19

Encouraged by Hilmar, I've decided to update the answer with all the steps necessary to calculate the Reverberation Time from a scratch. Presumably, it will be useful for others interested in this area. Obviously, it is the simplest approach because more advanced are definitely beyond a scope. In the beginning, you must obtain the impulse response of a room....


15

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_{-\...


13

Real-time low-latency partitioned convolution reverb with a long impulse response works by dividing the impulse response into unequally sized partitions. The shortest partitions (blocks) are at the beginning of the impulse response, and the partition length grows towards the end of the impulse response: Each partition length can be processed separately, ...


11

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 ...


11

Your professor is right, and you're almost right too. The filter is clearly an FIR filter, but because its frequency response can be expressed as a geometric series, a recursive implementation is possible. If you write the transfer function as a rational function you get $$H(z)=2\frac{1-z^{-12}}{1+z^{-2}}\tag{1}$$ which is almost the same as you got, apart ...


10

Actually, I think I see why. $$X(j\Omega) = |X(j\Omega)|e^{-j\theta(\Omega)}$$ $|X(j\Omega)|$ is purely real, and therefore if we take the IFT it is even and symmetric. $\theta(\Omega)= a\Omega$ since the phase is linear, so $e^{-ja\Omega}$ merely shifts the corresponding even and symmetric magnitude in the time domain, so the resulting impulse response ...


8

This is a nice question. I will try solving it using 2 approaches (Which are basically the same). The solution is the Least Squares Solution: $$ \hat{h} = \arg \min_{h} \frac{1}{2} \left\| h \ast x - y \right\|_{2}^{2} $$ We assume data is given in finite discrete form (As it is in practice). The convolution is done in valid mode (Like in MATLAB valid ...


8

Let me clarify some definitions first. There is no such thing as a "passive impulse response", there are only passive systems. If you like, you can call the impulse response of a passive LTI system "passive", but that's not common usage. From what I understand from your question, you probably mean by a "passive impulse response" an impulse response $h[n]$ ...


8

No. The impulse response and frequency response of an LTI system are related by the Fourier transform, which is one-to-one.


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 ...


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

It looks like your transfer function is correct, but there's a small mistake in your partial fraction expansion: $$H(s)=\frac{2}{(s+4)(s+2)}=\frac{1}{s+2}-\frac{1}{s+4}\tag{1}$$ The corresponding impulse response is $$h(t)=(e^{-2t}-e^{-4t})u(t)\tag{2}$$ The response to $x(t)=te^{-2t}u(t)$ is indeed most easily computed by solving the convolution integral:...


6

In the general case you have $$H(z)=\frac{P(z)}{Q(z)}$$ where $P(z)$ and $Q(z)$ are polynomials in $z$. If - as is the case in your example - $Q(z)$ just has one single term, $H(z)$ is definitely FIR, because you can simply divide each term of $P(z)$ by that respective power of $z$, and the number of terms of $H(z)$ equals the number of terms of $P(z)$. ...


6

You can find the impulse response Let's take the case of a discrete system. If $s[n]$ is the unit step response of the system, we can write $$s[n]= u[n]\ast h[n]$$ where $h[n]$ is the impulse response of the system and $u[n]$ is the unit step function. Now using commutative property you can write $$s[n]=h[n]\ast u[n]$$ Expanding convolution we get $$s[...


6

$h_0(t)$ is inverse Fourier transform of $H(jw)$. Your formula is OK, you can continue your calculation to practice your math manipulation, why not. To check the result, you can remark that $H(jw)$ is rotated $sinc$ function. And $sinc$ must be the Fourier transform of a rectangular box function.


6

This is a fundamental topic and it needs a lot of explaination. I suggest you use the fading channel object of MATLAB, especially the Channel Visualization Tool. There are lots of examples with visualization in this link. MATLAB channel OBJECT


6

A causal first-order IIR filter is characterized by the following difference equation: $$y[n]=b_0x[n]+b_1x[n-1]-a_1y[n-1]\tag{1}$$ with $x[n]$ the input signal, and $y[n]$ the output signal. The impulse response of that system can be computed via the $\mathcal{Z}$-transform or otherwise, and it turns out to be $$h[n]=b_0\delta[n]+(-a_1)^{n-1}(b_1-b_0a_1)u[...


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

The intuitive answer is that an impulse in time at t=0 contains all frequencies of equal magnitude, so applying an impulse to an LTI system is the same as applying all frequencies at once, thus the result is the response of the system to all frequencies, i.e., the frequency response. For a real world example, you can find the total frequency response of a ...


5

It comes down to latency vs. complexity. If your filter is 10 seconds long, you need to store the audio data of the last ten seconds and then you are able to calculate the current output audio sample with a latency of basically zero (ignoring the time required for calculations here) simply by doing: $$y[0] = \sum_{k=0}^{l} x[-k] \dot h[k]$$ where $l$ is ...


5

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 ...


5

Let me provide a method, for part a, applied only for finding the impulse response $h(t)$ of an LTI system characterised by an LCCDE of the form $ \sum_{k=0}^{N}{ a_k {{d^k y(t)}\over {dt^k}}} = \sum_{k=0}^{M}{ b_k {{d^k x(t)}\over {dt^k}}}$ by using the classical time domain approach. A method which is generally ignored. Note that the input $x(t) = \...


5

The convolution theorem ("Multiplication in time is Convolution in frequency") states: $$\mathcal{F}\{x\cdot y\}=\mathcal{F}\{x\}*\mathcal{F}\{y\}$$ where $x, y$ are two time-domain signals, $\cdot$ denotes element-wise multiplication and $*$ is the convolution operator. So, the convolution in the frequency domain, is carried out between the Fourier ...


5

Given that $$ u[n-k_1] = \sum_{k=k_1}^\infty \delta[n-k] $$ you can take the difference between two steps with different shifts ($k_1 < k_2$) and obtain: $$ u[n-k_1] - u[n-k_2] = \sum_{k=k_1}^{k_2-1} \delta[n-k] $$ Then the difference $$ 3u[n-2] - 3u[n-6] = 3\left(\delta[n-2] + \delta[n-3] + \delta[n-4] + \delta[n-5]\right) $$ which appears in ...


5

A 1D LTI system is completely characterized by the function $h(t)=T\{\delta(t)\}$ which is denoted as the impulse response of the system. Given an LTI system with impulse response $h(t)$ you can then find its frequency response as $H(j\omega)=\mathcal{F}\{h(t)\}$ where $\mathcal{F}$ stands for continuous time Fourier transform. When a system is not LTI but, ...


5

Because you don't have background in wireless communications, I will try to answer as simply as possible. What is the time unit of the converted CIR? Just time unit. It can be second, ms, us, etc. They are convertible, aren't they? How much seconds are they (the 30 impulse responses) apart? In the case that CSI-pilots are equally seperated, impulses ...


5

Complex channel coefficient is just a way to represent the independent real coefficients. You just need to generate h = [h_0, h_1, h_2] = [hR_0, hR_1, hR_2] + 1i * [hI_0, hI_1, hI_2]. The independence/correlation between coefficients depend on your model. And if I were not wrong, the number of element of h is the order of your MA model. The idea behind ...


5

The property $$f(t)\delta(t-t_0)=f(t_0)\delta(t-t_0)\tag{1}$$ is only valid for a function $f(t)$ that is continuous at $t=t_0$. Since the Dirac delta impulse $\delta(t)$ is not a function (it is a distribution) and since $\delta(t)$ is not continuous, property $(1)$ does not hold (and does not make sense) for $f(t)=\delta(t)$. The product $\delta(t)\cdot ...


5

This system $$ y(t) = t^2 x(t) $$ is not LTI and therefore does not have an impulse response of the form $h(t) = \mathcal{T}\{\delta(t)\}$. So your statement $h(t) = t^2 \delta(t)$ is not correct... Hope this solves your confusion.


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 $$\int_{-\infty}^{\infty}\delta(at+b)\phi(t)dt=\frac{1}{|a|}\int_{-\infty}^{\infty}\delta(t+b/a)\phi(t)dt\tag{2}$$ ...


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