Direction finding is the problem of locating a transmitter from the signal it emits: a radio beacon, a jammer, a lost aircraft’s emergency locator, or, in the L19 Lab, an HB100 hidden in a box. We do not control the transmitter, and we do not know its power or its antenna pattern. The one thing we can measure reliably is the direction its wave comes from, because that direction sets the times at which the wave reaches each element of an array.

A direction measured from one place is a bearing. A bearing alone says which way to look but not how far. A radar gets range from the delay of its own echo, but a direction finder only listens, so there is no delay to measure. The received power does not give range either, since it depends on the transmitter’s power and pattern, which are unknown. Distance comes instead from geometry: two bearings taken from two known positions cross at the transmitter. The first half of this lesson measures a bearing; the second half turns bearings into a position.

L18 started from this picture. A plane wave arriving from \(\theta_s\), measured from broadside, travels an extra \(d\sin\theta_s\) to reach each successive element, so neighboring elements see the wave with a phase difference of \(kd\sin\theta_s\), and element \(n\) sees it \(n\ kd\sin\theta_s\) ahead of element 0. In L18 we chose the steer angle and computed the ramp to command. Here the source chooses the ramp, and the source’s direction is the unknown.

If we could measure the phase on each element, the bearing would follow from L18’s inverse problem: difference neighboring phases, unwrap, and take

Some direction finders work this way. They are interferometers: two or more antennas, each with its own receiver, and a direct phase comparison between them. The PHASER cannot do it with eight elements, because the two ADAR1000s add their four elements each into one channel before anything is digitized (L17, “The Hybrid Beamformer”). The individual element phases never reach the computer. What reaches it is a sum, and the sum depends on the phase we command.

Steer the array to \(\theta_0\) and each element’s phase shifter delays its signal by \(n\ kd\sin\theta_0\). The arriving wave put a lead of \(n\ kd\sin\theta_s\) on element \(n\), so what is left on element \(n\) after the delay is \(n\psi\), with

The eight signals then add to \(\vert AF_N(\psi)\vert\), the array factor from L16 with \(\psi\) in place of its argument. It is largest, 1 after normalizing, at \(\psi = 0\), where the commanded delays exactly cancel the arrival lead and all eight signals add in phase. Any other steer angle leaves a residual ramp, and the sum falls off along the main lobe, through its nulls, and over the sidelobes.

The beam sweep steps \(\theta_0\) across the field of view and records the sum at each step. The figure shows what it records for a source at 30°: the curve is the array pattern, centered on the source. This is the same function the L18 Lab measured, read the other way, and the curve L16 drew with the grid and the noise floor in “The Measured Sweep”. There the source sat still and the curve showed where the beam pointed; here the beam moves and the peak shows where the source is.

At 10.525 GHz, \(\lambda = 28.5\ \text{mm}\) and the array’s half-power beamwidth at broadside is

It is tempting to conclude that a bearing is only good to about 13°, but the beamwidth limits something else. Finding the peak of one smooth lobe is a location estimate, and how much noise jitters the curve sets its precision, not how wide the curve is. The standard estimate for a beam that scans across a source is

The lobe’s slope sets it. Noise adds a random voltage to every sample of the sweep, about \(1/\sqrt{2\ \text{SNR}}\) of the peak voltage, and the slope of the lobe turns that voltage error into an angle error:

The constant \(k\) is that slope in units of \(1/\theta_\text{HP}\). Textbook trackers quote \(k \approx 1.6\); the course array’s lobe, normalized to 1 at its peak, has a slope of \(1.05/\theta_\text{HP}\) at its half-power points. With the array’s own slope, an SNR of 30 dB (a factor of 1000) gives

and the textbook constant gives 0.2°. Either way the noise term is a fraction of the 1.05° grid step at 30° (next frame). An HB100 two meters away is tens of decibels above the noise, so noise is not what limits the bearing.

The beamwidth sets resolution: the ability to tell two sources apart. Two sources closer than about one beamwidth produce lobes that overlap into a single peak somewhere between them, and the sweep reports one wrong bearing instead of two right ones. The figure shows both cases for two equal sources: 20° apart the sweep has two peaks with a dip between them, and 8° apart it has one peak midway between the sources. For the course array the first null sits \(\lambda/Nd = 0.254\) away in \(\sin\theta\), 14.7° at broadside, which is the usual measure of the separation two sources need. Only one HB100 is on during the fox hunt, so the sweep sees one source.

The sweep cannot set any steer angle it likes. The ADAR1000’s phase shifters have 7 bits, so the progressive phase moves in whole steps of \(360^\circ/128 = 2.8125^\circ\) (L18, “Phase-Shifter Resolution”). At the HB100’s frequency, \(kd = 176.8^\circ\), so one step of progressive phase moves the beam by a fixed 0.0159 in \(\sin\theta\). The same step in \(\sin\theta\) is a larger step in angle away from broadside, by the factor \(1/\cos\theta\) in the slide’s second line, the same stretching that broadens the steered beam in L18.

The Angle readout is the steer angle of the largest sample, so it can only land on the grid. Unless the source happens to sit on a grid point, the reported angle is the nearest one, off by up to half a step. In the figure the source is at 30.0° and Angle reads 29.5°. This is one reason to prefer a source within about ±30° of broadside: the grid step and the beamwidth grow as \(1/\cos\theta\), and the gain falls as \(\cos\theta\) (L18, “Scan Loss”). At 45° the step is 1.29° and the beam is about 18° wide, and 45° is where the PHASER starts to lose tracking.

The phase ramp depends only on the extra path between neighboring elements, and that path is \(d\) times the component of the arrival direction along the line of elements. Put \(\hat{\mathbf x}\) along the element row, \(\hat{\mathbf y}\) along broadside, and \(\hat{\mathbf z}\) up. A unit vector \(\hat{\mathbf u}\) toward a source at azimuth az, measured from broadside in the horizontal plane, and elevation el has a horizontal part of length \(\cos(\text{el})\), the adjacent side of the elevation angle, and a vertical part \(\sin(\text{el})\). The horizontal part splits into \(\cos(\text{el})\sin(\text{az})\) along the row and \(\cos(\text{el})\cos(\text{az})\) along broadside. In L18 the source was in the horizontal plane and the component along the row was \(\sin\theta_s\); here it is

That is where the cosine comes from: tipping the arrival direction out of the horizontal plane shortens its projection onto the row. Every direction on a cone around the row has the same projection, and so the same reading, which is why a line array measures a cone angle and not a floor-plan azimuth.

When the source is at the array’s height, el is zero and the two angles agree. Below it, the reading is pulled toward broadside, and at classroom distances the pull is not small. The figure shows it in two flat views. A kit on a table 0.9 m (3 ft) above a box on the floor 1.5 m (5 ft) out sees the box 31.0° down, so the arrival direction reaches out only \(\cos 31^\circ = 0.86\) of the way horizontally (side view). On the floor plan the box is still 30° off broadside, but the horizontal projection of its direction, the unit arrow toward the box with its vertical part dropped, \((\sin\text{az}\cos\text{el},\ \cos\text{az}\cos\text{el},\ 0)\), is only \(\cos 31^\circ = 0.86\) long. Its part along the row is \(0.86 \times 0.500 = 0.429\) instead of 0.500 (top view). The array cannot tell that shortened projection from a level source at the angle whose part along the row is also 0.429, \(\arcsin 0.429 = 25.4^\circ\), the dashed line in the top view, and so it reports 25.4°: a 4.6° error, three times the whole bearing budget. The cosine scales the sine of the angle, not the angle itself: \(30^\circ \times \cos 31^\circ = 25.7^\circ\) comes out close only because \(\sin\theta \approx \theta\) for small angles, and the gap grows off broadside. At 45° it reads 37.3°. At 3.0 m (10 ft) out the box is 16.7° down and 30° reads 28.6°. The error is zero at broadside, grows off it, and always pulls toward broadside. On a firing line that shows up mostly as depth: five stations 0.9 m above a box at \((0.25,\ 1.5)\ \text{m}\) put the uncorrected fix 0.25 m too far downrange.

Two things remove it. Putting the kits at the boxes’ height makes el zero. When that is not possible, undo it. With the height difference \(\Delta h\) known and a station’s horizontal range \(r\) to the fix, \(\text{el} = \arctan(\Delta h/r)\), and each reading converts back to a floor-plan angle with

The range comes from a first, uncorrected fix: correct each reading, solve again, and repeat once. In the example above one pass brings the fix within 0.04 m of the box and a second within 0.01 m.

Aiming the board does not help. Tipping it up or down toward the boxes turns it about the row and leaves the reading alone, because the cone is centered on the row. Raising one end of the row tilts the cone and adds the same kind of error, so keep the row level.

The cone has a back half as well, a source behind the array at the mirror angle, but the PHASER’s patches sit over a ground plane and receive almost nothing from behind, so a bearing is always in front of the array.

A whiteboard, a metal cabinet, or a person near the line of sight sends a second copy of the signal toward the array from a different direction. The array adds both, so the sweep is the sum of two patterns, one centered on the source and a weaker one centered on the reflector, with a relative phase set by the extra path the reflection travels. When the two directions are within about a beamwidth of each other, the lobes merge and the peak moves toward the reflection. The worst case over every phase, for a reflection 5° to 15° from the direct path, is a shift of 0.6° for a reflection 20 dB down, 2.0° at 10 dB down, and 3.2° at 6 dB down. The figure shows the −10 dB case at its worst, a reflection about 12° from the direct path.

A reflection more than about 20° away forms its own lobe and moves the main peak much less. It takes over the bearing if someone stands between the kit and the source: the direct lobe drops, the reflection’s lobe is now the largest, and Angle reports the direction of the whiteboard. A bearing several degrees off from the others most often comes from a reflection that has taken over the peak.

The Angle readout measures the angle from the array’s broadside, which is wherever the kit happens to face. To draw the bearing on a floor plan we need it in the room’s frame. Measure angles counterclockwise from a fixed room axis, seen from above. If broadside points along heading \(h\) and positive Angle is the counterclockwise side, the room bearing is

Which physical side of the kit is positive depends on which way the kit faces; the L18 Lab found it by moving the source toward element E8. Get it wrong and a source at 30° lands on the plan at −30°, a 60° error, so check it with one shot at a known position before trusting any bearing.

Each term of \(\beta = h + \theta_\text{est}\) carries an error, and the error in \(\beta\) is the two combined. The table is the budget for one bearing: one row per independent error, the term of the equation it enters through, where it comes from, and its one-sigma size. The heading error adds directly to the bearing error, and in the room it is the largest term. Laying the array’s edge along a taped line holds \(h\) to about 1°. The grid term is the worst case, half a step, rather than the 0.30° standard deviation of an error spread evenly over a 1.05° step, which keeps the budget conservative. The grid, multipath, and noise terms make up \(\sigma_\theta\), the error in the Angle readout itself, and all four are independent, so they add as a root sum of squares, the chain on the slide. Setting the heading by eye, about 3°, raises the total to 3.2°.

A bearing error is an angle, but the fix is a position, so what matters is how far the error moves the line where the source is. A line rotated by \(\sigma_\beta\) about the kit moves sideways, at range \(R\), by the cross-range error

At the budget’s 1.5° that is 0.05 m at 2 m; at \(\sigma_\beta = 2^\circ\), a little worse, it is 0.03 m at 1 m, 0.07 m at 2 m, and 0.14 m at 4 m. The error grows in proportion to range, which is why a closer station gives a tighter line. It is also all in one direction. A bearing says nothing about how far along the line the source is, so its uncertainty is a narrow strip, \(2\delta_\perp\) wide at one standard deviation and unbounded along its length. A second bearing’s strip is what limits the fix along the first one, and how well it does that depends on the angle between them, “The Crossing Angle” below.

On the L19 Lab’s firing range every station faces straight downrange, so all the stations share one heading and the bearing line takes a simpler form. Put the origin at the middle of the tape, the station 0 mark, with \(x\) along the tape and \(y\) downrange, and take Angle \(\theta\) positive toward the \(+x\) end of the tape. A station at \(x_i\) sits at \((x_i,\ 0)\) and faces \(+y\). A source at \((x,\ y)\) is \(y\) downrange of it and \(x - x_i\) across, the two legs of the dotted right triangle in the figure, and \(\theta_i\) is the angle at the station between downrange and the source, so

Solving for \(x\), a reading \(\theta_i\) at station \(x_i\) puts the source on the line

Every point on that line fits the reading, so one bearing is one equation in the two unknowns \(x\) and \(y\): it says which way to look, but not how far. A second bearing from another station is a second equation, and the two lines cross at the source.

Two bearings cross exactly, but three or more never quite meet: every bearing carries some error, so three lines form a small triangle. To pick the point that comes closest to all of them, we first need a way to say how far a point is from one line. Vectors make that one line of algebra.

Start from the previous frame’s triangle. From station \(i\) at \(\mathbf p_i = (x_i,\ 0)\) to the source, the legs are \(x - x_i\) across and \(y\) downrange, and dividing each by the hypotenuse gives \(\sin\theta_i\) and \(\cos\theta_i\). So the unit vector that points from the station along the bearing is

and turning it 90° clockwise gives the unit vector across the line, toward \(+x\),

Now take any point \(\mathbf x = (x,\ y)\) and the arrow \(\mathbf x - \mathbf p_i\) from the station to it. That arrow splits into a part along \(\hat{\mathbf u}_i\), how far down the line the point is, and a part along \(\hat{\mathbf n}_i\), how far off the line it is. The dot product with a unit vector picks out the part along it, so the point’s perpendicular miss is

the green segment in the figure, in meters and signed: positive on the \(+x\) side of the line. Setting \(e_i = 0\) gives \(\tan\theta_i = (x - x_i)/y\), the previous frame’s line, so the bearing line is exactly the set of points with no miss.

Another way to see the same arithmetic is as a change of coordinates. Subtracting \(\mathbf p_i\) moves the origin to station \(i\), and dotting with \(\hat{\mathbf u}_i\) and \(\hat{\mathbf n}_i\) reads the point’s coordinates on axes turned by \(\theta_i\), one along the bearing and one across it. In that frame \(\hat{\mathbf u}_i\cdot(\mathbf x - \mathbf p_i)\) is how far down range the point is and \(e_i\) is its cross-range coordinate, the same direction the cross-range error of “The Bearing Error Budget” measures. The bearing line is where that cross-range coordinate is zero, and least squares looks for the one point whose cross-range coordinate is nearly zero in every station’s frame at once.

With two bearings the crossing is exact. A third bearing adds a third line, and because every bearing carries some error, the three lines form a small triangle rather than meeting at a point. The figure magnifies it: no point lies on all three lines, so no point has all three misses \(e_i\) of the previous frame equal to zero. The fix has to be a compromise.

The usual compromise is least squares: choose the point that makes the sum of the squared misses, \(e_1^2 + e_2^2 + e_3^2\), as small as it can be. Squaring keeps a miss on either side of a line from canceling one on the other side, gives one smooth bowl with a single bottom, and counts a large miss more heavily than two small ones, which suits bearings whose errors are mostly small.

There is a physical picture of where the bottom of that bowl is. Tie the candidate point to each line with a spring that pulls it straight toward the line, along \(\hat{\mathbf n}_i\), with a force proportional to the miss \(e_i\). The point comes to rest where the pulls balance,

and that resting point is exactly the least-squares fix: the arrows in the figure are those pulls, and they add to zero. The picture also says what a bad bearing does. A reflection that bends one line pulls the fix toward it, and because the other lines pull back, that line ends up with the largest miss, which is how “Residuals and Outliers” below finds it.

Nobody solves this by hand in the fox hunt. The L19 Lab’s pre-lab has you write it as a short Python function around numpy.linalg.lstsq and test it on a supplied set of bearings; “The General Form” next shows what that call computes, and the two frames after it work an example.

Unweighted least squares treats every line as equally reliable. A bearing from a kit far from the source is less reliable in meters, as the cross-range error showed, so if the ranges differ a lot, weighting each term by \(1/r_i^2\) gives the better fix. In the fox hunt every bearing that counts is within about ±45° of broadside, so the ranges from the stations to one box differ by at most a factor of \(\sqrt{2}\), the weights by a factor of 2, and the unweighted form is close enough.

Each station’s bearing line is one linear equation in the unknown point. Its miss from “Distance From a Line”, \(e_i = (x - x_i)\cos\theta_i - y\sin\theta_i\), is zero on the line, and moving the known part to the right-hand side writes the line as

Write one such equation per station, one under the other, as on the slide. In matrix form this is \(\mathbf A\mathbf x = \mathbf b\):

\(\mathbf A\) holds the coefficients of \(x\) and \(y\), one row per station; each row is that station’s \(\hat{\mathbf n}_i\). \(\mathbf b\) holds the right-hand sides. With \(N\) stations \(\mathbf A\) is \(N\times 2\), and \(\mathbf A\mathbf x - \mathbf b\) is the list of misses \((e_1,\ e_2,\ \dots,\ e_N)\).

With two stations the two equations have one exact solution, the crossing of “A Two-Bearing Fix”. With three or more there are more equations than unknowns, and no point satisfies them all. The least-squares point satisfies two combinations of them instead, the normal equations,

They say the pulls of “Many Bearings: Least Squares” balance, and the next frame expands \(\mathbf A^{\mathsf T}(\mathbf A\mathbf x - \mathbf b)\) once to show it. They leave two equations in two unknowns, which always have one answer unless the lines are all parallel.

numpy.linalg.lstsq(A, b) solves the normal equations in one line, and the frames after the next build \(\mathbf A\) and \(\mathbf b\) for three real readings and solve them.

\(\mathbf A^{\mathsf T}\) is \(\mathbf A\) turned on its side: the rows of \(\mathbf A\), one per station, become its columns. Multiplying a matrix by a column vector takes each row of the matrix, multiplies it entry by entry with the vector, and adds. The first row of \(\mathbf A^{\mathsf T}\) holds the three \(\cos\theta_i\), so the first entry of the product is \(e_1\cos\theta_1 + e_2\cos\theta_2 + e_3\cos\theta_3\); the second row holds the three \(-\sin\theta_i\), which gives the second entry. Read by columns instead, the same product is each station’s \(\hat{\mathbf n}_i\) scaled by its own miss \(e_i\), added up, which is exactly the sum of the three arrows in the least-squares figure: each line pulls the point toward itself along \(\hat{\mathbf n}_i\), harder the more it misses.

At the least-squares fix those pulls cancel, so \(\mathbf A^{\mathsf T}(\mathbf A\mathbf x - \mathbf b) = 0\). Multiplying out, \(\mathbf A^{\mathsf T}\mathbf A\ \mathbf x - \mathbf A^{\mathsf T}\mathbf b = 0\), which is the normal equations. \(\mathbf A^{\mathsf T}\mathbf A\) is \(2\times 2\) and \(\mathbf A^{\mathsf T}\mathbf b\) has two entries whatever the number of stations, so the system always comes down to two equations in \(x\) and \(y\).

For those who have had calculus, the same equations come from asking where \(S = \sum_i e_i^2\) is smallest. \(S\) is a bowl over the floor, and at its bottom a small step in \(x\) or in \(y\) does not change it, so both slopes are zero. Each \(e_i = \cos\theta_i\ x - \sin\theta_i\ y - x_i\cos\theta_i\) is linear, so by the chain rule

which are the two entries of \(2\mathbf A^{\mathsf T}\mathbf e\). The bottom of the bowl and the balance of the pulls are the same point.

Stations at \(x_i = -0.75\), 0, and +0.75 m on the tape read Angle 34.6°, 8.8°, and −17.7°; each \(\theta_i\) is that station’s Angle readout, measured from broadside and positive toward the \(+x\) end of the tape, not a room bearing. Building the system takes three numbers per station. Its row of \(\mathbf A\) is \(\cos\theta_i\) and \(-\sin\theta_i\), the coefficients of \(x\) and \(y\) in its line equation, and its entry of \(\mathbf b\) is \(x_i\cos\theta_i\); station 1, for example, gives \(\cos 34.6^\circ = 0.823\), \(-\sin 34.6^\circ = -0.568\), and \(-0.75 \times 0.823 = -0.617\). Stack the three rows and the three entries and the system is the one on the slide: three equations, two unknowns.

To check by hand: \(\mathbf A^{\mathsf T}\mathbf A\) sums \(\cos^2\theta_i\), \(-\cos\theta_i\sin\theta_i\), and \(\sin^2\theta_i\) over the three stations, and \(\mathbf A^{\mathsf T}\mathbf b\) sums \(x_i\cos^2\theta_i\) and \(-x_i\cos\theta_i\sin\theta_i\); for the first entry, \(0.823^2 + 0.988^2 + 0.953^2 = 2.562\). The normal equations are

The determinant is \(2.562 \times 0.438 - 0.329^2 = 1.014\), and the \(2\times 2\) inverse gives

The box was at \((0.25,\ 1.50)\ \text{m}\), so the fix lands 0.01 m from it. The misses, \(\mathbf A\mathbf x - \mathbf b\), are −0.016, 0.028, and −0.015 m, all under each station’s own \(R_i\tan\sigma_\beta\) of about 0.04 m at 1.5°: three good bearings (the check in “Residuals and Outliers” below). In Python the whole frame is two lines: build A and b from the readings as on the previous frame, then numpy.linalg.lstsq(A, b).

Choosing where to stand is part of the measurement, because the geometry sets the error as much as the instrument does. On a firing line every bearing looks downrange, so the lines cross steeply only when the stations are far apart compared with the depth, and depth is the direction the fix measures worst.

Take two stations a baseline \(B\) apart and a box midway between them at depth \(y\), a range \(R\) from each. Each line moves sideways by \(\delta_\perp = R\sigma_\beta\), with \(\sigma_\beta\) in radians, and the two lines cross at \(\gamma\), with

The two lines sit symmetrically about the downrange direction, so the axes of the error ellipse point cross-range and downrange. The eigenvalues of \(\mathbf M\) are \(2\cos^2(\gamma/2)/\delta_\perp^2\) in cross-range and \(2\sin^2(\gamma/2)/\delta_\perp^2\) in depth, and the depth error is the inverse square root of the second:

With the taped-heading budget, \(\sigma_\beta = 1.5^\circ\), and the L19 Lab’s station marks 0.75 m apart, the end stations, 3 m apart, fix a box 2 m downrange to 0.08 m in depth, stations 1.5 m apart to 0.11 m, and neighboring stations, 0.75 m apart, to 0.20 m. In cross-range the same pairs fix it to 0.04 to 0.06 m. The depth error grows as \(R^2\) and falls as \(1/B\), so the stations farthest apart that still detect the fox give the best fix.

Spreading the stations has a limit: a station far to one side sees the box well off broadside. Within ±30° the grid step and the beamwidth stay near their broadside values. At ±45° the step is 1.29°, the beam is about 18° wide, the gain is 1.5 dB down (L18, “Scan Loss”), and the PHASER starts to lose tracking. A station that sees a box farther out than that gives no bearing worth using, and the HB100’s own beam, aimed only roughly at the line, may leave it too weak a tone to read anyway.

Two systematic errors from earlier in the lesson remain. The cone-angle error pulls every bearing toward broadside when the source is above or below the array. Put the kits at the boxes’ height, or correct each reading as “The Cone Angle” describes, and keep each element row level. A reflector near the line of sight shifts the peak or replaces it, so keep the path from each kit to the box field clear, people included. Finally, two bearings always cross somewhere, right or wrong. A third bearing is the only check that the first two are consistent.

The geometry also sets how close two boxes can be and still be told apart. A simulation of a row of boxes, with the fix assigned to the nearest box, names the right one 99% of the time at 0.75 m spacing with two kits 4 m out crossing at 90° and \(\sigma_\beta = 2^\circ\), and 93% of the time at 0.5 m. The L19 Lab’s boxes sit at least 0.75 m apart for the same reason.