9
The system
$$y[n]=y[n-1]+x[n]\tag{1}$$
is an ideal accumulator, i.e., it computes the cumulative sum of the input samples:
$$y[n]=\sum_{k=-\infty}^nx[k]\tag{2}$$
It is in a way analogous to a continuous-time integrator, but this doesn't mean that you will necessarily obtain an ideal integrator by transforming the discrete-time system to a continuous-time ...
8
Your original differentiator, which should be $x(n)-x(n-1)$, is called a "first difference" differentiator. That differentiator amplifies high-frequency noise. As a next step I suggest you try what's called the "central difference" differentiator defined by:
$$ \mathit{Diff} = \frac{x(n)-x(n-2)}{2} $$
which does not amplify high-frequency noise.
6
Your simple integrator is called a "Rectangular Rule" integrator. There are more complicated (and more accurate) integrators called "Trapazoidal Rule", "Simpson's Rule", and "Tick's Rule" integrators.
5
A step response test is an easy way to determine the bandwidth. Sum a small step into the control voltage of your oscillator (VCO or NCO), and measure the 90% to 10% fall time of the corrected response at the output of the loop filter as shown in this block diagram. Note that the loop will respond in such a way to completely cancel the injected offset, but ...
5
In a 2003 paper in French, "Estimation par maximum de vraisemblance de la dérivée d’un signal bruité. Application à la caractérisation de vérins pneumatiques" (Maximum likelihood estimation of the derivative of a noisy signal. Application to the characterization of pneumatic cylinders) [from the early GRETSI french-speaking conference on signal and image ...
3
Two suggestions to move forward:
Reduce $K_i$ to the point of an acceptable overshoot (this will provide the bottom line answer for comparison to the computations.
Do system identification (Bode plots) on the open loop system and individual components to isolate the difference between implementation and loop model; the computations were compared below ...
3
As the book's title is control theory; a controller is the general broad name given to any piece of hardware (or sowftware) that aims to control some processes, in your aerodynamics field probably the flight process; by menas of receiving signals from sensors and sending commands to propellers - actuators, and in between is the control hardware or software......
3
Well, the two systems differ only at low frequencies. In fact, if you define
$$
R_L = \dfrac{\tau_zs+1}{\tau_ps+1},\qquad R_I = \dfrac{\tau_zs+1}{\tau_ps}
$$
you have that $R_I(j\omega)\underset{\omega\to\infty}{\to}R_L(j\omega)$.
Therefore, you can expect that the behavior of the closed-loop system differs only at lower frequencies.
What does the PI do?
...
2
If you require two outputs, which belong to a movement in x and the other in y direction, you could start with the assumption, that both movements are independent an simply implement two PID controller, each caring for one direction.
It gets interesting at the point where you translate the output of these two controller to signals to the actuators. Here it ...
2
Switching between 2 PID controller is called gain-scheduling. There are various ways to implement gain scheduling. But basically, the idea is that the gain should change smoothly. You should not change suddenly. You could have one set of gains if T < T1, another set if T > T2 and use interpolation between the 2 sets if T1 < T < T2.
You can also ...
2
Considering the Laplace transform of a function $x(t)$ as :
$$X(s) = \mathcal{L}\{x(t)\} = \int_{-\infty}^{\infty} x(t) e^{-st} dt \tag{0}$$
then to find the following Laplace transform
$$\mathcal{L} \{ \int_{-\infty}^{t} e(\tau) d\tau \} \tag{1} $$
you can write the integral as a convolution of $e(t)$ with the unit-step function $u(t)$ as
$$\mathcal{L}\{ \...
2
1 There's a mistake in the PID connection. You must feed the quadrature component, i.e $U_q$ to the PID, not $U_d$. The setpoint of your PLL is $U_q = 0$ because you want your PLL to be in phase with your 3-phase input i.e. $U_d = 1, U_q = 0$.
2 - Perhaps there are hidden delays in the block you instantiated ?
3 - Notice the error equation of your PLL is ...
2
This is just a different flavor of what Dan Boschen suggests. In fact, the actual summing junction part is exactly what he's suggesting.
Insert a summing junction, and then inject a signal ($u_A$ in my diagram). For more work, you can get the frequency response directly. If you inject a sine wave at $u_A$ while you pick off $u_R$ and $y$, you can let it ...
2
I have heard that the word comes from "plant" as in "steam plant". But I don't know of any historical research that's been done to pin down the term.
I did just check my copy of "On Governors" by James Clerk Maxwell himself (that guy got around). At least in 1868, Maxwell wasn't using the word "plant" for "thing ...
1
Define "appropriate".
Yes, the output of the controller will be limited to between your action_max and action_min -- this is good. Since your output is also your integrator state, you'll be effecting anti-windup. If I'm not horribly mistaken, if your output is saturated, as soon as the error term starts pointing in the opposite direction of the ...
1
Following your edit, the first challenge in creating a controller is the task to create a stabilizing controller. After that, performance can be tuned. In order to find if the created controller stabilizes the closed-loop, the nyquist plot can be used. The nyquist stability criterion states that the amount of encirclements of the point -1 (hereby are counter-...
1
I'm a bit puzzled with your Bode plot, here's what I get with your transfer function and Matlab 2019b. My phase starts at almost -180 degrees while yours start at 0 degree. Is it possible that you've made a mistake somewhere in your analysis?
Your transfer function is a bit complex, you could simplify it by removing the fast poles and zeroes. For example, ...
1
You have 2 right-half plane zeroes : 0.012 and 18. The zero at +18 is "fast" and will not affect the performance much, but your slow zero at 0.012 will severely limit your performance. You can't cancel this right-half plane zero with a right-half plane pole in your controller, your controller output will be unbounded.
Is this homework or a real ...
1
"Plant" doesn't only refer to the living thing that has chlorophyll; it, in the context of control, refers to a technical system with in and output – some production line, a whole factory, a nuclear power plant.
As such, there's nothing more intuitive than the word itself: a plant is what is being controlled.
1
Your controller controls pitch and roll. However, in the graph you showed us we only see x,y and z.
Could you add the plot for pitch and roll? I'm not an expert in aircraft control, but I think you can have a constant pitch and roll and still move in the x,y,z plane.
I'll edit my answer as soon as you give more info.
Edit : You roll seems stable, you could ...
1
K is the gain parameter for which you will be at any particular point on the root locus. As you increase K, the closed loop poles will start from the open loop poles and move toward the open-loop zeros as depicted in the root locus showing all possible locations of closed loop poles versus gain K.
This is because the general form for the closed loop and open ...
1
The objective of a controller is to respond to the error, an example of a controller is the PID.
On the other hand the objective of a compensator is to change the original dynamics of the plant, examples of compensators are the lead, lag, and lag-lead compensators.
1
1 - Do you have some general idea of what your transfer function looks like? Or at least a frequency response? Could you perform some kind of step response? You could then try to fit an order-2 model on your step response and then try to tune your controller.
2 - You mention that your system is underdamped. I assume that you have underdamped stable poles (...
1
One important point you may be missing is assuming that the noise of the ADC is truly white over your sampling bandwidth- it isn't! You will have spurs from signal correlations and ADC non-linearities that will limit the noise floor you can achieve. This will be specified by the SFDR (spurious-free-dynamic-range) of the ADC for guidance as to what you may be ...
1
The PID can be tested in an open-loop setup, however any integral gain <> 0 will cause the PID output to run away, so you are likely to hit clamping (like control voltage maxima) with your test rig.
What really matters and needs to be tuned is the closed-loop behaviour. This includes your plants behaviour. Stability is prime goal, maximum overshoot and ...
1
Yes, the system should be an integrator.
What method did you use in the call to d2c.
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