How to turn S11 to time domain by Matlab

Hi,everyone.I run HFSS to get S11 for my circuit,HFSS also provide time domain for S11.I want to obtain the same time domain response by Matlab.However the result is not even close.Can anyone help~Thx!

The input signal is an impulse and I perform the frequency domain simulation from 0~7.5GHz.

And both attached files are the result from HFSS.

Here's my code:

load S11_re.txt -ascii

load S11_im.txt -ascii

freq=1e+9*S11_re(:,1);

S11=S11_re(:,2)+1i*S11_im(:,2);

TDR=ifft(S11);

Fs=2*max(freq);

Ts=1/Fs;

N=numel(TDR);

tvec=(0:(N-1))*Ts;

plot(tvec,abs(TDR))

[Mathieu NOE]

the frequency domain data must contain the phase also (we need to compute the ifft of a complex valued transfer function, here you provide only the modulus)

[Guan Hao]

@Mathieu NOE Sorry,I'm not that familiar with the time domain signal.

Thanks for your help ! The phase data of frequency domain has been uploaded.

I've shared the complex value of S11 and also modified my code.

Answer 1

[Mathieu NOE]

hello again

I am mot sure to understand the shape of the impulse response from your S11_time.txt file :

why only two positive peaks ? from the Bode plot (first figure) I was expecting some kind of damped oscillations - alike what we get from the ifft (nb the symmetrical computation, including negative and positive frequency data)

also I was unsure what is the time unit in the S11_time.txt file ?

% freq domain (transfer function)
load S11_re.txt -ascii % freq + S11_re
load S11_im.txt -ascii % freq + S11_im
freq=1e+9*S11_re(:,1);
frf=S11_re(:,2)+1i*S11_im(:,2);

figure(1),
subplot(2,1,1),plot(freq,abs(frf))
xlabel('freq (Hz)')
ylabel('amplitude')
title('Impulse Response (IR)');
subplot(2,1,2),plot(freq,180/pi*angle(frf))
xlabel('freq (Hz)')
ylabel('phase(°)')


% IR (impulse response) obtained with ifft method
if mod(length(frf),2)==0 % iseven
    frf_sym = conj(frf(end:-1:2));
else
    frf_sym = conj(frf(end-1:-1:2));
end
    
TDR = real(ifft([frf; frf_sym])); % NB we need the negative and positive frequency complex FRF
TDR = TDR(1:101); % truncation is possible if TDR decays fast enough

Fs=2*max(freq);
Ts=1/Fs;
N=numel(TDR);
tvec=(0:(N-1))*Ts;

% compare to impulse response generated by your software
load S11_time.txt -ascii % time + IR
time = S11_time(:,1);
IR = S11_time(:,2);
time = time/1e9; % assuming time data was given in nanoseconds

figure(2),
plot(tvec,TDR./max(TDR),'b',time,IR./max(IR),'r')
xlabel('time (s)')
ylabel('amplitude')
title('Impulse Response (IR)');
legend('IR from re/im data','IR from file');

[Guan Hao]

@Mathieu NOE Thanks for your help!The S11_time file is the magnitude of the signal,so it returns two positive peak.I'll send another file.

By the way,the time unit is nanosecond.

[Mathieu NOE]

this time it's ok - how can you obtain either two positive peaks or pos / neg ?

results with provided file are matching , I simply removed the header lines so load will work fine (see attached txt file)

all the best

% freq domain (transfer function)
load S11_re.txt -ascii % freq + S11_re
load S11_im.txt -ascii % freq + S11_im
freq=1e+9*S11_re(:,1);
frf=S11_re(:,2)+1i*S11_im(:,2);

figure(1),
subplot(2,1,1),plot(freq,abs(frf))
xlabel('freq (Hz)')
ylabel('amplitude')
title('Impulse Response (IR)');
subplot(2,1,2),plot(freq,180/pi*angle(frf))
xlabel('freq (Hz)')
ylabel('phase(°)')


% IR (impulse response) obtained with ifft method
if mod(length(frf),2)==0 % iseven
    frf_sym = conj(frf(end:-1:2));
else
    frf_sym = conj(frf(end-1:-1:2));
end
    
TDR = real(ifft([frf; frf_sym])); % NB we need the negative and positive frequency complex FRF
TDR = TDR(1:101); % truncation is possible if TDR decays fast enough

Fs=2*max(freq);
Ts=1/Fs;
N=numel(TDR);
tvec=(0:(N-1))*Ts;

% compare to impulse response generated by your software
load S11_time.txt -ascii % time + IR
% remove the header lines otherwise load will throw an error message
% ============================================================================================
% ANSOFT                                                                              04/18/24
% Terminal S Parameter Plot 2                                                         00:46:05
% 
% --------------------------------------------------------------------------------------------
%         Time [ns]                   St(1,1) []
                                              
       
time = S11_time(:,1);
IR = S11_time(:,2);
time = time/1e9; % assuming time data was given in nanoseconds

figure(2),
plot(tvec,TDR./max(TDR),'b',time,IR./max(IR),'r')
xlabel('time (s)')
ylabel('amplitude')
title('Impulse Response (IR)');
legend('IR from re/im data','IR from file');

[Guan Hao]

@Mathieu NOE The software can display S11 in many ways,I sent the absolute value in former file,so it shows two positive peak.If I display the original data,it shows one postive peak and one negative peak.

Thank you for your help,I do learn a lot about how to do ifft.

By the way,I think the software inputs a rectangular pulse with the width of Ts(Time step) and the magnitude is 1/Ts.So maybe I should create a rectangular pulse and convolution with the impulse response?

But I got a question,how do I know where the rectangular pulse locates on the axis of time?

% create input rectangular pulse
t = 0 : Ts : 2e-9; 
Amp=1/Ts; % amplitude of the pulse
u=[zeros(1,numel(0:Ts:Ts)) Amp*ones(1,numel(Ts:Ts:2*Ts)) zeros(1,numel((2*Ts)+(Ts):Ts:2e-9))]; 
plot(u)
TDR=real(ifft([frf; frf_sym])); % NB we need the negative and positive frequency complex FRF
TDR=TDR(1:numel(u)); % truncation is possible if TDR decays fast enough
TDR1=conv(TDR,u)

[Mathieu NOE]

yes, you need to display the data with positive and negative peaks , otherwise we are comparing oranges and apples.

for the convolution with an impulse , an impulse will not modify your IR as a "true" impulse has falt spectrum (infinite)

as you create a rectangular implse with some samples delay, the effect is to delay your IR with a bit or smoothing (less ringing) as the spectrum of a rectangualr impulse has decaying energy as we go into higher frequencies.

% freq domain (transfer function)
load S11_re.txt -ascii % freq + S11_re
load S11_im.txt -ascii % freq + S11_im
freq=1e+9*S11_re(:,1);
frf=S11_re(:,2)+1i*S11_im(:,2);

figure(1),
subplot(2,1,1),plot(freq,abs(frf))
xlabel('freq (Hz)')
ylabel('amplitude')
title('Impulse Response (IR)');
subplot(2,1,2),plot(freq,180/pi*angle(frf))
xlabel('freq (Hz)')
ylabel('phase(°)')


% IR (impulse response) obtained with ifft method
if mod(length(frf),2)==0 % iseven
    frf_sym = conj(frf(end:-1:2));
else
    frf_sym = conj(frf(end-1:-1:2));
end
    
TDR = real(ifft([frf; frf_sym])); % NB we need the negative and positive frequency complex FRF
TDR = TDR(1:101); % truncation is possible if TDR decays fast enough

Fs=2*max(freq);
Ts=1/Fs;
N=numel(TDR);
tvec=(0:(N-1))*Ts;

% compare to impulse response generated by your software
load S11_time.txt -ascii % time + IR
% remove the header lines otherwise load will throw an error message
% ============================================================================================
% ANSOFT                                                                              04/18/24
% Terminal S Parameter Plot 2                                                         00:46:05
% 
% --------------------------------------------------------------------------------------------
%         Time [ns]                   St(1,1) []
                                              
       
time = S11_time(:,1);
IR = S11_time(:,2);
time = time/1e9; % assuming time data was given in nanoseconds

figure(2),
plot(tvec,TDR./max(TDR),'b',time,IR./max(IR),'r')
xlabel('time (s)')
ylabel('amplitude')
title('Impulse Response (IR)');
legend('IR from re/im data','IR from file');



% create input rectangular pulse
t = 0 : Ts : 2e-9; 
Amp=1/Ts; % amplitude of the pulse
u=[zeros(1,numel(0:Ts:Ts)) Amp*ones(1,numel(Ts:Ts:2*Ts)) zeros(1,numel((2*Ts)+(Ts):Ts:2e-9))]; 
figure(3),
% plot(u)
TDR=TDR(1:numel(u)); % truncation is possible if TDR decays fast enough
TDR1=conv(TDR,u)
tt = (0:numel(TDR1)-1)*Ts;
plot(tvec(1:numel(u)),TDR./max(TDR),'b',tt,TDR1./max(TDR1),'r')
xlabel('time (s)')
ylabel('amplitude')
title('Impulse Response (IR)');
legend('TDR','TDR1');

[Guan Hao]

@Mathieu NOE Was it nearly impossible to reconstruct the data produced by the software?

I saw some rectangular shape on the peak which was produced by the software.

[Mathieu NOE]

you notice that the IR lacks a bit of time resolution

we would need to acquire data at a faster rate (and your frequency data must also contain more high frequency data)

the rectangular shape is simply due to a "relatively" low sampling rate - so as you would obtain after downsampling a sharp peak , result is a coarse peak with rectangular shaped peak

[Guan Hao]

@Mathieu NOE I see.Thank you for your help sir!

[Mathieu NOE]

my pleasure !

[Guan Hao]

@Mathieu NOE Hi,sir.Sorry to bother you again.I'm thinking that maybe there's a possibility to obtain the third little peak (Multi-reflection) by the former two large peak (Two huge reflection in my circuit) by doing correlation.I've tried for few days but I can't get my result correct.The data of S11 was attached as an txt file.

Thx a lot !!!

%%% IFFT
% Reconstruct HFSS signal
% freq domain (transfer function)
load onepeakhighcircuit_re_im.txt -ascii % freq + S11_re
freq=1e+9*onepeakhighcircuit_re_im(:,1);
frf=onepeakhighcircuit_re_im(:,2)+1i*onepeakhighcircuit_re_im(:,3);
% IR (impulse response) obtained with ifft method
if mod(length(frf),2)==0 % iseven
    frf_sym = conj(frf(end:-1:2));
else
    frf_sym = conj(frf(end-1:-1:2));
end
 
Fs=2*max(freq);
Ts=1/Fs;
TDR=real(ifft([frf; frf_sym])); % NB we need the negative and positive frequency complex FRF
TDR=TDR(1:101); % truncation is possible if TDR decays fast enough
N=numel(TDR);
tvec=(0:(N-1))*Ts;
plot(tvec,TDR/max(TDR),'LineWidth',2)
hold on
grid on
% Cross Correlation
v=3e8;
t1=0.1/v;  % First reflection ending time,0.1 is the physical length of my circuit
l=t1/Ts;   % Number of points
R1=TDR(1:1:l+1); % First peak TDR
R2=TDR(l+1:1:(2*l)+1); % Second peak TDR
[c,lags]=xcorr(R1,R2); % Obtain multi-reflection from correlation
c1=[TDR(1:1:15);c;TDR(29:40)]; % Except the correlation part,the rest should remain the same
Tvec=(0:numel(c1)-1)*Ts;
plot(Tvec,c1/max(TDR),'--','LineWidth',3)
xlabel('time (ns)')
ylabel('normalized amplitude')
set(gca,'XLim',[0 2e-9])
set(gca,'XTick',0:2e-9/4:2e-9)
set(gca,'XTickLabel',0:0.5:2)
legend('IFFT From S11','Correlation Model')

[Mathieu NOE]

hello

no problem

I had a look at your code and played a bit with it , but I am not sure what method / principle you are using here to "create" that 3rd peak

so I may not really help you unless there is a paper or publication that coule help me understand what you are trying to achieve

[Guan Hao]

@Mathieu NOE Thanks for your reply.

When a signal is traveling in the circuit , it will generate the reflection when the impedance varies. There're two greatest impedance deficit in the circuit that produce two peak of reflection.When I send a signal from input,first reflection occurs at the first impedance deficit and the second reflection occurs at the second impedance deficit,both of the reflections travel back to input port.However,the reflection produced by the second peak produced another reflection when it met the first impedance deficit and it bounced between these two impedance deficit.That's why the third peak was seen in the IR. (multi-reflection)

And the third peak was related to the first and second peak.That's the reason why I'm trying to do correlation to try to predict the third peak.

Hope this is helpful !!!

[Mathieu NOE]

tx for the explanation

before coding it, simply , how do you predict when the third peak should be time localised from the two first peaks ? what is the math behind this ?