3 views (last 30 days)
Show older comments
Muhammad Hamza on 16 Mar 2023
-
-
Link
Direct link to this question
https://matlabcentral.mathworks.com/matlabcentral/answers/1929955-how-to-find-angle-of-impedence
Commented: Muhammad Hamza on 4 Apr 2023
Accepted Answer: Muhammad Hamza
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.
7 Comments Show 5 older commentsHide 5 older comments
Show 5 older commentsHide 5 older comments
Mathieu NOE on 16 Mar 2023
Direct link to this comment
https://matlabcentral.mathworks.com/matlabcentral/answers/1929955-how-to-find-angle-of-impedence#comment_2664510
Edited: Mathieu NOE on 16 Mar 2023
hello
doing a impedance measurement (or simulation) is basically making a transfer function estimate between your current and voltage
you can use the matlab function tfestimate for that (Signal Processing Toolbox required)
Transfer function estimate - MATLAB tfestimate - MathWorks France
the test signal must cover the entire frequency range of interest (chirp or random signals are the two most common used)
the result is a complex FRF (frequency response function) , you will take the amplitude and phase of it (with abs and angle functions)
Muhammad Hamza on 20 Mar 2023
Direct link to this comment
https://matlabcentral.mathworks.com/matlabcentral/answers/1929955-how-to-find-angle-of-impedence#comment_2668915
- Angle.PNG
Hello Mathieu, Thank you for your explanation,I implemented this but unfortunately the same result,I got the amplitude correct but not the angle,Now I have analized the data and I noticed that I always got 0 or 180 angle in degrees Because of the real and non real values of my angle.Please see the attachment.
May be this is the reason that when I am trying to find the angle I get 0 or 180(depends on the data) at the point of my desired Impedence value.Else all the other values of my data are giving the correct angle.It is happening might be due to the fact that for Impedence it is dividing the peak voltage amplitude by peak current amplitude of all my 129 sample values and gives the correct angle and it gives angle 0 or 180 where i got the exact required Impedence value which is actually the division of peak values of both.
OR May be I should try to do signal reconstruction and try to find angle and impedence,If fft doesnot work for me in that case? Or else I come to know what i am doing wrong actually. Thanks in advance.
Mathieu NOE on 20 Mar 2023
Direct link to this comment
https://matlabcentral.mathworks.com/matlabcentral/answers/1929955-how-to-find-angle-of-impedence#comment_2668935
hello
I am still unsure what your are doing and how you do the phase computation
are you doing it by using angle(fft(Voltage)./fft(current)) ?
Muhammad Hamza on 20 Mar 2023
Direct link to this comment
https://matlabcentral.mathworks.com/matlabcentral/answers/1929955-how-to-find-angle-of-impedence#comment_2669170
Hi Mathieu,
Impedence=(VoltageAmplitude./CurrentAmplitude)
Theta=angle(FFTvoltage./FFTCurrent)
AND WITH tfestimate I am doing like this below
estimatetesting=tfestimate(MEANCURRENTWITHDELAY,MEANVOLTAGEWITHDELAY)
itsangle=angle(estimatetesting);
itsangleinDegrees=rad2deg(itsangle)
itsamplitude=abs(estimatetesting)
Mathieu NOE on 20 Mar 2023
Direct link to this comment
https://matlabcentral.mathworks.com/matlabcentral/answers/1929955-how-to-find-angle-of-impedence#comment_2669420
ok
and you get now the right angle ?
Muhammad Hamza on 20 Mar 2023
Direct link to this comment
https://matlabcentral.mathworks.com/matlabcentral/answers/1929955-how-to-find-angle-of-impedence#comment_2669500
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 on 20 Mar 2023
Direct link to this comment
https://matlabcentral.mathworks.com/matlabcentral/answers/1929955-how-to-find-angle-of-impedence#comment_2669580
Open in MATLAB Online
@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.
.
Sign in to comment.
Sign in to answer this question.
Accepted Answer
Muhammad Hamza on 4 Apr 2023
Hello Everyone,
My problem was solved sorry for informing late.
Many thanks to @Mathieu NOE @Star Strider @Swaraj
Best Regards,
Muhammad Hamza.
3 Comments Show 1 older commentHide 1 older comment
Show 1 older commentHide 1 older comment
Mathieu NOE on 4 Apr 2023
Direct link to this comment
https://matlabcentral.mathworks.com/matlabcentral/answers/1929955-how-to-find-angle-of-impedence#comment_2689764
Good news !!
Star Strider on 4 Apr 2023
Direct link to this comment
https://matlabcentral.mathworks.com/matlabcentral/answers/1929955-how-to-find-angle-of-impedence#comment_2689949
What was the solution?
Muhammad Hamza on 4 Apr 2023
Direct link to this comment
https://matlabcentral.mathworks.com/matlabcentral/answers/1929955-how-to-find-angle-of-impedence#comment_2690109
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.
Sign in to comment.
More Answers (1)
Swaraj on 4 Apr 2023
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)
0 Comments Show -2 older commentsHide -2 older comments
Show -2 older commentsHide -2 older comments
Sign in to comment.
Sign in to answer this question.
See Also
Categories
Signal ProcessingSignal Processing ToolboxSpectral AnalysisSpectral Measurements
Find more on Spectral Measurements in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
An Error Occurred
Unable to complete the action because of changes made to the page. Reload the page to see its updated state.
Select a Web Site
Choose a web site to get translated content where available and see local events and offers. Based on your location, we recommend that you select: .
You can also select a web site from the following list
Americas
- América Latina (Español)
- Canada (English)
- United States (English)
- Belgium (English)
- Denmark (English)
- Deutschland (Deutsch)
- España (Español)
- Finland (English)
- France (Français)
- Ireland (English)
- Italia (Italiano)
- Luxembourg (English)
- Netherlands (English)
- Norway (English)
- Österreich (Deutsch)
- Portugal (English)
- Sweden (English)
- Switzerland
- Deutsch
- English
- Français
- United Kingdom(English)
Asia Pacific
- Australia (English)
- India (English)
- New Zealand (English)
- 中国
- 日本Japanese (日本語)
- 한국Korean (한국어)
Contact your local office
