# Tag Info

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As you know, poles and zeroes are the respective roots of the denominator and the numerator of a complex transfer function. Once they are known, they can be placed on a map - the $s$-plane - depending on their real and imaginary components. The below drawing shows an example for a transfer function featuring one zero and three poles: The left-side of the ...

4

Bibo stability is all about systems external stability which is determined by applying the external input with zero initial condition (transfer function in other words) so if you check bibo stability of G(s) ,it would be bibo stable But Asymptotic stability is all about systems internal stability which can be determined by applying the non zero initial ...

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Consider an OLTF $$\small G(s)=\frac{s-a}{s+b}$$ with $\small a>0$ and $\small b>0$. The root locus would show the closed loop pole being attracted to the zero, hence pulled to the right and becoming slower, thereby reducing the bandwidth. A pure time delay can be modelled as a first order Pade approximant: $$\small e^{-sT}\rightarrow \frac{1}{1+Ts}... 3 A system with a cancelled pole and zero may still be unstable (just not in the BIBO sense) -- it can be mathematically shown that a system with such a cancellation may still misbehave due to its response to an initial condition - this is in contrast to a stable system with all poles in the LHP, where the contribution of the initial condition to the output ... 2 One way you can achieve "placeable zeros" is by using a dynamic feedback. As given in the question, this is what you achieved with constant feedback on three loops:$$G(z)=\frac{1}{z-1}\text{NTF(z)}=\frac{1}{\left(1+\frac{\alpha}{z-1}\right)\left(1+\frac{\beta}{z-1}\right)\left(1+\frac{\gamma}{z-1}\right)}\text{NTF(z)}=\frac{(z-1)^3}{(z-1+...

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I need to know its value You can estimate it from the data sheet: - So, it has an open voltage of typically 5.5 volts and it has a shorting current of 1.2 mA. This means it typically has an internal pull-up resistor of 5.5/0.0012 = 4583 ohms. This is a typical value and as you can see it might be 5.5/0.0022 (which equals 2500 ohms) or 6/0.0012 = 5000 ohms. ...

2

The right-half-plane zero (RHPZ) in a switching converter featuring an indirect energy-transfer scheme is present because of two things: the output current delivered to the load depends on $(1-D)$. For instance, in a boost converter operated in continuous conduction mode (CCM), the output current is the diode average current expressed as: $I_{out}=I_L(1-... 2 "Voltage-current dynamics" means the response of the motor current to applied voltage (ignoring, it appears, the back-emf generated by the motor -- presumably the controller is fast enough that it can ignore that). It looks like the author meant "current position dynamics" to mean the response of the vehicle position to motor current. ... 2 What you have drawn seems to be for negative feedback. This is the result for positive feedback. RootLocusPlot[k (s - 3)/(s^2 + 3 s + 2), {k, 0, 10}, FeedbackType -> "Positive"] 1 Here's a little help with the partial fraction decomposition: - \dfrac{1}{s}\cdot \dfrac{k}{s^2 + s +k}\hspace{1cm} =\hspace{1cm} \dfrac{1}{s}+ \dfrac{-s-1}{s^2 + s + k} Can you take it from here? If not try this inverse laplace solver. 1 Standard form is$G(s)=\large\frac{K_{ss}\omega_n^2}{s^2+2\zeta \omega_n s+\omega_n^2}$, where$\small K_{ss}\\$ is the steady state gain.

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The procedure is quite simple. When you have the BODE plot for the LOOP GAIN (open-loop transfer function) you only must shift the magnitude plot up or down until the desired phase margin is met. This shifting will not influence the phase plot. As a result of this shifting the frequency where the magnitude crosses the horizontal axis (0 dB) will be shifted ...

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Old question but regarding And in block diagram, there is only one solution for a control system design ? the answer is no, there's no unique representation in block diagrams of a control system. There're theorems on the algebra of block diagrams. Applying them you can manipulate block diagrams to obtain others just as valid; they're equivalent. This is ...

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Jeff is right. The system is not linear, it is affine. You can approach this by changing the system slightly to get a linear system. You can split the input into two parts, a constant nominal value and a perturbation from that nominal value: $$V_{IN} = \bar{V}_{IN} + \delta V_{IN}$$ As you can see, setting the nominal part to -5/m will ensure the system is ...

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