Phased Array Antenna: Calculating Intensity and Radiation Patterns with Four Antennas, Assignments of Quantum Physics

A problem solution for calculating the intensity and radiation patterns of a phased array antenna with four antennas. The problem involves antennas emitting coherently with the same intensity and wavelength, initially in phase. How the interference effects create directional antennas, with maxima only in the north and south directions due to the path length difference being an integer multiple of the wavelength. The document also discusses the effect of introducing a relative phase between adjacent antennas and calculates the intensity in each direction for the new phasor diagrams.

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Physics 214 Problem 5 Week 2
Phased Array
We can use interference effects to make a very directional antenna (either for transmitting or
receiving). As an example, consider the following situation: 4 radio antennas are equally spaced
in a line, separated by intervals of length d. The antennas all emit radiation coherently, with the
same intensity I1 and the same wavelength λ. Initially the antennas all emit in phase.
1. For concreteness, assume d = λ/4. Draw phasor diagrams corresponding to the net amplitude
far away from the antennas in the directions indicated. Calculate the intensity in each of these
directions. Hint: In considering radiation to the east, the signal from antenna #1 takes longer to
arrive, so its phasor will lag behind that of antenna #2, which will lag behind that of antenna #3,
etc.
N,S: φ = 2πδ/λ If we are far away, then in the up (N) or down (S) directions, all the path lengths
are essentially equal. Therefore, we have δ = 0 = φ:
Atot = 4A1 Itot = 16I1
E: φ = 2πδ/λ In this case, δ = r1 r2 = d = λ/4. Therefore, φ = 2π(λ/4)/λ = π/2:
Atot = 0 Itot = 0.
W: φ = 2πδ/λ In this case, δ = r1 r2 =d = −λ/4. Therefore, φ = 2π(−λ/4)/λ = −π/2:
Atot = 0 Itot = 0.
To the west, the antenna #2 signal lags the antenna #1 signal, so A2 will lag behind A1, and so
on.
Solution
A1
A2
A3
A4
A1
A2
A3
A4
pf3

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Physics 214 Problem 5 Week 2

Phased Array

We can use interference effects to make a very directional antenna (either for transmitting or

receiving). As an example, consider the following situation: 4 radio antennas are equally spaced

in a line, separated by intervals of length d. The antennas all emit radiation coherently, with the

same intensity I 1

and the same wavelength λ. Initially the antennas all emit in phase.

  1. For concreteness, assume d = λ/4. Draw phasor diagrams corresponding to the net amplitude

far away from the antennas in the directions indicated. Calculate the intensity in each of these

directions. Hint: In considering radiation to the east, the signal from antenna #1 takes longer to

arrive, so its phasor will lag behind that of antenna #2, which will lag behind that of antenna #3,

etc.

N,S: φ = 2πδ/λ If we are far away, then in the up (N) or down (S) directions, all the path lengths

are essentially equal. Therefore, we have δ = 0 = φ:

A

tot

= 4A

1

 I

tot

= 16I

1

E: φ = 2πδ/λ In this case, δ = r 1

  • r 2

= d = λ/4. Therefore, φ = 2π(λ/4)/λ = π/2:

A

tot

= 0  I

tot

W: φ = 2πδ/λ In this case, δ = r 1

  • r 2

=−d = −λ/4. Therefore, φ = 2π(−λ/4)/λ = −π/2:

A

tot

= 0  I

tot

To the west, the antenna #2 signal lags the antenna #1 signal, so A 2

will lag behind A 1

, and so

on.

Solution

A

1

A

2

A

3

A

4

A

1

A

2

A

3

A

4

  1. Since d < λ, there are no other principal maxima except along the N and S directions. Why?

The only way to get a principal maximum is if the path length difference is an integer multiple of

the wavelength. Since d is less than one wavelength, there’s only 1 principal maximum (on each

side, i.e., one N and one S).

Thus, with four antennas, the power is emitted primarily into two lobes (even though each

antenna individually is emitting into all directions). Similarly, if this array were acting as a

receiver, it would primarily detect signals from the North and South. By using more antennas,

the directionality would become sharper.

Now we use slightly more sophisticated electronics to introduce a relative phase Δφ = π/

between adjacent antennas, as shown below. That is, if the time-dependence of antenna 1 is

cos(ωt), the time-dependence of antenna 2 is cos(ωt + π/2), the time-dependence of antenna 3 is

cos(ωt + π), etc. We will see that this has the effect of changing the direction of the primary

radiation lobes.

  1. Draw new phasor diagrams corresponding to the net amplitude far away from the antennas in

the directions indicated. Calculate the intensity in each of these directions.

N,S: As before, δ = 0. But now each phasor is rotated counterclockwise by φ i

A

tot

= 0  I

tot

φ 2

− φ 1

= π/

A

1

A

2

A

3

A

4