Re: Apple video bandwidth question

[MdntTrain <js@cimmeri.com>]

> If we have a 14 MHz pixel clock and a $55 pattern it's a 7 MHz square
> wave that can be represented as a sum of the 7 MHz fundamental sine wave
> plus odd harmonics.
> Suppose we have a 7 MHz "bandwidth". Then by definition 7 MHz is a
> "corner" frequency that will be passed at -3 dB compared to low
> frequencies. If the bandwidth is say 5 MHz, if will have a level below
> that but definitely visible. Just blurry.
>
> -Alex.

Ok, I get it.   So if, hypothetically, we were putting out a $55
pattern as a sine wave instead of a square wave, in that case our
bandwidth would truly be 7mhz.  Likewise, a sine wave $55 pattern at
hires' 7mhz dot clock would truly be 3.5mhz bandwidth.  The more
square the wave (the more harmonics added to the fundamental), the
greater the theoretical (not practical) bandwidth requirement.

Thanks again for clearing up that mind mud.

 jS