501-C3:
The 27 was constructed of older commercial radios and lots of hard work.  Several compromises were made because funds just weren’t always available to buy the right stuff.  One example was the 75 ohm cable TV hard line that was used to feed the main transmitting antenna.  A couple of years ago the big break came when the LARC applied for and received the tax-exempt status of 501-C3.  Soon thereafter the Lowell Community Fund awarded the club a cash grant for the purchase of new equipment to really upgrade the 145.27 repeater.

Accomplished:

An open stub cut to the third harmonic of the operating frequency will pass the operating frequency and cancel odd harmonics.

Nulls are reflections that arrive at the feed point 180 degrees out of phase with the input signal and present very low impedance at their wavelength.

If a cable is terminated into its characteristic impedance, there will be no nulls (reflections).

If the cable is shorted, the first null will occur at the second harmonic of the ¼ wavelength.  Successive nulls will occur at every even multiple of ¼ wavelength. The span between any two nulls will be the second harmonic of ¼ wavelength.

In addition to providing lightening protection, a shorted ¼ wavelength stub will pass the operating frequency and cancel even harmonics.  One stub can be cut to “short-out” the even harmonic and can be cascaded with a second stub to eliminate odd harmonics.

The impedance of a device connected to one end of a ¼ WL section of cable (or odd multiple of ¼ wavelength) is transferred to the end of the section.  Therefore, a ¼ wavelength section of coaxial cable (or odd multiples) can be used to transfer the output Z of one device to the input Z of another.
 

The “Feet” calculation:
To determine the distance to a fault (open or short) sweep the feed-line into a “T” and use the following formula:

(492 x F) / VF = Distance (In feet) to a fault (open or short),
F = the frequency span between any two nulls in MHz.
VF is the velocity factor of the cable
 

Free space WL :
In feet:  ¼ wavelength = 245.9 / F   

Electrical length:
In feet:  ¼ wavelength = 245.9 / F x VF
In inches  (more accurate) ¼ wavelength = ((11808 / F) x VF ) / 4
 

Making cables:
Determine or verify the VF:
1. Put a connector on one end of a cable and leave the other end open.  Connect the cable to a “T” and sweep it with your spectrum analyzer – tracking generator.  
(Cable lengths from the generate and analyze ports to the “T” are not critical.)

2.  Find the lowest frequency null (the ¼ WL frequency)  ( F ). Record it.

3. Determine the free space ¼ WL (in inches)  = (11808 / F) / 4 Record it.

4.  Measure from the center of the “T” to the open end of the cable to determine the actual length of the cable in inches.   Record it.

5.  Determine the VF  VF  = Free space WL  /  Actual length Record it.
 
 
 

No Spectrum Analyzer?
To calculate the ¼ wavelength of a cable when the VF and length in inches are known:
F = (VF / (L” x 4) ) x 11808
 
 

Example:  Operating frequency: 45.65 MHz.

1.  Find the free space ¼ length:    11808 / 45.65 = 258.66  /  4 = 64.66”

2.  Measure the stub from the center of the “T” to the open end:   80.74”

3.  VF = 64.66 / 80.74 = .800

4.  Determine the electrical ¼ wavelength: = free space ¼WL x VF 64.66 x .8 = 51.728”

5.  Determine the length of the required cable:
Measure the approximate length of cable that is required to connect device A to device B, then round off the length to the next odd number of ¼ wavelengths.  For example 3. 
51.728” x 3 = 155.184”   (Actual length in inches, 3 - ¼ WL)
 
 
 

Tricks to cutting cables:
Measurements must include the following:
1. The distance to the center of a “T”
2. Any couplers or adapters (Heaven forbid)

“N” Connectors
Figure the length of the cable from the shoulder of one male pin to the shoulder of the next male pin.  This applies to any type of connector.  The length should calculated as if it were fully engaged with its mate.
 

Notes:
Use ¼ wavelength sections (or odd multiples) to couple the following devices 
Cavities in series or a duplexer
Transmitter and duplexer input, cavity and amplifier
Receiver and duplexer input, cavity and amplifier

The pass frequency of an open stub should be an even harmonic of the ¼ wavelength.
The pass frequency of a shorted stub should be an odd harmonic of the ¼ wavelength.

Cable length calculations for cavities and duplexers should always be based on the PASS frequency.

Velocity factor is also defined as the square root of e, where e = the dielectric constant.

Exercise:
Fabricate a stub of RG8U that would provide lightening protection and pass 145.27 MHz.
1. The first null will occur at the second harmonic of the ¼ wavelength.
145.270 x 2 = 290.54 MHz.  First null when shorted.
2. Calculate the electrical length (in inches) of a ¼ wavelength at 145.270 MHz.  (VF = .66)
(11808 / 145.270) = 81.283 x .66 = 53.64 / 4 = 13.41”  To the shorting point.
Allow the center conductor to extend a foot or so for the ground connection.
 
 
 

Velocity factors for popular 50 ohm cables

Cable	VF
RG58 (Polly)	.66
RG8 (Polly)	.659
RG8X	.84
LMR-400	.85
RG-142 (Teflon)	.695
RG-214	.66
3/8” Superflex  (Blue)	.801
¼” Hard	.841
½” Hard	.81
7/8”  Hard	.81

Matching a line to a load of different impedance:
Problem:  Match a 600-ohm line to an antenna with 72-ohm feed point impedance.
The Z of the section is determined by: The square root of (72 x 600) = 208 ohms
(The antenna should be tuned to resonance before the matching section is connected.)
 

A few theoretical notes:

Characteristic impedance: Z0 is determined by spacing between the conductors and the conductor size.

Calculate the free space wavelength when the frequency is known:
300 / F (MHz.) = 1WL (in meters) Example:  300 / 146.52 = 2.047 meters

Calculate the frequency when the wavelength is known
300 / WL = F (MHz.)  Example:  300 / 2.047 = 146.52 MHz. 

Conversions:
1 foot = .3047 meter
1 meter = 3.28 feet