5\quad\Longrightarrow\quad K=3\tag{2}$$ With$K=3$we obtain $$H(0)=\frac{K}{2+K}=\frac35\tag{3}$$ which leaves step response$C$as the only option. 2 The system is causal, provided that the recursion is forward; i.e., it's recursed for increasing$k$. Seeing that you are confused about causality tests, let me elaborate on it. Let's put the definition of causality from Oppenheim's Signals & Systems book : A system is causal if the output at any time depends only on values of the input at the present ... 2 You can use a window as an FIR filter (convolve with window in time), such that the zero crossings of the window transform are at$\omega_0 = 2\pi * 2k*200$to filter the even harmonics, and phase shift the window by transform$\frac{\pi}{2}$to filter the odd harmonics. For example, a rectangular window of length$M$will have zero crossings at$\omega = \...

1

You can use the manage/export functionality to convert your Audacity filter preset into a text file. The file contains frequency/value pairs plus some side info. It looks like Audacity builds an 8192 FFT grid and uses a B-spline to interpolate the frequency/value pairs over the entire grid. I assume that just do an inverse FFT plus some circular time ...

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If the system described by the transfer function $H(s)$ is stable, you can obtain its frequency response by substituting $s=j\omega$, and use the relation that you found: $$S_Y(\omega)=S_X(\omega)\big|H(j\omega)\big|^2\tag{1}$$ where $S_X(\omega)$ and $S_Y(\omega)$ denote the power spectra of the system's input and its output, respectively.

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