How to find Angle of Impedence?

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Muhammad Hamza
Muhammad Hamza 2023-3-16
评论： Muhammad Hamza 2023-4-4
Hello,
I am using fft to find amplitude of current and amplitude of voltage seperately.And then using angle command to determine angles of them respectively.But now I want to find Angle of Impedence as well,For that I am dividing the voltage amplitude by current amplitude and fortunately getting the correct impedence as well.But after that when i try to find the angle of Impedence I am not getting the correct angle.Could some body tell me why I am not getting the correct angle?.What is the problem or what should i need to do?. Thanks in advance.
For Example.
Impedence=(AmplitudeV/AmplitudeI);
Theta=angle(Impedence);
Is it the correct way to find Impedence using fft?
Please note that I am a new user of MATLAB,So apologies in advance if you feel that question is very basic.
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Muhammad Hamza
Muhammad Hamza 2023-3-20
No still not getting the correct angle.Tried both methods but same result,Getting the desired impedence but not the desired angle,Please see my above detailed explanation again. Thanks
Star Strider
Star Strider 2023-3-20
Ran in:
@Muhammad Hamza —
I am not certain what you are actually doing.
Consider something like this —
Fs = 1000; % Sampling Frequency (Hz)
t = linspace(0,10*Fs-1,10*Fs).'/Fs; % Assume Column Vectors
% Check = 1/(t(2)-t(1))
Fr = 250; % Signal Frequency
I = sin(2*pi*Fr*t); % Current Signal
V = 10*cos(2*pi*Fr*t); % Voltage Signal
Fn = Fs/2; % Nyquist Frequency (Hz)
L = numel(t); % Signal Length
NFFT = 2^nextpow2(L); % For Efficiency
FT_V = fft(V.*hann(L), NFFT)/L; % Voltage Fourier Transform (Windowed)
FT_I = fft(I.*hann(L), NFFT)/L; % Current Fourier Transform (Windowed)
Fv = linspace(0, 1, NFFT/2+1)*Fn; % Frequency Vector (One-Sided Fourier Transfomr)
Iv = 1:numel(Fv); % Index Vector (For Convenience)
figure
subplot(2,1,1)
plot(Fv, abs(FT_V(Iv))*2)
grid
ylabel('Magnitude')
subplot(2,1,2)
plot(Fv, rad2deg(angle(FT_V(Iv))))
grid
ylabel('Phase (°)')
sgtitle('Voltage')
figure
subplot(2,1,1)
plot(Fv, abs(FT_I(Iv))*2)
grid
ylabel('Magnitude')
subplot(2,1,2)
plot(Fv, rad2deg(angle(FT_I(Iv))))
grid
ylabel('Phase (°)')
sgtitle('Current')
FT_Z = FT_V ./ FT_I;
figure
subplot(2,1,1)
plot(Fv, abs(FT_Z(Iv))*2)
grid
ylabel('Magnitude')
subplot(2,1,2)
plot(Fv, rad2deg(angle(FT_Z(Iv))))
grid
ylabel('Phase (°)')
sgtitle('Impedence')
Windowing the Fourier transform reduces the amplitude of the resulting peaks by ½.
.

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采纳的回答

Muhammad Hamza
Muhammad Hamza 2023-4-4
Hello Everyone,
My problem was solved sorry for informing late.
Many thanks to @Mathieu NOE @Star Strider @Swaraj
Best Regards,
Muhammad Hamza.
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Muhammad Hamza
Muhammad Hamza 2023-4-4
There wee two things which I corrected,
1)The reference data I was working on was not accurate enough especially with floating numbers,Matlab by default showed me 4 digits after decimal so I have to change it to long g to get more precise results.
2)Another problem was solved by subtracting my angle values from 180,I guess the reason I found is that Matlab was returning me pi when I have negative real numbers and 0 when I have non negative real numbers.

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更多回答（1 个）

Swaraj
Swaraj 2023-4-4
You should take into consideration that there can be phase difference between the voltage and the current signals. One way is to use the complex Impedance Formula. It can be calculated as the ratio of the complex Voltage to Complex Current.
Z = V/I
Here V and I are Complex Numbers.
Then try using the angle function as “angle(Z)”.
Phase angle - MATLAB angle (mathworks.com)

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