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# Forza Horizon 4 Demo Available

This demo is a Play Anywhere title which means it’s available for download on both PC and Xbox One, the two platforms that *F**orza Horizon 4 *will launch for. The page to download the demo is right here if you’d like to get it for yourself. Forza Horizon 4 is scheduled to release next month on October 2 for Xbox One and PC. If you’re picking up the Ultimate Edition, however, you’ll gain access a few days earlier on September 28.

« NVIDIA Announces Tesla T4 Tensor Core GPU ·

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**Forza Horizon 3 gets patch that fixes stuttering -**10/17/2016 09:30 AM

Small note for the Forza Horizon 3 players out there. A new patch is available that should fix stuttering. Shortly after Forza Horizon 3, the new entry of the series developed by Playground Games, ha...

**Forza Horizon 3 Launch Trailer -**09/22/2016 11:53 AM

Two drivers have been released yesterday especially for this title from AMD and Nvidia. Microsoft has released the launch trailer for Forza Horizon 3. In Forza Horizon 3, players will be able to cust...

**Forza Horizon 3 Gone Gold (video) -**09/01/2016 08:54 AM

The worldwide release of Forza Horizon 3 is coming up on Sept. 27 (and four days earlier on Sept. 23 for Forza Horizon 3 Ultimate Edition owners), and today we crossed the latest milestone ahead of la...

**Fox2232**

Senior Member

Posts: 11810

Joined: 2012-07-20

**#5585182 Posted on: 09/16/2018 05:09 PM**

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.

**Noisiv**

Senior Member

Posts: 7885

Joined: 2010-11-16

**#5585214 Posted on: 09/16/2018 06:23 PM**

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.

**Fox2232**

Senior Member

Posts: 11810

Joined: 2012-07-20

**#5585356 Posted on: 09/17/2018 07:13 AM**

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:

**Noisiv**

Senior Member

Posts: 7885

Joined: 2010-11-16

**#5585443 Posted on: 09/17/2018 10:26 AM**

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

Having 2 pulses/signals that close together as in #31 leads to very high frequency. In that case well above regular sampling rates.

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).

Click here to post a comment for this news story on the message forum.

NoisivSenior Member

Posts: 7885

Joined: 2010-11-16

#5585162 Posted on: 09/16/2018 03:35 PMYes, 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.

y1+y2 = A*sin(W*t)) + A*sin (W*t - phase)

I have just done it. Principle of superposition

Q.E.DFact 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.

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