Question E9C14:
How does the radiation pattern of a horizontally-polarized antenna mounted above a long slope compare with the same antenna mounted above flat ground?  

Choices:  
A: The main lobe takeoff angle increases in the downhill direction  
B: The main lobe takeoff angle decreases in the downhill direction  
C: The horizontal beamwidth decreases in the downhill direction  
D: The horizontal beamwidth increases in the uphill direction    

Correct Answer is  B: The main lobe takeoff angle decreases in the downhill direction  

Correct answer: B

1) Why B is correct
- A long, uniform slope acts like a tilted ground plane. In the downhill direction the ground “falls away,” which makes the antenna’s effective height larger along that path, so the first elevation lobe is launched at a lower elevation angle (smaller takeoff angle).
- Two-ray model with a horizontal antenna over ground (reflection coefficient ≈ −1 at low angles) gives maxima when:
  $$2kh\sin\theta = (2m+1)\pi \quad (m=0,1,2,\dots)$$
  For the first lobe:
  $$\sin\theta_0 \approx \frac{\lambda}{4h}$$
  A slope by angle $\alpha$ replaces $\theta$ with $\theta \pm \alpha$ (downhill uses “+”):
  $$2kh\sin(\theta \pm \alpha) = (2m+1)\pi$$
  So for the first lobe:
  $$\theta_{0,\text{down}} \approx \arcsin\!\left(\frac{\lambda}{4h}\right) - \alpha$$
  Which is a smaller angle than over flat ground; thus the takeoff angle decreases downhill.

2) Why the other choices are incorrect
- A: “Takeoff angle increases downhill” — opposite of what happens; downhill decreases the angle by about the slope angle.
- C: “Horizontal beamwidth decreases downhill” — the slope primarily tilts the elevation pattern; it does not materially change azimuth (horizontal) beamwidth for a long, uniform slope.
- D: “Horizontal beamwidth increases uphill” — same issue as C; the main effect is a tilt in elevation, not a change in azimuth beamwidth.

3) Useful formulas / rules of thumb
- First elevation-lobe angle over flat ground (horizontal pol.):
  $$\sin\theta_0 \approx \frac{\lambda}{4h}$$
  Higher $h$ lowers $\theta_0$.
- Over a uniform slope of angle $\alpha$:
  $$\theta_{0,\text{down}} \approx \arcsin\!\left(\frac{\lambda}{4h}\right) - \alpha,$$
  $$\theta_{0,\text{up}} \approx \arcsin\!\left(\frac{\lambda}{4h}\right) + \alpha.$$
- Rule of thumb: A long, even slope tilts the elevation pattern by roughly the slope angle; downhill lowers and uphill raises the takeoff angle.