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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(...


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Adapted from an answer to a different question (as mentioned in a comment) in the hope that this question will not get thrown up repeatedly by Community Wiki as one of the Top Questions.... There is no "flipping" of the impulse response by a linear (time-invariant) system. The output of a linear time-invariant system is the sum of scaled and time-...


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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....


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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 ...


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Somewhere in the chain between the sound card analog input and the samples that you're plotting, there is certainly a lowpass filter. There is likely an analog anti-aliasing filter before the ADC; in addition, there are likely one or more lowpass filters applied during resampling processes either on the card or in the operating system's audio driver stack (...


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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 ...


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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_{-\...


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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

The Butterworth filter's frequency response is the result of specific formulas and its characteristic is the flat passband frequency response. Consequently, if the coefficients of the IIR filter are modified in any way, the filter might not maintain the "Butterworth" characteristics. In addition to the responses by "Hilmar" and "Jason R", maybe you could ...


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A somewhat visual complement to the other answers You are talking about systems that are linear and time invariant. Exponential functions have one peculiar property (and can be actually defined by it): doing a time translation results in the same function multiplied by a constant. So $$ e^{t-t_0}=e^{-t_0}e^t$$ The red exponential could as well be the ...


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


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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 ...


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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.


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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, ...


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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 $h[n]$ ...


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There is no magic bullet, I'm afraid. You can use an elliptic filter to independently control pass band ripple and stop band attenuation, however you will find that the decay rate is closely related to the steepness and overall bandwidth of the filter. You can make the filter decay drastically faster by reducing the filter order to 1, but then again the ...


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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

You can approach the problem using the state transition matrix by solving the standard non-homogeneous ODE in the first equation. The solution to $\dot{x}(t)=A x(t) + B u(t)$ is $$x(t)=x_0 e^{At}+\int_{0}^te^{A(t-t')}Bu(t')dt'$$ where $x_0=x(0)$. The quantity $e^{At}$ is called the state transition matrix (also the solution to the homogeneous ...


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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 ...


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If the impulse response is $h[n], n\geq 0$, then the step response $s[m]$ at any time $m≥0$ is $$s[m]=h[0]+h[1]+h[2]+\cdots +h[m].$$ Note that if the filter is a finite impulse response (FIR) filter so that $h[m] = 0$ for $m > M$, then the step response at $M$ is $$s[M] = h[0]+h[1]+h[2]+\cdots +h[M]$$ and remains at this value forever afterwards, that is, ...


6

It is important to realize that due to the LTI property the response to an impulse $\delta(n)$ describes the system completely, because any input signal can be written as a weighted sum of shifted impulses: $$x(n)=\sum_kx(k)\delta(n-k)\tag{1}$$ Let $\mathcal{T}\{\cdot\}$ denote the system operator, so the response to a shifted impulse is $$\mathcal{T}\{\...


6

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 ...


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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 $P(z)$. ...


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$h_0(t)$ 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.


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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


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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[...


5

The length of the impulse response is typically related to the frequency resolution of the channel transfer function. As a rule of thumb: the more detail there is in the frequency response, the longer the impulse response will be. In practice there are a few things you can do: If you have full access to a similar, you can simply measure it with a very long ...


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Here is one possible suggestion: use a second order microphone (e.g. cardioid microphone). The human voice is a point source and has strong reactive velocity near-field component which velocity mics are sensitive too. In plain English this means you get substantially more relative bass when the speaker is close by. You could leverage to compare the energy in ...


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If a linear time-invariant system has (deterministic) input $x(t)$ and output $y(t)$, then the cross-correlation function $R_{y,x}$ of the output and input has value $h \star R_{x,x}$ where $R_{x,x}$ is the autocorrelation function of the input signal and $h$ is the impulse response of the linear system. For completeness, the autocorrelation function $R_{y,...


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Ok. You are touching on the topic of deconvolution. You can do this most easily by transforming the data into the frequency domain, doing some basic math and then converting back to the time domain. For example for case 1, if you transform the signals you have $ H(f) = Y(f)/X(f) $ so $h[n] = IDFT[Y(f)/X(f)]$ You can do similar manipulations to handle ...


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