Below is my attempt at describing the needed maths for programming the DSP part of the Humanise.org project. (See http://humanise.org). Humanise.org is an open source project to create a high quality software oriented metal detector suitable for humanitarian demining. I'm trying to do this article with simple school type maths (no integration, no radians, no complex maths etc). In particular I will be trying to describe how a Phase Lock Loop (PLL) and how a Two Channel Lock-In Amplifier work. The actual PLL and Lock-In Amplifier is described in another article. This is just the maths behind them. I expect it could also be used as an introduction to DFT and then FFT. I looked for something as simply as this that still gave exact maths sufficient for it to be programmed from but although I found many attempts by others, I never liked any. By doing my own version, it means that I can directly host it instead of just linking to it and then finding later that what I had linked to is no longer there. I don't consider myself good at this type of maths so I'm looking for others to check out the maths. When I convert it to HTML I will convert a lot of the symbols to their proper symbols and get rid of all the * used as multiplication signs. I also hope to illustrate it with sine wave graphs etc. I'm not good at making diagrams so I'm hoping that others wanting to help in the Humanise.org will make some diagrams for it. Can you please read it over and tell me any mistakes that I have made. --------------------------------------------------------------------------------------------------- Note on AC/DC terminology: With Electronics we talk about currents being DC (Direct Current) or AC (Alternating Current) or a mixture of both. When converted to numbers and graphed on a typical X and Y axis graph, the DC is represented as a shift of the graph on the Y axis while the AC is a continuing alternation on the Y axis without any overall shift. Throughout the discussion and mathematics on this website we will continue to use the term DC and AC for similar actions on streams of numbers representing waveforms even though there is no actual current still involved. Similarly we assume that the typical X and Y axis graphs exist for each waveform with the Y axis showing the wave's Amplitude and the X axis being used for Time. Gentle Maths Version: Now lets try describing the maths behind the Phase Lock Loop and the Two Channel Lock-In Amplifier using nothing more than simple high school maths. It has been stated that the heart of all Fourier analysis is, amazingly, all based on the following single high school trigonometry formula for the product of two sines: sin(X1) * sin(X2) = 1/2 * cos(X1-X2) - 1/2 * cos(X1+X2) For sine waveforms, the above X1 and X2 are based on time and frequency. For simplicity, let us assume that Frequency is given to us in units of degrees per second where a complete cycle of the sine wave is 360 degrees. This is sometimes called Angular Velocity. We will use F to represent this Frequency. Let us assume that the sine table or sine lookup we use (sin for short) is based on degrees. Let us also assume that time represented by t is given in seconds. For these sine waves, the X1 and X2 above are as follows: (F * t) + Initial_Phase_Angle_In Degrees As F is an Angle per time (degrees per second), at each time point we multiply the Angle per time by the current time (in seconds) we get an angle (in degrees) which is what we need for looking up the the sin lookup. Unfortunately, frequencies are not normally specified as degrees per second. Rather, they are specified in cycles per second. This creates a mishmash between the units used for angle with the the sin lookup and the units used for angle per time in the frequency specification. Let us represent these cycles per second frequency by f. Then F = 360 * f With the sin lookup still being based on angles that have 360 units per waveform unit, X1 and X2 now needs to be as follows: (360 * f * t) + Initial_Phase_Angle_In_Degrees We have had to place in the factor of 360 to take account of the mishmash with 360 degrees in each cycle in the sin lookup but frequency being specified with whole cycles per second. Note that with each of the above definitions of X1 and X2, once we have a fixed input frequency and phase, the only thing that is varying is time. For frequencies, the initial equation relates the Amplitude of these waves to time. Now lets start using the initial equation with waveforms. sin(X1) * sin(X2) = 1/2 * cos(X1-X2) - 1/2 * cos(X1+X2) When this equation is used on waveforms, it tells us what happens when we have one sin(X1) waveform input plus another another sin(X2) waveform input and we multiply them together as they are being input. We are just multiplying the same point in time on one waveform to the same point in time of the other waveform. The resultant waveform is (1/2 * cos(X1-X2)) - (1/2 * cos(X1+X2)). Lets see what happens when X1 and X2 are different frequencies. Lets assume that these frequencies are f1 and f2 and that both of these are of an amplitude of 1. If we look at the first resultant term 1/2 * cos(X1-X2) With waveform X1 being a frequency f1 and X2 being a frequency f2, this will be equal to 1/2 * cos((360 * f1 * t) - (360 *f2 * t)) This is simplified to 1/2 * cos(360 * (f1-f2) * t) This is a waveform with a frequency that is the difference between the two input frequencies. As we know these frequencies are different (we made that assumption at the beginning) then this must produce a wave form that has a non zero frequency that is the difference between the two frequencies. As this is a standard sine waveform, it is symmetric about the X axis. This is, its an AC only waveform with an amplitude of one half (due to the 1/2 previous to the cos(X1-X2)). Similarly with the 1/2 * cos(X1+X2) term, the result 1/2 * cos((360 * (f1+f2) * t) + 2 * Initial_Phase_Angle) is a waveform that is the addition of the two frequencies. This produces a second waveform of half the amplitude of the input waveforms. In this case it has a frequency that is the sum of the two frequencies. As this is also a standard sine waveform, it is also symmetric about the X axis. The addition of two pure AC complete waveforms always produces a resultant waveform that is pure AC only. The word 'complete' is here used to indicate that we are looking at complete cycles, not just part of a cycle of the waveform. As can be seen from the above, so long as the frequencies are different, there is no DC offset in the result. Lets assume that X1 and X2 above are exactly equal. Start again with the initial equation: sin(X1) * sin(X1) = 1/2 * cos(X1-X1) - 1/2 * cos(X1+X1) As Cos(X1-X1) = Cos(0) = 1 Sin(X1) * Sin(X1) = 1/2 - (1/2 * cos(2 * X1)) In this case, the same input sine waves multiplied by each other produce a DC offset (equal to half the amplitude of the input sine waves) plus a wave at double the frequency i.e. (2 * X1) and half the amplitude of the input sine waves. One strange aspect of sine waves is that whether a wave is added or subtracted doesn't change its position with respect to the Y axis. That is, it doesn't give or change the wave's DC offset. As the wave is minus, it is equivalent to the wave being 180 degrees out of phase. That is, the following equation is equivalent to the last one: Sin(X1) * Sin(X1) = 1/2 + (1/2 * cos((2 * X1) + 180 degrees)) Lets assume that X1 and X2 above are equal except that X2 is phase shifted forward by 90 degrees. Sin(X1) * Sin(X1 + 90 degrees) = 1/2 * cos(0 - 90 degrees) - 1/2 * cos((2 * X1) + 90 degrees) As Cos(-90 degrees) = 0 Sin(X1) * Sin(X1 + 90 degrees) = - 1/2 * cos((2 * X1) + 90 degrees) = 1/2 * cos((2 * X1) - 90 degrees) In the case of waveforms this means that when input sine waves of the same frequency but with a phase offset of 90 degrees are multiplied by each other, a wave at double the frequency i.e. (2 * X1) and half the amplitude of the input sine waves is produced. As the wave is minus, it is equivalent to the wave being 180 degrees out of phase but the extra 90 degrees in ((2 * X1) + 90 degrees) then shifts it back 90 degrees. Note that this wave is symmetrical about the X axis. There is no DC offset. If we did the same as above but with X2 phase shifted back by 90 degrees (the same as phase shifted forward by 270 degrees) then we would get a similar result in that the resultant wave has no DC offset. Sin(X1) * Sin(X1 - 90 degrees) = - 1/2 * cos((2 * X1) - 90 degrees) Lets assume that X1 and X2 above are equal except that X2 is phase shifted by 180 degrees. This is the equivalent to being inverted. Starting at our initial equation we see that: sin(X1) * (-1 * sin(X1)) = -1 * ( 1/2 * cos(X1-X1) - 1/2 * cos(X1+X1)) = 1/2 * cos(X1+X1) - 1/2 * cos(X1-X1) Again, as cos(0) = 1 = (1/2 * cos(2*X1)) - 1/2 In the case of waveforms this means that when input sine waves that are equal except one is inverted from the other, are multiplied by each other, they produce a minus DC offset (equal to half the amplitude of the input sine waves) plus a wave at double the frequency i.e. (2 * X1) and half the amplitude of the input sine waves. From these equations we can see that we only obtain a DC offset when we multiply two waveforms if they are exactly at the same frequency. Even then, we only obtain the DC offset if the waves are not at 90 or 270 degrees phase difference from each other. The DC offset reaches a maximum positive value when the waves are exactly in phase with each other and reach a maximum negative value when the waves are 180 degrees out of phase. Also note that we have simplified the basic equation by leaving off the amplitudes. We have assumed that the amplitudes of the two input waves are both 1. Assume the amplitude of the sin(X1) input wave is A and the amplitude of the sin(X2) input wave is B then the original equation would be (A * sin(X1)) * (B * sin(X2)) = (A * B ) * (1/2 * cos(X1-X2) - 1/2 * cos(X1+X2)) If the amplitudes of the input sin(X1) and sin(X2) waves is not 1 then all the previous formulas can be corrected by simply multiply the results by a factor of (A * B). The previous equations should have explained the workings of the Phase Lock Loop. The following expands this to show the workings of the Two Channel Lock-In Amplifier. Lets see what happens if we create an X2 waveform that is the same as an input X1 except that it is phase shifted an angle Theta from X1. sin(X1) * sin(X1 + Theta degrees) = (1/2 * cos(X1-X1 - Theta degrees)) - (1/2 * cos(2*X1 + Theta degrees)) = (1/2 * cos(- Theta degrees)) - (1/2 * cos(2*X1 + Theta degrees)) The second term (1/2 * cos(2*X1 + Theta degrees) by itself is a pure sine wave that is symmetrical about the X axis. It has no DC component. If we are only interested in the resultant DC component then we can throw away this term. The first term gives us the value of the resultant waveform if we averaged it, filtering out the AC component. The DC component would be 1/2 * cos(- Theta degrees) = 1/2 * cos(Theta degrees) cos(Theta degrees) is a constant so we have a constant DC offset of 1/2 * cos(Theta degrees). If we perform this with an input wave amplitude of A and an internally created wave of amplitude B, and measure the resultant DC offset as Y1 then Y1 = 1/2 * A * B * cos(Theta degrees) If we now create an X3 waveform that is the same as X2 except that it is phase sifted by (Theta plus 90) degrees from X1 and use that to multiply against another copy of the input X1 waveform. sin(X1) * sin(X1 + (Theta + 90) degrees) = (1/2 * cos(X1-X1 - (Theta + 90) degrees)) - (1/2 * cos(2 *X1 + (Theta + 90) degrees)) Again we are only interested in the DC component so we throw away the second term. The first term gives us 1/2 * cos((Theta + 90) degrees) = 1/2 sin(Theta degrees) If we perform this with an input wave amplitude of A and an internally created wave of amplitude of B, (assume that X3 is the same amplitude as X2) and measure the resultant DC offset as Y2 then Y2 = 1/2 * A * B * sin(Theta degrees). If we now square Y1 and Y2, add these squares together and then take the square root (Square root of (Y1**2 + Y2**2)) = square root of ((1/2 * A * B * cos(Theta degrees))**2 + (1/2 * A * B * sin(Theta degrees))**2) = square root of (1/4 *A*A*B*B* (cos(Theta degrees)**2) + 1/4 *A*A*B*B* sin(Theta degrees)**2) = square root of (1/4 *A*A*B*B* (cos(Theta degrees)**2) + 1/4 *A*A*B*B* sin(Theta degrees)**2) = square root of (1/4 *A*A*B*B * ((sin(Theta degrees)**2) + (cos(Theta degrees)**2))) Note: sin(X)**2 + cos(X)**2 = 1 This is a trigonometrical identity that is easily derived from the main trigonometrical identity we have been using. Using this reduces the above to = square root of (1/4*A*A*B*B*) = 1/2 * A * B Consequently, if we know B then the Amplitude A can be calculated from A = 2 * (Square root (Y1**2 + Y2**2)) / B Using the above, if we have an input waveform X1 that contained a known frequency but we didn't know the phase or amplitude of this frequency, we can find the amplitude by creating two reference waveforms, both of Amplitude B and frequency the same as X1. The second internally created reference waveform would be created with a 90 degree phase shift from the first reference waveform. We also create a second copy of the input waveform. We now multiply the input waveform by the first reference waveform and average the result so that we measure the resultant Y1. We also multiply the copy of the input waveform by the second reference waveform and average the result so that we measure the resultant Y2. The amplitude A of the known frequency within the input waveform can now be calculated as A = 2 * (Square root (Y1**2 + Y2**2)) / B Lets denote by Y3 what would have been measured had the internally created reference waveform been exactly in phase with the input frequency. This gets rid of the 1/2 that's needed when comparing to the input waveform. The measured Y1 and Y2 are the sine and cosine components of the Y3 amplitude. They have the same relationship as the clock hand (of Y3 length) has to the X and Y axis or the same relationship as the hypotenuse (of Y3 length) of a right angle triangle. Consequently, we can find the true amplitude or "magnitude" as the hypotenuse of this right triangle by taking the square root of the sum of the squares of the sine and cosine components. This is called the "vector sum" of the components. Similarly, we can find the phase as the angle whose tangent is equal to the sine component divided by the cosine component. The sine and cosine components are often referred to as the "in-phase" and "quadrature" components. Some engineering and mathematicians prefer to use "real" and "imaginary" for the sine and cosine components. From this they create some very intimidating expressions using an imaginary number (square root of -1) in a style of maths called "complex notation". Also note that the first internally created reference frequency can be at any phase difference to the actual signal. Given any particular pure sinewave, without a DC component, at a particular frequency f1, it can be fully described by specifying its amplitude and phase. Alternatively, the sine wave can be described by specifying two amplitudes, these being the equivalent in-phase and quadrature components of that frequency. As we can change where (which phase angle) our first reference frequency is, there is effectively an infinite number of pairs of amplitudes that can be used as an equivalent to the phase plus amplitude pair. As long as we keep in mind where the the reference is in each case, all these pairs of amplitudes are equivalents. With VLF metal detectors, where we are both transmitting the frequency on one coil and receiving the resultant frequency on another coil, the sine wave being output is normally used as our reference (sine component) frequency for the primary received frequency. Consequently, in this case it clearly makes sense to call the sine wave component the in-phase component. Changes in the received amplitude of this component are normally caused by resistive property effects in the soil and anomalies near the coil. Note that resistive is here used to indicated properties that change the received voltage in phase with the signal like resistance does. It doesn't necessarily mean resistance itself. Changes in the received amplitude of the cosine component are normally caused by reactive (such as capacitance and inductance) property effects of the soil and anomalies near the coil. When programming using any of the above equations, you need to recognise the order in which the sine or cosine waves will be processed. The sine graphs with time increasing on the X axis are a representation of waves that are travelling from right to left. Generally, we expect things on paper to be represented from left to right so this works out to be opposite to what many people expect. To confirm that it is this direction note what happens if we place the sine or cosine graph onto a long tape with time in seconds marked as normal along the X axis. We cover over the tape except that we leave a small view port that allows us to see the tape at the zero time mark (normally the Y axis). We now fire a starters gun to mark our zero time point. Following the gun firing we pull the tape so that the time marks of 1 second, 2 seconds etc are viewable at our view port at 1 second, 2 seconds etc after our zero time. To do this, we have to pull the tape to the left so that the graphed sine wave is moving right to left. The conclusion here is that if you have a sine wave and a cosine wave being input to the processor, the cosine wave will reach the processor at 1/4 wave cycle ahead of the sine wave or 3/4 wave cycle after the sine wave. The equation cos(X) = sin(X + 90 degrees) indicates that cosine is before sine by 90 degrees. We should also point out the usefulness of the earlier discussed fact that multiplying two frequencies produces a sine wave that is the difference of the two frequencies plus a sine wave that is the sum of the two frequencies. As an example, Let us assume that we are interested in using an audio card input to look at an input that has a frequency of 300Khz. This is a higher frequency than current audio cards can handle correctly. To make it suitable we could create an oscillator that is for example a fixed 295khz. We then multiply this oscillator frequency to the 300Khz input. The result is a waveform at (300 + 295)Khz plus a wave form at (300 - 295)Khz. We can then use a low pass filter to filter out the 595Khz waveform leaving us with just a 5Khz waveform which is very easy for the audio card to handle. As the original 300Khz either changes in frequency or changes in amplitude, the 5Khz input to the audio card will similarly change. This principle is used in many applications such as the super-heterodyne receiver It also explains a common problem in many wave systems called Intermodulation Distortion. This distortion occurs when two sine waves of frequencies X1 and X2 in the input result in the creation of other unwanted frequency components. These frequencies include (X1+X2), (X1-X2), (2*X1-x2), (2*X2-X1), and generally (mX1 ± nX2) for integer m and n. Generally, the size of the unwanted frequency components falls rapidly as m and n increase. A similar explanation for some of the above is given at the following link: http://www.daqarta.com/eex01.htm
Please check my article on DSP basics
Started by ●March 17, 2009
Reply by ●March 17, 20092009-03-17
Are you assuming that you will only have metal mines or mines with a significant amount of metal? Dirk
Reply by ●March 17, 20092009-03-17
On 17 Mar, 18:34, bellda2...@cox.net wrote:> Are you assuming that you will only have metal mines or mines with a > significant amount of metal?If so, it would be a classic ostrich - I've seen it all too often in the past. People tend to focus on what they *can* find, ingnore all the cases they can *not* handle, and go on as if what they can do is sufficient. Of course, that might be OK in a comm system which is subjected to multipath propagation; most people would argue it's not, with mines involved. There are some applications where anything less than 100% success is unacceptable. Rune
Reply by ●March 17, 20092009-03-17
On Mar 18, 2:32�am, "Humanise" <k...@humanise.org> wrote:> Below is my attempt at describing the needed maths for programming the DSP > part of the Humanise.org project. (Seehttp://humanise.org). �Humanise.org > is an open source project to create a high quality software oriented metal > detector suitable for humanitarian demining. �I'm trying to do this article > with simple school type maths (no integration, no radians, no complex maths > etc). > > In particular I will be trying to describe how a Phase Lock Loop (PLL) and > how a Two Channel Lock-In Amplifier work. �The actual PLL and Lock-In > Amplifier is described in another article. �This is just the maths behind > them. �I expect it could also be used as an introduction to DFT and then > FFT. �I looked for something as simply as this that still gave exact maths > sufficient for it to be programmed from but although I found many attempts > by others, I never liked any. �By doing my own version, it means that I can > directly host it instead of just linking to it and then finding later that > what I had linked to is no longer there. > > I don't consider myself good at this type of maths so I'm looking for > others to check out the maths. �When I convert it to HTML I will convert a > lot of the symbols to their proper symbols and get rid of all the * used as > multiplication signs. �I also hope to illustrate it with sine wave graphs > etc. �I'm not good at making diagrams so I'm hoping that others wanting to > help in the Humanise.org will make some diagrams for it. > > Can you please read it over and tell me any mistakes that I have made. > > --------------------------------------------------------------------------------------------------- > Note on AC/DC terminology: > > With Electronics we talk about currents being DC (Direct Current) or AC > (Alternating Current) or a mixture of both. �When converted to numbers and > graphed on a typical X and Y axis graph, the DC is represented as a shift > of the graph on the Y axis while the AC is a continuing alternation on the > Y axis without any overall shift. > > Throughout the discussion and mathematics on this website we will continue > to use the term DC and AC for similar actions on streams of numbers > representing waveforms even though there is no actual current still > involved. �Similarly we assume that the typical X and Y axis graphs exist > for each waveform with the Y axis showing the wave's Amplitude and the X > axis being used for Time. > > Gentle Maths Version: > > Now lets try describing the maths behind the Phase Lock Loop and the Two > Channel Lock-In Amplifier using nothing more than simple high school > maths. > > It has been stated that the heart of all Fourier analysis is, amazingly, > all based on the following single high school trigonometry formula for the > product of two sines: > > sin(X1) * sin(X2) = 1/2 * cos(X1-X2) - 1/2 * cos(X1+X2) > > For sine waveforms, the above X1 and X2 are based on time and frequency. > > For simplicity, let us assume that Frequency is given to us in units of > degrees per second where a complete cycle of the sine wave is 360 degrees. > This is sometimes called Angular Velocity. �We will use F to represent this > Frequency. �Let us assume that the sine table or sine lookup we use (sin > for short) is based on degrees. �Let us also assume that time represented > by t is given in seconds. �For these sine waves, the X1 and X2 above are as > follows: > > � �(F * t) �+ Initial_Phase_Angle_In Degrees > > As F is an Angle per time (degrees per second), at each time point we > multiply the Angle per time by the current time (in seconds) we get an > angle (in degrees) which is what we need for looking up the the sin > lookup. > > Unfortunately, frequencies are not normally specified as degrees per > second. �Rather, they are specified in cycles per second. � This creates a > mishmash between the units used for angle with the the sin lookup and the > units used for angle per time in the frequency specification. �Let us > represent these cycles per second frequency by f. �Then > > � � F = 360 * f > > With the sin lookup still being based on angles that have 360 units per > waveform unit, X1 and X2 now needs to be as follows: > > � � (360 * f * t) + Initial_Phase_Angle_In_Degrees > > We have had to place in the factor of 360 to take account of the mishmash > with 360 degrees in each cycle in the sin lookup but frequency being > specified with whole cycles per second. > > Note that with each of the above definitions of X1 and X2, once we have a > fixed input frequency and phase, the only thing that is varying is time. � > For frequencies, the initial equation relates the Amplitude of these waves > to time. > > Now lets start using the initial equation with waveforms. > > sin(X1) * sin(X2) = 1/2 * cos(X1-X2) - 1/2 * cos(X1+X2) > > When this equation is used on waveforms, it tells us what happens when we > have one sin(X1) waveform input plus another another sin(X2) waveform input > and we multiply them together as they are being input. �We are just > multiplying the same point in time on one waveform to the same point in > time of the other waveform. �The resultant waveform is (1/2 * cos(X1-X2)) - > (1/2 * cos(X1+X2)). > > Lets see what happens when X1 and X2 are different frequencies. �Lets > assume that these frequencies are f1 and f2 and that both of these are of > an amplitude of 1. �If we look at the first resultant term > > �1/2 * cos(X1-X2) > > With waveform X1 being a frequency f1 and X2 being a frequency f2, this > will be equal to > > �1/2 * cos((360 * f1 * t) - (360 *f2 * t)) > > This is simplified to > > �1/2 * cos(360 * (f1-f2) * t) > > This is a waveform with a frequency that is the difference between the two > input frequencies. As we know these frequencies are different (we made that > assumption at the beginning) then this must produce a wave form that has a > non zero frequency that is the difference between the two frequencies. �As > this is a standard sine waveform, it is symmetric about the X axis. �This > is, its an AC only waveform with an amplitude of one half (due to the 1/2 > previous to the cos(X1-X2)). > > Similarly with the 1/2 * cos(X1+X2) term, the result > > �1/2 * cos((360 * (f1+f2) * t) + 2 * Initial_Phase_Angle) > > � is a waveform that is the addition of the two frequencies. �This > produces a second waveform of half the amplitude of the input waveforms. > In this case it has a frequency that is the sum of the two frequencies. �As > this is also a standard sine waveform, it is also symmetric about the X > axis. > > The addition of two pure AC complete waveforms always produces a resultant > waveform that is pure AC only. �The word 'complete' is here used to > indicate that we are looking at complete cycles, not just part of a cycle > of the waveform. �As can be seen from the above, so long as the frequencies > are different, there is no DC offset in the result. > > Lets assume that X1 and X2 above are exactly equal. �Start again with the > initial equation: > > sin(X1) * sin(X1) = 1/2 * cos(X1-X1) - 1/2 * cos(X1+X1) > > As Cos(X1-X1) = Cos(0) = 1 > > Sin(X1) * Sin(X1) = 1/2 - (1/2 * cos(2 * X1)) > > In this case, the same input sine waves multiplied by each other produce a > DC offset (equal to half the amplitude of the input sine waves) plus a wave > at double the frequency i.e. (2 * X1) and half the amplitude of the input > sine waves. > > One strange aspect of sine waves is that whether a wave is added or > subtracted doesn't change its position with respect to the Y axis. �That > is, it doesn't give or change the wave's DC offset. �As the wave is minus, > it is equivalent to the wave being 180 degrees out of phase. �That is, the > following equation is equivalent to the last one: > > Sin(X1) * Sin(X1) = 1/2 + (1/2 * cos((2 * X1) + 180 degrees)) > > Lets assume that X1 and X2 above are equal except that X2 is phase shifted > forward by 90 degrees. > > Sin(X1) * Sin(X1 + 90 degrees) = 1/2 * cos(0 - 90 degrees) - 1/2 * cos((2 > * X1) + 90 degrees) > > As Cos(-90 degrees) = 0 � > > Sin(X1) * Sin(X1 + 90 degrees) = - 1/2 * cos((2 * X1) + 90 degrees) > > � � � � � � � � � � � �= � 1/2 * cos((2 * X1) - 90 degrees) > > In the case of waveforms this means that when input sine waves of the same > frequency but with a phase offset of 90 degrees are multiplied by each > other, a wave at double the frequency i.e. (2 * X1) and half the amplitude > of the input sine waves is produced. �As the wave is minus, it is > equivalent to the wave being 180 degrees out of phase but the extra 90 > degrees in ((2 * X1) + 90 degrees) then shifts it back 90 degrees. �Note > that this wave is symmetrical about the X axis. �There is no DC offset. > > If we did the same as above but with X2 phase shifted back by 90 degrees > (the same as phase shifted forward by 270 degrees) then we would get a > similar result in that the resultant wave has no DC offset. > > Sin(X1) * Sin(X1 - 90 degrees) = - 1/2 * cos((2 * X1) - 90 degrees) > > Lets assume that X1 and X2 above are equal except that X2 is phase shifted > by 180 degrees. �This is the equivalent to being inverted. � Starting at > our initial equation we see that: > > sin(X1) * (-1 * sin(X1)) = -1 * ( 1/2 * cos(X1-X1) - 1/2 * cos(X1+X1)) > � � � � � � � � � � �= �1/2 * cos(X1+X1) - 1/2 * cos(X1-X1) > > �Again, as cos(0) = 1 > > � � � � � � � � � � �= �(1/2 * cos(2*X1)) - 1/2 > > In the case of waveforms this means that when input sine waves that are > equal except one is inverted from the other, are multiplied by each other, > they produce a minus DC offset (equal to half the amplitude of the input > sine waves) plus a wave at double the frequency i.e. (2 * X1) and half the > amplitude of the input sine waves. > > From these equations we can see that we only obtain a DC offset when we > multiply two waveforms if they are exactly at the same frequency. �Even > then, we only obtain the DC offset if the waves are not at 90 or 270 > degrees phase difference from each other. �The DC offset reaches a maximum > positive value when the waves are exactly in phase with each other and > reach a maximum negative value when the waves are 180 degrees out of > phase. > > Also note that we have simplified the basic equation by leaving off the > amplitudes. �We have assumed that the amplitudes of the two input waves are > both 1. �Assume the amplitude of the sin(X1) input wave is A and the > amplitude of the sin(X2) input wave is B then the original equation would > be > > (A * sin(X1)) * (B * sin(X2)) = (A * B ) * (1/2 * cos(X1-X2) - 1/2 * > cos(X1+X2)) > > If the amplitudes of the input sin(X1) and sin(X2) waves is not 1 then all > the previous formulas can be corrected by simply multiply the results by a > factor of (A * B). > > The previous equations should have explained the workings of the Phase > Lock Loop. �The following expands this to show the workings of the Two > Channel Lock-In Amplifier. > > Lets see what happens if we create an X2 waveform that is the same as an > input X1 except that it is phase shifted an angle Theta from X1. > > sin(X1) * sin(X1 + Theta degrees) = (1/2 * cos(X1-X1 - Theta degrees)) - > (1/2 * cos(2*X1 + Theta degrees)) > � � � � � � � � �= �(1/2 * cos(- Theta degrees)) - (1/2 * cos(2*X1 + Theta > degrees)) > > The second term (1/2 * cos(2*X1 + Theta degrees) by itself is a pure sine > wave that is symmetrical about the X axis. �It has no DC component. �If we > are only interested in the resultant DC component then we can throw away > this term. �The first term gives us the value of the resultant waveform if > we averaged it, filtering out the AC component. �The DC component would be > > 1/2 * cos(- Theta degrees) = 1/2 * cos(Theta degrees) > > cos(Theta degrees) is a constant so we have a constant DC offset of 1/2 * > cos(Theta degrees). �If we perform this with an input wave amplitude of A > and an internally created wave of amplitude B, and measure the resultant DC > offset as Y1 then > > � �Y1 = 1/2 * A * B * cos(Theta degrees) > > If we now create an X3 waveform that is the same as X2 except that it is > phase sifted by (Theta plus 90) degrees from X1 and use that to multiply > against another copy of the input X1 waveform. > > sin(X1) * sin(X1 + (Theta + 90) degrees) = (1/2 * cos(X1-X1 - (Theta + 90) > degrees)) - (1/2 * cos(2 *X1 + (Theta + 90) degrees)) > > Again we are only interested in the DC component so we throw away the > second term. �The first term gives us > > � � �1/2 * cos((Theta + 90) degrees) = 1/2 sin(Theta degrees) > > If we perform this with an input wave amplitude of A and an internally > created wave of amplitude of B, (assume that X3 is the same amplitude as > X2) and measure the resultant DC offset as Y2 then > > � Y2 = 1/2 * A * B * sin(Theta degrees). > > If we now square Y1 and Y2, add these squares together and then take the > square root > > (Square root of (Y1**2 + Y2**2)) > �= square root of �((1/2 * A * B * cos(Theta degrees))**2 + (1/2 * A * B * > sin(Theta degrees))**2) > �= square root of �(1/4 *A*A*B*B* (cos(Theta degrees)**2) �+ �1/4 > *A*A*B*B* sin(Theta degrees)**2) > �= square root of �(1/4 *A*A*B*B* (cos(Theta degrees)**2) �+ �1/4 > *A*A*B*B* sin(Theta degrees)**2) > �= square root of �(1/4 *A*A*B*B * ((sin(Theta degrees)**2) + (cos(Theta > degrees)**2))) > > � � �Note: sin(X)**2 + cos(X)**2 = 1 > � � �This is a trigonometrical identity that is easily derived from the > main trigonometrical identity we have been using. �Using this reduces the > above to > > �= square root of �(1/4*A*A*B*B*) > �= 1/2 * A * B > > Consequently, if we know B then the Amplitude A can be calculated from > > � A = 2 * (Square root (Y1**2 + Y2**2)) / B > > Using the above, if we have an input waveform X1 that contained a known > frequency but we didn't know the phase or amplitude of this frequency, we > can find the amplitude by creating two reference waveforms, both of > Amplitude B and frequency the same as X1. �The second internally created > reference waveform would be created with a 90 degree phase shift from the > first reference waveform. �We also create a second copy of the input > waveform. > > We now multiply the input waveform by the first reference waveform and > average the result so that we measure the resultant Y1. �We also multiply > the copy of the input waveform by the second reference waveform and average > the result so that we measure the resultant Y2. �The amplitude A of the > known frequency within the input waveform can now be calculated as > > � A = 2 * (Square root (Y1**2 + Y2**2)) / �B > > Lets denote by Y3 what would have been measured had the internally created > reference waveform been exactly in phase with the input frequency. �This > gets rid of the 1/2 that's needed when comparing to the input waveform. > The measured Y1 and Y2 are the sine and cosine components of the Y3 > amplitude. �They have the same relationship as the clock hand (of Y3 > length) has to the X and Y axis or the same relationship as the hypotenuse > (of Y3 length) of a right angle triangle. > > Consequently, we can find the true amplitude or "magnitude" as the > hypotenuse of this right triangle by taking the square root of the sum of > the squares of the sine and cosine components. This is called the "vector > sum" of the components. �Similarly, we can find the phase as the angle > whose tangent is equal to the sine component divided by the cosine > component. > > The sine and cosine components are often referred to as the "in-phase" and > "quadrature" components. Some engineering and mathematicians prefer to use > "real" and "imaginary" for the sine and cosine components. � From this they > create some very intimidating expressions using an imaginary number (square > root of -1) in a style of maths called "complex notation". > > Also note that the first internally created reference frequency can be at > any phase difference to the actual signal. �Given any particular pure > sinewave, without a DC component, at a particular frequency f1, it can be > fully described by specifying its amplitude and phase. �Alternatively, the > sine wave can be described by specifying two amplitudes, these being the > equivalent in-phase and quadrature components of that frequency. �As we can > change where (which phase angle) our first reference frequency is, there is > effectively an infinite number of pairs of amplitudes that can be used as > an equivalent to the phase plus amplitude pair. �As long as we keep in mind > where the the reference is in each case, all these pairs of amplitudes are > equivalents. � > > With VLF metal detectors, where we are both transmitting the frequency on > one coil and receiving the resultant frequency on another coil, the sine > wave being output is normally used as our reference (sine component) > frequency for the primary received frequency. �Consequently, in this case > it clearly makes sense to call the sine wave component the in-phase > component. �Changes in the received amplitude of this component are > normally caused by resistive property effects in the soil and anomalies > near the coil. � Note that resistive is here used to indicated properties > that change the received voltage in phase with the signal like resistance > does. �It doesn't necessarily mean resistance itself. Changes in the > received amplitude of the cosine component are normally caused by reactive > (such as capacitance and inductance) property effects of the soil and > anomalies near the coil. > > When programming using any of the above equations, you need to recognise > the order in which the sine or cosine waves will be processed. �The sine > graphs with time increasing on the X axis are a representation of waves > that are travelling from right to left. �Generally, we expect things on > paper to be represented from left to right so this works out to be opposite > to what many people expect. > > To confirm that it is this direction note what happens if we place the > sine or cosine graph onto a long tape with time in seconds marked as normal > along the X axis. �We cover over the tape except that we leave a small view > port that allows us to see the tape at the zero time mark (normally the Y > axis). �We now fire a starters gun to mark our zero time point. �Following > the gun firing we pull the tape so that the time marks of 1 second, 2 > seconds etc are viewable at our view port at 1 second, 2 seconds etc after > our zero time. �To do this, we have to pull the tape to the left so that > the graphed sine wave is moving right to left. > > The conclusion here is that if you have a sine wave and a cosine wave > being input to the processor, the cosine wave will reach the processor at > 1/4 wave cycle ahead of the sine wave or 3/4 wave cycle after the sine > wave. �The equation cos(X) = sin(X + 90 degrees) indicates that cosine is > before sine by 90 degrees. > > We should also point out the usefulness of the earlier discussed fact that > multiplying two frequencies produces a sine wave that is the difference of > the two frequencies plus a sine wave that is the sum of the two > frequencies. > > As an example, Let us assume that we are interested in using an audio card > input to look at an input that has a frequency of 300Khz. �This is a higher > frequency than current audio cards can handle correctly. �To make it > suitable we could create an oscillator that is for example a fixed 295khz. > We then multiply this oscillator frequency to the 300Khz input. �The result > is a waveform at (300 + 295)Khz plus a wave form at (300 - 295)Khz. �We can > then use a low pass filter to filter out the 595Khz waveform leaving us > with just a 5Khz waveform which is very easy for the audio card to handle. > As the original 300Khz either changes in frequency or changes in amplitude, > the 5Khz input to the audio card will similarly change. �This principle is > used in many applications such as the super-heterodyne receiver > > It also explains a common problem in many wave systems called > Intermodulation Distortion. This distortion occurs when two sine waves of > frequencies X1 and X2 in the input result in the creation of other unwanted > frequency components. �These frequencies include (X1+X2), (X1-X2), > (2*X1-x2), (2*X2-X1), and generally (mX1 � nX2) for integer m and n. > Generally, the size of the unwanted frequency components falls rapidly as m > and n increase. > > A similar explanation for some of the above is given at the following > link:http://www.daqarta.com/eex01.htmThat's not maths...that's kids stiff from school. For DSP you will need the Fourier transform,Fourier Series,Convolution integral and all the discrete equivalents. z-transform of course,power series expansions,Laplace transforms etc etc Not for amateurs. Hardy
Reply by ●March 17, 20092009-03-17
On Tue, 17 Mar 2009 10:59:56 -0700, Rune Allnor wrote:> On 17 Mar, 18:34, bellda2...@cox.net wrote: >> Are you assuming that you will only have metal mines or mines with a >> significant amount of metal? > > If so, it would be a classic ostrich - I've seen it all too often in the > past. People tend to focus on what they *can* find, ingnore all the > cases they can *not* handle, and go on as if what they can do is > sufficient. > > Of course, that might be OK in a comm system which is subjected to > multipath propagation; most people would argue it's not, with mines > involved. There are some applications where anything less than 100% > success is unacceptable. > > RuneUnfortunately, mine detection is _not_ 100%, yet it still must be done. No matter what else happens, the locals will go out and plow those fields eventually, because the possibility of getting blown up is better than the certainty of starving. Sooner or later all the mines will be 'detected' -- it's just a question of whether you have to use up one farmer's kid for each mine, or whether you use up one mine detection expert for each 100 mines. There are many mines that cannot be detected with a metal detector. Still, using a metal detector as _part_ of the effort isn't a bad thing if it can reduce the incidence of injury and death. -- http://www.wescottdesign.com
Reply by ●March 17, 20092009-03-17
On 17 Mar, 20:02, Tim Wescott <t...@seemywebsite.com> wrote:> On Tue, 17 Mar 2009 10:59:56 -0700, Rune Allnor wrote: > > On 17 Mar, 18:34, bellda2...@cox.net wrote: > >> Are you assuming that you will only have metal mines or mines with a > >> significant amount of metal? > > > If so, it would be a classic ostrich - I've seen it all too often in the > > past. People tend to focus on what they *can* find, ingnore all the > > cases they can *not* handle, and go on as if what they can do is > > sufficient....> There are many mines that cannot be detected with a metal detector. � > Still, using a metal detector as _part_ of the effort isn't a bad thing > if it can reduce the incidence of injury and death.Well, this and other problems can only be solved by a divid-and- conquer approach: First take the easy stuff (e.g. by metal detectors), then the harder stuff, after that the even harder stuff etec, until one is done. However, this requires a mind-set that any given methods has limited scope, and complementary methods exist. To my experience, there aren't a lot of people around (any at all?) who have that mind-set: Whatever method people come up with is seen as The One and Final Answer, and anything not covered by it is irrelevant. Rune
Reply by ●March 17, 20092009-03-17
On Tue, 17 Mar 2009 12:35:13 -0700, Rune Allnor wrote:> On 17 Mar, 20:02, Tim Wescott <t...@seemywebsite.com> wrote: >> On Tue, 17 Mar 2009 10:59:56 -0700, Rune Allnor wrote: >> > On 17 Mar, 18:34, bellda2...@cox.net wrote: >> >> Are you assuming that you will only have metal mines or mines with a >> >> significant amount of metal? >> >> > If so, it would be a classic ostrich - I've seen it all too often in >> > the past. People tend to focus on what they *can* find, ingnore all >> > the cases they can *not* handle, and go on as if what they can do is >> > sufficient. > ... >> There are many mines that cannot be detected with a metal detector. >> Still, using a metal detector as _part_ of the effort isn't a bad thing >> if it can reduce the incidence of injury and death. > > Well, this and other problems can only be solved by a divid-and- conquer > approach: First take the easy stuff (e.g. by metal detectors), > then the harder stuff, after that the even harder stuff etec, until one > is done. > > However, this requires a mind-set that any given methods has limited > scope, and complementary methods exist. To my experience, there aren't a > lot of people around (any at all?) who have that mind-set: Whatever > method people come up with is seen as The One and Final Answer, and > anything not covered by it is irrelevant. > > RuneIf you dig a bit you'll find that there's quite a bit of attention being paid to the land mine problem. People keep trying to get complaisant, but other people keep getting completely or partially blown up, which tends to focus attention on the deficiencies of any one method. -- http://www.wescottdesign.com
Reply by ●March 17, 20092009-03-17
Tim Wescott <tim@seemywebsite.com> wrote:> Unfortunately, mine detection is _not_ 100%, yet it still must be done. > No matter what else happens, the locals will go out and plow those fields > eventually, because the possibility of getting blown up is better than > the certainty of starving. Sooner or later all the mines will be > 'detected' -- it's just a question of whether you have to use up one > farmer's kid for each mine, or whether you use up one mine detection > expert for each 100 mines.I have wondered about using a pulsed RF source and a blast shield. The shield would keep the explosion fragments away from the user and the RF source. That probably depends on there being enough metal near the explosive charge to generate a spark, but not much more than that. The frequency would be chosen to penetrate the ground to a reasonable depth, and not be too hard to generate. (Maybe left over magnetrons from microwave ovens.)> There are many mines that cannot be detected with a metal detector. > Still, using a metal detector as _part_ of the effort isn't a bad thing > if it can reduce the incidence of injury and death.-- glen
Reply by ●March 17, 20092009-03-17
glen herrmannsfeldt wrote:> Tim Wescott <tim@seemywebsite.com> wrote: > >> Unfortunately, mine detection is _not_ 100%, yet it still must be done. >> No matter what else happens, the locals will go out and plow those fields >> eventually, because the possibility of getting blown up is better than >> the certainty of starving. Sooner or later all the mines will be >> 'detected' -- it's just a question of whether you have to use up one >> farmer's kid for each mine, or whether you use up one mine detection >> expert for each 100 mines. > > I have wondered about using a pulsed RF source and a blast > shield. The shield would keep the explosion fragments > away from the user and the RF source. That probably depends > on there being enough metal near the explosive charge to > generate a spark, but not much more than that. > > The frequency would be chosen to penetrate the ground to > a reasonable depth, and not be too hard to generate. > (Maybe left over magnetrons from microwave ovens.) > >> There are many mines that cannot be detected with a metal detector. >> Still, using a metal detector as _part_ of the effort isn't a bad thing >> if it can reduce the incidence of injury and death.Chain flails? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●March 17, 20092009-03-17
On 17 Mar, 21:49, Tim Wescott <t...@seemywebsite.com> wrote:> On Tue, 17 Mar 2009 12:35:13 -0700, Rune Allnor wrote: > > On 17 Mar, 20:02, Tim Wescott <t...@seemywebsite.com> wrote: > >> On Tue, 17 Mar 2009 10:59:56 -0700, Rune Allnor wrote: > >> > On 17 Mar, 18:34, bellda2...@cox.net wrote: > >> >> Are you assuming that you will only have metal mines or mines with a > >> >> significant amount of metal? > > >> > If so, it would be a classic ostrich - I've seen it all too often in > >> > the past. People tend to focus on what they *can* find, ingnore all > >> > the cases they can *not* handle, and go on as if what they can do is > >> > sufficient. > > ... > >> There are many mines that cannot be detected with a metal detector. > >> Still, using a metal detector as _part_ of the effort isn't a bad thing > >> if it can reduce the incidence of injury and death. > > > Well, this and other problems can only be solved by a divid-and- conquer > > approach: First take the easy stuff (e.g. by metal detectors), > > then the harder stuff, after that the even harder stuff etec, until one > > is done. > > > However, this requires a mind-set that any given methods has limited > > scope, and complementary methods exist. To my experience, there aren't a > > lot of people around (any at all?) who have that mind-set: Whatever > > method people come up with is seen as The One and Final Answer, and > > anything not covered by it is irrelevant. > > > Rune > > If you dig a bit you'll find that there's quite a bit of attention being > paid to the land mine problem. �I know. Once upon a time we tried to get funding for a project that used certain seismic waves I had analyzed as basis for a land mine 'sonar'. That is, use a seismic tranciever and use reflections to detect reflections from objects in the soil. There were two things that put me off the idea: - My boss presented the idea to the potential funders as The One And Final Solution for the problem, dismissing off hand other alternative approaches. - My boss dismissed off hand any problems with our approach, like rocks, lumps of clay, potatoes or other root fruits; any other inhomogeneites in the soil that would cause reflections. Once you've seen those kinds of things up close, you recognize the mindset just about everywhere.> People keep trying to get complaisant, > but other people keep getting completely or partially blown up, which > tends to focus attention on the deficiencies of any one method.Right. Which raises the question if one should use brute force instead, at least in agricultural areas. Like those vehicles with whirling chains at the front, which beat the mines so they detonate. Get something like that to be used on a regular tractor, and clear at least enough land so the locals can start growing their own food. Rune
