Conference cookie::notes$archive:cd_v1

    I have a Harmon Kardon cassette deck and receiver, both of which I'm
    very happy with.  I haven't tried their CD player though. They do build
    a lot of audio equipment with excellent price/performance.

    Ever thought about buying a Yamaha CDX-810U ?  It has 8X oversampling,
    excellent remote/programming capabilities and seems to be really
    well made. The model 910 has an eighteen bit digital filter which
    gives you 118 dB S/N ratio. (The 810 has a sixteen bit filter
    giving only 106 dB.) There is also a model 1110.
    
    I'm in Canada, so it wouldn't be useful to quote prices, except
    in terms of ratios. 
    
    Example: (Cdn. price of Y CDX-810u) / (Cdn. price of H-K 101) = 1.75
    
    I intend to buy an 810 myself in a couple of weeks.

    
    I used to own a H/K HD500, their first machine.
    
    It had (at the time) audibly better performance than Sony, Technics,
    etc.  and included many of the niceties you mentioned.  It came
    out before oversampling became popular so I replaced it this year
    with a Sony.  I would have bought another H/K except:
    
    1.  Too much money.  I got the HD500 at about 1/2 list price when
    the factory closed them out.  My budget was only $350 max so H/K
    was out for me this time.
    
    2.  No indexing!!!  I got a few classical CDs where they were set
    up as only one track and numerous indices.  The H/K could not jump
    to an index so you could only get to certain movements with audio
    fast forward (very tedious).

    On the other hand, it had real pluses:
    
    1. The most powerful headphone amp I have heard on a CD player.
     More volume than you could ever use, even if you like screamingly
    loud rock music.
    
    2.  A simple, uncluttered control interface.  Very simple to use.
    Yes, my Sony has more functions, some of which I find superfluous
    and all of which add considerable extra controls.


    The only caveat I would have with H/K is that if you are planning
    to buy from Natural Sound in Framingham, MA (don't know where you
    are located) I have had *extremely* bad experiences there with them
    over service of H/K stuff (the factory, however is excellent to
    deal with).  
    
    

    .3 is another example of paying more for stereo equipment and getting
    less.  The most expensive is not always the best.


    							-Paul

.2>    well made. The model 910 has an eighteen bit digital filter which
.2>    gives you 118 dB S/N ratio. (The 810 has a sixteen bit filter
.2>    giving only 106 dB.) There is also a model 1110.
    
    	Wait a minute!
    
    	If the disc is recorded with 16 bits, the S/N ratio will not
    increase if you decode the data with more than 16 bits.
    
    -mike z

    re .5
    
    Actually, it will.  I'm not completely up on the technical aspects of
    this, but many new CD players are using 18 and even 20 bit decoding to
    achieve better signal to noise ratios and lower distortion specs.  The
    increased specs are achieved by shifting the bits before they go
    through the filter.  Hence, 16 bits are shifted to achieve the 18 or 20
    bit codes.  The filter can then take advantage of this, although I'm
    not exactly sure how it does it.  It would be great if someone could
    fill in the huge gaps in my explanation.
    
    Yet another way to increase performance without using better quality
    parts.
    							Regards,
    							Dave 

    	After reading .2 again, I see that "18 bits" refers to the digital
    filter, not the D/A converter.
    
    	Now, if the recorded signal has a given S/N ratio, how can a
    filter increase that number?  It seems to me that the 1 LSB of
    noise will always be there.
    
    	So, are the CD manufacturers claiming that the signal coming
    out of the audio jack has a S/N ratio in excess of 10log(2^#bits)?

    -mike z

    re .7
    
    Correct me if I'm wrong, but 10log2^16 = 160.  So there aren't any
    filters that claim to have a better S/N ratio than the original
    recorded signal.  I believe that the 18 bit filters also have that 1
    LSB of noise, but when you shift the signal back to 16 bits, some of
    the noise is shifted out of the signal. 
    
    --Dave

    My understanding of the need to use more than 16 bits in the decoding
    or filtering process is to improve linearity at low levels. I don't
    remember reading anything that says the > 16 bits is used to increase
    the S/N ratio (maybe I'm reading the wrong rags).
    
    Ken....
    

I am not sure what the significance of 10*log(2^16) is, in this 
context, and why a previous reply took ln(2^16), rather than log. My 
own understanding is that calculation of the dynamic range (NOT S/N
ratio), is based on 20*log(2^16), which happens to be just over 96
dB for CD recordings. 

As to S/N ratio, it is a function of number of bits going into the
D/A. Therefore, if the preceding digital filter increases the number
of bits, then the S/N ratio will improve. This is true even though
these are not 'true bits' but simply an approximation --actually,
'prediction' is a more correct term-- of what the value would have
been. I believe each additional bit will improve the S/N ratio by
approximately 6 dB. 

All these S/N ratios are BEFORE the analog filter which follows the 
A/D. This filter can improve the final S/N ratio further, depending 
on its design. Over 100 dB for 16 bits and over 110 dB for 18 bits 
are quite plausible.

And, now: can anyone tell me how significant is the 100 dB to 110 dB 
improvement? Also, what's the point in adding 2 bits to A/D whose 
ACCURACY is only guaranteed to the 5th MSB and whose monotonicity 
is not even guaranteed?

-- Ron

.11>I am not sure what the significance of 10*log(2^16) is, in this 
    
    	When converting a voltage (or current) ratio from absolute
    numbers to a decibel scale, the equation is 10*log(V2/V1).
    
    	If V2=2^16 (maximum voltage) and V1=1 (mimimum voltage, 1
    LSB), that yields 100.9dB for 16 bits, 97dB for 14 bits.
    
    	If the input signal has a given S/N ratio, no amount of
    data interpolation should change it.
    
    	What am I doing wrong here?
    
    -mike z

    >    	If V2=2^16 (maximum voltage) and V1=1 (mimimum voltage, 1
    >LSB), that yields 100.9dB for 16 bits, 97dB for 14 bits.
     
    How do you get these numbers?  Are you taking the base-10 log? 
    I get 10*log(65536) = 48.16db, for 16-bits.
    
    Could someone clarify the difference between Dynamic Range and S/N
    ratio (at the 16-bit encoding level)?  Dynamic range is based on
    a power ratio rather than voltage, so we get 20log(65536) or about
    96dB.  Is S/N ratio based on voltage, so it's half of that?
    
    			Steve

	Damn VAX BASIC!  My calculator was at work so I popped into
    BASIC to take the log().
    
    	I just realized that in BASIC, log() is a NATURAL log, log10
    is what I should've done.
    
    	OK, now given 96dB, 20*log(2^16), (since we are talking about
    power here, not voltage - another mistake I made), how does
    translating to >16 bits achieve a better S/N ratio?
    
    -mike z

    I'm gonna stick my neck out here and claim that you *can't* increase
    your S/N ratio to be better than the theoretical limit imposed by
    the number of bits used when encoding the recording.  Of course
    most recordings don't get anywhere near the maximum voltage allowed
    by 16 bits, and the quiet bits on classical recordings must have
    a much poorer S/N ratio than the loud parts.  Is it perhaps that
    the 18-bit converters and such are used merely to prevent the worsening
    of the S/N during decoding?  You can lessen the worsening through
    the decoding process, but you can never do better than the theoretical
    limit imposed when the recording was first made.  96 dB is enough
    anyway really.  Your amps noise floor is bound to be a lot higher
    than the floor of your CD, and if not, it's because of cheap analog
    pre-amps in the output stage of the CD player, not because 'only'
    16 bits were used in the recording.
    
    			Steve

>    	What am I doing wrong here?

Here goes:

>    	When converting a voltage (or current) ratio from absolute
>    numbers to a decibel scale, the equation is 10*log(V2/V1).

No. 

When converting VOLTAGE (OR CURRENT) ratios, the equation is
20*log(V1/V2). 

When converting POWER ratios, the equation is 10*log(P1/P2). 

The above is easy to realize, if you remember that P1/P2=(V1/V2)^2
and that log(X^2)=2*log(X). 

Both S/N and dynamic range are given in VOLTAGES. Therefore,
S/N_ratio=20*log(V(signal)/V(noise)). 


>    	If V2=2^16 (maximum voltage) and V1=1 (minimum voltage, 1
>    LSB), that yields 100.9dB for 16 bits, 97dB for 14 bits.

As has already pointed out, one should use base 10, not natural 
base. And, yes, BASIC is an abortion on more counts than one...

    
>    	If the input signal has a given S/N ratio, no amount of
>    data interpolation should change it.

A lot depends on what you mean when you say 'input signal'. I
assumed that the signal input at the recording A/D carries zero
noise and that we were discussing only the noise generated by the
recording process itself. 

This noise is related to the sampling frequency and definitely CAN
BE REDUCED by either increasing the number of bits or by analog
filtering following the A/D. 
    
-- Ron 

.15>    I'm gonna stick my neck out here and claim that you *can't* increase
.15>    your S/N ratio to be better than the theoretical limit imposed by
.15>    the number of bits used when encoding the recording.
    
    	Exactly my point.  If the source has a S/N ratio of X, then
    even an ideal replication will still have a S/N ratio of X.

.16>When converting VOLTAGE (OR CURRENT) ratios, the equation is
.16>20*log(V1/V2). 
    
    	Whoops!  Absolutely right.  My mistake.
    
.16>A lot depends on what you mean when you say 'input signal'. I
.16>assumed that the signal input at the recording A/D carries zero
.16>noise and that we were discussing only the noise generated by the
.16>recording process itself. 
    
    	I assume that the recorded signal has 1 LSB of noise. You can
    call it round-off error if you prefer. It's the amount of random
    incorrectness in the recorded material. When playing it back it
    appears as a deviation from the real signal, so it's a source of
    noise.
    
    -mike z

.15>    I'm gonna stick my neck out here and claim that you *can't*
.15>    increase your S/N ratio to be better than the theoretical limit
.15>    imposed by the number of bits used when encoding the recording.

Yes, you can. Make that "the number of bits used when DECODING the
recording" to be correct. Once you 'oversample' and filter, you have
increased the effective number of bits. The decoding D/A has no way
of knowing --nor would it care had it known-- the number of bits
originally used for encoding. 

Another way to look at this (a bit heuristically, but still true) is
to say that filtering cuts noise and this is what the digital filter
does. 

Please note that distortion is harmonically related to the signal
while noise is not (in this case, the noise is harmonically related
to the sampling frequency). Thus, if the 'oversampling'/filtering
process does not produce the EXACT bits that were dropped from the
recording/encoded process, what you get is additional distortion,
but not noise. 


> .16>  A lot depends on what you mean when you say 'input signal'.
> .16>  I assumed that the signal input at the recording A/D carries
> .16>  zero noise and that we were discussing only the noise generated
> .16>  by the recording process itself. 
>     
>  I assume that the recorded signal has 1 LSB of noise. You can
>  call it round-off error if you prefer.

It is, in a way. To be very simplistic, it should be a maximum of
� 1/2 LSB (the absolute correct value cannot be more than 1/2 LSB
away from the encoded value). however, it turns out to be Q/SQR(12), 
where Q is the resolution (Q=1/2^n, where n is the number of bits). 
That's about 0.29 LSB.

If you're interested in this sort of thing, you can find a
simplified development of this equation in 'Introduction To Discrete
Systems' by Steiglitz. I am sure you can find a more rigorous
treatment, taking into account the statistical nature of the input
signal (which results in a much more elaborate formula but similar
numerical results) in 'Digital Signal Processing' by Oppenheimer.
Look for the section dealing with quantization noise (or 'error') in
either book. 

-- Ron