Propagation between Antennas

Knife-edge diffraction is the bending of a radio wave around a single sharp obstacle (a hill ridge, building edge) that blocks part of the line-of-sight path. It lets some energy reach a receiver in the "shadow" region, and the amount depends on how much of the first Fresnel zone is blocked.

Geometry

                     /\  obstacle edge
                    /  \
   Tx o------------h----\-------------o Rx
        <--- d1 ---->|<----- d2 ----->
   h = height of edge above the LOS line

Huygens' principle

Every point on the wavefront above the edge acts as a secondary source. These secondary wavelets add up (with different phases) behind the obstacle, so the field in the shadow is weak but not zero.

Fresnel–Kirchhoff diffraction parameter

The obstruction is described by the dimensionless parameter

v = h · √( 2(d1 + d2) / (λ·d1·d2) )

• v < 0: edge is below the LOS line (path clear)
• v = 0: edge just touches the LOS line
• v > 0: edge blocks the LOS line (receiver in shadow)

Diffraction loss

The received field is E = F(v)·E0, where E0 is the free-space field and F(v) is the complex Fresnel integral. The diffraction gain Gd = 20 log|F(v)| is read from a chart or approximated (Lee):

Key points

• When the edge just grazes the LOS (v = 0), the field is halved: a loss of 6 dB.
• For a clear path (v strongly negative) the field oscillates around E0 (±1 dB).
• Loss rises quickly in the shadow, roughly 20 log(0.225/v) for large v.
• For a negligible loss the first Fresnel zone should be at least about 60% clear of obstacles; this is used to fix tower heights in microwave link design.

The Friis transmission equation gives the power received by one antenna from another in free space in terms of the transmitted power, the two antenna gains, the wavelength and the distance. The path loss follows from it by setting both gains to unity.

Friis transmission equation

Consider a transmitter of power Pt feeding an antenna of gain Gt, and a receiving antenna of gain Gr (effective aperture Aer) at distance d in free space. Assume far field, matched polarisation, matched loads and no obstacles.

1. An isotropic source spreads Pt uniformly over a sphere of area 4πd², so the power density at distance d is P_iso = Pt / (4πd²) (W/m²)
2. A directive antenna with gain Gt increases the density in its main direction by Gt: P_d = Pt·Gt / (4πd²)
3. The receiving antenna collects the power falling on its effective aperture: Pr = P_d · Aer = Pt·Gt·Aer / (4πd²)
4. Effective aperture and gain are related by Aer = Gr·λ² / (4π). Substituting:

Pr = Pt·Gt·(Gr·λ²/4π) / (4πd²)

Pr = Pt·Gt·Gr·λ² / (4πd)²

Pr/Pt = Gt·Gr·(λ / 4πd)²

This is the Friis transmission equation. In terms of the two effective apertures (using Gt = 4πAet/λ²) it can also be written as Pr/Pt = Aet·Aer / (λ²d²).

In decibels:

Pr(dBW) = Pt(dBW) + Gt(dB) + Gr(dB) − 20 log(4πd/λ)

Path loss in free space

The free space path loss LFS is the ratio of transmitted to received power when both antennas are isotropic (Gt = Gr = 1), i.e. the loss due to spreading of the wave alone:

LFS = Pt/Pr = (4πd/λ)²

LFS(dB) = 20 log(4πd/λ)
        = 20 log(4π) + 20 log(d/λ)
        = 21.98 + 20 log(d/λ)
        ≈ 22 + 20 log(d/λ) dB

Using λ = c/f, with d in km and f in MHz:

LFS(dB) = 32.44 + 20 log d(km) + 20 log f(MHz)

(With f in GHz the constant becomes 92.44.) The loss grows by 6 dB each time the distance or the frequency is doubled.

Free space propagation is radio communication between a transmitter and receiver that have a clear, unobstructed line-of-sight path, with no reflection, refraction, absorption or diffraction from the earth, atmosphere or objects. The only effect on the signal is the spreading of energy over a larger and larger sphere, so the power density falls as 1/d². Satellite links and terrestrial microwave LOS links (with a clear first Fresnel zone) are close to free space.

Assumptions of the free space model

• Antennas are in each other's far field.
• The medium is lossless, homogeneous and isotropic (εr = μr = 1).
• Polarisations are matched and the antennas are matched to their loads.
• No obstacles in the first Fresnel zone, and no ground reflection.

Derivation of the Friis free space equation

Consider a transmitter of power Pt feeding an antenna of gain Gt, and a receiving antenna of gain Gr (effective aperture Aer) at distance d in free space. Assume far field, matched polarisation, matched loads and no obstacles.

1. An isotropic source spreads Pt uniformly over a sphere of area 4πd², so the power density at distance d is P_iso = Pt / (4πd²) (W/m²)
2. A directive antenna with gain Gt increases the density in its main direction by Gt: P_d = Pt·Gt / (4πd²)
3. The receiving antenna collects the power falling on its effective aperture: Pr = P_d · Aer = Pt·Gt·Aer / (4πd²)
4. Effective aperture and gain are related by Aer = Gr·λ² / (4π). Substituting:

Pr = Pt·Gt·(Gr·λ²/4π) / (4πd²)

Pr = Pt·Gt·Gr·λ² / (4πd)²

Pr/Pt = Gt·Gr·(λ / 4πd)²

This is the Friis transmission equation. In terms of the two effective apertures (using Gt = 4πAet/λ²) it can also be written as Pr/Pt = Aet·Aer / (λ²d²).

In decibels:

Pr(dBW) = Pt(dBW) + Gt(dB) + Gr(dB) − 20 log(4πd/λ)

This shows that the received power falls with the square of distance (6 dB per doubling of d) and, for fixed gains, with the square of frequency.

Free space path loss

The free space path loss LFS is the ratio of transmitted to received power when both antennas are isotropic (Gt = Gr = 1), i.e. the loss due to spreading of the wave alone:

LFS = Pt/Pr = (4πd/λ)²

LFS(dB) = 20 log(4πd/λ)
        = 20 log(4π) + 20 log(d/λ)
        = 21.98 + 20 log(d/λ)
        ≈ 22 + 20 log(d/λ) dB

Using λ = c/f, with d in km and f in MHz:

LFS(dB) = 32.44 + 20 log d(km) + 20 log f(MHz)

(With f in GHz the constant becomes 92.44.) The loss grows by 6 dB each time the distance or the frequency is doubled.

Example: At f = 10 GHz (λ = 0.03 m) and d = 50 km, LFS = 20 log(4π × 50000/0.03) = 146.4 dB.

Path loss is the reduction in power of a radio signal as it travels from the transmitting antenna to the receiving antenna, expressed as the ratio of transmitted power to received power (usually in dB). In free space it is caused only by the spherical spreading of the wave; over real paths, ground reflection, obstacles and the atmosphere add further loss.

Basic idea: spreading of power

           .-~~~-.      sphere of area 4πd²
         /         \
   Tx  (  * ------- )--> Rx (aperture Ae)
         \         /
           '-...-.

An isotropic antenna radiating Pt watts spreads it uniformly over a sphere of radius d:

P_iso = Pt / (4πd²)   W/m²

Effect of antenna gains

A transmitting antenna of gain Gt concentrates the power in its main beam:

P_d = Pt·Gt / (4πd²)

The receiving antenna extracts power in proportion to its effective aperture:

Pr = P_d · Ae ,   with  Ae = Gr·λ² / 4π

Pr = Pt·Gt·Gr·λ² / (4πd)²        (Friis equation)

In dB form:

Pr = Pt + Gt + Gr − 20 log(4πd/λ)     (dBW, dB)

Path loss (with antenna gains)

The overall transmission loss between the transmitter output and the receiver input is

L = Pt/Pr = (4πd/λ)² / (Gt·Gr)

L(dB) = 20 log(4πd/λ) − Gt(dB) − Gr(dB)

Free space (basic) path loss

With isotropic antennas (Gt = Gr = 1) we get the basic free space loss, which depends only on distance and wavelength:

LFS = (4πd/λ)²

LFS(dB) = 20 log(4π) + 20 log d − 20 log λ
        = 21.98 + 20 log(d/λ)

Putting λ = c/f = 3×10⁸/f:

LFS = 20 log(4π/3×10⁸) + 20 log d(m) + 20 log f(Hz)
    = −147.56 + 20 log d(m) + 20 log f(Hz)

With d in km and f in MHz:
LFS = 32.44 + 20 log d(km) + 20 log f(MHz)  dB

Path loss over a flat reflecting earth (two-ray model)

For a real LOS link above ground, the receiver gets a direct ray and a ground-reflected ray. For a perfectly reflecting ground (Γ ≈ −1) and d ≫ ht, hr, the path difference is Δ = 2ht·hr/d, and the received power becomes

Pr = Pt·Gt·Gr · (ht·hr)² / d⁴

L(dB) = 40 log d − 20 log ht − 20 log hr − Gt − Gr

So over a plane earth the loss rises at 40 dB per decade (12 dB per doubling of distance), compared with 20 dB per decade in free space, and it no longer depends on frequency.

Factors that add to path loss in practice

• Obstacles and diffraction (knife-edge loss, Fresnel zone blockage)
• Atmospheric absorption by oxygen and water vapour (above ~10 GHz)
• Rain attenuation (important above ~10 GHz)
• Multipath fading and shadowing in mobile links

Summary

Example: f = 6 GHz, d = 10,500 km gives LFS = 20 log(4π × 1.05×10⁷/0.05) = 188.4 dB.

Free space loss (LFS) is the loss in signal power between two isotropic antennas in free space, caused only by the spreading of the radiated energy over a growing sphere. It is not a loss of energy in the medium (free space is lossless); the receiving antenna simply intercepts a smaller fraction of the total power as distance increases.

Expression

From the Friis equation Pr = Pt·Gt·Gr·(λ/4πd)², putting Gt = Gr = 1:

LFS = Pt/Pr = (4πd/λ)²

LFS(dB) = 20 log(4πd/λ) = 22 + 20 log(d/λ)

LFS(dB) = 32.44 + 20 log d(km) + 20 log f(MHz)

Key features

• Increases by 6 dB each time distance doubles (20 dB per decade).
• Increases by 6 dB each time frequency doubles, because a higher-frequency isotropic antenna has a smaller effective aperture (Ae = λ²/4π).
• It is the reference loss used in every link budget: Pr = Pt + Gt + Gr − LFS − other losses.

Example: For d = 50 km and f = 10 GHz, LFS = 32.44 + 20 log 50 + 20 log 10000 = 146.4 dB.

The Friis transmission equation relates the power received by an antenna to the power transmitted by another antenna in free space, through their gains, the wavelength and the distance between them.

Derivation of the Friis transmission equation

Consider a transmitter of power Pt feeding an antenna of gain Gt, and a receiving antenna of gain Gr (effective aperture Aer) at distance d in free space. Assume far field, matched polarisation, matched loads and no obstacles.

1. An isotropic source spreads Pt uniformly over a sphere of area 4πd², so the power density at distance d is P_iso = Pt / (4πd²) (W/m²)
2. A directive antenna with gain Gt increases the density in its main direction by Gt: P_d = Pt·Gt / (4πd²)
3. The receiving antenna collects the power falling on its effective aperture: Pr = P_d · Aer = Pt·Gt·Aer / (4πd²)
4. Effective aperture and gain are related by Aer = Gr·λ² / (4π). Substituting:

Pr = Pt·Gt·(Gr·λ²/4π) / (4πd²)

Pr = Pt·Gt·Gr·λ² / (4πd)²

Pr/Pt = Gt·Gr·(λ / 4πd)²

This is the Friis transmission equation. In terms of the two effective apertures (using Gt = 4πAet/λ²) it can also be written as Pr/Pt = Aet·Aer / (λ²d²).

In decibels:

Pr(dBW) = Pt(dBW) + Gt(dB) + Gr(dB) − 20 log(4πd/λ)

The equation is valid only in the far field (d ≥ 2D²/λ), for matched polarisation and impedance, and for an unobstructed path. If not, extra factors (polarisation loss factor, reflection loss 1 − |Γ|²) are multiplied in.

Free space loss LFS = 22 + 20 log(d/λ)

Free space loss is the ratio Pt/Pr between two isotropic antennas (Gt = Gr = 1):

LFS = Pt/Pr = (4πd/λ)²

Taking 10 log of both sides:

LFS(dB) = 10 log (4πd/λ)²
        = 20 log(4πd/λ)
        = 20 log(4π) + 20 log(d/λ)
        = 20 × 1.0992 + 20 log(d/λ)
        = 21.98 + 20 log(d/λ)
        ≈ 22 + 20 log(d/λ)  dB

Hence proved. Here d and λ must be in the same units. In practical units, LFS = 32.44 + 20 log d(km) + 20 log f(MHz) dB.

We use the Friis equation in dB form. The radiated power is taken as 150 W (the "150 dB" in the question is read as 150 W, the usual value in this problem).

Given

• d = 50 km = 5×10⁴ m, f = 10 GHz
• Gt = 25 dB, Gr = 20 dB, Pt = 150 W

Step 1: Wavelength

λ = c/f = 3×10⁸ / 10×10⁹ = 0.03 m

Step 2: Free space loss

4πd/λ   = 4π × 5×10⁴ / 0.03 = 2.094×10⁷
LFS(dB) = 20 log(2.094×10⁷) = 146.42 dB

(Check: 32.44 + 20 log 50 + 20 log 10000 = 32.44 + 33.98 + 80 = 146.42 dB.)

Step 3: Transmitted power in dB

Pt = 10 log 150 = 21.76 dBW

Step 4: Received power

Pr = Pt + Gt + Gr − LFS
   = 21.76 + 25 + 20 − 146.42
   = −79.66 dBW
   = −49.66 dBm
Pr = 10^(−7.966) W ≈ 1.08×10⁻⁸ W = 10.8 nW

Answer: Pr ≈ −79.66 dBW (−49.66 dBm), i.e. about 10.8 nW.

(If the transmitted power were literally 150 dBW, the result would simply be 150 + 45 − 146.42 = 48.58 dBW; such a power is unrealistic, so 150 W is used.)

Path loss and received power

Given: f = 10 GHz, Gt = Gr = 25 dB, d = 100 km. The transmitted power is not given, so the received power is found in terms of Pt and also for an assumed Pt = 1 W (0 dBW).

Wavelength:

λ = c/f = 3×10⁸ / 10¹⁰ = 0.03 m

Free space path loss:

4πd/λ   = 4π × 10⁵ / 0.03 = 4.189×10⁷
LFS(dB) = 20 log(4.189×10⁷) = 152.44 dB
Check:  32.44 + 20 log 100 + 20 log 10000
      = 32.44 + 40 + 80 = 152.44 dB

Net link loss including antenna gains:

L = LFS − Gt − Gr = 152.44 − 25 − 25 = 102.44 dB

Received power:

Pr(dBW) = Pt(dBW) + 25 + 25 − 152.44 = Pt(dBW) − 102.44

For Pt = 1 W:
Pr = −102.44 dBW = −72.44 dBm ≈ 5.7×10⁻¹¹ W

Answer: Path loss = 152.44 dB (102.44 dB after the 50 dB of antenna gain); Pr = Pt − 102.44 dB, i.e. −102.44 dBW ≈ 57 pW for Pt = 1 W.

Derivation of the Friis transmission equation

Consider a transmitter of power Pt feeding an antenna of gain Gt, and a receiving antenna of gain Gr (effective aperture Aer) at distance d in free space. Assume far field, matched polarisation, matched loads and no obstacles.

1. An isotropic source spreads Pt uniformly over a sphere of area 4πd², so the power density at distance d is P_iso = Pt / (4πd²) (W/m²)
2. A directive antenna with gain Gt increases the density in its main direction by Gt: P_d = Pt·Gt / (4πd²)
3. The receiving antenna collects the power falling on its effective aperture: Pr = P_d · Aer = Pt·Gt·Aer / (4πd²)
4. Effective aperture and gain are related by Aer = Gr·λ² / (4π). Substituting:

Pr = Pt·Gt·(Gr·λ²/4π) / (4πd²)

Pr = Pt·Gt·Gr·λ² / (4πd)²

Pr/Pt = Gt·Gr·(λ / 4πd)²

This is the Friis transmission equation. In terms of the two effective apertures (using Gt = 4πAet/λ²) it can also be written as Pr/Pt = Aet·Aer / (λ²d²).

In decibels:

Pr(dBW) = Pt(dBW) + Gt(dB) + Gr(dB) − 20 log(4πd/λ)

Effective height (effective length) of an antenna is the height of an imaginary antenna carrying a uniform current equal to the actual feed-point current that would radiate the same field in the direction of maximum radiation. It converts the real, non-uniform current distribution into an equivalent uniform one.

Definition

h_e · I0 = ∫₀^h I(z) dz      (transmitting)

h_e = V_oc / E               (receiving)

where I0 is the base current, I(z) the current along the antenna, V_oc the open-circuit voltage induced and E the incident field. The effective height is always less than the physical height because current tapers to zero at the tip.

Typical values

Use in propagation

• Ground wave field strength from a vertical antenna: E = 120π·h_e·I/(λd) V/m. A larger h_e gives a stronger field, which is why MF broadcast towers use top loading to make the current nearly uniform.
• In space-wave (LOS) links, "effective antenna height" also means the height of the antenna above the average (effective) ground level of the reflecting area, not above the local foot of the tower. This height is used in d = 4.12(√ht + √hr) km and in the plane-earth path loss formula.

Fresnel zones are a series of ellipsoids around the direct line between a transmitter and receiver, such that a path via any point on the boundary of the nth zone is longer than the direct path by nλ/2.

          _..--~~~~~~~~--.._     1st Fresnel zone
   Tx o<-------- d1 --->|<--- d2 ------->o Rx
          ~~--..______..--~~
                        ^ radius r1 here

Radius of the nth zone

At a point that is d1 from the transmitter and d2 from the receiver:

rn = √( n·λ·d1·d2 / (d1 + d2) )

The first zone is largest at mid-path: r1(max) = ½√(λd), where d = d1 + d2.

Significance

• Secondary wavelets from adjacent zones arrive with a phase difference of 180°, so odd zones add and even zones subtract. Most of the energy reaching the receiver travels inside the first Fresnel zone.
• If an obstacle blocks part of the first zone, diffraction loss appears even when the optical LOS is clear. At grazing (obstacle touching LOS) the loss is about 6 dB.
• Design rule: keep at least 60% of the first Fresnel zone (0.6r1) clear of terrain and buildings. This sets tower heights for microwave links, with earth bulge (k = 4/3) also included.

Example: For d = 20 km at 6 GHz (λ = 0.05 m), r1 at mid-path = ½√(0.05 × 20000) = 15.8 m.

Friis transmission equation

Consider a transmitter of power Pt feeding an antenna of gain Gt, and a receiving antenna of gain Gr (effective aperture Aer) at distance d in free space. Assume far field, matched polarisation, matched loads and no obstacles.

1. An isotropic source spreads Pt uniformly over a sphere of area 4πd², so the power density at distance d is P_iso = Pt / (4πd²) (W/m²)
2. A directive antenna with gain Gt increases the density in its main direction by Gt: P_d = Pt·Gt / (4πd²)
3. The receiving antenna collects the power falling on its effective aperture: Pr = P_d · Aer = Pt·Gt·Aer / (4πd²)
4. Effective aperture and gain are related by Aer = Gr·λ² / (4π). Substituting:

Pr = Pt·Gt·(Gr·λ²/4π) / (4πd²)

Pr = Pt·Gt·Gr·λ² / (4πd)²

Pr/Pt = Gt·Gr·(λ / 4πd)²

This is the Friis transmission equation. In terms of the two effective apertures (using Gt = 4πAet/λ²) it can also be written as Pr/Pt = Aet·Aer / (λ²d²).

In decibels:

Pr(dBW) = Pt(dBW) + Gt(dB) + Gr(dB) − 20 log(4πd/λ)

Path loss in free space

With isotropic antennas (Gt = Gr = 1):

LFS = Pt/Pr = (4πd/λ)²
LFS(dB) = 20 log(4πd/λ) = 22 + 20 log(d/λ)
        = 32.44 + 20 log d(km) + 20 log f(MHz)

Line-of-sight distance d = 3.57(√ht + √hr) km

Because of the curvature of the earth, a straight ray from an antenna of height h touches the earth at the radio horizon.

          Tx
          |\   ht
          | \
          |  \  d1
   -------+---*  <- tangent point
     \    |   /
    R \   |  /  R
       \  | /
         O  (centre of earth)

From the right-angled triangle (radius R to the tangent point is perpendicular to the ray):

(R + ht)² = R² + d1²
d1² = 2R·ht + ht²
d1 ≈ √(2R·ht)          (ht ≪ R)

With R = 6370 km = 6.37×10⁶ m and h in metres:

d1 = √(2 × 6.37×10⁶ × ht) m
   = 3569√ht m
   = 3.57√ht km

Similarly for the receiving antenna, d2 = 3.57√hr km. The total LOS distance is

d = d1 + d2 = 3.57(√ht + √hr) km   (ht, hr in m)

This is the optical (geometric) horizon. Allowing for standard atmospheric refraction with effective earth radius kR (k = 4/3), the radio horizon becomes d = 4.12(√ht + √hr) km.

Transmission loss

Transmission loss is the ratio (in dB) of the power fed to the transmitting antenna to the power available at the receiving antenna terminals: L = 10 log(Pt/Pr). In free space it is the spreading loss reduced by the antenna gains, L = 20 log(4πd/λ) − Gt − Gr. With isotropic antennas it equals the basic free space loss LFS = 20 log(4πd/λ). Real links add losses from obstacles, ground, atmosphere, rain and feeders.

Derivation of the Friis transmission equation

Consider a transmitter of power Pt feeding an antenna of gain Gt, and a receiving antenna of gain Gr (effective aperture Aer) at distance d in free space. Assume far field, matched polarisation, matched loads and no obstacles.

1. An isotropic source spreads Pt uniformly over a sphere of area 4πd², so the power density at distance d is P_iso = Pt / (4πd²) (W/m²)
2. A directive antenna with gain Gt increases the density in its main direction by Gt: P_d = Pt·Gt / (4πd²)
3. The receiving antenna collects the power falling on its effective aperture: Pr = P_d · Aer = Pt·Gt·Aer / (4πd²)
4. Effective aperture and gain are related by Aer = Gr·λ² / (4π). Substituting:

Pr = Pt·Gt·(Gr·λ²/4π) / (4πd²)

Pr = Pt·Gt·Gr·λ² / (4πd)²

Pr/Pt = Gt·Gr·(λ / 4πd)²

This is the Friis transmission equation. In terms of the two effective apertures (using Gt = 4πAet/λ²) it can also be written as Pr/Pt = Aet·Aer / (λ²d²).

In decibels:

Pr(dBW) = Pt(dBW) + Gt(dB) + Gr(dB) − 20 log(4πd/λ)

From this, the transmission loss is

L = Pt/Pr = (4πd/λ)² / (Gt·Gr)
L(dB) = 32.44 + 20 log d(km) + 20 log f(MHz) − Gt − Gr

The received power falls as 1/d² (6 dB per doubling of distance). Doubling the frequency adds 6 dB of free space loss, but if dish antennas of fixed size are used, both gains rise by 6 dB, so high-frequency links with dishes actually gain overall.

Maximum power is received when the antennas are matched and aligned in free space, so the Friis equation applies directly. The receiving gain is found from the dish size and efficiency.

Given

• d = 15.5 km, f = 3.4 GHz, Pt = 250 W
• Gt = 25 dB = 10^2.5 = 316.23
• Receiving dish D = 6.4 m, η = 0.75

Step 1: Wavelength

λ = c/f = 3×10⁸ / 3.4×10⁹ = 0.08824 m

Step 2: Effective aperture and gain of the receiving dish

Ae = η·πD²/4 = 0.75 × π × 6.4² / 4 = 24.13 m²
Gr = 4π·Ae/λ² = 4π × 24.13 / 0.08824²
   = 3.894×10⁴  = 45.90 dB

Step 3: Received power (Friis, aperture form)

Pr = Pt·Gt·Ae / (4πd²)
   = 250 × 316.23 × 24.13 / (4π × (15500)²)
   = 1.9075×10⁶ / 3.0191×10⁹
   = 6.32×10⁻⁴ W

Check in dB

LFS = 20 log(4πd/λ) = 126.88 dB
Pt  = 10 log 250    = 23.98 dBW
Pr  = 23.98 + 25 + 45.90 − 126.88 = −31.99 dBW
    = −1.99 dBm ≈ 0.632 mW

Answer: Maximum received power ≈ 6.32×10⁻⁴ W = 0.632 mW (−31.99 dBW, −1.99 dBm).

Under free space conditions the Friis equation gives the path loss and received power.

Given

• Gt = Gr = 40 dB, f = 10 GHz, d = 90 km, Pt = 10 kW

Step 1: Wavelength

λ = c/f = 3×10⁸ / 10¹⁰ = 0.03 m

Step 2: Free space (transmission path) loss

4πd/λ = 4π × 9×10⁴ / 0.03 = 3.770×10⁷
LFS   = 20 log(3.770×10⁷) = 151.53 dB

Check: 32.44 + 20 log 90 + 20 log 10000
     = 32.44 + 39.08 + 80 = 151.53 dB

Net loss after the antenna gains: 151.53 − 40 − 40 = 71.53 dB.

Step 3: Received power

Pt = 10 log(10⁴) = 40 dBW
Pr = Pt + Gt + Gr − LFS
   = 40 + 40 + 40 − 151.53
   = −31.53 dBW = −1.53 dBm
Pr = 10^(−3.153) W ≈ 7.04×10⁻⁴ W

Answer: Path loss = 151.53 dB (71.53 dB including antenna gains); received power ≈ −31.53 dBW = 0.704 mW.

Using the Friis transmission equation in dB form.

Given

• d = 0.5 km = 500 m, f = 1 GHz
• Gt = 25 dB, Gr = 20 dB, Pt = 150 W

Step 1: Wavelength

λ = c/f = 3×10⁸ / 10⁹ = 0.3 m

Step 2: Free space loss

4πd/λ = 4π × 500 / 0.3 = 2.094×10⁴
LFS   = 20 log(2.094×10⁴) = 86.42 dB

Check: 32.44 + 20 log 0.5 + 20 log 1000
     = 32.44 − 6.02 + 60 = 86.42 dB

Step 3: Transmit power in dBm

Pt = 10 log(150 / 1 mW) = 10 log(1.5×10⁵) = 51.76 dBm

Step 4: Received power

Pr = Pt + Gt + Gr − LFS
   = 51.76 + 25 + 20 − 86.42
   = 10.34 dBm
Pr = 10^1.034 mW ≈ 10.8 mW

Answer: Received power ≈ 10.34 dBm (about 10.8 mW).

The Friis transmission equation gives the power received by an antenna from a transmitting antenna in free space, in terms of the transmitted power, antenna gains, wavelength and separation.

Pr = Pt·Gt·Gr·(λ / 4πd)²
Pr(dBW) = Pt + Gt + Gr − 20 log(4πd/λ)

Derivation in brief

1. Power density from an antenna of gain Gt at distance d: Pt·Gt/(4πd²).
2. Power picked up by the receiving antenna = density × effective aperture Aer.
3. Aer = Gr·λ²/4π. Substituting gives the equation above.

Assumptions / conditions

• Far-field separation (d ≥ 2D²/λ).
• Free space: no obstacles, reflections or absorption.
• Matched polarisation and impedances; otherwise multiply by the polarisation loss factor and (1 − |Γt|²)(1 − |Γr|²).

Uses

• Link budget of satellite and microwave LOS links.
• Defines the free space loss LFS = 20 log(4πd/λ) = 22 + 20 log(d/λ) dB.
• Antenna gain measurement (two- and three-antenna methods).

Received power falls as 1/d², i.e. 6 dB for every doubling of distance.

Using the Friis transmission equation Pr = Pt·Gt·Gr·(λ/4πd)².

Given

• f = 100 MHz, d = 0.75 km = 750 m, Pt = 120 W
• Gt = 30 dB = 1000, Gr = 25 dB = 316.23

Step 1: Wavelength

λ = c/f = 3×10⁸ / 10⁸ = 3 m

Step 2: Free space loss

4πd/λ = 4π × 750 / 3 = 3141.6
LFS   = 20 log(3141.6) = 69.94 dB

Step 3: Received power (ratio form)

Pr = 120 × 1000 × 316.23 × (1/3141.6)²
   = 3.795×10⁷ / 9.870×10⁶
   = 3.84 W

Check in dB

Pt = 10 log 120 = 20.79 dBW
Pr = 20.79 + 30 + 25 − 69.94 = 5.85 dBW = 35.85 dBm
10^0.585 = 3.84 W

Answer: Received power ≈ 3.84 W (5.85 dBW).

Derivation of free space path loss

Free space path loss is the ratio of transmitted to received power between two isotropic antennas in free space, caused by spherical spreading of the wave.

1. An isotropic antenna radiating Pt produces power density at distance d: P_d = Pt / (4πd²)
2. An isotropic receiving antenna has effective aperture Ae = λ²/4π.
3. Received power:

Pr = P_d · Ae = Pt/(4πd²) · λ²/(4π) = Pt·(λ/4πd)²

4. The path loss is therefore

LFS = Pt/Pr = (4πd/λ)² = (4πdf/c)²

LFS(dB) = 20 log(4πd/λ)
        = 22 + 20 log(d/λ)
        = 32.44 + 20 log d(km) + 20 log f(MHz)
        = 92.44 + 20 log d(km) + 20 log f(GHz)

With real antennas of gains Gt and Gr, the received power is Pr = Pt + Gt + Gr − LFS (dB), which is the Friis equation.

Numerical: f = 6 GHz, d = 10,500 km

λ = c/f = 3×10⁸ / 6×10⁹ = 0.05 m
4πd/λ = 4π × 1.05×10⁷ / 0.05 = 2.639×10⁹
LFS = 20 log(2.639×10⁹) = 188.43 dB

Check: 92.44 + 20 log 10500 + 20 log 6
     = 92.44 + 80.42 + 15.56 = 188.43 dB

Answer: Free space path loss ≈ 188.4 dB. (This is typical of a satellite uplink at C-band.)

Fresnel zones

Fresnel zones are concentric ellipsoids around the direct Tx–Rx line. A path through any point on the boundary of the nth zone is longer than the direct path by nλ/2.

          _..--~~~~~~~~--.._     1st zone
   Tx o<------- d1 ------>|<----- d2 ---->o Rx
          ~~--..______..--~~

Radius of the nth zone at distances d1 and d2 from the ends:

rn = √( n·λ·d1·d2 / (d1 + d2) )

• Waves from successive zones differ in phase by 180°, so most useful energy travels inside the first zone.
• Blocking part of the first zone causes diffraction loss even if the optical path is clear. Links are designed to keep at least 0.6r1 clear.

Knife-edge diffraction

When a single sharp obstacle (ridge, building edge) blocks the path, the wave bends over the edge into the shadow region (Huygens' principle: each point on the wavefront above the edge acts as a new source). The obstruction is measured by the Fresnel–Kirchhoff parameter

v = h·√( 2(d1 + d2) / (λ·d1·d2) )

where h is the height of the edge above the LOS line (negative if below).

The two ideas are linked: v = √2·h/r1, so the loss depends on how much of the first Fresnel zone the edge covers.