IF WE LET R BE THE RADIUS OF A SPHERE THAT CONTAINS THE ANTENNA AT A CERTAIN WAVELENGTH AND K BE 2*PI/WAVELENGTH, THEN WE CAN SAY ...
STATEMENT 1.
D = 2*K*R + (K*R)^2 IS A BOUNDARY BETWEEN NORMAL ANTENNAS AND SUPER-GAIN
ANTENNAS. (P. 225)
THIS IS PROBABLY TRUE AND ALLOWS UNCOUPLING OF THE GAIN AND BANDWIDTH.
LOOK AT A FEW CASES.
K*R >> 1 D = (K*R)^2 = 4*PI/(WAVELENGTH)^2 * PI*R^2
OR D = (AREA OF ANTENNA)/(EFF. AREA OF ISOTROPIC RADIATOR)
NOTE: THIS SHOULD BE THE CASE FOR HORNS AND PARABOLOIDS.
K*R = 1 D = 3 R = WAVELENGTH / (2*PI)
K*R << 1 D = 2*K*R ? ? ?
HOW CAN R BE VERY SMALL IF D = 1.5 FOR SMALL DIPOLES AND LOOPS AND BOTTOMS
OUT AT D = 1 FOR THE ISOTROPIC CASE ?
SEE BELOW.
NOTE: FOR SMALL DIPOLE ARRAYS (E.G. YAGIS) OF 2 TO 4 ELEMENTS, K*R IS
ABOUT 1 AND D IS ABOUT 3.
STATEMENT 2.
IF BW IS THE FRACTIONAL BANDWIDTH OF A SMALL ANTENNA OF A CERTAIN EFFCIENCY
AND Q = 2*PI*(MAX. ENERGY STORED)/(ENERGY DELIVERED IN ONE CYCLE), THEN ...
BW = 2*(K*R)^3/(EFFICIENCY) (MATCHED SOURCE) OR
Q = 1/(K*R)^3 ANTENNA ALONE (FIRST MODE DOMINATES)
K*R << 1 D = 1.5 (SIN THETA VARIATION ONLY)
EXAMPLE: SMALL LOOP
R = 1 METER F = 1.8 MHZ W.L. = 167 METERS
SO K*R = 0.037
THEN: Q = 18778 BW = 0.0001 => 0.187 KHZ FOR 100% EFF.
OR: BW = 0.01 => 18.7 KHZ FOR 1% EFF.
NOTE: THIS IS -20 DBI GAIN.
EXAMPLE: SHORT DIPOLE
R = 5 METERS F = 1.8 MHZ W.L. = 167 METERS
SO K*R = 0.185
THEN: Q = 157 BW = 0.0127 => 22.7 KHZ FOR 100% EFF.
OR: BW = 0.127 => 227 KHZ FOR 10% EFF.
NOTE: THIS IS -10 DBI GAIN. (MY TEE ANTENNA)
THIS IS A GOOD REASON WHY SHORT DIPOLES AND SMALL LOOPS HAVE SUCH NARROW BANDWIDTHS FOR GOOD EFFICIENCY OR WIDER BANDWIDTHS FOR POORER EFFICIENCY.
NOTE:
THIS DOES NOT SAY HOW TO REALIZE A CERTAIN PERFORMANCE ... JUST THAT IT
IS POSSIBLE ... OR NOT ... TO BUILD AN ANTENNA WITHIN A GIVEN VOLUME WITH
A GIVEN PERFORMANCE. IT SHOULD KEEP PEOPLE FROM ASKING THE IMPOSSIBLE.