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The impulse response and frequency response are two attributes that are useful for characterizing linear time-invariant (LTI) systems. They provide two different ways of calculating what an LTI system's output will be for a given input signal. A continuous-time LTI system is usually illustrated like this: In general, the system $H$ maps its input signal $x(... 16 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 ... 14 One approach would be to use the frequency-domain least-squares (FDLS) method. Given a set of (complex) samples of a discrete-time system's frequency response, and a filter order chosen by the designer, the FDLS method uses linear least-squares optimization to solve for the set of coefficients (which map directly to sets of poles and zeros) for the system ... 13 Transfer function estimation is usually implemented slightly differently than the method you describe. Your method computes $$\left\langle \frac{\mathcal{F}[y]}{\mathcal{F}[x]} \right\rangle$$ where$\langle$angle brackets$\rangle$represent averages taken over data segments, and a windowing function is applied to each data segment before taking the ... 13 Bang on something sharply once and plot how it responds in the time domain (as with an oscilloscope or pen plotter). That will be close to the impulse response. Get a tone generator and vibrate something with different frequencies. Some resonant frequencies it will amplify. Others it may not respond at all. Plot the response size and phase versus the ... 12 Welcome to Signal Processing! You're absolutely right. You cannot simply average DFT magnitudes and phases separately, especially phases. Here's a simple demonstration: Let$z = a+bi$. By definition, magnitude$|z|$and phase$\angle z$of$z$are: $$|z| = \sqrt{a^2 + b^2}$$ $$\angle z = \tan ^{-1} \left( \frac{b}{a} \right)$$ Average$z$of two complex ... 12 A single spike (as you call it) appears theoretically only for a infinite-length sinusoid. Since your signal is 100 samples length, it is not infinite. You actually multiplied your infinite signal with a window which has a value of 1 over 100 samples, and 0 elsewhere. Since multiplication in time domain is equivalent to convolution in the frequency domain, ... 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 $$H(s)=\frac{k}{(s-s_{\infty,1})(s-s_{\infty,2})\ldots (s-s_{\... 10 The impulse response is the response of a system to a single pulse of infinitely small duration and unit energy (a Dirac pulse). The frequency response shows how much each frequency is attenuated or amplified by the system. The frequency response of a system is the impulse response transformed to the frequency domain. If you have an impulse response, you ... 10 Let h(t) denote the impulse response of an LTI system. Then, for any input x(t), the output is$$y(t) = \int_{-\infty}^\infty h(\tau)x(t-\tau)\,\mathrm d\tau.$$In particular, the response to the input x(t) = \exp(j2\pi ft) is$$\begin{align} y(t) &= \int_{-\infty}^\infty h(\tau)\exp(j2\pi f(t-\tau))\,\mathrm d\tau\\ &= \exp(j2\pi ft)\int_{-\... 8 For a start, any non-linear system will not have an easily-identifiable frequency response. So, it's really a nonsensical question. I intend no offense; nonsensical questions are often the most enlightening! However one way to try to answer your question is to assume that the LTI filter involved is the mean (rather than the median) of the windowed data. ... 8 No. The impulse response and frequency response of an LTI system are related by the Fourier transform, which is one-to-one. 7 Not an easy job. A loudspeaker produces a complicated 3 dimensional sound field, which heavily interacts with the environment. Even at a distance of a about 3 feet the reflected energy is significantly bigger than the direct energy at most frequencies. Typical loudspeaker measurements are On axis response in an anechoic chamber Full 3D response in an ... 7 If you know that the system is linear and time invariant, the easiest method (assuming that you have no noise added in the process) is to let the system act on an impulse function. The Fourier transform of the output is the frequency response of the system. 7 Yes, you can do this with an LMS equalizer which uses the Wiener-Hopf equation to determine the least squared solution to the filter that would compensate for your channel, using the known transmit and receive sequences. The channel is the unknown being solved, and the tx and rx sequences are known. BOTTOM LINE: Here is the Matlab function with error ... 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})}{(... 7 The magnitude of that complex exponential is 1. Recall from complex algebra: any complex number can be expressed as z = r e^{j \phi} where |z|=r is its magnitude and \arg z = \phi is the argument. Using this note that |e^{-j\Omega \lambda}| = 1 $$which is why it "disappeared". 6 My colleagues have had great results with vector fitting: Vector Fitting is a robust numerical method for rational approximation in the frequency domain. It permits to identify state space models directly from measured or computed frequency responses, both for single or multiple input/output systems. The resulting approximation has guaranteed stable poles ... 6 Shortly, we have two kind of basic responses: time responses and frequency responses. Time responses test how the system works with momentary disturbance while the frequency response test it with continuous disturbance. Time responses contain things such as step response, ramp response and impulse response. Frequency responses contain sinusoidal responses. ... 6 Not right. Your basic assumption that all lowpass filters can be represent through your formula is wrong. What you have is a first order IIR low pass but there are many more options for a low pass filter, such as higher order IIR filter of different types (butterworth, bessel, chebycheff 1+2, elliptic) and plus various FIR filters as well. 6 As Dilip pointed out in the comment above, you can get the impulse response using the inverse Fourier transform. However, a slightly easier method might be to use the Laplace domain instead; it's more amenable to easy inverse transforming via transform tables. First, recall that the frequency response is really just the s-plane transfer function evaluated ... 6 Consider a liner discrete-time system. Assume we can define it in terms of an input-output relation as follows (you can assume a more general model but it is enough for our purpose):$$a_0y[n]+a_{1}y[n-1]+\cdots+a_{N}y[n-N]=b_0x[n]+b_{1}x[n-1]+\cdots+b_{M}x[n-M]\tag{1}$$When the coefficients \{a_i\} and \{b_i\} are constant, we call it a finite-order ... 6 Yes for 2D signals you can take a 2D fft, and if the 2D signal is represented in the time domain, then its fft is represented in the frequency domain. 2D FFT's have many other interesting applications, for example image creation in synthetic aperture radar (SAR), where an inverse 2D FFT of radar reflections results in the creation of an image. If your ... 6 This has absolutely nothing to do with causality. The frequency response of a real-valued filter (i.e., one with a real-valued impulse response) is (conjugate) symmetric, i.e., the negative frequencies are redundant. That's why it is sufficient to show the frequency response at non-negative frequencies only. You can easily see that symmetry as follows. The ... 5 I'm not entirely satisfied by Itamar Katz's answer, so here's my explanation. The DFT of an N length complex signal, x[n]=e^{\imath 2\pi f n/N} is$$X[k]=\mathcal{F}\{x[n]\}=\frac{e^{\imath 2\pi (f-k)}-1}{e^{\imath 2\pi (f-k)/N}-1}$$So, the power or the magnitude squared response is given by$$\left\vert X[k]\right\vert^2 = \left(\frac{\sin\left(\pi(... 5 In previous sections of the book, the fact that a discrete-time signal's spectrum is periodic may have been mentioned. It can be described formally as follows: $$X(e^{j\omega})=\frac1{T} \sum_{k=-\infty}^{\infty}X_C\biggr(j\biggr(\frac{\omega}{T}-\frac{2\pi k}{T}\biggr)\biggr)$$ beingX_C(j\omega)$the Fourier Transform of the continuous signal, and$T=1/...

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The analytic way is to substitute the variable $z$ by $e^{j\omega}$ to get the frequency response $H(\omega)$ (with $\omega = \frac{2 \pi f}{F_s}$) - that is to say, the frequency response is the $z$ transform evaluated on the unit circle. Note that matlab has a built-in function for plotting the frequency response straight from filter coefficients (freqz), ...

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FDLS requires a causal frequency response. Your prototype frequency response has zero phase everywhere, which is most definitely not causal. An IIR filter order of 50 is humongous. When FDLS has too many poles and zeroes available, it "tries" to cancel excess poles with excess zeroes. Unfortunately, due to numerical limitations, the cancellation is often ...

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What you're looking for is called a pruned DFT. In principle, it is possible to calculate a subset of outputs from a DFT using fewer mathematical operations. In practice, however, existing highly-optimized FFT implementations like FFTW are designed for full-output transforms. You'll find in many cases, unless you're only concerned with a very small ...

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You're definitely on the right track. The way you're trying to solve the problem is the best and simplest. You just need to realize that you need to evaluate the magnitude and phase of the frequency response just for one frequency, namely the frequency of the sinusoidal input signal: y[n]=\left|H(e^{j\omega_0})\right|\sin\left(n\omega_0+\phi(\omega_0)\...

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