You're welcome. As I am not from Structural engineering background but rather Mechanical, I cannot guide you accurately but will try to answer generally. If higher are modes are dominant in your structure, it could be either that higher modes are really contributing to the overall deformation or the lower modes have been suppressed by choice of boundary conditions (ex if the structure became too stiff). For example, consider a plate with too many accelerometers mounted on it. This certainly increases the system stiffness to such an extent that the lower modes are unable to make an impact. So the important thing to note here is that accelerometers need to be mounted in such a way that it does not alter the physical properties of plate (do not add mass or increase stiffness). Higher modes usually have complex shapes and are more localized. In contrast, lower modes have simple shapes and are more globalized. Also the excitation frequency range is important. Usually broad band excitation is given so as to check how the system responds overall. Certain frequencies will get excited which indicates resonance. But if the excitation frequency range is limited to high frequencies then the lower frequencies were not excited in the first place. So ensure all possible frequencies of interest are excited. These are the possible scenarios I could think of.

@@SrinathSrinivasan Thank you very much for your response Sir. Will try to follow your advice and see if there are improvements

There are many parameters to consider while designing an Anechoic/Semi-anechoic chamber. The most important is the cut-off frequency of the chamber which depends on the chamber dimensions and the wedge height. The lower the cut-off frequency, the larger should be the room size and the wedge height. Wedges help achieve near free field condition not only by absorbing high frequencies but also by trapping the lower frequencies due to their tapered shape. The wedge material must be a good absorber (ex. fiberglass, foam) with 0.99 absorption coefficient over a defined frequency range.

I had done some examples of mode shapes in general. You can watch them here : ru-vid.com/group/PLTkzvdFCLGkjrGqSw_C5J7iduYqwkVcZj

@@SrinathSrinivasan Thanks!!

The problem here is not with echoes but rather reverberation. Reverberation occurs when multiple reflections of the same sound arrives at slightly different times at the microphone creating one prolonging sound. As the room had flat walls and tiled floors, there was more opportunity for sound waves to undergo reflections. A reverberant environment does cause sound to be unintelligible and can increase the perceived amplitude of sound. However, it does not alter the frequency content of the sound since the environment (room) is not producing any sound but only reflecting already created sound.

Both PSD and frequency axis are in logarithmic scale. The graph therefore is a log-log graph.

For FEM (Finite Element Method) Simulation, one of the prerequisite is, the object of interest has to be discretized into finite particles called elements. In this video, the plate was divided into 100 elements. Larger the number of elements, better the results but also greater is the computational cost.

Yes.

@@SrinathSrinivasan thanks

Octave analysis is performed by passing a signal through many bandpass filters. While this is good for longer sounds, it may not be the best approach when it comes to analyzing short sounds where both time and frequency information are important. If we use FFT/Octave approach there are some things to consider. First the frequency resolution is fixed because the duration of signal is very low (ex. Door close sound). Higher frequency resolution can be obtained only when duration of signal is longer. To learn more about time-frequency relation check the link in video description. Second, this approach uses fixed frequency resolution over the entire spectrum, which also means that time resolution is fixed. The wavelet analysis overcomes this by using multi-resolution windowing so there is no further dependency on time and frequency. Hence one can achieve good time and frequency resolution at the same time.

@@SrinathSrinivasan ​Very thanks for your answer ! But in fact my question was not what i intended in my mind... I was thinking about iir bandpass, (Generally i use original signal minus 2th allpass.) From what i understand, wavelet use fir ? But what is the difference if we replace them with those bandpass iir filter ? would the latency be worse or the same ?

Yes, a wavelet can be considered as an FIR filter as it has a finite impulse response in the time domain without recursive components. Now merely comparing FIR and IIR with respect to latency, IIR has less latency as you probably know that IIR filter being recursive in nature uses past outputs as input resulting in fewer calculations compared to FIR which is non-recursive. The less latency of an IIR comes at an expense of phase issues. These can be rectified by using All pass filters but if higher order All pass are required then again it means more computation which can increase latency.

​@@SrinathSrinivasan Thanks !! I'm just testing my first FIR wavelet at 110 hz with 4096 tap. For the phase this is exact but note that at the exact frequency of interest, the phase could be exact, yet it degrade around that frequency. (depending of the type of bandpass, as a combination of lp and hp will maybe not do this ) I just test to see the difference, with only one bandpass (that is original signal minus a 2th allpass) the phase doesn't degrade so much around the frequency, but we have another problem, the envelope of the signal is extremely smoothed as 5X longer than the original, where the Fir have approximately a 1.7X longer envelope. Now i could use more bandpass filter in series with lower coefficient to get a less degraded envelope, But now, the phase is degraded even more around the frequency.. To get almost the same envelope elongation with IIR, i need to use 8 bandpass in series. But now the latency is even greater than with the FIR !... Well, this is a little approximate because i have some tool to visually see if the filter have the same response but it's not ultra precise, also, the IIR have a less sharp attenuation... But it give me an idea, using a IIR filter that degrade the envelope but have less latency and try to stop the detection with a FIR or more filter in series. But maybe i did'nt use the right windows for the wavelet, here a Hamming windows ?

Thank you. The horizontal axis for window function should rather be lower, please ignore it.

The well number 0 is present in QRD of any order. When installing QRD assuming N7, it is recommended to use arrays of N7 as using only one will not give meaningful results. So either one can leave the well '0' on one side or split the well '0' evenly along the ends. The latter option ensures that one does not need to worry about symmetry during final assembly. I do not know which manufacturer excludes well '0' altogether but there is a well known QRD manufacturer - RPG Acoustical systems. Their QRD model QRD-734 as mentioned in their website offers well '0' split along both ends. Out of curiosity I used QRDude software to check if there is any difference in keeping well '0' as it is or splitting it along ends and there was no observable difference. You can check the findings in the link in the video description now. Finally my thoughts, even if a manufacturer were to neglect well'0' altogether, they still have to use end-fins on both sides to cover the diffuser sides. In that process, they have just added an approximate well '0' on both sides since the end-fin will have some finite thickness.

Oh thank you very much for your detailed explanation! Now it makes sense... It was my mistake, never noticed that most likely those looking "simmetric" in fact have larger sides to compensate... I prefer the simmetric design, unless is equally good!

Sampling rate (in the time domain) is how many samples are being acquired in a given time. In this video, I considered two cases wherein the signal was being acquired with different sampling rates (2500Hz and 25000Hz). Now sampling rate has a consequence in the frequency domain which is bandwidth. One has to sample the signal at least 2.5 times than the highest frequency of interest. Hence in acoustic recordings, as our frequencies of interest lies within the human hearing range (20Hz - 20KHz), we usually sample at 48KHz. In other words a signal sampled at 48KHz would result in a bandwidth of up to 20KHz in the frequency domain. If the signal was only sampled at 24KHz then the bandwidth is now reduced to around 12KHz because the current sampling frequency is too slow to accurately sample the fast oscillating high frequency waves. To answer the other question: If your sine wave of 5Hz is sampled at 10Hz and 100Hz, the bandwidth in frequency domain is different in each case although you will only see 5Hz result in the spectrum. If you had an additional sine wave of say 20Hz then you cannot sample it with 10Hz as it is less than 20Hz, it will fail to capture it.

A transfer function is basically output/input. FRF is a measure of how the transfer function varies with respect to frequency.

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