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Which step response matches the system transfer function

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 ...
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4 votes

Which step response matches the system transfer function

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. ...
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How do bode plots work with unstable systems work?

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 ...
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3 votes
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bode plot, zero at 0Hz

$$H(s) = \frac{R C s}{R C s + 1}$$ 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 ...
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3 votes
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Matlab freqz and custom implementation differences

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, ...
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Why if I place the zero of the lead compensator at lower frequencies, do I obtain a resonance peak?

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 ...
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2 votes

Analyze stability of a closed-loop system with Bode

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 ...
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2 votes
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Finding transfer function from bode plot

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 ...
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2 votes
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Bode plot of the transfer function with unstable pole

If you pop in a very small s you get $$G(s) = \frac{(s+1)}{s(\frac{s}{10} - 1)} \approx\frac{1}{-s} = \frac{j}{\omega}.$$ That has indeed a phase of +90 or -270 ...
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2 votes

Differences between MATLAB and WolframAlpha and AstromMurray (minimun phase)

Whatever you are doing in Matlab appears to be wrong. Matlab has no problem getting the correct result if you code it correctly ...
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1 vote

Bode plot of the transfer function with unstable pole

The transfer function phase is the numerator phase minus the numerator phase The phase of the numerator can be expressed as $atan(\omega)$ For the denominator, it is a bit more complex. You need to ...
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1 vote

Z domain Transfer function to magnitude

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 ...
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1 vote

Stabilizing an unstable system and removing oscillations from step response

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) ...
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1 vote

How to deal with "weird" phase plots in bode diagram when designing a controller

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 ...
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1 vote

How to deal with "weird" phase plots in bode diagram when designing a controller

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 ...
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1 vote
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Creating Bode Plot from Experimental Data

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 ...
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1 vote
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What is the frequency of the Gaussian white noise model in control systems?

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 ...
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1 vote

Bode plot from from step response data

First of all - I'm not familiar with Python and didn't check your code. 1a) For FFTs it is the usual that the frequency resolution equals the inverse length of your data record. 1b) As far as I ...
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Why is my MATLAB's bode plot wildly off?

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, ...
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1 vote

Analyze stability of a closed loop system with Bode

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 ...
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  • 316
1 vote
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Slow and fast filter

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 ...
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1 vote
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Phase wrapping in Bode plot

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 ...
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1 vote
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Find the error in phase angle and dB gain Function Transfer

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 ...
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1 vote

Matlab - Bode plot of Lag Filter + Integrator

You should see that figure(2) is the only process that avoids some errors. Note on figure (1) the magnitude overlaps; you can add a constant say bode(lag_daccu_d,lagaccu+.0001,pts); to split them bu ...
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Matlab - Bode plot of Lag Filter + Integrator

I tried it in Octave, there's definitely a glitch like you said. I tried it with "Tustin" instead of "zoh", same result. However I was puzzled by your high sampling frequency. Your lag controller ...
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1 vote

How to make bode plot when output signal changes amplitude?

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 ...
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1 vote
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Could someone please explain how to get frequency response of a given Bode Plot?

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 ...
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1 vote

Could someone please explain how to get frequency response of a given Bode Plot?

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 ...
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What is the frequency response function for the Bode Plot pictured here?

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 $...
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  • 4,842
1 vote

Derive Bode Diagram from difference equation

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\...
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