AVS › AVS Forum › Audio › Audio theory, Setup and Chat › Very basic question about equalisation and cutting of discs
New Posts  All Forums:Forum Nav:

Very basic question about equalisation and cutting of discs

post #1 of 7
8/10/12 at 3:06pm
Thread Starter 
Everywhere I look the explanation given as to why de-emphasis of low frequencies is necessary is that low frequencies create bigger modulations and in the interest of extanding amount of music that can be stored we de-emphasize the lower frequencis. But how is frequency relevant here, isn't it just the amplitude of a sound wave that determines the amplitude of the groove on the disc?
Reply
Reply

AVS Top Picks

post #2 of 7
8/10/12 at 5:16pm
NO; it is proportional to the product of the signal voltage (voltage amplitude) times the period or time length of one cycle of the wave. Lower frequencies require the cutting head to keep moving at the same rate in one direction for a longer time, so the resulting groove is wider as the frequency is lowered (for any given amplitude).**


The thing you need to understand is that the instantaneous voltage output of the cartridge depends on the RATE at which the stylus is moving (not its physical displacement from rest). The cartridge is an electromagnetic generator, and the rate of stylus motion determines the output voltage.

To cut bass tracks on a vinyl record that will make the cartridge put out the needed voltage, the low-frequency excursion of the groove would be huge, since the master cutting head would have to maintain a high rate of movement for up to 25 milliseconds in each direction. The resulting grooves would be so wide on the record surface that a music track with low bass might cut the playing time to 3 or 4 minutes on each side of a 12-inch record. Another issue is that it would be very hard to make a cartridge that would track LF grooves that wide.

The only solution to this is to de-emphsize (reduce the signal voltage to the cutter head) at the lower frequencies. The lower the frequency, the worse the problem, so a roll-off filter is used when cutting the record. This is one of the functions of an RIAA equalization filter.

The phono preamp must then boost the low frequencies coming from the phono cartridge back up to their original relative level for playback using an inverse RIAA filter.

The RIAA filter also boosts the high frequencies during recording. This establishes a high signal-to-noise ratio on the record surface. When the high frequencies are de-emphasized by the RIAA filter during playback, the background noise is also reduced.

You should read the Wikipedia entry on "RIAA EQUALIZATION" which explains this in specific detail.

**- (for similar reasons, the physical excursion of a speaker cone is inversely proportional to frequency)


Quote:
Originally Posted by Lawrence001 View Post

Everywhere I look the explanation given as to why de-emphasis of low frequencies is necessary is that low frequencies create bigger modulations and in the interest of extanding amount of music that can be stored we de-emphasize the lower frequencis. But how is frequency relevant here, isn't it just the amplitude of a sound wave that determines the amplitude of the groove on the disc?

Edited by commsysman - 8/10/12 at 5:54pm
Reply
Reply
post #3 of 7
8/10/12 at 5:56pm
Quote:
Originally Posted by Lawrence001 View Post

Everywhere I look the explanation given as to why de-emphasis of low frequencies is necessary is that low frequencies create bigger modulations and in the interest of extanding amount of music that can be stored we de-emphasize the lower frequencis. But how is frequency relevant here, isn't it just the amplitude of a sound wave that determines the amplitude of the groove on the disc?

Phono cartridges as a rule generate a voltage that is proportional to the side-to-side and up-and-down velocity of the stylus. They basically generate an electrical signal by moving a magnet inside a coil of wire or vice-versa, so the voltage generated by motion goes up with the speed of the motion of the stylus. The cartridge is velocity sensitive, not amplitude sensitive.

This is how magnetic cartridges, which rule the world of quality phono cartridges, work.

There are also crystal, ceramic, and strain gauge cartridges that are position sensing. Crystal and ceramic cartridges are used in very cheap phonographs, including some modern low end turntables. They have a well-deserved reputation as groove busters.
Reply
Reply
post #4 of 7
8/11/12 at 6:11am
Thread Starter 
Quote:
Originally Posted by commsysman View Post

NO; it is proportional to the product of the signal voltage (voltage amplitude) times the period or time length of one cycle of the wave. Lower frequencies require the cutting head to keep moving at the same rate in one direction for a longer time, so the resulting groove is wider as the frequency is lowered (for any given amplitude).**
The thing you need to understand is that the instantaneous voltage output of the cartridge depends on the RATE at which the stylus is moving (not its physical displacement from rest). The cartridge is an electromagnetic generator, and the rate of stylus motion determines the output voltage.
To cut bass tracks on a vinyl record that will make the cartridge put out the needed voltage, the low-frequency excursion of the groove would be huge, since the master cutting head would have to maintain a high rate of movement for up to 25 milliseconds in each direction. The resulting grooves would be so wide on the record surface that a music track with low bass might cut the playing time to 3 or 4 minutes on each side of a 12-inch record. Another issue is that it would be very hard to make a cartridge that would track LF grooves that wide.
The only solution to this is to de-emphsize (reduce the signal voltage to the cutter head) at the lower frequencies. The lower the frequency, the worse the problem, so a roll-off filter is used when cutting the record. This is one of the functions of an RIAA equalization filter.
The phono preamp must then boost the low frequencies coming from the phono cartridge back up to their original relative level for playback using an inverse RIAA filter.
The RIAA filter also boosts the high frequencies during recording. This establishes a high signal-to-noise ratio on the record surface. When the high frequencies are de-emphasized by the RIAA filter during playback, the background noise is also reduced.
You should read the Wikipedia entry on "RIAA EQUALIZATION" which explains this in specific detail.
**- (for similar reasons, the physical excursion of a speaker cone is inversely proportional to frequency)

I understand that a low frequency will cut a larger wave across the record then a high frequency, but isn't it the case that the surface area taken up is ultimately the same. For ex. if you have two seconds of a high frequency sound you are going to get a lot of zig-zaging which is going to take up the same amount of surface area as two seconds of a low frequency sound which is going to cut a fewer number of smaller waves, but the displacement from the center is the same (provided the sound waves have the same amplitude) and therefore they take up the same amount of surface area, whether it's one big throw of the stylus or many small ones, the displacement is the same. So even tho an individual period of a low frequency wave takes up more time than an individual period of a high frequency wave, they ultimately take up the same surface area since the high frequency one is going to generate more waves on the disc in the same time period. I know I'm misunderstanding something here.
Reply
Reply
post #5 of 7
8/11/12 at 6:33am
Yes, you are missing something. As said above, output voltage and ultimately volume is proportional to stylus speed, not displacement.

Imagine low frequency A and high frequency B. Both have the same output level, so the stylus must move at the same speed. That means the "slope" if the groove is the same in both cases. Both grooves start out looking exactly the same on the record. The stylus moves the same speed in both. However, high frequency B has a shorter wavelength, so the groove makes its turn and heads back to center... while low frequency A has a longer wavelength and its groove just keep on going, cutting a wider path before it makes it's turn back to center.

What you described is essentially what the RIAA does. By making the low frequency groove the same width as the high frequency one, the slope of the groove is necessarily shallower. And again, since output is related to slope, the inverse RIAA mist be applied to bring the level back to it's original level.

Try it out. Draw two sine waves of the same peak amplitude, one short and high frequency, the other longer lower I'm frequency. Now look at the slope at the zero crossing. High frequency has a higher slope at the same peak amplitude. But remember, unlike acoustic waveforms, the stylus output is proportional to slope, not amplitude, of the groove. The higher frequency would be same volume if you were looking at microphone measurements of sound. But the higher frequency is louder if you are looking at a record groove sans RIAA.
Reply
Reply
post #6 of 7
8/11/12 at 6:35am
In the absence of EQ in the recording process, applying a sine wave voltage of varying frequency but constant amplitude to the cutting lathe input would result in a groove displacement that's inversely proportional to frequency. This is because the cutting head cuts a groove for which the velocity is proportional to the voltage applied to the cutting head. Thus the amplitude of the velocity would be constant with frequency in this case. However, displacement is the integral of velocity with respect to time. It is this relationship (the integral) that makes the displacement inversely proportional to frequency in the un-equalized case when a sine wave voltage of constant amplitude vs. frequency is applied to the cutting lathe input.

I would normally avoid referencing Stereophile for factual information because they are generally unreliable, but there is actually a very good Stereophile article on RIAA equalization by an engineer named Keith Howard here. He goes into much more detail than is possible in a forum post.

Edit: Oops, I see Bigus beat me to it smile.gif.
Edited by rock_bottom - 8/11/12 at 6:51am
Reply
Reply
post #7 of 7
8/11/12 at 8:32am
I don't know where to begin; everything you said here is totally wrong (except the last sentence).


Quote:
Originally Posted by Lawrence001 View Post

I understand that a low frequency will cut a larger wave across the record then a high frequency, but isn't it the case that the surface area taken up is ultimately the same. For ex. if you have two seconds of a high frequency sound you are going to get a lot of zig-zaging which is going to take up the same amount of surface area as two seconds of a low frequency sound which is going to cut a fewer number of smaller waves, but the displacement from the center is the same (provided the sound waves have the same amplitude) and therefore they take up the same amount of surface area, whether it's one big throw of the stylus or many small ones, the displacement is the same. So even tho an individual period of a low frequency wave takes up more time than an individual period of a high frequency wave, they ultimately take up the same surface area since the high frequency one is going to generate more waves on the disc in the same time period. I know I'm misunderstanding something here.
Reply
Reply
New Posts  All Forums:Forum Nav:
  Return Home
  Back to Forum: Audio theory, Setup and Chat

AVS Top Picks

AVS › AVS Forum › Audio › Audio theory, Setup and Chat › Very basic question about equalisation and cutting of discs