15

Step response and transfer function are interchangeable via Laplace transform (step response -> differentiate to get impulse response -> laplace -> transfer function), but you can't run a Laplace transform on a graph pulled out of a datasheet... For example, the user of a LDO will usually be interested in things like voltage sag/overshoot when load current ...


10

Another way to look at your question is when would you use PI control with the P term 0. The answer is basically "Whenever you think you can get away with it.". This main risk with only integral control is oscillation or large overshoots due to windup. If the output is low for a while, for example, then the integral term gets ever larger. If this happens ...


8

Your control is quite good despite if you feel in other way. I do think that a P-control with feed forward LUT (Look Up Table) would solve 98% of your need. Try this way: For each temperature setpoint you create a LUT table with output % that maintain the setpoint temperature at the steady state. If this setpoint is just 50 deg. C, then you don't even need ...


7

An impulse, i.e. hit it with a hammer, is not very friendly, particularly in systems that have mechanical bits and pieces. A perfect impulse cannot be generated, so you have to tailor the pulse duration to the system. A pulse that gives meaningful response data needs to be relatively strong, and that means high amplitude (strength = area = height x ...


6

A practical oscillator has to be designed taking into account component tolerances. In your example the ratio of RG to RF needs to be such that the gain round the loop is unity. Because of the tolerance in RG, RF and the capacitors there has to be some extra gain designed in so that in the worst case the gain is at least unity. In the typical case this ...


5

why increasing the "gain" of your system will cause instability If you have a servo control mechanism and you set a demand, the servo should rotate (or move) to the position demanded and all is good. However, if you have too much gain, the servo rapidly moves off in the direction needed but overshoots the target due to the momentum built up in the rapid ...


5

where E(s) is the error (and also the signal carried forward directly from the summing node), R(s) input and C(s) output. The error is "demand" minus "output" and the output and the demand are desired to be equal hence, the "thing that does the math" is a subtractor: - Picture from here. This definition has the slightly unsettling effect of yielding a ...


5

The graph with the P-control response looks quite decent already. Based on the P-control response, you just need a little integral action to take out the DC error. (unless you also want to boost the closed loop bandwidth...) You can use that P-control setup to characterize the open loop response even better, by feeding it sine wave inputs, superimposed over ...


4

Usually clipping limits the amplitude, but if you want to build a low-distortion oscillator, then every component must stay in its linear range. Clipping is not allowed. Which means we have a problem: self-excited sustained oscillations are only possible by means of a marginally stable system The Wien Bridge oscillator is an example of a solution. It ...


4

Any oscillator needs that little bit extra gain to start. Oscillations build cycle after cycle. Your example circuit is powered with a +5v supply. Oscillation peak-to-peak amplitude certainly cannot exceed that supply voltage, so peaks are clipped - there's your non-linearity that keeps amplitude from growing forever. It is not a particularly good oscillator,...


4

It's eminently possible for an open loop system to have exactly the same TF as a closed loop system: \$ \frac {1}{1+s}\$ could be the TF of a simple series RC circuit with R=1, C=1, or it could be a unity feedback closed loop system with an integrator, \$\frac{1}{s}\$, in the forward path. Or it could be something completely different. It's impossible to ...


4

The question assumes there is one damping factor i.e. the transfer function is dominated by a 2nd order system. In practice there could be several interacting 2nd order systems so care has to be taken here. Assuming it is dominated by a single 2nd order system, this diagram may be useful: - This picture allows you to calculate \$\zeta\$ by looking at the ...


4

In simply words: Regulator (controller) is a device that controls the object in closed loop on the basis of difference (error of regulation) between measurements of object's output and external steering signal, regulator tries to reduce error to zero. Compensator only change object characteristics (transfer function), e.g. phase of output signal. ...


4

You could use Newton's method. Yeah I know, it's not the "root locus method", but hey, it works. Let \$F(s)=3s^4+10s^3+21s^2+24s+30\$ Clean it up by dividing everything by 3 so our largest s is alone Let \$f(s)=\dfrac{F(s)}{3}=s^4+\frac{10}{3}s^3+7s^2+8s+10\$ then \$f'(s)=4s^3+10s^2+14s+8\$ Now we have everything we need to start converging onto the ...


4

Working in arctan is quite difficult. Instead, use complex notation, and this will give the answer, \$\omega =\large \frac{\sqrt{5}}{2}\$ rad/sec. ... In response to the OP's comment ... Thus, ignore the numerator gain (10) as this does not affect the phase angle, and work from: $$G(s)=\frac{(s+5)}{s(s^2 +s +1)}$$ Let \$s\rightarrow j\omega\$ : $$G(j\omega)=\...


3

It's apparent that a first-order block cannot have an oscillatory step response. How come? A first order system will reach a state of equilibrium in the fullness of time and the energy transfers between components will also reach equilibrium and remain stable in the fullness of time. An underdamped second order system will not have the component ...


3

The pfd (with charge pump) generates current pulses of fixed amplitude ICPICP like it is described for example here. For small phase deviations the length of these current pulses is proportional to the phase difference of the input signals. So the pfd output current is clearly not proportional to the phase error. Since the duty cycle is proportional to the ...


3

You should do some revision on complex numbers, e.g. the magnitude of a complex number, \$(a+jb)\$, is \$\sqrt{ (a^2+b^2)}\$ and not \$\sqrt{(a+jb)^2}\$, which is meaningless in this context - you should recognise that \$\sqrt {(a+jb)^2} = a+jb\$. Similarly, the phase angle of \$(a+jb)\$ is \$arctan (\frac{b}{a})\$. Magnitude and phase angle are scalar ...


3

What are the benefits of I control over PI control, and when should it be used / not be used? Just as with any other forms of control, the answer is simple - you should use the form that best suits the physical characteristics of the system you are trying to control. There is no generic "this one is the best" answer. Let's review the properties of I-only ...


3

To ensure that any feedback system is stable, the loop gain needs to have a phase difference less than 180 degrees when its magnitude is unity, in other words it needs to ave sufficient phase margin. Having a right half plane zero is non-favorable in this sense. It increases your loop gain and increases your phase difference. Consequently, your phase margin ...


2

One reason for using an integrator in the forward path of a closed-loop system (if integration is not present, inherently) is that it gives unity 'DC gain' or 'steady-state gain'. That is, the controller's I term forces the system steady-state error signal to zero for a step input. And, of equal importance, the I term also gives complete rejection of steady-...


2

Expanding on Ben's answer, let's presume we construct a PI with the integral replaced with a single pole low-pass filter. The TF for our controller is $$C(s) = K_p + \frac{K_f}{s+k} = \frac{K_ps+kK_p+K_f}{s+k},$$ so you see you've essentially turned your PI into a Lead-Lag compensator with pole/zero combo $$p = -k,\ \ z = -\left(k+\frac{K_f}{K_p}\right).$$ ...


2

"Single-pole lowpass filter", typically. The difference is that the lowpass filter has finite gain (1/k) at low frequencies, while the integrator theoretically has infinite gain at DC.


2

(Answering this despite broad scope because this is clearly a student capstone project... I'm treating this as a question about how a university student should approach capstone project management.) As with any design, first identify the project requirements, and try to break down the whole project into smaller design units. This is still a pretty big ...


2

The key of these stability questions is Nyquist´s stability criterion (which - in this case - can be expressed also with Barkhausens`s oscillation condition): LOOP GAIN OF UNITY. That means: A circuit with feedback will become unstable (oscillation or transistion to saturation) if at a certain frequency the total phase shift of the loop gain function ...


2

Take the simple example of an op-amp with dominant pole compensation. Open loop it has the approximate transfer function of Ao/(1+s*tau). Put resistive feedback around it and you can get an amplifier with whatever gain you want, and a single pole roll off at the gain-bandwidth divided by the gain. (Of course within GBW limits etc.) Now of course it's ...


2

Write it in the form \$\dfrac{K}{s(1+s/\omega_1)(1+s/{\omega_2})}\$, then you have the breakpoints and the correct gain term.


2

First check the transfer function for poles and zeros. In this case the numerator is a constant, so there are no zeros. The poles are the zeros of the denominator. By inspection we see that they are at s=0, s=-0.1, s=-100. For each pole we get additional -20dB/dec and additional -90 degrees phase shift. The gain starts to change at the corresponding ...


2

In the first calculation you correctly calculated the gain before the second pole to be 48000/100 On the second example the gain before the second pole is 48000/(100*0.1). So it is 20dB higher. This compensates with the higher attenuation of 20dB for the additional pole, so at w=1 the gain is unchanged. To avoid these issues it is better to use a ...


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