# Tag Info

6

The Kalman Filter is basically a framework to fuse 2 things: Measurement. Dynamic Model (Dynamic in the sense we can predict next value from a current value). In your case the model is composed of two things: Model which connects ${T}_{ext}$ and ${W}_{fuel}$ to ${T}_{in}$. Measurement of ${T}_{ext}$. Now, the reason you estimate ${T}_{ext}$ is ...

6

On each individual device, the speaker output can get subtracted from the microphone before it gets sent to other locations. This prevents others from hearing themselves through your microphone. When using two devices in within audible range of each other, the devices cannot subtract the speaker audio from the microphone audio because the information path ...

5

As you pointed out, there are many state-space realizations of one particular transfer function. The reason is that a transfer function only represents the input-output behavior of a system (observable and controllable dynamics) and not the internal states. That being said, you can directly write state-space realizations from a transfer function with the so-...

5

The final value of the step response is the DC gain of the closed-loop transfer function, which is generally different from the open-loop DC gain. Assuming unity gain feedback, the feed-forward transfer function $G(s)$ equals the open-loop transfer function, and the closed-loop transfer function is given by $$C(s)=\frac{G(s)}{1+G(s)}\tag{1}$$ The final value ...

4

Open loop gain at DC is -3dB or .707 and 0 degrees. We don’t know the forward gain but assuming it is the open loop gain, the closed loop gain would be $.707/(1+.707)= .4148$, matching the first plot. (With 60 degrees of phase margin however I would have expected a response closer to the third plot, but as explained in the comments this is due to my ...

4

The transfer function is $H(s) = \frac{16.94s + 579.5}{s^2 + 507.2s + 1224}$ This transfer function has 2 poles, one slow pole at -2.4248 and a fast pole at -504.7752. The function has a slowish zero at -34.2. Good news, your poles and zero are all in the left-half plane. It is much easier to control a system with zeroes and poles in the left-half plane ...

3

That depends on what you want to plot: The response of the continuous time system or the response of a sampled discrete time system, which will always depend on the choice of sample rate. Your system is not band-limited, so it can't be sampled without some amount of aliasing. Do I just evaluate s at iω If you just want to look at the approximate frequency ...

3

The Explanation from @ScienceGeyser provides a good explanation to the phenomenon. There are two more things to address the question on how is this phenomenon avoided. The feedback read by the microphone is not identical to the audio sent to the speakers. There is the physical response from the speakers, the acoustic of the device and the environment, if ...

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

I would say that the difference is not about time domain or frequency domain, is more about using transfer function or state space equations. The state equation for a system with a single input and a single output is $$\sum_{n=0}^{N} a_n \left(\frac{d}{d t}\right)^n y(t) = \sum_{n=0}^{N} b_n \left(\frac{d}{d t}\right)^n u(t)$$ The state space representation ...

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 ...

2

This is an interesting question since both squared error and absolute error are convex functions, so they are both going to give the optimal solution when minimized. My intuition is that the $\ell_2$-norm (sum of squared values of error) converges to zero more quickly than the $\ell_1$-norm (sum of absolute values of error) when the search direction is right....

2

A transfer function describes an LTI system. As such, the given system can be described by a transfer function. However, if there are non-zero initial conditions, the system is no longer linear because there's a contribution in the output that does not depend on the input signal but only on the initial conditions. Consequently, the transfer function cannot ...

2

The closed loop poles are the roots of the polynomial $$D(s)=s^2+2s+2+K\tag{1}$$ and, according to the root locus plot, they are $s_{1,2}=-1\pm 2j$. Consequently, we get $$2+K=|1+2j|^2=5\quad\Longrightarrow\quad K=3\tag{2}$$ With $K=3$ we obtain $$H(0)=\frac{K}{2+K}=\frac35\tag{3}$$ which leaves step response $C$ as the only option.

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

You need to properly select a time vector for the step function. I suspect that Matlab has trouble automatically selecting a good time vector for small delays. try this close all; clear all; clc; s = tf('s'); sys_retard = exp(-2E-6*s); sys_retard_pade = pade(sys_retard, 3); T = linspace(0, 10*2E-6, 1000); figure(1); output1 = step(sys_retard, T); output2 ...

2

Hard limits on what the output actuator can do has got to be the most common control non-linearity there is. The search term that'll get you hits on academic papers is "actuator saturation". But the moment I get outside linear controller design everything gets nebulous; discussions of a dozen different techniques with no discussion of how to ...

2

Person driving car. Deer entering road. Person turning the steering wheel sharply to avoid deer. Person turning the steering wheel sharply to get back on track, overshooting in their response. Car going for a lithobreaking with the local roadside geologic environment.

1

If you have the symbolic math toolbox, then you should be able to follow the instructions in Chapter 3 here. I tried to do this for your example: syms t tau a = 10; omega = 2*pi*100; f = a*exp(-a^2/(4.0*t))/(2.0*sqrt(pi)*t^(3.0/2.0)); g = exp(1i*omega*t); z = int(subs(f,tau)*subs(g,t-tau),tau,-inf,inf); z = simplify(z); figure(1) ezplot(f) but the ...

1

You can model the effect of the various delays and still work in the s-domain. Your DAC can be modeled as a zero-order-hold with a T period. In the s-domain you could model the zero-order-hold by a $\frac{T}{2}$ delay or $e^{\frac{-Ts}{2}}$. While this is not technically accurate, it should be good enough. Secondly, your software must have some kind of ...

1

You need to solve (6.3) for $G_R$, here is the step-by-step: $G_w = \frac{G_R(z) G_P(z)}{1 + G_R(z) G_P(z)}$, multiply both sides by $1 + G_R(z) G_P(z)$ to get: $G_R(z)G_w(z)G_P(z) - G_R(z)G_P(z) = -G_w$, now solve for $G_R(z)$: $G_R(z) = -\frac{G_w(z)}{G_w(z)G_P(z) - G_P(z)} = -\frac{G_w(z)}{G_P(z)(G_w(z) -1)} = \frac{1}{G_P(z)}\frac{G_w(z)}{1 - G_w(z)}$

1

Only the first criterion is correct when referring to input-output stability. The second criterion is just a way to compute the poles, as mentioned in a comment. All realizable continuous-time systems using lumped elements have poles, which doesn't contradict the stability requirement. Poles in the right half-plane (RHP) make a causal system unstable. Note ...

1

Fixed point processing is not for the faint of heart. It typically requires managing scaling factors, clipping point, signal to noise ratios, underflow, limit cycles, rounding behavior at each single point of your processing chain. Your best shot at dealing with different data ranges is to rigorously use the Q-number format for each data, state and ...

1

For $|\zeta| \le 1$, let $\zeta= \cos\theta$, so $\theta=\mathrm{arccos}\,\zeta$ \begin{align*}g(s) &= K\frac{\omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2}\\ \\ &= K\frac{\omega_n^2}{s^2 + 2\omega_n s \cos\theta+ \omega_n^2(\cos^2\theta +\sin^2\theta)}\\ \\ &= K\frac{\omega_n^2}{(s + \omega_n \cos\theta)^2+ \omega_n^2\sin^2\theta}\\ \\ &... 1 From the root locus plot, the open loop transfer function has a real-valued pole at s_0=0 and a complex conjugate pole pair s_1=-1.5+ 2j and s_1^*. There's also a zero at s=-2. Consequently, G(s) is given byG(s)=\frac{s+2}{s(s-s_1)(s-s_1^*)}$$The closed-loop transfer function is$$H(s)=\frac{kG(s)}{1+kG(s)}=\frac{k(s+2)}{s(s-s_1)(s-s_1^*)+k(s+2)...

1

First, typically when you're exerting a force on something and getting a position, the acceleration varies instantaneously with force. A more or less universal equation of motion for a single-axis linear system would be $m \ddot x = f_v(\dot x) + f_p(x)$. For a mass-spring-damper system, it'd be $m \ddot x = b \dot x + k x$. So you can easily express that ...

1

The problem with your example is that $\infty\cdot 0$ isn't necessarily equal to zero. The only way to judge what is happening in the limit $K\to\infty$ is to divide the original equation by $K$: $$\frac{D(s)}{K}+N(s)=0\tag{1}$$ Now it is obvious that for $K\to\infty$ the actual value of $D(s)$ is irrelevant, as long as it is finite. Consequently, the only ...

1

No such thing as a single frequency of the noise. That's exactly why it's called white; it has power in all frequency ranges, but not at a single frequency. Finally, is there a frequency-domain representation of Gaussian white noise? Yes, a constant power spectral density for all frequencies. That's like white light (which contains also a continuum of all ...

1

Assuming $P$ is the transfer function of your process and $C$ the transfer function of your controller. The closed-loop transfer function of a standard control loop, with the controller in the normal path, is $$G(s) = \frac{C(s)P(s)}{1+C(s)P(s)}$$ while the transfer function of your alternative controller is $$G_{alt}(s) = \frac{P(s)}{1+C_{alt}(s)P(s)}$$ ...

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