Análisis del sintonizador FM de Neuros

Para una frecuencia Freq expresada en unidades de 0.1 Mhz el algoritmo para obtener el canal es el siguiente:

• Chan = MaxChan - ((MaxFreq-Freq)*6)

Por ejemplo, tomando como tope de dial MaxFreq=1080 (MaxChan=7122)

• Chan = 7122 - ((1080-Freq)*6)

Para un canal Chan el algoritmo para obtener la frecuencia es el siguiente:

• Freq = MaxFreq - ((MaxChan - Chan) / 6)

Por ejemplo, tomando como tope de dial MaxFreq=1080 (MaxChan=7122)

• Freq = 1080 - ((7122 - Chan) / 6)

Conociendo MaxFreq se puede obtener MaxChan empleando estas ecuaciones, y viceversa.
Si deseamos ampliar el límite superior de recepción (MaxFreq) deberemos partir de una base conocida,
por ejemplo:

• OldMaxFreq = 1080 ; OldMaxChan = 7122
• NewMaxFreq = 2200 ; NewMaxChan = (NewMaxFreq - OldMaxFreq)*6 + OldMaxChan = 13842

Problema:
Se ofrecen seis canales para cada 100kHz.
100 no es divisible entre 6.

El tuner ofrece 1230 posiciones para 2050 posibles canales (en intervalos de 10kHz) en la gama estándar FM.

It's bad english translation time! (It wasn't bad at all!)

An Analysis of the Neuros FM tuner

The tuner offers 1230 channels (in intervals of 16.666 kHz) in the 20.50 MHz range of standard FM, which is 87.5 to 108 MHz.

For a frequency (Freq) expressed in units of 0.1 Mhz the algorithm to obtain the channel is:

Chan = MaxChan - ((MaxFreq - Freq)*6)

For example, taking the top of dial MaxFreq = 1080 (108.0 MHz) and (MaxChan = 7122)

Chan = 7122 - ((1080 - Freq)*6)

For a channel (Chan) the algorithm to obtain the frequency is:

Freq = MaxFreq - ((MaxChan - Chan)/6)

For example, taking the top of the dial MaxFreq = 1080 (108.0 MHz) and (MaxChan = 7122)

Freq = 1080 - ((7122 - Chan)/6)

Knowing MaxFreq, MaxChan can be obtained using these equations, and vice versa.

If we wish to extend the upper limit of reception (MaxFreq), we must start our calculations from a previously-known base, for example:

• OldMaxFreq = 1080 ; OldMaxChan = 7122
• NewMaxFreq = 2200 ; NewMaxChan = (NewMaxFreq - OldMaxFreq)*6 + OldMaxChan = 13842

Problem: Each channel is one sixth of 100 kHz - but 100 is not evenly divisible by 6, So the resulting channels are at 16.666 kHz intervals.

Feel free to clean my translation up.

-- GarBage - 16 Oct 2004

Glad to assist!

-- PhilSalkie - 24 Nov 2004