Antecedentes de satisfacción del cliente

sin²θ sin θdθ (3.69)

On performing the integration over θ, the total radiated power is P_rad = 1

2|I0|²36.54 (3.70)

Equating this to the power dissipated in an equivalent resistor the radiation resistance of a monopole is

R_rad = 36.54 Ω (3.71)

The radiation resistance of a quarter wavelength long monopole place above a ground plane is half that of a half-wave dipole radiating in free space.

The radiation intensity has a maximum value along θ = π/2, and is given by

U_max= η 2

I₀ 2π

² (3.72)

The maximum directivity is calculated by dividing U_max by the average radiation intensity

D = 4πUmax

P_rad = 3.284 (3.73)

Thus, the directivity of a quarter wave monopole above a ground plane is equal to twice that of a half-wave dipole radiating in free space. The maximum directivity occurs along the ground plane and the radiation is vertically polarized.

3.4 Small Loop Antenna

Let us now study another type of antenna called a loop antenna constructed using thin wires. The loop antenna can take different shapes such as a circle, a square, a triangle, etc. The radiation characteristics of a loop antenna depend on the size and shape of the antenna. The circular loop antenna is one of the easiest to construct as well as to analyse.

Consider a single turn circular loop antenna of radius a, symmetrically placed about the origin on the x-y plane, carrying a current Ie(x^, y^, z^) as shown in Fig. 3.8. The wire diameter is assumed to be very small and, hence, we assume the current Ie(x^, y^, z^) to be along the wire. The magnetic vector

z

I0

I0

a

P (x, y, z) P (r, , )

' d'

dl' = ad' Q (a, ') a _

Q (x', y', z') 

' = ^¯/2

R

r

x

y

Fig. 3.8 Geometry of a small loop antenna

potential is an integral over the current distribution given by A = μ

4π



c

I_e(x^, y^, z^)e^−jkR

R dl^ (3.74)

where the path of integration is along the loop. The distance, R, from the source point (x^, y^, z^) to the field point (x, y, z), is given by

R =



(x− x^)²+ (y− y^)²+ (z− z^)² (3.75) Because of the circular shape of the curve, it is convenient to represent the source point in cylindrical coordinates and, as usual, we use spherical coordinates for the field point. To convert the field point (x, y, z) to (r, θ, φ) coordinates, we use the transformation equations

x = r sin θ cos φ (3.76)

y = r sin θ sin φ (3.77)

z = r cos θ (3.78)

which when substituted into Eqn (3.75) yield R =



(r²− 2r(x^sin θ cos φ + y^sin θ sin φ + z^cos θ) + x^2+ y^2+ z^2) (3.79) If the observation point is far away from the antenna, we may expand the above expression using the binomial series and neglect the terms involving

1/r and its higher powers. With this simplification, R is approximated as R r − (x^sin θ cos φ + y^sin θ sin φ + z^cos θ) (3.80) To convert the source point coordinates (x^, y^, z^) to cylindrical coordinates, we can use the transformation equations

x^= a cos φ^ (3.81)

y^= a sin φ^ (3.82)

z^= 0 (3.83)

Substituting the above into Eqn (3.80) and simplifying we get the approxi-mate expression for R as

R r − a sin θ cos(φ − φ^) (3.84) This approximate expression for R is used in the phase term of the vector potential. For the amplitude, the R in the denominator of Eqn (3.74) is replaced by r. The elemental length dl^ along the loop in cylindrical coordi-nates is given by

dl^ = adφ^ (3.85)

For a loop with a circumference which is small compared to the wave-length, we can assume the current to be constant over the loop and, hence, the current element, Iedl^= aφI0adφ^, can be expressed in rectangular coor-dinates as

I_edl^ = (−axI₀sin φ^+ a_yI₀cos φ^)adφ^ (3.86) where I₀ is the amplitude of the current.

Substituting this expression for the current and the far-field approxima-tion for R in the vector potential equaapproxima-tion [Eqn (3.74)], we get

A = μ

4πI₀e^−jkr r



C

)−axsin φ^+ a_ycos φ^*ejka sin θ cos(φ−φ^)adφ^ (3.87)

Now, to convert this expression into spherical coordinates, the unit vec-tors a_x and a_y are written in terms of spherical coordinates. In spherical coordinates the unit vectors a_x and a_y are given by

ax= arsin θ cos φ + a_θcos θ cos φ− aφsin φ (3.88) ay = arsin θ sin φ + aθcos θ sin φ + aφcos φ (3.89)

Substituting these into Eqn (3.87) and simplifying we get three individual components of the vector potential in the spherical coordinate system as

A_r = μ Since the current is φ-symmetric, the vector potential is also φ-symmetric.

Therefore, it is sufficient to evaluate the above integrals at any one value of φ. Without loss of generality, we choose φ = 0.

Let us first consider the evaluation of A_φ. Splitting the limits of integration into two parts Substituting φ^ = φ^+ π in the second integral and using the expansion e^jφ= cos φ + j sin φ, Aφ can be simplified to Since the circumference of the loop is small compared to the wavelength, ka 1 and we can further approximate sin(ka sin θ cos φ^) ka sin θ cos φ^. Thus we have

Aφ μ

4I0jka²sin θe^−jkr

r (3.95)

Using the even and odd properties of the integrands in Eqns 3.90 and 3.91, we can show that A_r= 0 and A_θ= 0 (see Example 3.9). Since the vector potential A has only the a_φ component, in the far-field region the E and H fields computed using the relationships given by Eqns (3.30) and (3.31) result in E_φand H_θ components alone. These are given by

Eφ= ηa²k²

The radiation pattern has a null along the axis of the loop (θ = 0^◦) and the maximum is in the θ = 90^◦ plane. The fields have the same power pattern as that of a Hertzian dipole (see Fig. 2.3).

EXAMPLE3.9

Show that A_r given by Eqn (3.90) is equal to zero.

Solution: Since the current is φ independent, we can evaluate the integral in Eqn (3.90) at φ = 0. The integral is given by

I =

 _2π

φ^=0sin(−φ^)ejka sin θ cos(−φ^)dφ^ Substituting φ^ = ψ− π

I =

 _π

ψ=−πsin(π− ψ)ejka sin θ cos(π−ψ)dψ and simplifying the integrand

I =

 _π

ψ=−πsin(ψ)e−jka sin θ cos(ψ)dψ

Expanding the exponential term using Euler’s formula (Kreyszig 1999) I =

 _π

ψ=−πsin(ψ) cos[ka sin θ cos(ψ)]dψ

− j

 _π

ψ=−πsin(ψ) sin[ka sin θ cos(ψ)]dψ

cos ψ is an even function of ψ and, therefore, cos(ka sin θ cos ψ) is also an even function of ψ. Since sin ψ is an odd function, the integrand in the first integral is an odd function of ψ. Similarly, sin(ka sin θ cos φ) is also an even function. Therefore, the integrand of the second integral is also an odd function of ψ. Using the property of definite integrals

 _a

−af (x)dx = 0 if f (x) is an odd function of x we get

I = 0 which implies that A_r= 0.

Using the procedure outlined in Section 3.1 we can now compute the radiation characteristics of a small loop antenna. For the sake of continuity, we may repeat some of the steps in this section. First, let us compute the time-averaged power density given by

S = 1

2Re(E× H^∗) = 1

2ηa_r|Eφ|² (3.98) Substituting the value of E_φ from Eqn (3.96), this reduces to

S = a_rS = a_rη 2

a²k²|I0| 4r

₂

sin²θ (3.99)

Thus, the radiation intensity is

U (θ, φ) = r²S = η 2

a²k²|I0| 4

₂

sin²θ W/sr (3.100) The total radiated power, P_rad is obtained by integrating the radiation intensity over 4π solid angle of the sphere

P_rad =

 _2π

φ=0

 _π

θ=0

U (θ, φ) sin θdθdφ (3.101) Substituting the value of U from Eqn (3.100) and performing the integration (see Example 3.10), we obtain

P_rad= 10π²a⁴k⁴|I0|² W (3.102) Equating this power to the power dissipated in an equivalent resistance carrying the same current

10π²a⁴k⁴|I0|² = 1

2|I0|²R_rad (3.103) we get the radiation resistance of a loop antenna as

R_rad = 20π²k⁴a⁴ (3.104) Let L_A= πa² be the area of the loop and L_C = 2πa be the circumference of the loop. We can show that the radiation resistance can be written as

R_rad = 20π²

L_C λ

₄

= 31171L²_A

λ⁴ = 320π⁶

a λ

₄

(3.105)

The radiation resistance of a small loop is generally very small and is difficult to match to the source. The radiation resistance can be increased by having more turns in the loop. For example, the radiation resistance of a single turn loop of radius 0.05λ is 1.92 Ω. If the loop is made of N turns and is carrying the same input current, I₀, the loop current would be N I₀. Hence the field strength of the multi-turn loop would be N times that of a single turn loop. Replacing I0 by N I0 in the left hand side of Eqn (3.103), while keeping the same I0 on the right hand side, we get R_rad, N² times that of a single turn loop

R_rad = 20π²N² If the antenna considered in Example above has 10 turns, the radiation resistance becomes 192 Ω. If the loss resistance of a single turn loop is R_loss, for an N -turn loop the loss resistance increases only by a factor of N . Therefore, the multi-turn loop has a higher radiation efficiency compared to a single turn loop.

EXAMPLE3.10

Prove the relation given by Eqn (3.102).

Solution: Substituting the expression for U from Eqn (3.100) into Eqn (3.101)

Substituting the limits and simplifying P_rad = 2π120π

EXAMPLE3.11

What is the total power radiated by a small circular loop of radius 0.5 m carrying a current of 10 A at 15 MHz? If the loop is symmetrically placed at the origin and in the x-y plane, calculate the magnitude of the electric field intensity in the x-y plane at a distance of 10 km.

Solution: The wavelength of the 15 MHz electromagnetic wave propagating in free space is

λ = c

f = 3× 10⁸

15× 10⁶ = 20 m The propagation constant is

k = 2π λ = 2π

20 rad/m

Substituting a = 0.5 m, I₀= 10 A, and k = π/10 rad/m into Eqn (3.102) P_rad = 10π²a⁴k⁴|I0|²= 10π²0.5⁴

π 10

₄

|10|²= 6.01 W From Eqn (3.96)

|Eφ|θ=90^◦ = ηa²k²I₀ 4r

=

376.73× 0.5²×

π 10

₂

× 10 4× 10 × 10³

= 2.32 mV/m EXAMPLE3.12

A large circular loop having a circumference of 1λ is placed in the x-y plane symmetrically about the origin. Derive an expression for the electric field of the wave radiated along the z-direction. Show that the wave is circularly polarized if the current on the loop is a travelling wave given by

I = I0e^{j(ωt−kaφ)}

where ω is the angular frequency, k is the propagation constant of the current and is equal to the propagation constant in free space, a is the radius of the loop, and φ^ is the angle measured from the x-axis.

Solution: For a loop of circumference λ, [a = λ/(2π)]

ka = 2π λ

λ 2π = 1

Therefore, the current in the loop reduces to I = I₀e^−jφe^jωt

The current phasor can be represented as a vector in the Cartesian coordi-nate system [see Eqn (3.86)]

I_e=−axI sin φ^+ a_yI cos φ^

Expressing a_x and a_y in spherical coordinates and simplifying

I_e= I₀e^−jφ[a_rsin θ sin(φ− φ^) + a_θcos θ sin(φ− φ^) + a_φcos(φ− φ^)]

In the far-field region only the transverse components of E and H exist, therefore, it is sufficient to compute A_θ and A_φ components of the vector potential [see Eqns (3.30) and (3.31)]. Substituting the expression for the current distribution, ka = 1, and the far-field approximations to R into Eqn (3.74) for the vector potential, we can write the transverse components of A as

A_θ = μ 4πI0

e^−jkr r cos θ

 _2π

φ^=0sin(φ− φ^)ej sin θ cos(φ−φ^)e^−jφadφ^ A_φ= μ

4πI₀e^−jkr r

 _2π

φ^=0cos(φ− φ^)ej sin θ cos(φ−φ^)e^−jφadφ^

For the fields along the z-axis we can further simplify these expressions by substituting θ = 0.

A_θ = μ

4πI₀e^−jkr r

 _2π

φ^=0sin(φ− φ^)e^−jφadφ^ A_φ= μ

4πI₀e^−jkr r

 _2π

φ^=0

cos(φ− φ^)e^−jφadφ^

Substituting (φ^− φ) = t, we have dφ^= dt and the limits of integration

−φ to 2π − φ. Hence we can write

A_θ =− μ

4πI0ae^−jkr r e^−jφ

 _2π−φ

t=−φ sin(t)e^−jtdt A_φ= μ

4πI₀ae^−jkr r e^−jφ

 _2π−φ

t=−φ cos(t)e^−jtdt

Consider the integral X =

 _2π−φ

t=−φ sin(t)e^−jtdt Using the identity

sin t = e^jt− e^−jt 2j the above integral can be written as

X = 1 2j

 _2π−φ

t=−φ [e^jt− e^−jt]e^−jtdt = 1 2j

 _2π−φ

t=−φ [1− e^−j2t]dt

Performing the indicated integration, substituting the limits and simplifying X = 1

2j



t +e^−j2t 2j

_2π−φ

t=−φ

=−jπ Similarly, we can show that

Y =

 _2π−φ

t=−φ cos(t)e^−jtdt = π

Therefore, the components of the vector potential reduce to A_θ=−μ

4πI₀ae^−jkr

r e^−jφ(−jπ) A_φ= μ

4πI₀ae^−jkr r e^−jφπ

In the far-field region the electric field vectors can be computed from the magnetic vector potential using Eqn (3.30)

E =−jωAt=−jω(aθA_θ+ a_φA_φ) and the components of the electric field are given by

Eθ=−jωAθ= ωμ

4 I0ae^−jkr r e^−jφ Eφ=−jωAφ=−jωμ

4 I0ae^−jkr r e^−jφ Therefore, the electric field can be written as

E = ωμ

4 I0ae^−jkr

r e^−jφ(a_θ− jaφ) This represents a right circularly polarized wave.

Exercises

3.1 Let A_r(r, θ, φ), A_θ(r, θ, φ), and A_φ (r, θ, φ)be the components of a magnetic vector potential,A. In the far-field region these can be expressed as

A_r(r, θ, φ) = A_r0(θ, φ)e^−jkr/r A_θ(r, θ, φ) = A_θ0(θ, φ)e^−jkr/r A_φ(r, θ, φ) = A_φ0(θ, φ)e^−jkr/r whereA_r0(θ, φ),A_θ0(θ, φ), andA_φ0(θ, φ) are independent ofr. Substituting this into B =∇ × A and taking only the radia-ted field components show that

H = jk

μ[a_θA_φ(r, θ, φ)− aφA_θ(r, θ, φ)]

Further, show that this can be expressed as H =−jω

ηa_r× At

where A_t= [a_θA_θ(r, θ, φ) + a_φA_φ(r, θ, φ)]. Using this result and

E = 1

jω∇ × H show that

E =−jωAt

3.2 Derive an expression for the vector effec-tive length of a z-oriented short dipole antenna of lengthl, located at the origin.

Assume a triangular current distribution on the dipole.

3.3 Calculate the radiation resistance of a short dipole (with a triangular current dis-tribution) of length 0.3 m operating at 100 MHz. If the total resistance of the an-tenna is 2.2Ω, calculate the maximum ef-fective area and the radiation efficiency of the dipole. Calculate the open circuit volt-age induced at the terminals of the dipole

if it is oriented along thez-direction and the incident electric field at the dipole is Eⁱ = (a_x4 + a_y3 + a_z5)V/m.

Answer: 1.97Ω, 0.9597 m², 89.5%

3.4 Derive expressions for the radiated elec-tric and magnetic fields of a thin dipole that supports a triangular current distri-bution, located at(x^, y^, z^)and oriented along the (a)x-direction, (b) y-direction, and (c)z-direction.

3.5 Repeat Problem 3.4 replacing the trian-gular current distribution by a sinusoidal current distribution.

3.6 On performing the integration, show that Eqn (3.45) reduces to Eqn (3.46).

3.7 Calculate the radiation efficiency of a half-wave dipole if the loss resistance is 1 Ω. What is its maximum effective aperture if the frequency of operation is 145 MHz?

Answer: 98.65%, 0.552 m² 3.8 Two half-wave dipoles operating at 2.4 GHz are used to establish a wire-less communication link. The antennas are matched to the transmitter and the re-ceiver, respectively. The maximum trans-mit power is 100 mW and for reliable com-munication the received power has to be at least −80 dBm. Calculate the maxi-mum possible distance over which reliable communication can be established using this system.

Answer: 1.63 km 3.9 If one of the dipoles in Problem 3.8 is re-placed by a circularly polarized antenna having the same gain as that of the half-wave dipole, calculate the distance over which the communication link can be es-tablished.

Answer: 1.15 km

3.10 Calculate the maximum value of the elec-tric field intensity at a distance of 1 km from a semi-circular loop of radius 10 cm placed in front of an infinitely large, perfect electrical conductor as shown in Fig. 3.9.

The loop carries a constant current of 4 A at 100 MHz.

Infinite perfect electric conductor

Circular loop

y I

x

z

10 cm

Fig. 3.9 Geometry of a semi-circular loop in front of an infinitely large, perfect electrical conductor

Answer: 16.5 mV/m

3.11 IfR_rad,1andR_loss,1are the radiation and loss resistance of a one-turn small loop, derive an expression for the radiation efficiency of an N-turn loop of identical radius and, hence, show that an N-turn loop is more efficient.

3.12 Calculate the radiation resistance of a 20-turn, 1 m diameter loop antenna oper-ating at 10 MHz. If the loss resistance of a one-turn loop is 1Ω, calculate the radia-tion efficiency. Assume a constant current in the loop.

Answer: 32.16%

In document UNIVERSIDAD CATÓLICA LOS ÁNGELES DE CHIMBOTE FACULTAD DE CIENCIAS CONTABLES, FINANCIERAS Y ADMINISTRATIVAS. ESCUELA PROFESIONAL DE ADMINISTRACIÓN (página 41-46)