The \( \alpha -\beta -\gamma \) filter

Example 1 – Weighting the gold

At the time \( n \), the estimate \( \hat{x}_{n,n} \) would be the average of all previous measurements:

The dynamic model in this example is static (or constant) since the weight of gold doesn't change over time, therefore \( \hat{x}_{n+1,n}= \hat{x}_{n,n} \).

Although the above equation is mathematically correct, it is not practical for implementation. In order to estimate \( \hat{x}_{n,n} \) we need to remember all historical measurements; therefore, we need a large memory. We also need to recalculate the average repeatedly if we want to update the estimated value after every new measurement. Thus, we need a more powerful CPU.

It would be more practical to keep the last estimate only (\( \hat{x}_{n-1,n-1} \)) and update it after every new measurement. The following figure exemplifies the required algorithm:

• Estimate the current state based on the measurement and prior prediction.
• Predict the next state based on the current state estimate using the Dynamic Model.

We can modify the averaging equation for our needs using a small mathematical trick:

\( \hat{x}_{n-1,n-1} \) is the estimated state of \( x \) at the time \( n - 1\), based on the measurement at the time \( n-1 \).

Let's find \( \hat{x}_{n,n-1} \) (the predicted state of \( x \) at the time \( n \)) , based on \( \hat{x}_{n-1,n-1} \) (the estimation at the time \( n - 1 \)). In other words, we would like to extrapolate \( \hat{x}_{n-1,n-1} \) to the time \( n \).

Since the dynamic model in this example is static, the predicted state of \( x \) equals the estimated state of \( x \): \( \hat{x}_{n,n-1} = \hat{x}_{n-1,n-1} \).

Based on the above, the estimate of the current state \( \hat{x}_{n,n} \), can be written as follows:

The above equation is one of the five Kalman filter equations. It is called the State Update Equation. It means the following:

The factor \( \frac{1}{n} \) is specific to our example. We will discuss the vital role of this factor later, but right now, I would like to note that in "Kalman Filter language," this factor is called the Kalman Gain. It is denoted by \( K_{n} \). The subscript \( n \) indicates that the Kalman Gain can change with every iteration.

The discovery of \( K_{n} \) was one of Rudolf Kalman's significant contributions.

Before we get into the guts of the Kalman Filter, we use the Greek letter \( \alpha _{n} \) instead of \( K_{n} \).

So, the State Update Equation looks as follows:

The term \( \left( z_{n}- \hat{x}_{n,n-1} \right) \) is the "measurement residual," also called innovation. The innovation contains new information.

In this example, \( \frac{1}{n} \) decreases as \( n \) increases. In the beginning, we don't have enough information about the current state; thus, the first estimation is based on the first measurement \( \frac{1}{n}|_{n=1} = 1 \). As we continue, each successive measurement has less weight in the estimation process, since \( \frac{1}{n} \) decreases. At some point, the contribution of the new measurements will become negligible.

Let's continue with the example. Before we make the first measurement, we can guess (or rough estimate) the gold bar weight simply by reading the stamp on the gold bar. It is called the Initial Guess, and it is our first estimate.

The Kalman Filter requires the initial guess as a preset, which can be very rough.

Estimation algorithm

The following chart depicts the estimation algorithm that is used in this example.

Now we are ready to start the measurement and estimation process.

The numerical example

Iteration Zero

Initialization

Our initial guess of the gold bar weight is 1000 grams. The initial guess is used only once for the filter initiation. Thus, it won't be required for successive iterations.

\[ \hat{x}_{0,0}=1000g \]

Prediction

The weight of the gold bar is not supposed to change. Therefore, the dynamic model of the system is static. Our next state estimate (prediction) equals the initialization:

\[ \hat{x}_{1,0} = \hat{x}_{0,0}=1000g \]

First Iteration

Step 1

Making the weight measurement with the scales:

\[ z_{1}= 996g \]

Step 2

Calculating the gain. In our example \( \alpha_{n}= \frac{1}{n} \) , thus:

\[ \alpha_{1}= \frac{1}{1}=1 \]

Calculating the current estimate using the State Update Equation:

\[ \hat{x}_{1,1}= \hat{x}_{1,0}+ \alpha _{1} \left( z_{1}- \hat{x}_{1,0} \right) = 1000+1 \left( 996-1000 \right) = 996g \]

Step 3

The dynamic model of the system is static; thus, the weight of the gold bar is not supposed to change. Our next state estimate (prediction) equals to current state estimate:

\[ \hat{x}_{2,1}= \hat{x}_{1,1}=996g \]

Second Iteration

After a unit time delay, the predicted estimate from the previous iteration becomes the prior estimate in the current iteration:

\[ \hat{x}_{2,1}=996g \]

Step 1

Making the second measurement of the weight:

\[ z_{2}= 994g \]

Step 2

Calculating the gain:

\[ \alpha _{2}= \frac{1}{2} \]

Calculating the current estimate:

\[ \hat{x}_{2,2}= \hat{x}_{2,1}+ \alpha_{2} \left( z_{2}- \hat{x}_{2,1} \right) =996+\frac{1}{2} \left( 994-996 \right) = 995g \]

Step 3

\[ \hat{x}_{3,2}= \hat{x}_{2,2}=995g \]

Third Iteration

\[ z_{3}= 1021g~~~~~~~~~~~~~~~~~~~~~~~~~~ \alpha_{3}= \frac{1}{3} \] \[ \hat{x}_{3,3}=~ 995+\frac{1}{3} \left( 1021-995 \right) =1003.67g \] \[ \hat{x}_{4,3}=1003.67g \]

Fourth Iteration

\[ z_{4}= 1000g~~~~~~~~~~~~~~~~~~~~~~~~~~ \alpha_{4}= \frac{1}{4} \] \[ \hat{x}_{4,4}= 1003.67+\frac{1}{4} \left( 1000-1003.67 \right) =1002.75g \] \[ \hat{x}_{5,4}=1002.75g \]

Fifth Iteration

\[ z_{5}= 1002g~~~~~~~~~~~~~~~~~~~~~~~~~~ \alpha_{5}= \frac{1}{5} \] \[ \hat{x}_{5,5}= 1002.75+\frac{1}{5} \left( 1002-1002.75 \right) =1002.6g \] \[ \hat{x}_{6,5}=1002.6g \]

Sixth Iteration

\[ z_{6}= 1010g~~~~~~~~~~~~~~~~~~~~~~~~~~ \alpha_{6}= \frac{1}{6} \] \[ \hat{x}_{6,6}= 1002.6+\frac{1}{6} \left( 1010-1002.6 \right) =1003.83 \] \[ \hat{x}_{7,6}=1003.83g \]

Seventh Iteration

\[ z_{7}=983g~~~~~~~~~~~~~~~~~~~~~~~~~~ \alpha_{7}= \frac{1}{7} \] \[ \hat{x}_{7,7}= 1003.83+\frac{1}{7} \left( 983-1003.83 \right) =1000.86g \] \[ \hat{x}_{8,7}=1000.86g \]

Eighth Iteration

\[ z_{8}=971g~~~~~~~~~~~~~~~~~~~~~~~~~~ \alpha_{8}= \frac{1}{8} \] \[ \hat{x}_{8,8}= 1000.86+\frac{1}{8} \left( 971-1000.86 \right) =997.125g \] \[ \hat{x}_{9,8}=997.125g \]

Ninth Iteration

\[ z_{9}=993g~~~~~~~~~~~~~~~~~~~~~~~~~~ \alpha_{9}= \frac{1}{9} \] \[ \hat{x}_{9,9}= 997.125+\frac{1}{9} \left( 993-997.125 \right) =996.67g \] \[ \hat{x}_{10,9}=996.67g \]

Tenth Iteration

\[ z_{10}=1023g~~~~~~~~~~~~~~~~~~~~~~~~~~ \alpha_{10}= \frac{1}{10} \] \[ \hat{x}_{10,10}= 996.67+\frac{1}{10} \left( 1023-996.67 \right) =999.3g \] \[ \hat{x}_{11,10}=999.3g \]

We can stop here. The gain decreases with each measurement. Therefore, the contribution of each successive measurement is lower than the contribution of the previous measurement. We get pretty close to the true weight, which is 1000g. If we were making more measurements, we would get closer to the true value.

The following table summarizes our measurements and estimates, and the chart compares the measured values, the estimates, and the true value.

Results analysis

The following chart compares the true, measured, and estimated values.

The estimation algorithm has a smoothing effect on the measurements and converges toward the true value.

The required number of measurements

What should be the number of measurements? Should we stop after 10, 100, or maybe 1000 measurements?

The required number of measurements depends on the desired estimation precision. We will have the answer to this question once we've gone through the book's second part. However, the inquisitive reader can sneak a peek at the Appendix J of the book.

Example summary

In this example, we've developed a simple estimation algorithm for a static system. We have also derived the state update equation, one of the five Kalman Filter equations. We will revise the state update equation in the next chapter.

Example 2 – Tracking the constant velocity aircraft in one dimension

Thus, the velocity is a derivative of the range:

For discrete measurements:

The radar sends a track beam in the direction of the target at a constant rate. The track-to-track interval is \( \Delta t \).

Two motion equations describe the system dynamic model for constant velocity motion:

According to these equations, the aircraft range at the next tracking cycle equals the range at the current track cycle plus the target velocity multiplied by the track-to-track interval. Since we assume constant velocity in this example, the velocity at the next cycle equals the velocity at the current cycle.

The above system of equations is called a State Extrapolation Equation (also called a Transition Equation or a Prediction Equation) and is also one of the five Kalman filter equations. This system of equations extrapolates the current state to the next state (prediction).

We have already used the State Extrapolation Equation in the previous example, where we assumed that the weight at the next state equals the weight at the current state.

The State Extrapolation Equations depend on the system dynamics and differ from example to example.

There is a general form of the State Extrapolation Equation in matrix notation. We will learn it later.

In this example, we use the above equations specific to our case.

Now we are going to modify the State Update Equation for our example.

The \( \alpha - \beta \) filter

Let the radar track-to-track ( \( \Delta t \) ) period be 5 seconds. Assume that at time \( n - 1 \), the estimated range of the UAV (Unmanned Aerial Vehicle) is 30,000m, and the estimated UAV velocity is 40m/s.

Using the State Extrapolation Equations, we can predict the target position at time \( n \):

\[ \hat{x}_{n,n-1}= \hat{x}_{n-1,n-1}+ \Delta t\hat{\dot{x}}_{n-1,n-1}=30000+5\times40=30200m \]

The target velocity prediction for time \( n \):

\[ \hat{\dot{x}}_{n,n-1}= \hat{\dot{x}}_{n-1,n-1}=40m/s \]

However, at time \( n \), the radar measures range ( \( z_{n} \) ) of 30,110m and not 30,200m as expected. There is a 90m gap between the expected (predicted) range and the measured range. There are two possible reasons for this gap:

• The radar measurements are not precise
• The aircraft velocity has changed. The new aircraft velocity is: \( \frac{30,110-30,000}{5}=22m/s \)

Which of the two statements is true?

Let us write down the State Update Equation for the velocity:

The value of the factor \( \beta \) depends on the precision level of the radar. Suppose that the \( 1 \sigma \) precision of the radar is 20m. The 90 meters gap between the predicted and measured ranges most likely results from a change in the aircraft velocity. In this case, we should set the \( \beta \) factor to a high value. If we set \( \beta \) = 0.9, then the estimated velocity would be:

\[ \hat{\dot{x}}_{n,n}= \hat{\dot{x}}_{n,n-1}+ \beta \left( \frac{z_{n}-\hat{x}_{n,n-1}}{ \Delta t} \right) =40+0.9 \left( \frac{30110-30200}{5} \right) =23.8m/s \]

On the other hand, suppose that the \( 1 \sigma \) precision of the radar is 150m. Then the 90 meters gap probably results from the radar measurement error. In this case, we should set the \( \beta \) factor to a low value. If we set \( \beta \) = 0.1, then the estimated velocity would be:

\[ \hat{\dot{x}}_{n,n}= \hat{\dot{x}}_{n,n-1}+ \beta \left( \frac{z_{n}-\hat{x}_{n,n-1}}{ \Delta t} \right) =40+0.1 \left( \frac{30110-30200}{5} \right) =38.2m/s \]

If the aircraft velocity has changed from 40m/s to 22m/s, we see this after 10 track cycles (running the above equation 10 times with \( \beta \) = 0.1). If the gap has been caused by measurement error, then the successive measurements would be in front or behind the predicted positions. Thus on average, the target velocity would not change.

The State Update Equation for the aircraft position is similar to the equation that was derived in the previous example:

Unlike the previous example, where the \( \alpha \) factor is recalculated in each iteration ( \( \alpha _{n}= \frac{1}{n} \) ), the \( \alpha \) factor is constant in this example.

The magnitude of the \( \alpha \) factor depends on the radar measurement precision. For high precision-radar, we should choose high \( \alpha \), giving high weight to the measurements. If \( \alpha = 1 \), then the estimated range equals the measured range:

\[ \hat{x}_{n,n}= \hat{x}_{n,n-1}+ 1 \left( z_{n}- \hat{x}_{n,n-1} \right) = z_{n} \]

If \( \alpha =0 \) , then the measurement has no meaning:

\[ \hat{x}_{n,n} = \hat{x}_{n,n-1}+ 0 \left( z_{n}- \hat{x}_{n,n-1} \right) = \hat{x}_{n,n-1} \]

So, we have derived a system of equations that composes the State Update Equation for the radar tracker. They are also called \( \alpha - \beta \) track update equations or \( \alpha - \beta \) track filtering equations.

Estimation Algorithm

The following chart depicts the estimation algorithm that is used in this example.

Unlike the previous example, the Gain values \( \alpha \) and \( \beta \) are given for this example. For the Kalman Filter, the \( \alpha \) and \( \beta \) are replaced by Kalman Gain, which is calculated at each iteration, but we learn it later.

Now we are ready to start a numerical example.

The numerical example

Consider an aircraft moving radially toward (or away from) a radar in a one-dimensional world.

The \( \alpha - \beta \) filter parameters are:

• \( \alpha =0.2 \).
• \( \beta =0.1 \).

The track-to-track interval is 5 seconds.

Iteration Zero

Initialization

The initial conditions for the time \( n=0 \) are given:

\[ \hat{x}_{0,0}=30000m \] \[ \hat{\dot{x}}_{0,0}=40m/s \]

Prediction

The initial guess should be extrapolated to the first cycle using the State Extrapolation Equations:

\[ \hat{x}_{n+1,n}= \hat{x}_{n,n}+ \Delta t\hat{\dot{x}}_{n,n} \rightarrow \hat{x}_{1,0}= \hat{x}_{0,0}+ \Delta t\hat{\dot{x}}_{0,0} =30000+5 \times 40=30200m \] \[ \hat{\dot{x}}_{n+1,n}= \hat{\dot{x}}_{n,n} \rightarrow \hat{\dot{x}}_{1,0}= \hat{\dot{x}}_{0,0} =40m/s \]

First Iteration

In the first cycle ( \( n=1 \) ), the initial guess is the prior estimate:

\[ \hat{x}_{n,n-1} = \hat{x}_{1,0}=30200m \] \[ \hat{\dot{x}}_{n,n-1} = \hat{\dot{x}}_{1,0}=40m/s \]

Step 1

The radar measures the aircraft range:

\[ z_{1}= 30171m \]

Step 2

Calculating the current estimate using the State Update Equation:

\[ \hat{x}_{1,1} = \hat{x}_{1,0}+ \alpha \left( z_{1}- \hat{x}_{1,0} \right) =30200+0.2 \left( 30171-30200 \right) = 30194.2m \] \[ \hat{\dot{x}}_{1,1}= \hat{\dot{x}}_{1,0}+ \beta \left( \frac{z_{1}-\hat{x}_{1,0}}{ \Delta t} \right) = 40+0.1 \left( \frac{30171-30200}{5} \right) = 39.42m/s \]

Step 3

Calculating the next state estimate using the State Extrapolation Equations:

\[ \hat{x}_{2,1}= \hat{x}_{1,1}+ \Delta t\hat{\dot{x}}_{1,1} =30194.2+5 \times 39.42=30391.3m \] \[ \hat{\dot{x}}_{2,1}= \hat{\dot{x}}_{1,1} =39.42m/s \]

Second Iteration

After a unit time delay, the predicted estimate from the previous iteration becomes the prior estimate in the current iteration:

\[ \hat{x}_{2,1}=30391.3m \] \[ \hat{\dot{x}}_{2,1}= 39.42m/s \]

Step 1

The radar measures the aircraft range:

\[ z_{2}= 30353m \]

Step 2

Calculating the current estimate using the State Update Equation:

\[ \hat{x}_{2,2} = \hat{x}_{2,1}+ \alpha \left( z_{2}- \hat{x}_{2,1} \right) =30391.3+0.2 \left( 30353-30391.3 \right) = 30383.64m \] \[ \hat{\dot{x}}_{2,2}= \hat{\dot{x}}_{2,1}+ \beta \left( \frac{z_{2}-\hat{x}_{2,1}}{ \Delta t} \right) = 39.42 + 0.1 \left( \frac{30353-30391.3}{5} \right) = 38.65m/s \]

Step 3

Calculating the next state estimate using the State Extrapolation Equations:

\[ \hat{x}_{3,2}= \hat{x}_{2,2}+ \Delta t\hat{\dot{x}}_{2,2} =30383.64+5 \times 38.65=30576.9m \] \[ \hat{\dot{x}}_{3,2}= \hat{\dot{x}}_{2,2} =38.65m/s \]

Third Iteration

\[ z_{3}= 30756m \] \[ \hat{x}_{3,3}= 30576.9+0.2 \left( 30756 -30576.9 \right) = 30612.73m \] \[ \hat{\dot{x}}_{3,3}= 38.65+0.1 \left( \frac{30756 - 30576.9}{5} \right) = 42.2m/s \] \[ \hat{x}_{4,3}= 30612.73+5 \times 42.2=30823.9m \] \[ \hat{\dot{x}}_{4,3}= 42.2m/s \]

Fourth Iteration

\[ z_{4}= 30799m \] \[ \hat{x}_{4,4}= 30823.9+0.2 \left( 30799-30823.9 \right) = 30818.93m \] \[ \hat{\dot{x}}_{4,4}= 42.2+0.1 \left( \frac{30799-30823.9}{5} \right) = 41.7m/s \] \[ \hat{x}_{5,4}= 30818.93+5 \times 41.7=31027.6m \] \[ \hat{\dot{x}}_{5,4}= 41.7m/s \]

Fifth Iteration

\[ z_{5}= 31018m \] \[ \hat{x}_{5,5}= 31027.6+0.2 \left( 31018 -31027.6 \right) = 31025.7m \] \[ \hat{\dot{x}}_{5,5}= 41.7+0.1 \left( \frac{31018 -31027.6}{5} \right) = 41.55m/s \] \[ \hat{x}_{6,5}= 31025.7+5 \times 41.55 = 31233.4m \] \[ \hat{\dot{x}}_{6,5}= 41.55m/s \]

Sixth Iteration

\[ z_{6}= 31278m \] \[ \hat{x}_{6,6}= 31233.4+0.2 \left( 31278 -31233.4 \right) = 31242.3m \] \[ \hat{\dot{x}}_{6,6}= 41.55+0.1 \left( \frac{31278 -31233.4}{5} \right) = 42.44m/s \] \[ \hat{x}_{7,6}= 31242.3+5 \times 42.44 = 31454.5m \] \[ \hat{\dot{x}}_{7,6}= 42.44m/s \]

Seventh Iteration

\[ z_{7}= 31276m \] \[ \hat{x}_{7,7}= 31454.5+0.2 \left( 31276 -31454.5 \right) = 31418.8m \] \[ \hat{\dot{x}}_{7,7}= 42.44+0.1 \left( \frac{31276 -31454.5}{5} \right) = 38.9m/s \] \[ \hat{x}_{8,7}= 31418.8+5 \times 38.9 = 31613.15m \] \[ \hat{\dot{x}}_{8,7}= 38.9m/s \]

Eighth Iteration

\[ z_{8}= 31379m \] \[ \hat{x}_{8,8}= 31613.15+0.2 \left( 31379 -31613.15 \right) = 31566.3m \] \[ \hat{\dot{x}}_{8,8}= 38.9+0.1 \left( \frac{31379 -31613.15}{5} \right) = 34.2m/s \] \[ \hat{x}_{9,8}= 31566.3 + 5 \times 34.2 = 31737.24m \] \[ \hat{\dot{x}}_{9,8}= 34.2m/s \]

Ninth Iteration

\[ z_{9}= 31748m \] \[ \hat{x}_{9,9}= 31737.24+0.2 \left( 31748 -31737.24 \right) = 31739.4m \] \[ \hat{\dot{x}}_{9,9}= 34.2+0.1 \left( \frac{31748 -31737.24}{5} \right) = 34.4m/s \] \[ \hat{x}_{10,9}= 31739.4+5 \times 34.4=31911.4m \] \[ \hat{\dot{x}}_{10,9}=34.4m/s \]

Tenth Iteration

\[ z_{10}= 32175m \] \[ \hat{x}_{10,10}= 31911.4+0.2 \left( 32175 -31911.4 \right) = 31964.1m \] \[ \hat{\dot{x}}_{10,10}= 34.4+0.1 \left( \frac{32175 -31911.4}{5} \right) = 39.67m/s \] \[ \hat{x}_{11,10}= 31964.1+5 \times 39.67 = 32162.45m \] \[ \hat{\dot{x}}_{11,10}= 39.67m/s \]

The following table summarizes our measurements and estimates.

Results analysis

The following chart compares the true values, measured values, and estimates.

Our estimation algorithm has a smoothing effect on the measurements and converges toward the true value.

Using high \( \alpha \) and \( \beta \)

The following chart depicts the true, measured, and estimated values for \( \alpha = 0.8 \) and \( \beta = 0.5 \).

The "smoothing" degree of this filter is much lower. The "current estimate" is very close to the measured values, and predicted estimate errors are relatively high.

So, shall we always choose low values for \( \alpha \) and \( \beta \) ?

The answer is NO. The value of \( \alpha \) and \( \beta \) should depend on the measurement precision. If we use high-precision equipment, like laser radar, we would prefer a high \( \alpha \) and \( \beta \) that follow measurements. In this case, the filter would quickly respond to a velocity change of the target. On the other hand, if measurement precision is low, we prefer low \( \alpha \) and \( \beta \). In this case, the filter smoothes the uncertainty (errors) in the measurements. However, the filter reaction to target velocity changes would be much slower.

Example summary

We've derived the \( \alpha - \beta \) filter state update equation. We've also learned the State Extrapolation Equation. We've developed an estimation algorithm for a one-dimensional dynamic system based on the \( \alpha - \beta \) filter and solved a numerical example for a constant velocity target.

Example 3 – Tracking accelerating aircraft in one dimension

The following chart depicts the target range, velocity and acceleration vs. time.

As you can see from the chart, the aircraft velocity is constant for the first 20 seconds and then grows linearly. The range grows linearly for the first 20 seconds and then grows quadratically.

We are going to track this aircraft with the \( \alpha - \beta \) filter that was used in the previous example.

The numerical example

Consider an aircraft moving radially toward (or away from) a radar in a one-dimensional world.

The \( \alpha - \beta \) filter parameters are:

• \( \alpha =0.2 \)
• \( \beta =0.1 \)

The track-to-track interval is 5 seconds.

Iteration Zero

Initialization

The initial conditions for the time \( n=0 \) are given:

\[ \hat{x}_{0,0}=30000m \] \[ \hat{\dot{x}}_{0,0}=50m/s \]

Prediction

The initial guess shall be extrapolated to the first cycle using the State Extrapolation Equations:

\[ \hat{x}_{n+1,n}= \hat{x}_{n,n}+ \Delta t\hat{\dot{x}}_{n,n} \rightarrow \hat{x}_{1,0}= \hat{x}_{0,0}+ \Delta t\hat{\dot{x}}_{0,0} =30000+5 \times 50=30250m \] \[ \hat{\dot{x}}_{n+1,n}= \hat{\dot{x}}_{n,n} \rightarrow \hat{\dot{x}}_{1,0}= \hat{\dot{x}}_{0,0} =50m/s \]

Iterations 1-10

All filter iterations are summarized in the following table:

Results analysis

The following charts compare the true, measured, and estimated values for the range and velocity for the first 75 seconds.

You can see a constant gap between true or measured values and estimates. The gap is called a lag error. Other common names for the lag error are:

• Dynamic error
• Systematic error
• Bias error
• Truncation error

The lag error appears during the acceleration period. A significant lag error can result in the target loss. The lag error is unacceptable in certain applications, such as missile guidance or air defense.

Example summary

We've examined the lag error caused by target acceleration.

Example 4 – Tracking accelerating aircraft with the \( \alpha -\beta -\gamma \) filter

Results analysis

The following charts compare the true, measured, and estimated values for the range, velocity, and acceleration for the first 50 seconds.

Example summary

As you can see, the \( \alpha -\beta -\gamma \) filter with dynamic model equations that include acceleration can track the target with constant acceleration and eliminate the lag error.

But what happens in the case of a maneuvering target? The target can suddenly change the flight direction by making a maneuver. The dynamic model of the target can also include a jerk (changing acceleration). In such cases, the \( \alpha -\beta -\gamma \) filter with constant \( \alpha -\beta -\gamma \) coefficients produces estimation errors and, in some cases, loses the target track.

The Kalman filter can handle uncertainty in the dynamic model. It is our next topic, right after the summary.

Summary of the \( \alpha -\beta -(\gamma) \) filter

I hope to write a tutorial about some of these filters. But this tutorial is about the Kalman Filter, which is the topic of the following examples.