I am trying to make an indicator out of the --straight line-- linear regression pinescript drawing tool included in tradingview (not their regression channel tool or any type least squares moving average/moving regression which plots a continuous curved line). I wish to see one straight line through the chart prices defined in the custom time period input. The plot of this straight line will have acustom start time and length to define the end (or even better an end time instead f length). Next, I wish to add a rate of return feature by taking the total price move of the straight regression line from beginning to end, divided by # of time periods between the start and end date. I have tried the code seen here but I am getting a curved line, not straight. Then for the next step, I am not sure how to get the start and end price point of the regression line to divide by the total time period.
//#version=4
study("lreg ", shorttitle="run-ave", overlay=true)
timeYear = input(2022, title="Year", minval=1991, maxval=2100, type=input.integer)
timeMonth = input(1, title="Month", minval=1, maxval=12, type=input.integer)
timeDay = input(04, title="Day", minval=1, maxval=31, type=input.integer)
timeHours = input(9, title="Hours", minval=0, maxval=23, type=input.integer)
timeMinutes = input(30, title="Minutes", minval=0, maxval=59, type=input.integer)
timeSeconds = input(0, title="Seconds", minval=0, maxval=59, type=input.integer)
length = input(14)
// Initilization of variables only once
var delta = 0
// start time at 0 from a particular time interval
if(year == timeYear and month == timeMonth and dayofmonth == timeDay and hour == timeHours and minute == timeMinutes and second == timeSeconds)
delta := 0
// Count number of bars
if(year >= timeYear and month >= timeMonth and dayofmonth > timeDay)
delta += 1
plotchar(delta, title="days passed from startdate", color=color.green, char='')
// rate of change and moving average of ROC
source = input(close, "Source")
float lreg = na
float lregg = 0
if delta >= 1
lreg:= linreg(close, length, offset=0)
plot(lreg)
Related
I implemented a 3d linear kalman filter to track the position of object. here are the details of the KF.
my state is 3d position only and my input(control vector) is 3d velocity (vx,vy,vz). and measurement is 3d position.
dt = time interval in state update
std_vx,std_vy,std_vz - standard deviation in control input e.g. 2 mtrs/sec
std_cooperative_pos_x,std_cooperative_pos_y,std_cooperative_pos_z - standard deviation of measurement error e.g. around 40 mtrs each axis
A = numpy.matrix([[1,0,0],[0,1,0],[0,0,1]]) # State transition matrix
B = numpy.matrix([[dt,0,0],[0,dt,0],[0,0,dt]]) # control input marix
H = numpy.matrix([[1,0,0],[0,1,0],[0,0,1]])
Q = numpy.matrix([[std_vxstd_vxdt, 0, 0],[0, std_vystd_vydt, 0],[0, 0 ,std_vzstd_vzdt]] ) # = Ex error in input , process error
R = numpy.matrix([[std_cooperative_pos_x2,0,0],[0,std_cooperative_pos_y2,0],[0,0,std_cooperative_pos_z**2]]) # measurement co- variance matrix
P = numpy.matrix([[1,0,0],[0,1,0],[0,0,1]])
here are my few questions :
how much the time interval that is dt value matters in the performance of filter. as i want dt such as 10 sec , or 20 sec .....or upto 200 secs. for dt values upto 10 sec i am able to get some improvement i my track but beyond that its not working . I am doing something wrong or its natural .
i implemented a linear filter is that right as to me combination of postion and velocity over time appears to be linear only .
thanks in advance for your time and help.
The process noise matrix 'Q' is a covariance matrix - it must contain variances rather than standard deviations. You have:
Q = numpy.matrix([[std_vxstd_vxdt, 0, 0],[0, std_vystd_vydt, 0],[0, 0 ,std_vzstd_vzdt]] ) # = Ex error in input , process error
You want:
Q = numpy.matrix([[std_vxstd_vx*std_vxstd_vx * dt *dt, 0, 0], ...
As you can see, the process noise matrix grows with dt^2. So the time interval is very important if it is the time between measurements. You need only one propagation step between measurements for this filter. There is no advantage to multiple steps (a smaller dt) unless you are approximating non-linear state propagation or there's more measurements available.
I have an acceleration data for X-axis and time vector for it. I determined the peaks more than threshold and now I should find the FFT for every peak.
As result I have this:
Peak Value 1 = 458, index 1988
Peak Value 2 = 456, index 1990
Peak Value 3 = 450, index 12081
....
Peak Value 9 = 432, index 12151
To find these peaks I used the peakfinder script.
The command [peakLoc, peakMag] = peakfinder(x0,...) gives me location and magnitude of peaks.
Also I have the Time (from time data vector) for each peak.
So what I suppose, that I should take every peak, find its width (or some data points around the peak) and make the FFT. Am I right? Could you help me in that?
I'm working in Octave and I'm new here :)
Code:
load ("C:\\..patch..\\peakfinder.m");
d =dlmread("C:\\..patch..\\acc2.csv", ";");
T=d(:,1);
Ax=d(:,2);
[peakInd peakVal]=peakfinder(Ax,10,430,1);
peakTime=T(peakInd);
[sortVal sortInd] = sort(peakVal, 'descend');
originInd = peakInd(sortInd);
for k = 1 : length(sortVal)
fprintf(1, 'Peak #%d = %d, index%d\n', k, sortVal(k), originInd (k));
end
plot(T,Ax,'b-',T(peakInd),Ax(peakInd),'rv');
and here you can download the data http://www.filedropper.com/acc2
FFT
d =dlmread("C:\\..path..\\acc2.csv", ";");
T=d(:,1);
Ax=d(:,2);
% sampling frequency
Fs_a=2000;
% length of FFT
Length_Ax=numel(Ax);
% number of lines of Fourier spectrum
fft_L= Fs_a*2;
% an array of time samples
T_Ax=0:1/Fs_a: Length_Ax;
fft_Ax=abs(fft(Ax,fft_L));
fft_Ax=2*fft_Ax./fft_L;
F=0:Fs_a/fft_L:Fs_a/2-1/fft_L;
subplot(3,1,1);
plot(T,Ax);
title('Ax axis');
xlabel('time (s)');
ylabel('amplitude)'); grid on;
subplot(3,1,2);
plot(F,fft_Ax(1:length(F)));
title('spectrum max Ax axis');
xlabel('frequency (Hz)');
ylabel('amplitude'); grid on;
It looks like you have two clusters of peaks, so I would plot the data over three plots: one of the whole timeseries, one zoomed in on the first cluster, and the last one zoomed in on the second cluster (note I have divided all your time values by 1e6 otherwise the tick labels get ugly):
figure
subplot(3,1,1)
plot(T/1e6,Ax,'b-',peakTime/1e6,peakVal,'rv');
subplot(3,1,2)
plot(T/1e6,Ax,'b-',peakTime(1:4)/1e6,peakVal(1:4),'rv');
axis([0.99*peakTime(1)/1e6 1.01*peakTime(4)/1e6 0.9*peakVal(1) 1.1*peakVal(4)])
subplot(3,1,3)
plot(T/1e6,Ax,'b-',peakTime(5:end)/1e6,peakVal(5:end),'rv');
axis([0.995*peakTime(5)/1e6 1.005*peakTime(end)/1e6 0.9*peakVal(5) 1.1*peakVal(end)])
I have set the axes around the extreme time and acceleration values, using some coefficients to have some "padding" around (the values of these coefficients were obtained through trial and error). This gives me the following plot, hopefully this is the sort of thing you are after. You can add x and y labels if you wish.
EDIT
Here's how I would do the FFT:
Fs = 2000;
L = length(Ax);
NFFT = 2^nextpow2(L); % Next power of 2 from length of Ax
Ax_FFT = fft(Ax,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2+1);
% Plot single-sided amplitude spectrum.
figure
semilogx(f,2*abs(Ax_FFT(1:NFFT/2+1))) % using semilogx as huge DC component
title('Single-Sided Amplitude Spectrum of Ax')
xlabel('Frequency (Hz)')
ylabel('|Ax(f)|')
ylim([0 300])
giving the following result:
I need to output a 32bit bit-stream onto a pin in verilog. I know verilog has the streaming operators pack and unpack but I do not believe they will do what I want it to do.
I have 32x512 FIFO RAM in which data is stored. Data for the variable "I" stored on the first 32 bits and the data for variable "Q" is stored on the next 32 bits (the rest of FIFO saves data in this alternating fashion). I need to continually get a 32bit stream off the FIFO RAM and output the 32bit data stream onto a pin. My FIFO has three output signals(a signal for the 32 bit data stream(32_data), a signal to say when the FIFO is empty (32_empty), and a signal to say when the FIFO is full(32_full)) My sudo code is the following (It's sudo code because I know how to do everything else but the part I need help with and I wanted to keep it simple for understanding):
process # posedge clock
begin
if (32_empty != 1) then //if the FIFO has data
if (32_full == 1) then //if the FIFO is full, then we lose data (for testing purposes to know if I need to make the RAM bigger
PIN_1 <= 1; //output onto a pin that the FIFO is full
PIN_2 <= 0; //clear pin 2 from outputting data for "I"
PIN_3 <= 0; //clear pin 3 from outputting data for "Q"
else if (en_Q == 0)
(stream 32bit data for variable "I" onto pin 2) //variable "I" output//HELP-This is where I need help figuring out how to stream the output, 32_data, onto a pin
en_Q <= ~en_Q; // toggle en_Q so next 32bit stream will be for "Q"
else if (en_Q ==1)
(stream 32bit data for variable "Q" onto pin 3) //variable "Q" output//HELP-This is where I need help figuring out how to stream the output, 32_data, onto a pin
en_Q <= ~en_Q; // toggle en_Q so next 32bit stream will be for "I"
end
If you could help me with figuring out how to stream a 32 bit data stream onto a pin, that would be great!
Thanks in advance
I have added the suggestion. Could I put the data on the pins with a for loop? The following is my code segment and the bottom part is the shift register and outputting to a pin:
`// Wires and registers related to data capturing
wire capture_clk;
reg [31:0] capture_data;
wire capture_en;
reg [4:0] slowdown;
wire capture_full;
reg capture_open;
reg capture_open_cross;
reg capture_has_been_full;
reg capture_has_been_nonfull;
reg has_been_full_cross;
reg has_been_full;
// Data capture section
// ====================
always #(posedge capture_clk)
begin
if (capture_en)
capture_data <= user_w_write_32_data; // Bogus data source
// The slowdown register limits the data pace to 1/32 the bus_clk
// when capture_clk = bus_clk. This is necessary, because the
// core in the evaluation kit is configured for simplicity, and
// not for performance. Sustained data rates of 200 MB/sec are
// easily reached with performance-oriented setting.
// The slowdown register has no function in a real-life application.
slowdown <= slowdown + 1;
// capture_has_been_full remembers that the FIFO has been full
// until the file is closed. capture_has_been_nonfull prevents
// capture_has_been_full to respond to the initial full condition
// every FIFO displays on reset.
if (!capture_full)
capture_has_been_nonfull <= 1;
else if (!capture_open)
capture_has_been_nonfull <= 0;
if (capture_full && capture_has_been_nonfull)
capture_has_been_full <= 1;
else if (!capture_open)
capture_has_been_full <= 0;
end
// The dependency on slowdown is only for bogus data
assign capture_en = capture_open && !capture_full &&
!capture_has_been_full &&
(slowdown == 0);
// Clock crossing logic: bus_clk -> capture_clk
always #(posedge capture_clk)
begin
capture_open_cross <= user_r_read_32_open;
capture_open <= capture_open_cross;
end
// Clock crossing logic: capture_clk -> bus_clk
always #(posedge bus_clk)
begin
has_been_full_cross <= capture_has_been_full;
has_been_full <= has_been_full_cross;
end
// The user_r_read_32_eof signal is required to go from '0' to '1' only on
// a clock cycle following an asserted read enable, according to Xillybus'
// core API. This is assured, since it's a logical AND between
// user_r_read_32_empty and has_been_full. has_been_full goes high when the
// FIFO is full, so it's guaranteed that user_r_read_32_empty is low when
// that happens. On the other hand, user_r_read_32_empty is a FIFO's empty
// signal, which naturally meets the requirement.
assign user_r_read_32_eof = user_r_read_32_empty && has_been_full;
assign user_w_write_32_full = 0;
// The data capture clock here is bus_clk for simplicity, but clock domain
// crossing is done properly, so capture_clk can be an independent clock
// without any other changes.
assign capture_clk = bus_clk;
async_fifo_32x512 fifo_32 //FIFO created using Xilinx FIFO Generator Wizard
(
.rst(!user_r_read_32_open),
.wr_clk(capture_clk),
.rd_clk(bus_clk),
.din(capture_data),
.wr_en(capture_en),
.rd_en(user_r_read_32_rden),
.dout(user_r_read_32_data),
.full(capture_full),
.empty(user_r_read_32_empty)
);
reg Q_en = 1'b0; //starting value is 0 because first 32bit is I
reg [31:0] data_outI;
reg [31:0] data_outQ;
reg I;
reg Q;
integer counter;
always #(posedge bus_clk) begin
if(Q_en == 1'b0) begin //To output to I signal
data_outI <= user_r_read_32_data;
for (counter = 0; counter < 32; counter = counter + 1) begin //output to pins
I = data_outI[0];
data_outI <= (data_outI >> 1);
Q = data_outQ[0];
data_outQ <= (data_outQ >> 1);
end
Q_en <= ~Q_en;
end
else if(Q_en == 1'b1) begin //To output to Q signal
data_outQ <= user_r_read_32_data;
for (counter = 0; counter < 32; counter = counter + 1) begin //output to pins
I = data_outI[0];
data_outI <= (data_outI >> 1);
Q = data_outQ[0];
data_outQ <= (data_outQ >> 1);
end
Q_en <= ~Q_en;
end
end
assign PS_GPIO_ONE_I = I; //Assign Pin I
assign PS_GPIO_TWO_Q = Q; //Assign Pin Q
`
Basically, you'd want to do something like the following:
Fetch an item from the fifo into a 32 bit register (lets call data)
Each clock cycle, put the lsb of data onto the pin, and then right shift data by one value.
Keep repeating this shifting for 32 clock cycles, until all of the data has been shifted out.
Toggle the value of en_Q, and fetch another 32 bit item.
You should be able to make a small state machine that can handle this sequence. You can't shift out all 32 bits in a single clock cycle as you have done in your pseudo-code, unless you also have a 32x clock available, and that would likely be a more complicated design.
Currently I'm having trouble creating multiple format in SSRS for one Axis. The current situation is that I have a measure that could bring a maximum value of 1.2M (Currency) but an average of 500K depending the period. Originally I have the labelformat property set to 0,,M, but this doesn't bring the correct scale when the value is less than a million.
This are the current solution I have tried:
=IIF(Sum(Fields!Current_Year.Value, "CustomerSales") > Sum(Fields!Last_Year.Value, "CustomerSales"),
IIF(Sum(Fields!Current_Year.Value, "CustomerSales") < 1000000, "0,K", "0,,M"),
IIF(Sum(Fields!Last_Year.Value, "CustomerSales") < 1000000, "0,K", "0,,M"))
Public Function LabelCustomFormat(ByVal CY As Integer, ByVal LY As Integer) As String
If CY > LY Then
If(CY < 1000000) THEN
Return "0,K"
Else
Return "0,,M"
End If
Else
IF(LY < 1000000) THEN
Return "0,K"
Else
Return "0,,M"
End If
End If
End Function
Is it possible to have the different scale in one Axis of a Chart? If so, please help.
Well after trying to make a dynamic scale in the axis without success, I left the label format with the expression 0,,M and set the interval to a 1M instead of 500K.
I'm trying to compute the Fourier coefficients for a waveform using MATLAB. The coefficients can be computed using the following formulas:
T is chosen to be 1 which gives omega = 2pi.
However I'm having issues performing the integrals. The functions are are triangle wave (Which can be generated using sawtooth(t,0.5) if I'm not mistaking) as well as a square wave.
I've tried with the following code (For the triangle wave):
function [ a0,am,bm ] = test( numTerms )
b_m = zeros(1,numTerms);
w=2*pi;
for i = 1:numTerms
f1 = #(t) sawtooth(t,0.5).*cos(i*w*t);
f2 = #(t) sawtooth(t,0.5).*sin(i*w*t);
am(i) = 2*quad(f1,0,1);
bm(i) = 2*quad(f2,0,1);
end
end
However it's not getting anywhere near the values I need. The b_m coefficients are given for a
triangle wave and are supposed to be 1/m^2 and -1/m^2 when m is odd alternating beginning with the positive term.
The major issue for me is that I don't quite understand how integrals work in MATLAB and I'm not sure whether or not the approach I've chosen works.
Edit:
To clairify, this is the form that I'm looking to write the function on when the coefficients have been determined:
Here's an attempt using fft:
function [ a0,am,bm ] = test( numTerms )
T=2*pi;
w=1;
t = [0:0.1:2];
f = fft(sawtooth(t,0.5));
am = real(f);
bm = imag(f);
func = num2str(f(1));
for i = 1:numTerms
func = strcat(func,'+',num2str(am(i)),'*cos(',num2str(i*w),'*t)','+',num2str(bm(i)),'*sin(',num2str(i*w),'*t)');
end
y = inline(func);
plot(t,y(t));
end
Looks to me that your problem is what sawtooth returns the mathworks documentation says that:
sawtooth(t,width) generates a modified triangle wave where width, a scalar parameter between 0 and 1, determines the point between 0 and 2π at which the maximum occurs. The function increases from -1 to 1 on the interval 0 to 2πwidth, then decreases linearly from 1 to -1 on the interval 2πwidth to 2π. Thus a parameter of 0.5 specifies a standard triangle wave, symmetric about time instant π with peak-to-peak amplitude of 1. sawtooth(t,1) is equivalent to sawtooth(t).
So I'm guessing that's part of your problem.
After you responded I looked into it some more. Looks to me like it's the quad function; not very accurate! I recast the problem like this:
function [ a0,am,bm ] = sotest( t, numTerms )
bm = zeros(1,numTerms);
am = zeros(1,numTerms);
% 2L = 1
L = 0.5;
for ii = 1:numTerms
am(ii) = (1/L)*quadl(#(x) aCos(x,ii,L),0,2*L);
bm(ii) = (1/L)*quadl(#(x) aSin(x,ii,L),0,2*L);
end
ii = 0;
a0 = (1/L)*trapz( t, t.*cos((ii*pi*t)/L) );
% now let's test it
y = ones(size(t))*(a0/2);
for ii=1:numTerms
y = y + am(ii)*cos(ii*2*pi*t);
y = y + bm(ii)*sin(ii*2*pi*t);
end
figure; plot( t, y);
end
function a = aCos(t,n,L)
a = t.*cos((n*pi*t)/L);
end
function b = aSin(t,n,L)
b = t.*sin((n*pi*t)/L);
end
And then I called it like:
[ a0,am,bm ] = sotest( t, 100 );
and I got:
Sweetness!!!
All I really changed was from quad to quadl. I figured that out by using trapz which worked great until the time vector I was using didn't have enough resolution, which led me to believe it was a numerical issue rather than something fundamental. Hope this helps!
To troubleshoot your code I would plot the functions you are using and investigate, how the quad function samples them. You might be undersampling them, so make sure your minimum step size is smaller than the period of the function by at least factor 10.
I would suggest using the FFTs that are built-in to Matlab. Not only is the FFT the most efficient method to compute a spectrum (it is n*log(n) dependent on the length n of the array, whereas the integral in n^2 dependent), it will also give you automatically the frequency points that are supported by your (equally spaced) time data. If you compute the integral yourself (might be needed if datapoints are not equally spaced), you might calculate frequency data that are not resolved (closer spacing than 1/over the spacing in time, i.e. beyond the 'Fourier limit').