# Tag Info

10

I did my signal processing Ph.D. in a control systems department. My take is that signal processing is open loop; control systems close the loop. Apart from that, the mathematics behind both are very similar. It's the applications that are generally very different.

8

Both draw on Linear System Theory (a.k.a. "Signals and Systems"). So also does Communications Systems and Linear Electric Circuits, Electronic Circuits,and Distributed Networks (a.k.a. Transmission Lines). Both worry about system stability. Poles have to be inside the unit circle. DSP is actually broader than either Controls or Communications. Control ...

5

Two principles here: When dealing with a differential equation, you define intermediate state variables so everything is in terms of first derivatives. This system is nonlinear, so the state-space equations won't be in terms of matrices. Applying these principles, we define a state vector: $$\mathbf x = [x_1, x_2]^T,$$ where: $$x_1 = y \\ x_2 = \dot y$$...

4

There's a fairly simple distinction. Signal processing is a set of tools that can be used for control engineering. Control engineering is about making something move how you want it to move. Some of the tools of signal processing will help with that (and some won't; backward filtering doesn't happen in real-time without a TARDIS). Signal processing is ...

4

Like you mentionned, you cannot cancel a right-half-plane zero (or a zero outside the unit circle) by placing a pole on it. A unstable pole in your compensator will make the command of your controller unbounded (i.e. it will reach infinity). There are no ways to cancel a right-half-plane zero. It's sometimes possible to "remove" a right-half-plane zero by ...

3

Again I refer you to the Murray's book. Which is free by the way http://www.cds.caltech.edu/~murray/books/AM08/pdf/fbs-limits_18Aug2019.pdf From the book "Furthermore a zero is said to be “slow” if its magnitude is smaller than the intended closed loop bandwidth."

3

Intuitively (and I leave it as an exercise for you to work it out with math), it's because you have to push harder on the plant to get it to move fast. Consider a bowl of honey and a spoon -- moving the spoon around in the bowl slowly takes little effort, but moving it fast requires a lot of effort, because honey is viscous. It may be most informative to ...

3

A system with simple distinct poles on the imaginary axis (and note that the origin is on the imaginary axis) and no poles in the right half-plane is called marginally stable. If you have poles with multiplicity greater than $1$ on the imaginary axis, or if there are poles in the right half-plane, then the system is unstable. For discrete-time systems, the ...

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

The requirement, for causal, real-time system implementations (where time is the independent parameter) that continuously minimize an output error with respect to a reference criterion, distinguishes the control systems discipline. You could search MIT Open Courseware, such as https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-30-feedback-control-...

2

Control engineering are often taught in similar or even same courses of study, up the masters' degrees. In the general system modeling approach, where inputs ($I$) and outputs ($O$) are related through systems ($S$), I would say that, for a target $O$, they either work on $S$ or $I$: control engineers tend to put (strong) contraints on outputs of a system, ...

2

$$H(s) = {K*\frac{s-z}{s-p}}\\ \\$$ when |z| < |p|, you have a lead compensator. It adds phase between a certain band, and can help you improve your phase margin. It can also increase your bandwidth. You typically use it to improve your transient as you mentionned. You can think of it as a PD controller cascaded with a low-pass filter. The gain K is ...

2

How can a stable system have a RHP pole? It can't. Now, I have that if I do the root locus I have: (image left out of quote) so I have a pole in the RHP. This is where you misunderstand. The word "locus" means "The set of all points whose coordinates satisfy a given equation or condition". The traces in that root locus show all possible locations ...

2

That's a simple process, where $y(k)=u(k-3)$, as the $z^{-1}$ blocks just perform a delay by one sample. $x_1(k)$, $x_2(k)$ and $x_3(k)$ are just arbitrary names. In other words: output will be exactly like input, only delayed by three samples.

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1 - It can be a good idea to put a filter $C_r(s)$ in order to smooth the reference. For example, if you have a speed regulator, and the reference jumps from 100 km/h to 200 km/h, having a low-pass filter will smooth the actual reference fed to the closed-loop. It could prevent excitation of unmodeled dynamics, actuator wind-up, etc. That being said if $C_r(... 2 You could get away with defining it as such (or extending the concept to the pole with the most positive real part). But typically the concept of a dominant pole is applied to systems that are truly stable, and thus actually settle out. A system with singular poles on the imaginary axis (and no poles in the right-half plane) is considered "metastable", ... 2 Bandwidth in control system usually mean open-loop bandwidth and is usually defined as the frequency where the open loop transfer function gain crosses the 0 dB gain. The open loop transfer function is defined as$C(s)P(s)$. Usually, but not always, processes have some kind of low-pass transfer function. Therefore, to increase the bandwidth your controller ... 2 The first compensator$\frac{3s+1}{0.1s + 1}$has a high-frequency gain of 30. The second compensator$\frac{6s+1}{0.1s + 1}$has a high-frequency gain of 60. The difference is 6.02 dB, it looks pretty close to the difference between the green and red line. 2 1 - How do I analyze the stability of the last case? AFAIK, I cannot use the margin and allmargin Matlab functions as the delay is not in the open loop transfer function ($G_{comp}G_{process}$). In both cases the open-loop transfer function is$1+G_{comp}G_{process}e^{-Ts}$The open-loop transfer function is the transfer function from the output of the ... 1 As near as we can tell by experiment, causality is nature's way of doing its thing. Causality says that if you have a system$y(t) = h\left(x(t), t\right)$, and it is causal, then$y(t_0)$is dependent only on values of$x(t)$for$t < t_0*. Causality doesn't say you can't combine systems into larger systems (as a rather pertinent example, you can ... 1 If a RHP zero imposes limitation on the bandwidth in the frequency domain, it also imposes limitation in the time domain. This is due to time-frequency duality. For example, if you want to have a really fast step response, a slow RHP zero will prevent you from reaching that goal. I really recommand Murray's book for more information http://www.cds.caltech.... 1 The transmission zeros of a MIMO system are obtained by solving \begin{align} \dot{x}^* = A\,x^* + B\,u^*, \\ y = C\,x^* + D\,u^*, \end{align} forx^*$and$u^*$such that$y = 0$. By using the Laplace variable$s$for the time derivative that system of equation can also be written as $$\begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} A - s\,I &... 1 or the transfer function from the input 𝑢 to the output, how can I do this ? Is that not what your G is? If you want to find the transfer function from just one element of u to the output, then either delete the columns of B that don't pertain to that element of u and get your transfer function, or just look at the column of the transfer function ... 1 You already have the answer S+T = 1, the relation must hold true at all frequencies. Therefore if your sensitivity function S has a high-pass characterisitic, the complementary sensitivity function T will have low-pass characteristic. As for bandwidth, your curves make sense. The red sensitivity function has a larger bandwidth than the green ... 1 Hi I am confused about the implementation of AGC in Simulink example. I am unable to relate the given implementation to the algorithm given by MATLAB. Kindly help me in understanding the reason behind the difference in algorithm and implementation. Best Regards Sunny 1 From the diagram in the Algorithms section of the documentation you can see how the different quantities are computed: Note that z in the diagram is an estimate of the output power.^1 The error signal e is computed by comparing the reference value A to \ln(z). So if you choose$$A=\ln(P)\tag{1}$$then the average output power will be adjusted to ... 1 This is kind of hand-wavy, but you can look at this from two different perspectives: One, you can look at z^{-1} as a "back-step" operator; i.e. if X(z) = \mathcal{Z}\lbrace x_n \rbrace, then (with a few 'i's left undotted and 't's uncrossed) \frac{X(z)}{z} = \mathcal{Z}\lbrace x_{n-1} \rbrace. You can also look at s as a derivative operator: if X(... 1 You're right, the given pole and zero are wrong. They should be$$s_0=-\frac{1}{\tau}$$and$$s_{\infty}=-\frac{1}{\beta\tau}$$because for$s=s_0$the numerator becomes zero, and for$s=s_{\infty}\$ the denominator becomes zero.

1

This looks like it might be able to be written as a dynamic programming problem. Bellman's equation looks very similar to the way you have expressed your control problem.

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