Absolutely great optimized game , runs very good on all hardware. This game it's very multithreaded optimized and uses all my 16 cores at 25% equally. It runs as a charm on nvidia or amd video cards as as well as Intel or AMD cpus. Job well done by Turn 10 and microsoft studio. Forza Horizon 4 Post # 38 a lot of testing done already.
I have a crappy monitor so 1080P on a GTX1070 R5 1600X 16GB 3200 https://imgur.com/a/JLlf1lF It's actually a bit smoother than FH3
WTF. Why would you ever try and use 192khz as default? Even the highest of high end A/V receivers can't handle that. It's very niche. 48khz is what you need for games, anything more is not supported by 99% of games.
WTF. Why would you ever try and use 44khz as default? Even the highest of high end A/V receivers can't handle that. It's very niche. 11khz is what you need for games, anything more is not supported by 99% of games. - - - - But on serious note. At what sampling rate do you want to mix few dozen of sounds when each can start with microsecond offset from each other. Since microsecond offset implies time unit of 1 milionth of second, you want 2MHz sampling rate to prevent loss of information. There is difference between output quality and mixing quality. You could as well say that older person does not hear sounds above 16kHz, so they should use 32kHz sampling rate, right? Wrong!
Works perfectly on 4K with 8700K and 1080...Very impressed at how smooth and great it looks and plays...real nice.
https://s8.**********/8dk5ugyf5/FH4.png These results seemed too good to be true but I ran the benchmark 3 times in a row and was getting similar results everytime. For some reason I had to re-enable audio enhancements for my audio device in order to get in-game sound to work, not sure what's up with that but aside from that, amazing demo..
You are confusing frequency domain behavior of a system with its time domain response. Please get yourself acquainted with the sampling theorem. A sampled waveforms contains ALL the information without any distortions, when the sampling rate exceeds twice the highest frequency contained by the sampled waveform. Do note that the theorem is not dealing with approximations. It states a necessary condition for the faithful representation, ie. a perfect reconstruction of the signal.
Now think. I have shown mere example of mixing 2 audio signals. Which may be exactly same, but they have certain delay. FYI Having two 1Hz signals mixed with 0.5s delay makes them 2Hz signal requiring double sampling rate. to prevent loss of information. Take this 1Hz signal gain and mix it with another 1Hz signal, this time with 0.001s delay. Yes, now you have tiny 1kHz "bzzz" which takes 0.001s from start till end, but you need already 2kHz sampling rate. So, when you have half dozen of signals each at same sampling rate, mere time delay is going to increase requirements on sampling rate to prevent information loss. Now imagine 2 wave forms one at 10Hz, one at 11Hz. What sampling rate you need to combine them without information loss? For output device, nobody really cares, but mixing... That's why there is still analog mixing in studios.
So you are willingly choosing to simply disregard a well established theoretical work and follow your intuition instead. OK lets follow the intuition: No sh1t? That's exactly what Nyquist theorem says: 2 * 1Hz = 2Hz sampling rate needed. But that's not all you had said in your previous post, and it's not what you're saying all along. You're saying a tiny time shift somehow requires a sampling rate increase. The smaller the time shift, the bigger sampling rate increase needed: Forget everything and just think from a informational point of view how stewpid would be that the one time introduction of a constant (a timeshift,a one dimensional scalar) should require a stupendous amount of the sampling rate increase. ---> It doesn't. It's just a time shift ie: Omega*t -> Omega (t+ TimeShift)
Again: Nyquist sampling theorem is NOT an approximation. It states that in order to perfectly reconstruct the signal you need twice the bandwidth of the highest frequency.
Yes, I am saying that if you have freaking 2 signals shifted by 0.00000000001s, there is no way to mix them without loss of information of F*ing 44kHz sampling rate. Simple as that is, that's reality. Fact is that you know words, but do not understand model. that one beat mixed with another delayed by 0.001s really makes it 1000Hz signal. Have beat mixed with another in 0.000001s, and it is 1MHz. Moment you do not have control over delays nor source sampling rates of each signal, you need to go up. A lot as you could have realized from 10 & 11Hz mixup.
y1+y2 = A*sin(W*t)) + A*sin (W*t - phase) I have just done it. Principle of superposition Q.E.D You are confusing FFT original domain frequency with its representation in the frequency domain. But even if you are not acquainted with FFT or sampling theorem it should be intuitively obvious that adding phase shifted identical waves should not be expensive. It should not be expensive from the informational or the computational or the storage needed point of view; you're merely introducing a simple constant and then adding the waves. In other words: it's cheap as f*k and it does not require stupendous increase of the needed bandwidth
Well, do yourself a favor. Take one of those lovely math modeling applications. Way you write hints that you have some at hand or accessible. Take simple small part of song. Do SRC from 44.1kHz to 44.101kHz (by 1Hz up). There should be no loss, no distortion right? Then repeat by 1Hz again and again till you are at 44.2kHz. Finish by SRC to 48kHz for comfortable playback on any player. Compare the two. Or compare them in mathematical way. Take that original 44.1kHz and do SCR directly to 48kHz. If there is no loss of information then 100step conversion will be exactly same as 1 step conversion.
To what purpose? Besides, I have no idea what you've just said. Your idea that adding time shifted signal on top of the identical signal leads to explosion of the required bandwidth, eventually going to infinity as the time shift approaches zero, ie the idea that adding two almost identical signals would lead to requirement of infinite bandwidth, is in disagreement with every logic and intuition. Not to mention with the fundamental theorem of information theory, with superposition principle, with sampling theorem etc etc. Adding two identical harmonics which are infinitesimally time shifted results in the harmonic of exactly the same frequency, and almost double the amplitude! A sin(kx - ωt) + A sin(kx - ωt + φ) = 2A cos (φ /2) sin (kx - ωt + φ/2) ~ 2A sin(kx - ωt) More generally adding two identical harmonics which are time shifted, results in the harmonic of the same frequency, and the amplitude modified by a factor of 2*cos (φ /2). Not to explosion of the required bandwidth.
What you described is loss of information. You doubled the amplitude instead of recording 2 separate pulses. Remember that two 1Hz signals from before? At that low frequency, you still understood that you either increase sampling rate appropriately or you have loss of information. At higher frequency, you kindly just throw information out of window. Why is that? Well, that friend you quote so much did explain it to you in easy to understand words. Having 2 pulses/signals that close together as in #31 leads to very high frequency. In that case well above regular sampling rates. Here, we two have small fight. That's due to your theoretical knowledge (while wast) being in way of practical use:
The dude forgot to do his filtering. Page 19: Filtering to avoid aliasing Page 21: Over sampling http://www.atomhard.byethost24.com/pub/Sampling_Theory.pdf Pulses? Lets not move the goalpost and let us stay with the your original example of 1Hz signal: A sin(kx - ωt) + A sin(kx - ωt + φ) = B sin (kx - ωt + φ/2) B = 2A cos (φ /2) See? What we get is a phase shift (φ/2) and a modulated amplitude ( 2A cos (φ /2) ) , but the frequency of the resulting wave stays the same (ω).
No kind of filtering can get you out of this. Quite contrary, filter preventing filtering in this situation would remove deviating frequency and then you would be left with nothing. What that video shown to you is damn simple. It is how sine wave running at 10Hz gets captured if you have 21 or 22 equally spaced capture points over time (sampling at 21 or 22Hz). In less than minute this video explains clearly that to capture and reconstruct 10Hz signal you need sampling rate to be 20Hz (2x of sampled content), but at same time it is not sufficient for proper representation of all other frequencies. He could as well keep 20Hz sampling rate and change sampled 10Hz sine to 9,5Hz. Effect would be same. = = = = But hey, I bet you can make 10Hz sine perfectly clear on 21 or 22 sampled points equally distributed over time within one second. (sarcasm, if missed)
Did you even read the comments? Nvm my link. They are laughing at him... I have explained this several times already. Sampling theorem is not an approximation. We are done here.
