For the last time, PLEASE: clipping does NOT blow speakers/subwoofers

[PV Audio 5,000+ posts]

Ok, so I got curious.
When I went home for lunch, I took my "high end" stock Infinity speaker that I took out of my Dodge. I hooked it up to a AAA battery. I only had the patience to hold it there for about two minutes. Whatever the case, in that time, I didn't smell any burning voice coil like I would have expected, so I have to admit defeat on this one.

School was a little while ago, but I understood that DC of any sort would hurt a coil. But that doesn't make sense since you have inductive circuits in the DC world. Whatever the case, if I had remembered correctly, then I believe a clipped wave would damage the voice coil, but this is obviously not the case. I stand by the rest of my words, minus the part about the voice coil getting damaged.

I'm sorry if I caused any confusion. And thanks for the arguements.

Murray

Son, that's the point of this thread: to dispel common rumors and myths. //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif
And by son, I call everyone that even though I know you're older than me.

 

[Haunz 10+ year member]

Familiar with the Fourier Transform? You can easily break that square wave down in to its AC components, with no DC present.
The question arises: is it not DC during the zero slope portions of the signal? This is a question of semantics where we must first define a window. If you accept a very small window in which there is no alternating polarity, then yes, we could say this is representative of DC. However, logic says this is pointless. Imagine, if you will, a square wave with a frequency of 100 Hz. In order to accept this as DC, we would have to use a window of no more than 1/200th of a second (as anything larger would have a change in polarity).

Not to mention that durring that 5ms window a speaker's cone would still be oscillating with the harmonics superimposed on the 100 Hz fundamental...

Interestingly enough, clipping an asymmetrical signal can produce DC voltage from an amplifier's output... This is explained in some detail on Rod Elliott's website...

 

[maldecido33 10+ year member]

When discussing 'more area under the curve' in a waveform, the key is to realize that this displays more power over time.

Because a sin wave is a function of time, ie f(x)=sin(x)

That is, of course, it's simplest form.

 

[maldecido33 10+ year member]

This will probably confuse people, but I'd like to point out that the majority of clipped signals are not square waves. A square wave is an ideally clipped (oxymoron anyone?) signal.

 

[WhoSayWho? 5,000+ posts]

Yes. The easiest way to think of it is how I described it initially. The clipped signal is no longer a sine wave; it has become a square wave. Imagine it as literally square, pi on the base, 1 tall on the sides. That has more area than a sine curve from 0 to pi, as the area when the function is increasing to pi/2 isn't filled, as well as the area from pi/2 to pi. The square wave does have that area filled. Obviously a true clipped signal doesn't have 90 degree angles, but the concept is still the same.

Ww, that sounds pretty bad. No wonder clipped signals blow so many subs.

 

[DBfan187 5,000+ posts]

I don't understand how that is not DC. It is not alternating between positive and negative. That's why it shows up as a straight line on the scope. I know it's not as bad as sending straight DC, since there is still AC between the DC lines, but it still can't be good for the speaker.
I guess DC might not push the speaker to it's full limit, so that might not mechanically damage the speaker, but I know sending straight DC into a coil that is meant for AC is damaging. It heats up a lot more than alternating the current. I don't remember the exact physics behind it though.

That is a great write up from Fosgate. I can only assume they are correct, it just doesn't make sense to me.

Murray

If it changes in time, it's not DC. //content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif

 

[PV Audio 5,000+ posts]

If it changes in time, it's not DC. //content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif

Lighthouse for the save again. That's what I was trying to illustrate before: DC only has a polarity and amplitude (we're thinking the most rudimentary sense), whereas AC has polarity, amplitude and frequency, which is period/time.

 

[Haunz 10+ year member]

Lighthouse for the save again. That's what I was trying to illustrate before: DC only has a polarity and amplitude (we're thinking the most rudimentary sense), whereas AC has polarity, amplitude and frequency, which is period/time.

I can hook a switch up to a battery and turn it on and off at 1hz..... does that make it AC ?

The clear distinction is that a square wave is made up of an alternating fundamental and odd order harmonics.... the point should be that a square wave is a collection of alternating signals...... which are all sine waves.....

 

[PV Audio 5,000+ posts]

I can hook a switch up to a battery and turn it on and off at 1hz..... does that make it AC ?



The clear distinction is that a square wave is made up of an alternating fundamental and odd order harmonics.... the point should be that a square wave is a collection of alternating signals...... which are all sine waves.....

Okay smarty pants, we're talking about a continuous function here. //content.invisioncic.com/y282845/emoticons/biggrin.gif.d71a5d36fcbab170f2364c9f2e3946cb.gif

 

[mil81 10+ year member]

If it changes in time, it's not DC. //content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif

But what I was talking about was the portion of the signal that was clipped. If you took a snap shot of only that part of the wave, you will see a DC signal. That means for that short portion of time your speaker is seeing DC. Now we have already established that that will not necissarily harm the speaker, but it is however DC voltage for that period of time. This is because, like you said, there is no change in voltage over that period of time.

Regardless, this is off topic, since I have been proven wrong in that clipping will not harm the speaker.

Murray

 

[PV Audio 5,000+ posts]

But what I was talking about was the portion of the signal that was clipped. If you took a snap shot of only that part of the wave, you will see a DC signal. That means for that short portion of time your speaker is seeing DC. Now we have already established that that will not necissarily harm the speaker, but it is however DC voltage for that period of time. This is because, like you said, there is no change in voltage over that period of time.
Regardless, this is off topic, since I have been proven wrong in that clipping will not harm the speaker.

Murray

That's a terrible argument though. That's like saying that if you pick a time period small enough, the function of my heart on the EEG stops beating. In reality, you know it doesn't, and because the waveform looks like something for a fraction of a second does not mean it is that actual thing.

 

[DBfan187 5,000+ posts]

But what I was talking about was the portion of the signal that was clipped. If you took a snap shot of only that part of the wave, you will see a DC signal. That means for that short portion of time your speaker is seeing DC. Now we have already established that that will not necissarily harm the speaker, but it is however DC voltage for that period of time. This is because, like you said, there is no change in voltage over that period of time.
Regardless, this is off topic, since I have been proven wrong in that clipping will not harm the speaker.

Murray

No, there is NO DC present, it's all AC. If an amp outputs DC, then that amp is broken.

read http://www.audiogroupforum.com/csforum/showthread.php?t=4332

 

[mil81 10+ year member]

That's a terrible argument though. That's like saying that if you pick a time period small enough, the function of my heart on the EEG stops beating. In reality, you know it doesn't, and because the waveform looks like something for a fraction of a second does not mean it is that actual thing.

Yes, there is a period of time where your heart is sitting there not doing anything. It doesn't mean you're dead, it just means for that time period, you're heart is not beating. That is a great arguement for what I am saying.

Look at what happens to the speaker at that point in time. The reason it sounds funny, is because the speaker isn't following it's smooth sine wave anymore. It has an interuption in the sine wave that pushes the speaker straight out. Then the speaker struggles to get back into it's sine wave path.

Come on guys, you are supposed to be Electrical Engineering students. What does DC look like on an Oscilloscope? It is a straight line. Much like what you are seeing when your signal gets clipped. Exactly like that actually.

I don't know how you can argue this.

Murray

 

[ExpoSport 10+ year member]

5 day late response to the first couple pages...but LOL @ the CS major people. so true...i'm in a computer programming major (wouldnt think it by seeing me...no anime shirts or WoW talk) and there are some odd lookers there. one guy has a ****** lookin moustache and a **** rat tail like whoa. couple hippie lookin people.

sry for the off topic post...but i was readin and it made me lol //content.invisioncic.com/y282845/emoticons/biggrin.gif.d71a5d36fcbab170f2364c9f2e3946cb.gif

 

[Lakota 5,000+ posts]

When you are describing distortion that has been added to the original recording (presumably on purpose), the amplifier will simply reproduce the sound, it will not make the amplifier automatically output a square waveform.

That's what I was trying to say, maybe there was a bit of misunderstanding. An amplifier simply takes in a small signal and amplifies it into a larger one of the same characteristics. If the output signal is different in any aspect besides the amplitude, then there is distortion. The gain should be linear.