GE MASTR II Self-Quieting Problems

Help!!!
My converted mobile self-quiets!

By Scott Zimmerman N3XCC

Problem:
This problem occurs most often when a UHF MASTR II mobile is being converted for repeater service, but it can happen in other radios and other bands as well. The problem is that when the radio is converted and tuned in a certain band of frequencies, an active transmitter causes the receiver to completely quiet.

Background information:
In any receiver, there are many 'sensitive spots'; frequencies that the receiver can hear if the signals are strong enough. These sensitive spots are found at every harmonic of the first receiver local oscillator fundamental crystal frequency (F_lo1), plus and minus the second local oscillator frequency (F_lo2). Written as a formula: (Harmonic * F_lo1)+ F_lo2 and (Harmonic * F_lo1) - F_lo2. These two formulas are for high-side and low-side injection respectively. The job of a radio's front end is to allow the desired receive frequency through to its hot spot and reject all other frequencies.

The term Low-Side Injection (LSI) is used to describe the design condition where the first Local Oscillator's (L.O.) frequency is lower than the desired receive frequency. The other side of this coin is High-Side Injection (HSI). This is when the first L.O. is higher in frequency than the frequency we are trying to receive.

An Example of Injection Sides:
In a UHF MASTR II, for example, the receiver uses a 27 times multiplier. To calculate the crystal frequency for a specific operating frequency, you simply follow one of the formulas below, where F_o is the desired receive frequency and F_c is the crystal frequency needed:

Low-Side injection: F_c = (F_o - 11.2 MHz) / 27

High-Side injection: F_c = (F_o + 11.2 MHz) / 27

The factory normally uses low-side injection in the 450-512 MHz frequency splits.

Example #1: Low-side Injection on 448.600
Say we wanted a receiver to receive on 448.600 MHz, with low-side injection, we would follow the above formula for LSI and subtract 11.2 MHz from 448.600 MHz. This would tell us we need a first L.O. frequency of 437.400 MHz. We would then divide this number by 27 to get a crystal frequency of 16.2 MHz.

Example #2: High-side Injection on 448.600
We could also use high-side injection in our receiver. We would follow the above formula for HSI and add 11.2 MHz to 448.600 MHz and get a first L.O. frequency of 459.800 MHz. We would then divide this number by 27 to get a crystal frequency of 17.0296 MHz.

As you can see above a crystal oscillating at 16.2 MHz -OR- 17.0296 MHz will make a receiver that will receive our desired receiver input of 448.600 MHz. This is of course assuming that the local oscillator multiplier chain will tune correctly at both frequencies.

More Background Theory:
Now, suppose we use 471.000 MHz as our desired receive frequency using low-side injection. We would follow the above formula for HSI and subtract 11.2 MHz from 471.000 and get 459.800 MHz. The same frequency as in the HSI example above! We divide by 27, and again come up with 17.0296 MHz as our crystal frequency.

What the example above tells us is that our crystal cut for 17.0296 MHz will work to receive on both 471.000 and 448.600 MHz. The only thing that determines which of these two frequencies our receiver will pay attention to is the radio's front-end filtering.

Just as the math described above shows that our receiver will receive either side of our desired LO frequency, it will also receive on either side of a different harmonic of the crystal frequency (an undesired LO frequency).

Suppose we use our 16.2 MHz crystal from our first example above, but we operate it at the 13th harmonic. 16.2 MHz times 13 equals 210.600 MHz. Our receiver without any front-end filtering would receive 11.2 MHz above and below this frequency. Doing the math, we find those frequencies to be 199.400 and 221.800 MHz. This phenomenon will happen at EVERY multiple of the receive crystal. So as you can see, a receiver without any front-end filtering will receive signals on a large number of frequencies. This is one reason why a radio with a poor front end suffers from intermod due to pagers and other out of band transmitters.

If you wish to play with multipliers and high/low injection numbers, here's a 56kB Excel spread sheet file you can download and experiment with.

Why is this relevant to our problem?
Just as every receiver has "sensitive spots" as shown above, every transmitter has "hot spots". Transmitter "hot spots" are harmonics of the transmit crystal that leak out of the exciter circuitry.

An Example of Tx Hotspots:
On a UHF MASTR II, the transmitter multiplier is 36. To calculate the crystal frequency needed for a particular transmit frequency, you simply divide the transmit frequency by 36. So for a transmit frequency of 443.600, the output frequency of the first examples above, you would divide 443.600 MHz by 36. This works out to a crystal frequency of 12.322222 MHz. Just for kicks, let's see what the 18th harmonic of that frequency is. 12.322222 x 18 equals 221.800. Hey, that's the same frequency as the HSI sensitive spot of the 13th harmonic of our receive crystal!!!

If you wish to play with transmit multipliers to find hotspots, here's a 96kB Excel spread sheet file you can download and experiment with.

Problem defined:
When low-side injection is used on a GE MASTR II radio, the receiver has a sensitive spot at the same frequency as the 18th harmonic of the transmitter crystal.

The Solution:
When ordering receive crystals for your UHF MASTR II project, order them for high-side injection. High-side injection also has a secondary benefit when converting to the ham band buy putting the first LO frequency closer to the original commercial first LO operating frequency. This allows the receiver to operate in the ham band without the need of additional component changes, and usually results in performance that equals published specifications.