Error/Mismatch in determing impedance using fft in simulink imported data.

Question

Md. Golam Zakaria 2024-5-22

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2121586-error-mismatch-in-determing-impedance-using-fft-in-simulink-imported-data

评论： Md. Golam Zakaria 2024-5-28

采纳的回答： Sayan

voda.slx

在 MATLAB Online 中打开

Hello Everyone

I need to calculate the impedance of a more or less complex circuit using FFT. I have implemented the circuit in simulink and saved the voltage and current in the workapace using To Workspace block. However, after performing fft , I didnt quite get the expected result. Now to check if my code was right, I was testing it on a simple circuit. I have implemented a simple RC circuit . The model is attached. To determine impedance I am using the theory Z(w)=fft(voltage)/fft(current). R=Real (Z(w), X=Imag(Z(w)).

Now, The snapshot of the circuit--

The code I am using-

clc
% clear
Time = out.tout;
Load_Voltage = out.V_L;
Load_Current = out.I_L;
Fs = 1 / mean(diff(Time));
N = length(Load_Voltage);
fft_Load_Voltage = fft(Load_Voltage);
fft_Load_Current = fft(Load_Current);
fft_Load_Voltage_single_sided = fft_Load_Voltage(1:N/2+1);
fft_Load_Current_single_sided = fft_Load_Current(1:N/2+1);
frequencies = Fs * (0:(N/2)) / N;
frequency_of_interest =13.56e6;
[~, index] = min(abs(frequencies - frequency_of_interest));
Z1 = fft_Load_Voltage_single_sided(index) / fft_Load_Current_single_sided(index);
R1 = real(Z1);
X1 = imag(Z1);
fprintf('R: %.2f ohms, X: %.2f ohms\n', R1, X1);

I have set the value of the resistor as 30 and the value of capacitor such as it would exibit -50 ohm at 13.56 Mhz, which is the driving frequency. But the results are off. The result I get is: R: 25.79 ohms, X: -37.52 ohms.

What is the problem here? Is it the code or anything with the simulink model. However if its just a resistive circuit the code gives correct result. This a very important step in a much bigger project. It is crucial.

Your help will be really appreciated.

Thank You.

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Pat Gipper 2024-5-22

编辑：Pat Gipper 2024-5-22

The model you attached and the picture that you show is a series RL circuit, with R = 50 Ohms and L = 0.587 uH.

Md. Golam Zakaria 2024-5-23

@Pat Gipper Fixed it. Sorry for that.

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Answer 1

Sayan 2024-5-28

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https://ww2.mathworks.cn/matlabcentral/answers/2121586-error-mismatch-in-determing-impedance-using-fft-in-simulink-imported-data#answer_1464286

voda.slx

Hi Md. Golam,

You do not need to use the powergui block here, as the block is only required in the case of solving Specialized Power System blocks. I could resolve the issue by setting the solver type to "Fixed-step". The "Variable-step" solver does not compute the value of the signal at a regular time interval. For FFT analysis, a fixed-step size ensures uniform sampling intervals, which is a prerequisite for standard FFT algorithms. I have got the expected value of resistance and capacitive impedence, as shown below.

I am attaching the updated model here. You can know more about the comparison between "Fixed-step" and "Variable-step" solver in the following documentation.

https://www.mathworks.com/help/simulink/ug/compare-solvers.html

Hope this resolves your issue.

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Sayan 2024-5-28

编辑：Sayan 2024-5-28

resample internally takes care of the interpolation. Please look at the part of the documentation where an example is shown to resample a non uniformally spaced data.

Md. Golam Zakaria 2024-5-28

@Sayan I have tried interpolation on the original concept of mine and its working. I will surely try resampling . Maybe it will give me more accurate results. Thank you for this. You have helped me big time.

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Answer 2

Pat Gipper 2024-5-28

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https://ww2.mathworks.cn/matlabcentral/answers/2121586-error-mismatch-in-determing-impedance-using-fft-in-simulink-imported-data#answer_1464486

@Md. Golam Zakaria You can use the variable step solver and force it to output at fixed intervals by putting a "Sample time" value other than -1 into the "To Workspace" blocks of your simulation as shown in the image below. I tend to use a workspace variable instead of a fixed number for flexibility.

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