I am trying to simulate a LQG controller for CART wind turbine. I have read an article that has done this and checked the calculations and theory step by step.
This is the model:
In which A,B,C are state space matrices, Kf feedback gain, Kk kalman gain, w process noise and v is measurement noise which are gaussian noise.
The corresponding block diagram is:
My simulink simulation:
initial conditions are assumed to be zero and it is set in integrator block. The states are x1 rotor speed, x2 drive train torsion spring force, x3 generator speed.
The operating work at which the linear turbine model is obtained is as
below so I set input to 42rpm:
The output should be like this:
but what I get is this:
I can't find out why. Is it a problem with simulink because I have done the same thing as the article did or am I doing something wrong?
Related
For a university project, I want to train a (simulated) robot to hit a ball given the position and velocity. The first thing to try is policy gradients: I have a parametric trajectory generator. For every training position, I feed the position through my network, send the trajectory to the simulator and get a reward back. I now can use that as the loss, sample the gradient, feed it back and update the weights of my network so that it does better next time.
Therefore, goal is to learn the mapping from position to trajectory weights. When using all-star compute graph libraries like Theano and Tensorflow (or Keras), the I have the problem that I do not know how to actually model that system. I want to have standard fully connected layers first, then the output are my trajectory weights. But how do I actually calculate the loss so that it can use the backprop?
In a custom loss function, I would ignore/not specify the true labels, run the simulator and return the loss it gives. But from what I read, you need to return a Theano/Tensorflow function which is symbolic. My loss is quite complicated, so I do not want to move it from simulator to network. How can I implement that? Problem then is to differentiate that loss, as I might need to sample to get that gradient.
I've had a similar problem some time ago.
There was a loss function which relied heavily on optimized C code and third-party libraries.
Porting this to tensorflow was not possible.
But we still wanted to train a tensorflow graph to create steering signals from the current setup.
Here is an
ipython notebook which explains how to mix numerical and analytical derivatives
https://nbviewer.jupyter.org/gist/lhk/5943fa09922693a0fbbbf8dc9d1b05c0
Here is a more detailed description of the idea behind it:
The training of the graph is an optimization problem, so you will definitely need the derivative of the loss.
The challenge is to mix the analytical derivative in tensorflow and the numerical derivative of your loss.
You need this setup
Input I
output P
Graph G maps I to P, P = G(I)
add a constant of the same shape as P, P = C * G(I)
Loss function L
Training the tensorflow graph works with backpropagation.
For every parameter X in the graph, the following derivative is computed
dL/dX = dL/dP * dP/dX
The second part of that, dP/dX comes for free by just setting up the tensorflow graph. But we still need the derivative of the loss.
Now there's a trick.
We want tensorflow to update X based on the correct gradient dL/dP * dP/dX
but we can't get tensorflow to compute dL/dP, because that's not a tensorflow graph.
we will instead use P~ = P * C,
the derivative of that is dP~ / dX = dP/dX * C
So if we set C to dL/dP, we get the correct gradient.
We simply have to estimate C with a numerical gradient.
This is the algorithm:
set up your graph, multiply the output with a constant C
feed 1 for the constant, compute a forward pass, get the prediction P
compute the loss at P
compute the numerical derivative of P
feed the numerical derivative as C, compute the backward pass, update the parameters
I was studying for a signals & systems project and I have come across this code on high and low pass filters for an audio signal on the internet. Now I have tested this code and it works but I really don't understand how it is doing the low/high pass action.
The logic is that a sound is read into MATLAB by using the audioread or wavread function and the audio is stored as an nx2 matrix. The n depends on the sampling rate and the 2 columns are due to the 2 sterio channels.
Now here is the code for the low pass;
[hootie,fs]=wavread('hootie.wav'); % loads Hootie
out=hootie;
for n=2:length(hootie)
out(n,1)=.9*out(n-1,1)+hootie(n,1); % left
out(n,2)=.9*out(n-1,2)+hootie(n,2); % right
end
And this is for the high pass;
out=hootie;
for n=2:length(hootie)
out(n,1)=hootie(n,1)-hootie(n-1,1); % left
out(n,2)=hootie(n,2)-hootie(n-1,2); % right
end
I would really like to know how this produces the filtering effect since this is making no sense to me yet it works. Also shouldn't there be any cutoff points in these filters ?
The frequency response for a filter can be roughly estimated using a pole-zero plot. How this works can be found on the internet, for example in this link. The filter can be for example be a so called Finite Impulse Response (FIR) filter, or an Infinite Impulse Response (IIR) filter. The FIR-filters properties is determined only from the input signal (no feedback, open loop), while the IIR-filter uses the previous signal output to control the current signal output (feedback loop or closed loop). The general equation can be written like,
a_0*y(n)+a_1*y(n-1)+... = b_0*x(n)+ b_1*x(n-1)+...
Applying the discrete fourier transform you may define a filter H(z) = X(z)/Y(Z) using the fact that it is possible to define a filter H(z) so that Y(Z)=H(Z)*X(Z). Note that I skip a lot of steps here to cut down this text to proper length.
The point of the discussion is that these discrete poles can be mapped in a pole-zero plot. The pole-zero plot for digital filters plots the poles and zeros in a diagram where the normalized frequencies, relative to the sampling frequencies are illustrated by the unit circle, where fs/2 is located at 180 degrees( eg. a frequency fs/8 will be defined as the polar coordinate (r, phi)=(1,pi/4) ). The "zeros" are then the nominator polynom A(z) and the poles are defined by the denominator polynom B(z). A frequency close to a zero will have an attenuation at that frequency. A frequency close to a pole will instead have a high amplifictation at that frequency instead. Further, frequencies far from a pole is attenuated and frequencies far from a zero is amplified.
For your highpass filter you have a polynom,
y(n)=x(n)-x(n-1),
for each channel. This is transformed and it is possble to create a filter,
H(z) = 1 - z^(-1)
For your lowpass filter the equation instead looks like this,
y(n) - y(n-1) = x(n),
which becomes the filter
H(z) = 1/( 1-0.9*z^(-1) ).
Placing these filters in the pole-zero plot you will have the zero in the highpass filter on the positive x-axis. This means that you will have high attenuation for low frequencies and high amplification for high frequencies. The pole in the lowpass filter will also be loccated on the positive x-axis and will thus amplify low frequencies and attenuate high frequencies.
This description is best illustrated with images, which is why I recommend you to follow my links. Good luck and please comment ask if anything is unclear.
Lets say I have a mass-spring-damper system...
here is my code (matlab)...
% system parameters
m=4; k=256; c=1; wn=sqrt(k/m); z=c/2/sqrt(m*k); wd=wn*sqrt(1-z^2);
% initial conditions
x0=0; v0=0;
%% time
dt=.001; tMax=2*pi; t=0:dt:tMax;
% input
F=cos(t); Fw=fft(F);
% impulse response function
h=1/m/wd*exp(-z*wn*t).*sin(wd*t); H=fft(h);
% convolution
convolution=Fw.*H; sol=ifft(convolution);
% plot
plot(t,sol)
so I can successfully retrieve a plot, however I am getting strange responses I also programmed a RK4 method that solves the system of differential equations so I know how the plot SHOULD look like, and the plot I am getting from using FFT has an amplitude of a like 2 when it should have an amplitude of like .05.
So, how can I solve for the steady state response for this system using FFT. I want to use FFT because it is about 3 orders of magnitude faster than numerical integration methods.
Keep in mind I am defining my periodic input as cos(t) which has a period of 2*pi that is why I only used FFT over the time vector that spanned 0 to 2*pi (1 period). I also noticed if I changed the tMax time to a multiple of 2*pi, like 10*pi, I got a similar looking plot but the amplitude was 4 rather than 2, either way still not .05!. maybe there is some kind of factor I need to multiply by?
also I plotted : plot(t,Fw) expecting to see one peak at 1 since the forcing function is cos(t), yet I did not see any peaks (maybe I shouldn't be plotting Fw vs t)
I know it is possible to solve for the steady state response using fourier transform / fft, I am just missing something! I need help and understanding!!
The original results
Running the code you provided and comparing the result with the RK4 code posted in your other question, we get the following responses:
where the blue curve represents the FFT based implementation, and the red curve represents your alternate RK4 implementation. As you have pointed out, the curves are quite different.
Getting the correct response
The most obvious problem is of course the amplitude, and the main sources of the amplitude discrepancies of the code posted in this question are the same as the ones I indicated in my answer to your other question:
The RK4 implementation performs a numeric integration which correctly scales the summed values by the integration step dt. This scaling is lacking from the FFT based implementation.
The impulse response used in the FFT based implementation is consistent with the driving force being scaled by the mass m, a factor which was missing from the RK4 implementation.
Fixing those two issues results in responses which are a little closer, but still not identical. As you probably found out given the changes in the posted code of your other question, another thing that was lacking was zero padding of the input and of the impulse response, without which you were getting a circular convolution rather than a linear convolution:
f=[cos(t),zeros(1,length(t)-1)]; %force f
h=[1/m/wd*exp(-z*wn*t).*sin(wd*t),zeros(1,length(t)-1)]; %impulse response
Finally the last element to ensure the convolution yields good result is to use a good approximation to the infinite length impulse response. How long is long enough depends on the rate of decay of the impulse response. With the parameters you provided, the impulse response would have died down to 1% of its original values after approximately 11*pi. So extending the time span to tMax=14*pi (to include a full 2*pi cycle after the impulse response has died down), would probably be enough.
Obtaining the steady-state response
The simplest way to obtain the steady-state response is then to discard the initial transient. In this process we discard a integer number of cycles of the reference driving force (this of course requires knowledge of the driving force's fundamental frequency):
T0 = tMax-2*pi;
delay = min(find(t>T0));
sol = sol(delay:end);
plot([0:length(sol)-1]*dt, sol, 'b');
axis([0 2*pi]);
The resulting responses are then:
where again the blue curve represents the FFT based implementation, and the red curve represents your alternate RK4 implementation. Much better!
An alternate method
Computing the response for many cycles waiting for the transient response to die down and extracting the remaining samples corresponding
to the steady state might appear to be a little wasteful, despite the fact that the computation is still fairly fast thanks to the FFT.
So, let's go back a little and look at the problem domain. As you are probably aware,
the mass-spring-damper system is governed by the differential equation:
where f(t) is the driving force in this case.
Note that the general solution to the homogeneous equation has the form:
The key is then to realize that the general solution in the case where c>0 and m>0 vanishes in the steady state (t going to infinity).
The steady-state solution is thus only dependent on the particular solution to the non-homogenous equation.
This particular solution can be found by the method of undetermined coefficients, for a driving force of the form
by correspondingly assuming that the solution has the form
Substituting in the differential equation yields the equations:
thus, the solution can be implemented as:
EF0 = [wn*wn-w*w 2*z*wn*w; -2*z*wn*w wn*wn-w*w]\[1/m; 0];
sol = EF0(1)*cos(w*t)+EF0(2)*sin(w*t);
plot(t, sol);
where w=2*pi in your case.
Generalization
The above approach can be generalized for more arbitrary periodic driving forces by expressing the driving force as a
Fourier Series (assuming the driving force function satisfies the Dirichlet conditions):
The particular solution can correspondingly be assumed to have the form
Solving for the particular solution can be done in a way very similar to the earlier case. This result in the following implementation:
% normalize
y = F/m;
% compute coefficients proportional to the Fourier series coefficients
Yw = fft(y);
% setup the equations to solve the particular solution of the differential equation
% by the method of undetermined coefficients
k = [0:N/2];
w = 2*pi*k/T;
A = wn*wn-w.*w;
B = 2*z*wn*w;
% solve the equation [A B;-B A][real(Xw); imag(Xw)] = [real(Yw); imag(Yw)] equation
% Note that solution can be obtained by writing [A B;-B A] as a scaling + rotation
% of a 2D vector, which we solve using complex number algebra
C = sqrt(A.*A+B.*B);
theta = acos(A./C);
Ywp = exp(j*theta)./C.*Yw([1:N/2+1]);
% build a hermitian-symmetric spectrum
Xw = [Ywp conj(fliplr(Ywp(2:end-1)))];
% bring back to time-domain (function synthesis from Fourier Series coefficients)
x = ifft(Xw);
A final note
I purposely avoided the undamped c=0 case in the above derivation. In this case, the oscillation never die down and the general solution to the homogeneous equation does not have to be the trivial one.
The final "steady-state" in this case may or may not have the same period as the driving force. As a matter of fact, it may not be periodic at all if the period oscillations from the general solution is not related to the period of the driving force through a rational number (ratio of integers).
can someone help me with another matlab project:
Is it possible to create a simple low pass filter like in RC circuits? For instance if we create a sine wave like y=10*sin(2*pift).
Can i just filter the signal with a cutoff frequency to be able to see how the filtered signal looks like(for the amplitude decay)?
So when i enter some information for example f = cutoff and plot the filtered signal this must reduce the amplitude somewhere at 30% from the input signal ?
I've been looking for the functions like butter and cheby but they seem to do other kind of filtering like removing the noise from a signal. I just want some simple plots(input and output) which will show the principle of RC, RL low and high pass filters.
An RC circuit is a fist-order filter. To approximate a first order hardware filter, I generally use a IIR filter. A butterworth filter is usually my first choice for IIR, but for a first-order response, it doesn't really matter.
Here's an example with the cutoff frequency being the same as the signal frequency, so the filtered signal should be 3 dB down...
%first, make signal
fs = 1000; %your sample rate, Hz
dur_sec = 1; %what is the duration of your signal, seconds
t_sec = ([1:dur_sec*fs]-1)/fs; %here is a vctor of time
freq_Hz = 25; %what frequency do you want your sign wave
y = sin(2*pi*freq_Hz*t_sec); %make your sine wave
%Second, make your filter
N = 1; %first order
cutoff_Hz = 25; %should be 3dB down at the cutoff
[b,a]=butter(N,cutoff_Hz/(fs/2),'lowpass'); %this makes a lowpass filter
%Third, apply the filter
y_filt = filter(b,a,y);
%Last, plot the results
figure;
plot(t_sec,y,t_sec,y_filt);
xlabel('Time (sec)');
ylabel('Amplitude');
ylim([-1 1]);
legend('Raw','Filtered');
title(['1st-Order Filter with Cutoff at ' num2str(cutoff_Hz) ' Hz']);
As an alternative to using the built-in filter design functions such as butter, you could choose model the circuit itself. A simple first order RC (or RL) circuit would result a first order differential equation. For an RC circuit, you'd then integrate the equation through time, given your sine wave as stimulation. That would work fine, but could be more of a hassle, depending upon your background.
For a simple, first-order hardware filter that is properly buffered by an op-amp on either end of the RC, I think that you'd find that the result using the first order butter filter is going to be awfully close (the same?) as modeling the circuit. The butter filter is way easier to implement in software (because i just gave you the code above), so I'd go that route.
When you move to 2nd order hardware filters, however, that's where you have to be more careful. You have a few options:
1) Continue to model your 2nd order hardware using one of the built-in filter functions. A second order butter is trivial to implement (alter the N in the code above), but this might not model the specific hardware filter that you've created. You'll have to choose the right kind of IIR filter to match the architecture of your hardware filter.
2) If you haven't picked your architecture for your hardware filter, you could choose the architecture to follow one of the canonical filter types, just so that it is easy to model via butter, cheby1, or whatever.
3) You could go back to modeling the circuit with differential equations. This would let you model any filter circuit, whether it followed a canonical type or not. You could also put non-linear effects in here, if you wanted.
For the first order RC filter, though, I think that any of the built-in filter types will be a decent enough model for an RC filter. I'd suggest that you play with my example code above. I think that it will satisfy your need.
Chip
I want to estimate the time of arrival of GPR echo signals using Music algorithm in matlab, I am using the duality property of Fourier transform.
I am first applying FFT on the obtained signal and then passing these as parameters to pmusic function, i am still getting the result in frequency domain.?
Short Answer: You're using the wrong function here.
As far as I can tell Matlab's pmusic function returns the pseudospectrum of an input signal.
If you click on the pseudospectrum link, you'll see that the pseudospectrum of a signal lives in the frequency domain. In particular, look at the plot:
(from Matlab's documentation: Plotting Pseudospectrum Data)
Notice that the result is in the frequency domain.
Assuming that by GPR you mean Ground Penetrating Radar, then try radar or sonar echo detection approach to estimate the two way transit time.
This can be done and the theory has been published in several papers. See, for example, here:
STAR Channel Estimation in DS-CDMA Systems
That paper describes spatiotemporal estimation (i.e. estimation of both time and direction of arrival), but you can ignore the spatial part and just do temporal estimation if you have a single-antenna receiver.
You probably won't want to use Matlab's pmusic function directly. It's always quicker and easier to write these sorts of functions for yourself, so you know what is actually going on. In the case of MUSIC:
% Get noise subspace (where M is number of signals)
[E, D] = eig(Rxx);
[lambda, idx] = sort(diag(D), 'descend');
E = E(:, idx);
En = E(:,M+1:end);
% [Construct matrix S, whose columns are the vectors to search]
% Calculate MUSIC null spectrum and convert to dB
Z = 10*log10(sum(abs(S'*En).^2, 2));
You can use the Phased array system toolbox of MATLAB if you want to estimate the DOA using different algorithms using a single command. Such as for Root MUSIC it is phased.RootMUSICEstimator phased.ESPRITEstimator.
However as Harry mentioned its easy to write your own function, once you define the signal subspace and receive vector, you can directly apply it in the MUSIC function to find its peaks.
This is another good reference.
http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=1143830