

Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
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.
Typology: Assignments
1 / 3
This page cannot be seen from the preview
Don't miss anything!


d d d
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.
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:
tot
1
tot
1
E: φ = 2πδ/λ In this case, δ = r 1
= d = λ/4. Therefore, φ = 2π(λ/4)/λ = π/2:
tot
tot
W: φ = 2πδ/λ In this case, δ = r 1
=−d = −λ/4. Therefore, φ = 2π(−λ/4)/λ = −π/2:
tot
tot
To the west, the antenna #2 signal lags the antenna #1 signal, so A 2
will lag behind A 1
, and so
on.
1
2
3
4
1
2
3
4
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.
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
tot
tot
φ 2
− φ 1
= π/
1
2
3
4