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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 ...

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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 ...

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The frequency response can exist, even if there are poles in the right half-plane, as long as there are no poles on the $j\omega$-axis. You cannot use the unilateral Laplace transform, instead you must use the bilateral Laplace transform (with the lower integration limit at $-\infty$), or, simply, the Fourier transform. The corresponding time domain ...

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Several corrections: This code does not make sense: precision = fs/n; w = linspace(0,pi-precision/2,n); f = w/pi*fs/2; Your precision should be the resolution in frequency domain (I think). But, now consider n=fs. Then precision=1 (which is fine), but w ranges only from (0...pi-0.5), i.e. you do not span the entire unit circle. The corrected code is (you ...

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$$H(s) = \frac{R C s}{R C s + 1}$$ If $R C s$ is much lower than 1 (i.e., $R C s \ll 1$, in Math), then you can make an approximation: $$H(s) \simeq \frac{R C s}{1}$$ Basically, the effect of the $R C s$ in the denominator becomes insignificant. So, just pick a radian frequency well below $1 / R C$ and compute the magnitude of $H(s)$ at that point -- mark ...

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The slope of the lowpass is 20dB/decade or 6dB/octave, that means it's a simple first order lowpass filter. At the corner frequency the gain is -3dB. Same of the phase. It goes from 0 to -90, so it's a first order filter. At the corner frequency, it's 45 degrees. Looking at both phase and level we conclude that it's a first order lowpass with $f_c = 1000 rad/... 2 The first compensator$\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 I thought that, seeing the Bode plots one could tell if the closed-loop system would be stable if the 0 dB crossing occured at a lower frequency than the −180° crossing. The usual formulation is to say that the phase margin at the 0dB crossing is > 0. Matlab adds a vertical line to the phase diagram at the frequency where 0dB is. The length of that line is ... 1 The transfer function $$H(z)=\frac{1}{1-z^{-2}}\tag{1}$$ has two poles on the unit circle (at$z=\pm 1$). Consequently, the corresponding frequency response (i.e., the Fourier transform of the impulse response) doesn't exist as an ordinary function.$^1$Nevertheless, it is possible to evaluate the magnitude of$H(z)$on the unit circle$z=e^{j\omega}$if we ... 1 The OP has a mistake in the numerator vector as there is an additional space between the minus sign and the 1.027e14 coefficient which will cause that to operate as an expression (-2.58e13 - 1.027e14) instead of negate the sign. Had the system been correct, the resulting closed loop system with the compensator and gain chosen is underdamped. This can be ... 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 Assuming you have an input signal$ u = A cos(2\pi ft) $and you measure an output signal$y = B cos(2\pi ft + \theta) $The$\frac{B}{A}$ratio is the gain and$\theta$is the phase shift for frequency$f$. You could simply demodulate$y$by multiplying by$x = cos(2\pi ft) - jsin(2\pi ft)$$$z(t) = y*x = \frac{B}{2}(cos(4\pi ft) + cos(\theta) + j(sin(4\pi ... 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 I suspect that the reason for the difference is the number of samples used in the plot, since the extreme points are likely going to infinity (or very large numbers) such that if less points are used, those large values are simply missed in the computation. If the OP uses the same number of points in each, then the same result should be achieved- and for all ... 1 1) To determine if a system is stable by looking at a bode plot, generally, you just need to look at two things: Gain Margin and Phase Margin. First find the frequency at which you phase plot reaches -180 degrees. In this case, it is 18.6kHz (I am inferring this from the pictures you sent because the plot covers up this crossing). Then find the gain at that ... 1 The "Error" would be the actual phase and amplitude relative to the asymptotic plot shown. You computed the gain and phase, so now compare this to the asymptotic lines given by the plot to determine the error. 1 In the simplest terms, the slowness or fastness of a filter refers to its time domain characterization. Specifically a fast filter will have a short impulse response so that the output goes to steady state as quick as possible in time, whereas a slow filter will have a long impulse response, whose steady state output occurs after a longer delay, so that it ... 1 There is not really a difference between wrapped and unwrapped phase, it is mainly more visually pleasing when it is continues. The only way this could occur in an actual transfer function is when you would have two complex conjugate unstable poles and two complex conjugate stable zeroes with the same complex and real part, but the real part with opposite ... 1 You are probably still seeing other frequency components from the transient of the system. I assume you that you simulated your system for a fixed number of periods of the excitation signal. If this is the case then it would also explain why you only see the transient at high frequencies. This would be because a fixed number of high frequency periods take ... 1 In the most typical sense the Bode plot is actually a, very good, piece-wise linear approximation to a logarithmically plotted (on both x and y axes) nonlinear curve given by the frequency response magnitude of a linear time invariant system (mainly analog) with a rational transfer function which is described in your question by:$$ H(\omega) = \frac{10}{1 + ... 1 If your question is how to use this graph to find the transfer function the answer is simple : you can't. First of all, this graph give only half of the information: you have a gain plot but not the phase plot, and 2 systems with different TF could have the same gain profile. Second of all, this is obviously an asymptotic graph, as it presents a ... 1 Your Bode plot is a straight line. As we are on a semilog-scale, the equation is given by $$y = m \log(x) + b$$ In this case,$y$would be the magnitude of the frequency response$|H(j\omega)|$in$\mathrm{dB}$, and$x$would be the frequency$\omega$. Taking two points of the plot and solving the$2\times2$equation system should give you the function ... 1 You can evaluate the frequency response by evaluating$H(z)$in the unit circle:$H(e^{j\theta})$, where$\theta$is the discrete-time frequency. If you prefer a frequency in Hz, then use$\theta = 2\pi/f_s$(i.e. the sampling frequency corresponds to$\theta = 2\pi\$). The above expression will give you the frequency response as a complex number, ...

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