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LabVIEW Graphical Approach Using the VI from Figure 1, replace the CD Bode VI with the CD Gain and Phase Margin VI, found in the Frequency Response section of the Control For the example system, the controlled system - often referred to as the plant - is a first order system with a transfer function: G(s) = Gdc/(st + 1) We will Therefore, our bandwidth frequency will be the frequency corresponding to a gain of approximately -7 dB. Step Input (R(s) = 1 / s): (3) Ramp Input (R(s) = 1 / s^2): (4) Parabolic Input (R(s) = 1 / s^3): (5) When we design a controller, we usually useful reference

Figure 2: Bode Plots in LabVIEW Note the axes of the plots in Figure 2. Here is a simulation you can run to check how this works. The only input that will yield a finite steady-state error in this system is a ramp input. The system returned: (22) Invalid argument The remote host or network may be down. https://www.facstaff.bucknell.edu/mastascu/eControlHTML/Design/Perf1SSE.htm

Finding the phase margin is simply the matter of finding the new cross-over frequency and reading off the phase margin. The phase margin is defined as the change in open loop phase shift required to make a closed loop system unstable. Please try the request again.

It helps to get a feel for how things go. We can also read off the **plot that for** an input frequency of 0.3 radians, the output sinusoid should have a magnitude about one and the phase should be shifted by In this lesson, we will examine steady state error - SSE - in closed loop control systems. Steady State Error In Control System Problems We choose to zoom in between time equals 39.9 and 40.1 seconds because that will ensure that the system has reached steady state.

Figure 5:BodePlotof a System with Gain(Download) LabVIEW MathScript Approach If you used m-file code to model the system, enter the command bode(100*sys) into the MathScript Window. Steady State Error Matlab K = 37.33 ; s = tf('s'); G = (K*(s+3)*(s+5))/(s*(s+7)*(s+8)); sysCL = feedback(G,1); t = 0:0.1:50; u = t; [y,t,x] = lsim(sysCL,u,t); plot(t,y,'y',t,u,'m') xlabel('Time (sec)') ylabel('Amplitude') title('Input-purple, Output-yellow') In order to Be able to specify the SSE in a system with integral control. https://www.cds.caltech.edu/~murray/amwiki/index.php/FAQ:_What_is_steady_state_error%3F Let's use these concepts to design a controller for the following system: Figure 13: A Closed-Loop System In this system, Gc(s) is the controller, and G(s) is: The design must meet

LabVIEW Graphical Approach Increase the input signal frequency to 3, using the front panel control of your VI. How To Reduce Steady State Error However, at steady state we do have zero steady-state error as desired. Root Locus Controls Tutorials Menu State Space Back to Top Bookmark & Share Share Downloads Attachments: freq_step-resp-no-controller.vi freq_step-resp-pi-controller.vi frequency_bode-plot-gain.vi freq_gain-and-phase-margin.vi frequency_bode-pi-controller.vi frequency_bode-plot.vi frequencyresp_tutorial_vis.zip frequency_lsim.vi Ratings Rate this document Select a Let's place the **zero at 1 for** now and see what happens.

Create indicators for Magnitude Plot, Phase Plot, and Gain and Phase Margins. Just extend the low frequency line to the w=1 line. Steady State Error Example w = 0.3; num = 1; den = [1 0.5 1]; sys = tf(num,den); t = 0:0.1:100; u = sin(w*t); [y,t] = lsim(sys,u,t); plot(t,y,t,u) Result We must keep in mind that Steady State Error Wiki Therefore, we can get zero steady-state error by simply adding an integrator (a pole at the origin).

Then we can apply the equations we derived above. see here The difference between the measured constant output and the input constitutes a steady state error, or SSE. Now suppose you added a gain of 100. To confirm this, look at the Bode plots in Figure 4, find where the curve crosses the -40dB line, and read off the phase margin. Determine The Steady State Error For A Unit Step Input

The system returned: (22) Invalid argument The remote host or network may be down. Sinusoidal inputs with frequency greater than Wbw are attenuated (in magnitude) by a factor of 0.707 or greater (and are also shifted in phase). In LabVIEW, the graphical approach is best, so that is the approach we will use. http://interopix.com/steady-state/steady-state-error-unit-ramp-response.php You will get **a grade on a 0 (completely** wrong) to 100 (perfectly accurate answer) scale.

To be able to measure and predict accuracy in a control system, a standard measure of performance is widely used. Steady State Error Pdf That is especially true in computer controlled systems where the output value - an analog signal - is converted into a digital representation, and the processing - to generate the error, However, it has certain advantages, especially in real-life situations such as modeling transfer functions from physical data.

From the equation that relates Ts*Wbw to damping ratio, we find that Ts*Wbw ~ 21. Therefore, we ignore the transient response when we look at these plots. Remember that we are looking at the open-loop Bode plots. Steady State Error Control System Example Enter your answer in the box below, then click the button to submit your answer.

Looking at the plot, we find that it is approximately 1.4 rad/s. We have: E(s) = U(s) - Ks Y(s) since the error is the difference between the desired response, U(s), The measured response, = Ks Y(s). In essence we are no distinguishing between the controller and the plant in our feedback system. Get More Info The steady-state error can be read directly off the Bode plot as well.

Try several gains and compare results using the simulation. Yes No Submit This site uses cookies to offer you a better browsing experience. Generated Sun, 30 Oct 2016 05:18:35 GMT by s_hp106 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.10/ Connection Therefore, we can solve the problem following these steps: (8) (9) (10) Let's see the ramp input response for K = 37.33 by entering the following code in the MATLAB command

For a SISO linear system with state space dynamics with a stable matrix (eigenvalues have negative real part), the steady state error for a step input is given by In the Your cache administrator is webmaster. If G(s) is the open loop transfer function of a system and w is the frequency vector, we then plot G(j*w) versus w. There are two ways of solving this problem: one is graphical and the other is numerical.

And we know: Y(s) = Kp G(s) E(s). Figure 11:Linear Simulation ofa System(Download) Note that the output (white) tracks the input (red) fairly well; it is perhaps a few degrees behind the input as expected. Your grade is: Problem P4 What loop gain - Ks Kp G(0) - will produce a system with 1% SSE? The first thing we need to find is the damping ratio corresponding to a percent overshoot of 40%.

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