ECE 635 Spring 15 Prof. David R. Jackson ECE Dept. Notes 13 1
Overview In this set of notes we perform the algebra necessary to evaluate the p factor in closed form (assuming a thin substrate) and to simplify the final result.
Approximation of p From Notes 1 we have p = / / F( θ) φ+ G( θ) cos φ A( θφ, ) θ dθdφ F( θ) φ+ G( θ) cos φ A(, ) θ dθ dφ L cos kx W A( θφ, ) = WL c k y L kx k k x y = k = k θcosφ θφ 3
Assume h/λ Then we have (see Notes 11): ( ) F( θ ) jµ cosθ kh r j G( θ) ( n1 θ) kh ε r Also assume that n1 = εµ r r >> θ This implies that the patch is fairly small (high permittivity substrate) or that the angles of significant radiation are small. Then ( θ ) ( µ r cosθ) G( θ ) ( jµ ) kh F j kh r
We then have ( ) ( ) F + G r kh + φ ( θ) cos φ ( θ ) µ cos θ φ cos φ Therefore p / / ( + ) cos θ φ cos φ A( θφ, ) θ dθdφ ( + ) cos θ φ cos φ A(, ) θ dθ dφ Note: The p factor is now only a function of the patch dimensions not the substrate. 5
The patch array factor is or L cos kx W A( θφ, ) = WL c k y L kx L cos k A( θφ, ) WL c k L kx or x W = y L cos k A( θφ, ) A(, ) c k L kx x W = y A Recall: (,) = WL 6
In the denominator of the p expression we have (after cancelling the A(,) term): / / ( + ) = ( + ) cos θ φ cos φ θ dθ dφ cos θ 1 θ dθ 1 = + 1 3 = 3 7
Hence L / cos kx 3 W p ( cos θ φ cos φ) c k y θ dθ dφ + L kx Ug / / / ( ) = ( ) dθ dφ dθ dφ and factoring out a ( /) -, we have L / / cos kx 3 W p ( cos θ φ cos φ) c k y θdθ dφ + L 1 k x 8
Next, use [Abromowitz & Stegun] x 1+ ax x + ax cos x 1+ bx + bx x= k y x= k x W L where for a a b b x =.1665 =.761 =.967 =.375 Note: These are not Taylor series, but are approximations that are more uniformly accurate over the entire range. 9
The coe term may thus be approximated as cos x 1 bx + + bx 1+ x + x 1 x where we have used use a Taylor series for 1 x 1 We then have (keeping terms up to x ) cos x 16 1+ x b + x b b + + + 1 x 1
Define c b 16 c b b The numerical values are c. 91153 c. 7 7 88 1 Then cos x 1 x 1+ cx + cx 11
We then have p / / 3 + ( cos θ φ cos φ) kw kw 1 + a θφ + a θφ kl 1 cos + c θ φ θ dθ dφ 1
Neglect the following terms: We then have a, aa, c p / / 3 + ( cos θ φ cos φ) kw kw 1 + a ( ) θφ + a + a θφ + c θ φ θdθ dφ kl 1 cos 13
Next, we also neglect the following terms: a a c, c We then have p / / 3 + ( cos θ φ cos φ) kw kw 1 + a ( ) θφ + a + a θφ kl kw kl + c θcosφ + ac θφ θcosφ θ dθ dφ 1
Expanding, we have p / / 3 = { φ cos θ θ + cos φ θ kw 3 kw 3 a + a φ θ cos θ + φ cos φ θ kw 6 5 kw 5 ( a a) ( a a) + + φ θ cos θ + + φ cos φ θ kl 3 c cos φ φ cos θ + kl 3 θ c cos φ θ + kw kl 5 kw kl 5 + ac cos cos cos φ φ θ θ + ac φ φ θ dθ dφ } 15
All of the φ integrals may now be done in closed form: φdφ = φcos φdφ = 16 cos φdφ = 6 φdφ = 5 3 φdφ = 3 16 φcos φdφ = 3 cos φdφ = 3 16 φcos φdφ = 3 16
All of the θ integrals may also be done in closed form: θ dθ = 1 5 θ dθ = 8 15 3 θ dθ = 3 θcos θ dθ = 3 15 cos θθ dθ = 1 3 θcos θ dθ = 5 8 15 17
This yields p 3 = 1 kw 3 kw { + 1+ a + a 3 16 15 16 3 kw 5 8 kw 8 + ( a + a) ( a a) + + 3 15 3 15 kl kl 3 + c c + 16 15 16 3 kw kl 8 kw + ac ac + 3 15 kl 8 3 15 } 18
Simplifying, we obtain p = + a 1 1 kw ( ) 3 + + 56 ( a ) ( ) a kw 1 + c 5 kl ( ) 1 + ac kw kl 7 ( ) ( ) a a c =.1665 =.761 =.91153 19