Relationships Occurring With Sinusoidal Points March 11, 2002 by Andrew Burnson I have found that when a sine wave of the form f(x) = Asin(bx+c) passes through three points, several relationships are formed between the points and the sinusoidal function. The original goal was to be able to determine A, b, and c from these three points. Not only have I solved this problem, but have discovered what defines a sine wave to be a sine wave. So to start off let P, Q, and R represent these three unique points. The first restriction that must be made is that Py, Qy, and Ry can not all be zero. Later this restriction will be simplified, however. The purpose of this restriction is that if all are points have a y-coordinate of zero, then it is impossible to determine any characteristics of the sine wave because there would be an infinite number of solutions for it. The function of the sine wave that we are trying to match to these three points will be defined as f(x) = Asin(bx+c). In this function A is the amplitude, -c/b is the phase shift, and 2π / b is the period of the graph. The ultimate goal is to solve for A, b, and c, which will in turn lead to proof of some interesting relationships. Because we know that f(x) passes through the three points, the points coordinates can be substituted in. So, we know that: f(px) = Py f(qx) = Qy f(rx) = Ry Using substitution again: Py = Asin(bPx + c) Qy = Asin(bQx + c) Ry = Asin(bRx + c) With three unknowns and three equations, a system of equations should be able to handle the task of solving for A, b, and c. So, the first thing to do would be to solve for A in terms of b and c since A is readily available: - 1 -
This leads to the first known relationship. It may seem trivial, but it could come in handy if b and c are already known, or have already been solved for using the upcoming equations. Relationship 1.1 Amplitude from a point and normalized sine wave If P is a point with coordinates (x, y) and a function of the form f(x) = Asin(bx+c) passes through this point, then if and only if sin(bx+c) 0. The amplitude of the sine wave is A. Now, by substituting A in for every possible combination, three more equations can be generated: From the trigonometric angle addition identities, we know that: So, the equation will now be expanded so that it can be worked further. I will now only show work for just the first equation above. The others are identical in form, so after the first equation has been worked, the others will be updated to reflect the same derivations. So now, using the angle addition identity, the first equation can be rewritten as: And now using cross-multiplication the equation can be written as such: Then, by distributing the Py and Qy we can solve for c after a little more work: - 2 -
This solution results in the next equation. Relationship 1.2 Frequency-relative phase shift from two points If points P and Q have coordinates (x1, y1) and (x2, y2), respectively, and a sine wave of the form f(x) = Asin(bx+c) passes through points P and Q, where b > 0, A > 0, and, then, where c/b is the phase shift of the sine function. Now, since the other two equations that we had to begin with were of the same format. They would hence be derived in the exact same way. So if the other two equations were to be worked out in the same fashion, then we would have the following three equations, in total: Now, using substitution, three more equations could be created through the substitution of tan(c). But, I have found that no matter what equations you choose, you will end up with the same solution for c. This is reassuring because there should only be one solution for c since it is the last thing we are solving for. So, the following work will substitute tan(c) between the first two equations above: - 3 -
Now using cross-multiplication we get the following (large brackets indicate the equation being written on more than one line): And now by multiplying everything out: Now, the y-coordinates of the points can be factored by bringing everything to one side of the equation: But since [ out of the equation to result in the following: ] = 0, the factor Py Ry will simply drop The equation right now looks terribly complex, but if we remember the trigonometric angle subtraction identity then it can be greatly simplified. This identity states that. There are three instances where this identity can be utilized in the above equation, so it can now be simplified to the following: - 4 -
Also, Qy can be factored out and removed: And now b can be factored out a few times: However, as things stand right now, one can not solve for b algebraically. The reason for this is that there is a given varying conditions there can be anywhere from zero to infinite solutions to the equation. When we defined the points P, Q, and R, it was implied that they could exist anywhere and not necessarily even be in the same order on the x-axis. For example, point R could be before point P relative to the positive x-axis. This ambiguity is one of the things that is preventing the equation from being solved. So we will first make the restriction that. However, this still is not enough to solve the equation. The problem lies in the spacing of the x-coordinates of the three points. If the x-coordinate distance between points P and Q is not equal to the x-coordinate distance between points Q and R, then the number of solutions will vary from zero to an infinite number. So, the restriction PQ = QR must be made in order to ensure this. This may seem like a cumbersome restriction, but all it actually does is emphasize the importance of having a sample rate. When dealing with signal processing, for example, there is a rate at which any signal is sampled. All the points that would be contained in the sampled signal, are spaced evenly. All samples are assigned to a specific portion of time and the amount of time each sample is assigned to remains constant. So for sampled data, the new restriction will not hinder things in any way. Therefore, let z = the duration of each sample. In our case, it will be the distance between each point and the following point. And since a duration represents a physical amount of something z > 0 which also automatically includes the previous order restriction that. So, because of this sampling restriction we now know the following: Qx = Px + z Rx = Px + 2z From now on, I will refer to these types of points as sampled points. This new limitation may seem trivial, but watch what happens when Qx and Rx are substituted in: Now, the negative angle identities can be used to simplify this further: - 5 -
Here is where the real magic happens. The double angle identity,, can now be used: Now sin(bz) can be factored out: Normally, we would not just divide out sin(bz) from the equation, since it could contain valuable solutions. But the only solutions it contains are where. These are periodic solutions that are not only meaningless, but are invalid as you will see later in Relationship 3.1. So, sin(bz) can be safely divided out from the equation to leave us with: And now, b can finally be solved for: Finally, we have solved for b. This leads to the next equation. Relationship 1.3 Period of a sine wave passing through three sampled points If and only if a sine wave of the form f(x)=asin(bx+c) passes through the points P, Q, and R, where,, z > 0, and, then, where is the period of the sine wave. - 6 -
However, this is not the only thing that this equation says. In fact, this equation practically defines what a sine wave is. Because there is a inverse cosine taken, there are restrictions on both the domain and range. The domain of the inverse cosine function is [-1, 1]. The range is [0, π]. At first this came as a shock to me because these extremely narrow restrictions seemed to come out of no where. But several more relationships can now be proven. Since the domain is [-1, 1], this means that: Obviously, there are conditions such that this inequality would be false. There have not been any restrictions on the y-coordinates except that Qy cannot be zero. And yet, this new restriction still exists for some reason. It is my conclusion therefore, that if there ever are three sampled points that do not obey this inequality, then there is no sine wave of the form f(x)=asin(bx+c) that could possibly pass through the points. In the case of a complex signal with multiple frequencies present, this means that eventually these equations will fail since no one frequency or period can be established. It s like trying to determine the color of a painting, it simply cannot be done. So, it can be assumed then that Relationship 1.3 will only work with pure tone waves containing only one frequency. The inequality can be simplified somewhat to the following: By doing this we remove the restriction happens. When Qy = 0, it causes. I will attempt to explain why this. Mathematical proof of this is rather complicated, so instead I will use a diagram instead to show what is happening. The only time Qy = 0 is when the sine wave crosses the x-axis. So the sample before Q, which is P is going to have the reflected y-coordinate of the sample after Q, which is R. - 7 -
So, if Qy = 0 then. So, a sine wave can definitely pass through these points. The problem is that an infinite number of sine waves can pass through these points. This is why there was the restriction of back in Relationship 1.3. The equation could not solve for b without it because there would be this possibility for an infinite number of solutions. But now when it comes down to just determining if any sine wave could possibly pass through these points, the restriction is no longer present. This results in the following relationship. Relationship 2.1 Determining whether three points could be part of a sine wave If there are three sampled points, P, Q, and R, where Qy 0, then one function of the form f(x)=asin(bx+c) will contain these points if and only if. If there are three sampled points, P, Q, and R, where Qy = 0 and at least one sine wave of the form f(x)=asin(bx+c) passes through these points, then there are an infinite number of sine waves in the form f(x)=asin(bx+c) that will contain these points. This would be very useful for determining whether a certain set of points contains a sine wave. But, the only problem with this is that if you have many more points there is no guarantee that each three point set belongs to the same unique sine wave function. However, say you have three other sampled points, say, J, K, and L along with the first set of points P, Q, and R. Well, if both sets of points belong to the same frequency waveform, then the period, b, will remain constant. This leads to the following statement: This in turn can be simplified since each side of the equation is so similar: - 8 -
So, what this says is that if b remains constant, then the ratio of the sum of the two outerpoints y-coordinates and the middle point y-coordinate will remain constant. This leads to the following relationship: Relationship 2.2 Determining whether six points belong to the same period wave If there exist two sets of points each containing three sampled points the set P, Q, and R, and the set J, K, and L where QyKy 0 such that a sine wave of the form f(x) = Asin (bx+c) contains points P, Q, and R, and a sine wave of the form g(x) = Dsin(ex+h) contains points J, K, and L, then b=e if and only if. The reason why there are two functions defined, f and g, is that it is possible that two different waves could have the same frequency but could have different amplitudes and phase shifts. So, this was provided for the general case. But when f and g represent the same function, very interesting things happen. When the two sets of points overlap each other, a new relationship is produced. For example, if you allow point J to have the same coordinates as point Q and point K be equivalent to point R (see diagram), then the following equation will result after the substitutions have been made: - 9 -
Now, at first this does not seem to accomplish anything. As things are right now, there are four consecutive samples of some kind of sine wave. But, what if the last point, point L was unknown. By simply solving the equation for Ly you would be able to determine the coordinate of the sample point that would follow the set of three samples that was given. So, solving this equation for Ly: But, things can be made even easier if for example you wanted to keep calculating the following samples. Earlier, I discussed how for any amount of sampled points, if b is constant, then the ratio of the sum of the two outer-point s y-coordinates to the middle point s y-coordinate will remain constant. From now on I will refer to this as the sine wave constant since no matter what points you choose on a sine wave it will always be the same. This is the next relationship. Relationship 2.3 The sine wave constant If there are three sampled points P, Q, and R and only one function of the form f(x)=asin (bx+c) passes through these points, then k, the sine wave constant is defined to be: The sine wave constant practically defines what a sine wave is. It states that sine wave is all the points such that from any three sampled points, the ratio of the sum of the outer points y-coordinates to the middle point y-coordinate is constant. So, getting back to the problem at hand if k has already known, then using our original equation,, we could more easily solve for Ly: - 10 -
This equation proves to be more beneficial than the other more complicated solution for Ly, because now only two consecutive samples are needed in order to construct the following consecutive sample. This results in the next relationship. Relationship 2.4 Determining the next sample point on a sine wave given two sample points and the sine wave constant If there exist three sampled points, J, K, and L, appearing consecutively and a sine wave of the form f(x)=asin(bx+c) with a sine wave constant of k passes through these points, then. This is a very interesting relationship because it can compute the unknown sine of an angle using only one multiplication and one subtraction so long as you have the sine wave constant and the sine of two previous angles. In theory, one could derive an entire trigonometric lookup table if he or she knew the sines of just three small angles. This idea leads to the next relationship. Relationship 2.5 Computing the sine of an angle given the sines of just three angles If m is a positive real number designating the angle sample rate, and where x is any real number then,. To put this to the test, we will find where m=1. We know that This will of course simplify to the following: So this means that in the final equation: - 11 -
Checking this with a calculator, you can see for yourself that this relationship really exists. Then using,, and k, one could find, and then so on as long as accuracy permits. Now that the domain restrictions of the original inverse cosine function have been covered, there are a few things to point out about the range restrictions. To refresh, The range of the inverse cosine function is [0, π] which means: Using substitution we also know that Asin(bz+c). We know that:. Now let p = the period of the sine wave Now, if s = the sample rate of a signal and f = the frequency of the sine wave Asin(bz+c), then we also know that: Then substituting p in: Since we know how bz is restricted, we can also find out how the frequency will be restricted: and - 12 -
and Of course we know that is not valid since f 0 since the frequency must be positive. So it will be corrected to be. When the other inequality is simplified we get the following: and So what this states is that any sampled sine wave has a maximum frequency of half the sample rate. This is a simpler and more narrow proof of the Nyquist frequency theorem which states that a sine wave can only be reconstructed properly if it is sampled at twice its own frequency. This leads to the following set of relationships. Relationship 3.1 Frequency of a sine wave If s is the sample rate per unit of time, and f is the frequency of a sine wave of the form f (x)=asin(bx+c) which passes through three sampled points where z is the sample duration, then and. Also, the limitations on b are. - 13 -