VI. Local Properties of Radiation

Transcription

1 VI. Local Properties of Radiation Kirchhoff-Huygens Approximation Error for Shadowing by a Circular Window Relation to Fresnel Zone September 3 3 by H.L. Bertoni 1

2 Kirchhoff-Huygen Approximation y α dz dy r α E V E H Field Point (x,y,x) z Secondary Source Plane x = x Scalar approximation for fields incident in the y z plane jkr in jke EVH, ( x, y, z) EVH, (, y, z )( cosα + cosα ) dy dz 4πr September 3 3 by H.L. Bertoni

3 Example of Plane Wave Incidence - To show that the Kirchhoff-Huygen s method works correctly - For simplicity assume wave propagates along x ( α = ) ( ) = in E x, y, z E e jkx E in y r = x + ( ρ ) Without loss of generality' dθ ρ α take the field point to be on the xaxis ( x,, ). z x jkr jkr jke Ex E r dy dz jk e (,, ) = ( 1+ cos α) ' = E ( 1 + cos α) ρ dρ dθ' 4π 4π r jkr jk e = E ( 1 + cosα) ρ dρ r September 3 3 by H.L. Bertoni 3 π

4 Approximation for Integration Over ρ Re exp( jkr ) ρ r ρ r = x + ( ρ') For large ρ, subsequent half cycles of integrand cancel. Primary contribution to integral comes from vicinity of ρ =. x+3λ x+λ x+λ x ρ September 3 3 by H.L. Bertoni 4

5 Approximation for Integration Over ρ - cont. For x >> ρ, cosα 1 so that ( ) = Ex,, jke exp jk x + ρ ρ dρ Let u = jk x + ( ρ ), then du = jk and x + ρ ( ) = = x ρ dρ u jkx Ex,, E e du Ee, which is the plane wave field. jkx ( ) + ρ ( ) ( ) September 3 3 by H.L. Bertoni 5

6 Local Properties of Propagation Quantify the region about a ray through which the fields propagate y ( ) = Source at d,, with ZI 1 jkr in e E = f ( α ) r Without loss of generality, take ( ) field point at s,, along x axis. What region in the y z plane is responsible for wave propagation from the source to observation point? z x= d r α r α x=s x To answer the question, assume plane x = is opaque with transparent hole and examine the transmitted field. Question is then: 1. What location gives minimum distortion?. What is the minimum size of the hole for a given distortion? September 3 3 by H.L. Bertoni 6

7 How to find the Hole Position and Size By symmetry, hole must be centered on the x - axis and be circular. To create the hole, assume the plane has transmission coefficient T( ρ ). where ρ = ( y ) + ( z ), with T( )= 1and T( )=. To avoid diffraction that can confuse results, use continuously varying function for T( ρ ). September 3 3 by H.L. Bertoni 7

8 How to find the Hole Position and Size - cont. ρ ( ) Simple dependence for T ρ is w [ ] T( ρ )= exp ( ρ ) / w. 1/e 1 Then π jkr e Es f T jk e jkr (,, )= ( α ) ( ρ )( cosα+ cos α ) r r d θρ ' d ρ 4π T(ρ ) E in pattern da spherical wave September 3 3 by H.L. Bertoni 8

9 Evaluation of Integral for E(s,,) Integration over θ gives π so that E(s,,) = e r' - jkr' jkr f( α') T( ρ')(cosα+ cos α') jk e ρ' dρ' r Ray optical regime when d, s>> λ. Primary contribution to integral comes from the vicinity of ρ = ( )= () cosα+ cosα = and f α f. exp{ jk( r' + r Es (,, ) = jkf ( ) T( ρ') )} ρ' dρ' rr ' September 3 3 by H.L. Bertoni 9

10 Evaluation of Integral for E(s,,) - cont. For ds, >> λ we may further approximate rand r' using : In exponent ( ) + ( ) r = s + ρ s ρ s ( ) + ( ) r = d + ρ d ρ d In denominator r s and r d Then Es (,, ) = jkf ( ) exp ( ρ' ) w exp jk ( s d) or { } + + jk( d + s) 1 w (,, ) = jkf ( ) exp + Es e ds ( ρ') s + jk d s ( ρ ) ρ dρ ds + ( ρ') ρ' dρ' d sd 1 + Let ρ' = u + jk d s w ds September 3 3 by H.L. Bertoni 1 u

11 Evaluation of Integral for E(s,,) - cont. jk e Es (,, ) = f( ) ds jk d s exp u udu 1 + ( ) + w ds 1 Since exp u udu exp u Es,, f ( ) = ( ) e jk( d+ s) d + s jk( d+ s) ( ) = { ( )} = 1 sd 1 j kw s + d ( ) 1 Field at (s,,) in the absence of the window (w = ) Correction due to the presence of the window sd λsd Let ε = be a measure of error. kw s + d πw s d ( ) = ( + ) Then for wlarge, ε will be small and correction will be 1 + j ε. September 3 3 by H.L. Bertoni 11

12 Window Size in Terms of Fresnel Zones w Fn r r d s Definition of Fresnel ellipse ( r + r) ( d+ s)= nλ / for d, s, r d w d w / d >> λ = +( ) + ( ) Fn Fn r s w s w / s = +( ) + ( ) Fn Fn Fresnel zone ellipse is approximately wf = nλ / or d s 1 w F n wf n nλds/( d s). Substituting into the expression for ε gives ε. nπ w ( ) = + = If w= w, then ε= 1/ nπ and the amplitude correction is F n ε 1 1+ jε = 1+ ε 1+ = 1+. If n = 1, the error is 5%. n π September 3 3 by H.L. Bertoni 1 n

13 Fresnel Zone Size at Cellular Frequencies w Fn d s s+ d = R ( w ) = nλds/ ( d+ s) F n Maximum width when d = s= R/, in which case the full width is w F n = nλr If n = 1, the error is 5% and the maximum width of the Fresnel zone for R = 1 km is : at f = 9 MHz, λ = 1/ 3 m and w = m at f = 18. GHz, λ = 1/ 6m and w = 19. m September 3 3 by H.L. Bertoni 13 F1 F1

14 Fresnel Zone for Incident Plane Wave Near one end of a link from a distant source w Fn s For d >> s, then for n = 1, ( w ) = λds/( d + s) λs and F 1 w = F 1 λs September 3 3 by H.L. Bertoni 14