# Tag Info

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When talking about modeling, there are two things that usually get modeled: 1. the guitar amp, and 2. the speaker cabinet. Only the latter is modeled by an impulse response, which means that the cabinet is simply represented by an LTI system and implemented by convolution. This is of course an approximation but it works fairly well. You can find a lot of ...

8

If you're an EE student, you will have encountered the term LTI System (or you certainly will soon enough!): A system that, no matter the absolute time, outputs, given the same input, the same output; if you scale the input by a factor, the output is scaled by the same factor. Linear, time-invariant, so to speak. LTI systems can be applied to time-domain ...

6

The DC value is simply the mean. Since the signal is periodic you only need to take the mean of one period. This can be simply done with $$h(t)=\left\{\begin{matrix} 1/T, 0 < t < T\\ 0, {\rm elsewhere} \end{matrix}\right.$$ Clearly (1) and (2) are different. (1) will not converge, so that doesn't work (2) is the best way for a function that's ...

5

First of all it seems useful to establish what we mean by an equation like $$\delta(at+b)=\frac{1}{|a|}\delta(t+b/a),\qquad a\neq 0\tag{1}$$ Since the Dirac impulse $\delta(t)$ is a distribution, Eq. $(1)$ only makes sense if interpreted as $$\int_{-\infty}^{\infty}\delta(at+b)\phi(t)dt=\frac{1}{|a|}\int_{-\infty}^{\infty}\delta(t+b/a)\phi(t)dt\tag{2}$$ ...

5

A constant impulse response won't work because if the input signal has a non-zero DC component, the output will blow up. Note that the input signal has frequency components at DC and at integer multiples of $1/T$, the latter being its fundamental frequency. So you simply need a filter that retains the DC component and filters out all integer multiples of $1/... 5 Yes, that's correct. The impulse response of a cascade of two systems is the convolution of the individual Impulses responses. Convolution is commutative, so it doesn't matter in which order you convolve. Impulse responses are ONLY defined for linear time invariant systems, so this will work for systems like equalizers or filters but NOT for things like ... 4 You're almost there; you just need to connect a few dots. Let your$\frac{10}{5 + i 2 \pi f} = G(f)$. Then$g(t) = 10 u(t) e^{-5 t}$. Now we get into that exponent part. Your$h(t) = g(t - t_0)$is correct, but you're applying it incorrectly. You need to apply it to the part that I've labeled as$g(t)$:$h(t) = g(t - t_0) = 10 u(t - t_0) e^{-5 (t - t_0)}...

4

Because the impulse response completely characterizes an LTI system. The reason is an arbitrary input $x(t)$ can be written as an infinite sum of time shifted and scaled Delta functions: $$x(t) = \int_{-\infty}^{+\infty}x(\tau)\delta(t-\tau)d\tau$$ or alternatively, using the convolution operator, $$x(t) = x(t)*\delta(t)$$ Properties of linearity and time ...

3

Your proof is correct, and the result is that the system is time varying because the response to $x(t-t_0)$ does not in general equal a delayed response to the input $x(t)$. Of course, for the special case $t_0=2\pi k$ the delayed response to $x(t)$ equals the response to $x(t-t_0)$, but for time-invariance this equality must hold for any $t_0$.

3

Front/back is actually very hard too, at least for stationary simulation without head tracking. The main reason is simple: Left/Right is done by looking at the differences between the two ear signals, i.e. interaural level differences and interaural time differences. If a source is located to the left, the sound will arrive earlier on the left ear and will ...

3

(Adapted from this answer on dsp.SE) The reason that the impulse response (also called the unit pulse response for discrete-time systems) determines the output for arbitrary input $x$ to an LTI system is that The output of a linear time-invariant system in response to input $x$ is the sum of scaled and time-delayed versions of the impulse response. ...

3

I would do the following, first design a filter that notch's out just the DC. A first attempt would say to place a zero at z = 1, or at the unit circle on the real axis where real part = 1, the transfer function for this is given as $$H_{1}(z) = 1 - z^{-1} \tag{1}$$, the frequency response of this transfer function is shown below in the first figure. ...

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. $$... 3 The impulse response characterizes an LTI system completely, in the sense that the response to any input can be computed from the input and from the impulse response by convolution. That's a consequence of the system being linear and time-invariant. If we define stability as bounded-input bounded-output (BIBO) stability then stability can be determined from ... 3 I have no access to your audio files so I've downloaded: IR from here (mono/r1_omni.wav) - it's a really long one Anechoic recording from here (operatic-voice/mono/singing.wav) Resampled voice signals: Final convolved signal: As for your questions: 1. As you did the plot of IR in logarithmic scale it's clearly visible that towards its end there is ... 2 For an LTI system with impulse response h(t) and transfer function H(s)=\mathcal{L}\{h(t)\}, an input signal x(t)=e^{st} (note: no multiplication with the unit step function u(t)), results in an output$$\begin{align}y(t)&=(x\star h)(t)\\&=\int_{-\infty}^{\infty}h(\tau)x(t-\tau)d\tau\\&=\int_{-\infty}^{\infty}h(\tau)e^{s(t-\tau)}d\tau\\&...

2

[While commenting on Matt's answer, I tried to find a different path. I somehow failed to do so, but it is written, so] A folk (and false) interpretation of the Dirac $\delta(t)$ is that this is would be a function (false, in the classic sense, it cannot be evaluated; it should be understood as an application or operator on other functions, and called ...

2

It is a little late, but i'm also working on convolution reverb at the moment. If it is still of interest, you can use my code. Simply call the function convolution_reverb and pass the paths to the two audio files (audio and impulse response, both need to be .wav files), as well as the name for the result file to be created. import numpy as np from wave ...

2

If you’re looking for modeling the amplifier itself, convolution will not provide a complete model for the internal processes. However, convolution is the basis for a number of cab modeling products. I have a line 6 helix that I use frequently. A dry guitar doesn’t sound great. A dry guitar through an amp model sounds bad. A dry guitar through an amp and ...

2

I assume that this is a discrete-time problem where the maximum-length sequence is a pseudorandom sequence $x[k]$ of $\pm 1$ values and the noise is a sequence $n[k]$ of independent identically distributed (iid) zero-mean random variables with variance $\sigma^2$. Then, $\sum_{k=0}^{N-1} x[k]n[k]$ is also a sum of $N$ iid random variables and its variance ...

2

First of all, I wouldn't worry too much about the speaker response since it is relatively flat and the microphone has a much bigger roll-off. Since you've captured the frequency response using sweep, why not to skip the whole part of designing the filter that mimics the frequency response and use the original impulse response? I don't know what kind of ...

2

Firstly, for a response to be classified as the response to an impulse. The input first must qualify as an impulse to the system. For ex: a quick clap could be the impulse input to measure the sound impulse response of the room. In your case, hitting the hammer at point E must be a "good enough" impulse input for all the materials and lengths. The issue ...

2

Kronecker Delta used mainly in context of discrete time signals is defined as $$\delta_{ij}$$ is 1 only when $i=j$ or $i-j=0$, $0$ otherwise. So alternatively it is also written as $\delta[i-j]$. This value is exactly 1 and finite (refer below for the distinction with Dirac Delta) The unit impulse is just the kronecker Delta with $j=0$ hence we only refer ...

2

It's actually good that you couldn't prove it because the claim is wrong. Note that in discrete time we have $\delta[n]=1$ for $n=0$ and $\delta[n]=0$ for $n\neq 0$. So if $a\neq 0$, you simply get $\delta[an]=\delta[n]$ (note that $a$ must be an integer for this to make sense). If $a=0$ then $\delta[an]=1$ (for all $n$).

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I shall provide more details if I am correct. IMO if there is not root on the domain of integration, and here I suppose that $t\in \mathbb{R}$, then the argument never vanishes. Then, in an engineer fashion, one could say (but $t\mapsto \delta(t)$ is not a function): $$\delta(t^2+t+1) \equiv 0\,.$$ In more precise words, I would consider that for any ...

2

What is the best practice (and code solution) for detecting the onset of an impulse? ... The waveforms for these pulses do not necessarily go to zero between impulses, in fact, they rarely do. The signal goes close to zero, but not precisely. (I expect this has something to do with ambient noise around the signal, but am not certain . . .) Both of these ...

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A kernel is a more general concept, but an impulse response is a special case of a kernel. One usage of the term kernel is to describe an integral transform: $$y(t)=\int_{-\infty}^{\infty}x(\tau)K(\tau,t)d\tau\tag{1}$$ The function $K(\tau,t)$ is called the kernel of the integral transform. If you compare $(1)$ to the convolution y(t)=\int_{-\infty}^{\...

2

Clearly, for negative values of $t$, the system needs to know the future in order to determine its output. Hence, the system can't be causal. Since the system is also time-varying (show it!), its response to an impulse doesn't say much about its general behavior, unlike it would be the case for a linear time-invariant (LTI) system. So the given system's ...

1

For a discrete-time system if the impulse response is IIR and right sided, then the ROC will be outside of the largest pole. Else if the impulse response is IIR and left sided, then the ROC will be inside of the smallest pole. For all FIR (finite inpulse response) systems, the ROC will be all $z$ except possibly zero and/or infinity. For the continuous-...

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