Antenna Theory Exam No. 1 October 9, 2000

Transcription

1 ntenna Theory Exa No. 1 October 9, 000 Solve the following 4 probles. Each proble is 0% of the grade. To receive full credit, you ust show all work. If you need to assue anything, state your assuptions clearly. Reasonable assuptions that are necessary to solve the proble will be accepted. In all probles assue properties of free space (ε F/, µ H/, η Ω) 1. one wavelength, thin wire dipole antenna is given. ssuing a peak current of 1 at 600 MHz calculate: a. The radiated power of the antenna. b. The radiation resistance of the antenna. c. The directivity and axiu directivity of the antenna Solution: Find the far-field electric and agnetic field intensities. Calculate the tie averaged power density and integrate it over a sphere of radius R to obtain the radiated power. Fro radiated power and current, calculate the radiation resistance. Fro radiated power and power density calculate the radiation intensity and then the directivity of the antenna. Fro Eqs. (18.85) and (18.86): The tie averaged poer density is: E θ j R e j βr cos (βl/)cosθ cos(βl/) H φ ji 0 R e jβr cos (βl/)cosθ cos(βl/) θ φ ji 0 R e jβr cos (βl/)cosθ cos(βl/) j R e jβr cos (βl/)cosθ cos(βl/) R I 0 η cos (βl/)cosθ cos(βl/) 8 R ith Lλ and β/λ we get βl/. P av R I 0 η cos cosθ R Integrating over a spherical surface at radius R: s P av.ds φ0 θ0 I 0 η cos cosθ + 1 R dφ I 0 η 8 R 8 θ0 I 0 η 4 θ0 cos cosθ + 1 cos cosθ + 1 The integral is the sae as in Eq. (18.9) and, fro Table, equals 3.318:

2 b. Fro Eq. (18.36): I 0 η I 0 R rad R rad I Ω c. The definition of directivity is given in Eq. (18.50) and that of radiation intensity in Eq. (18.46): D(θ,φ) U(θ,φ) /4 4U(θ,φ) U(θ,φ) P av R I 0 η cos cosθ sr D(θ,φ) 4U(θ,φ) I 0 η cos cosθ + 1 cos cosθ For axiu directivity we note that the axiu value of the expression in brackets occurs at θ/ and equals. d half wavelength antenna is given. The axiu agnetic field intensity of the antenna in the far field is easured and found to be equal to 1 / (peak). ssuing free space and no losses, calculate the radiated power of the antenna. Radiation is at a 1 wavelength. Solution: Given the agnetic field intensity in the far field we can iediately obtain the electric field intensity and therefore the tie averaged power density. Integrating over a sphere of radius R gives the radiated power. Method : Fro Eq. (18.98): H φ ji cos 0 jβr e cosθ R The electric field intensity is therefore (Eq. (18.97)): φj1e jβr cos cosθ The power density is therefore: E θjηe jβr cos cosθ cos (/)cosθ Rη The radiated power is:

3 η cos (/)cosθ R dφ ηr cos (/)cosθ 1.18ηR R φ0 θ0 θ0 Notes: is the value of the integral taken fro Table.. The radiated power depends on where the agnetic field intensity was easured. Method : n alternative ethod is to note fro the expression for H that I 0 R 1 I 0 R Now, since we know that the radiation resistance of a half wavelength dipole is and using Eq. (18.36) we have: I 0 R rad (R) R 3. Three half wavelength dipoles are placed parallel to each other and are separated a distance h1. The three dipoles are fed with identical currents equal to (peak) at 1. GHz. Find: a. The total radiated power of the array. b. The directivity of the array. Solution: Use the expressions for an n eleent array with n3. a. The electric field intensity of a 3 eleent array of half wavelength dipoles ay be written directly fro Eqs. (18.13) and (18. 97): where in this case, E θ j R e j βr cos (/)cosθ e jψ sin(3ψ/) sin(ψ/) ψ βhcosφ (/λ)(4λ)cosφ 4cosφ where the fact that the wavelength equals 0.5 (λc/f / ) was used to write h4λ. Siilarly, for the agnetic field intensity we write: H φ ji 0 R e jβr cos (/)cosθ The tie averaged power density is therefore: The radiated power is: R cos (/)cosθ sin(3ψ/) 4 R sin(ψ/)

4 1 cos (/)cosθ sin(3ψ/) R dφ cos (/)cosθ sin(3ψ/) 4 φ0 θ0 R sin(ψ/) θ0 sin(ψ/) cos (/)cosθ sin(6cosφ) cos (/)cosθ sin(6cosφ) 40 θ0 sin(cosφ) θ0 sin(cosφ) b. The definition of directivity is given in Eq. (18.50) and that of radiation intensity in Eq. (18.46): D(θ,φ) 4 P av R D(θ,φ) U(θ,φ) /4 4U(θ,φ) U(θ,φ) P av R 4 cos (/)cosθ sin(6cosφ) 4 sin(cosφ) cos (/)cosθ sin(6cosφ) 40 θ0 sin(cosφ) sr θ0 cos (/)cosθ sin(6cosφ) sin(cosφ) cos (/)cosθ sin(6cosφ) sin(cosφ) 4. a. Calculate the ratio between the radiation resistance of a λ and a λ/ dipole, at any given frequency. b. Calculate the ratio between the axiu power density of a λ and a λ/ dipole. c. Based on the previous two responses which antenna is better overall?. ssue lossless antennas Solution: Calculate the radiated power for an arbitrary long antenna (Eq ) and evaluate it for a λ and a λ/ antenna. The ratio is then found. In (b), the sae relation is used. a. Fro Eq. (18.93): R rad η θ θ0 cos (L/λ)cosθ cos(l/λ) The integral is given in Table for Lλ and for Lλ/ as 4.37 and ith these: R rad (λ/) 1.18η, R rad(λ) 4.37η R rad (λ/) 1.18η, R rad (λ) R rad (λ/) 4.37η / 1.18η b. The tie averaged power density for an arbitrary long antenna is given in Eq. (18.89): cos (βl/)cosθ cos(βl/) P av 8 R

5 For the half wavelength dipole we have: cos (/)cosθ P av 8 R The axiu value occurs at θ/ and equals: For the λ dipole: P av ax 8 R P av cos cosθ 1 8 R The axiu value occurs when cos(cosθ) 1. This occurs at θ60. P av ax 8 R 1 1 sin60 8 R 3/ R P av (λ) ax P av (λ/) ax c. Based on the inforation in (a) and (b), the λ antenna is uch better. However, other aspect coe into play including rsadiation resistance, sidelobes and the like.