Question E9C03:
What type of radiation pattern is created by two 1/4-wavelength vertical antennas spaced 1/2-wavelength apart and fed in phase?  

Choices:  
A: Omni-directional  
B: Cardioid  
C: A figure-eight broadside to the axis of the array  
D: A figure-eight end-fire along the axis of the array    

Correct Answer is  C: A figure-eight broadside to the axis of the array  

Correct answer: C — A figure-eight broadside to the axis of the array

Why C is correct:
- Two identical 1/4-wave verticals spaced $d = 0.5\lambda$ and fed in phase ($\psi = 0$) form a classic 2-element broadside array.
- In the azimuth plane, the array factor for two elements is:
  $$AF(\phi) = 2\cos\left(\frac{k d \cos\phi + \psi}{2}\right), \quad k=\frac{2\pi}{\lambda}$$
  With $d=0.5\lambda$ and $\psi=0$: $k d = \pi$, so
  $$AF(\phi) = 2\cos\left(\frac{\pi}{2}\cos\phi\right)$$
- Broadside (perpendicular to the line between the antennas) is at $\phi=90^\circ$: $\cos\phi=0 \Rightarrow AF=2$ (maximum).
- Along the axis ($\phi=0^\circ$): $\cos\phi=1 \Rightarrow AF=2\cos(\pi/2)=0$ (nulls).
- This produces a bidirectional “figure-eight” pattern broadside to the array axis.

Why the other choices are incorrect:
- A: Omni-directional  
  Not omni: there are clear nulls along the array axis and maxima broadside. A single vertical is omni in azimuth; a spaced pair is not.
- B: Cardioid  
  A cardioid (single deep null, predominantly unidirectional) typically requires about $0.25\lambda$ spacing with a specific phase shift (often near $90^\circ$) or unequal amplitudes. In-phase at $0.5\lambda$ gives symmetric bidirectional, not cardioid.
- D: Figure-eight end-fire along the axis  
  End-fire needs a progressive phase shift to cancel the path difference so maxima occur along the axis. For $d=0.5\lambda$, end-fire requires $\psi \approx \pm\pi$ (180°), not $\psi=0$.

Useful formulas and rules of thumb:
- Path difference to a far point at angle $\phi$ from the array axis: $\Delta = d\cos\phi$; corresponding phase difference $k\Delta$.
- Two-element array factor:
  $$AF(\phi) = 2\cos\left(\frac{k d \cos\phi + \psi}{2}\right)$$
- Broadside vs. end-fire:
  - Fed in phase ($\psi=0$) → broadside maxima.
  - Fed with $\psi \approx -k d$ → end-fire toward $+\phi=0^\circ$.
- With $d=0.5\lambda$:
  - In-phase ($\psi=0$) → nulls on-axis, maxima broadside (this question).
  - $180^\circ$ out of phase ($\psi=\pi$) → end-fire along the axis.