# Discussion Forum

## Impulse Response of a SAW Sensor

 Topics: 4.2, SAW, Impulse Response
RSS feed   |   Turn on email notifications   |   1 Reply   Last post: September 18, 2015 6:13pm UTC

Mohamed Ashour

November 13, 2012 9:53pm UTC

Impulse Response of a SAW Sensor

Hi,

I've been working on a SAW Simulation for several months, i've been good using frequency response analysis approaches however i want to understand how this paper works

I've replicated the geometry and the conditions i used a step of 0.1ns and a total runtime of 200ns i've deconvoluted the output using matlab using this code

I = ifft(fft(v)./fft(i) ,'symmetric')
y=I;
Vin = 0.5;
Samplingtime = 0.1e-9
L=length(y);
NFFT = 2^nextpow2(L);
Y = fft(y,NFFT)/L;
Fs = 1/(Samplingtime)
f = Fs/2*linspace(0,1,NFFT/2+1);
%plot(f,2*abs(Y(1:NFFT/2+1))); %Plots Spectrum
Vf=2*abs(Y(1:NFFT/2+1));
Vf= Vf./Vin;
title('Insertion Loss')
xlabel('Frequency (Hz)')
ylabel('Insertion Loss(dB)')
plot(f,20*log10(Vf)); %Plots Insertion Loss

however i still can't get the same results as in the paper any suggestions?

thanks

Zaid Tareq

September 18, 2015 6:13pm UTC in response to Mohamed Ashour

Re: Impulse Response of a SAW Sensor

your input signal function is probably deferent then the one in the paper, i could not download the ppr but i suppose in the ppr the input is an impulse function (+Vi,-Vi), may be you used sine function V_in=A*sin(2*pi*f0*t), you will have different result in this case. i know you probably dont need the answer after three years but this may be helpful for some one els :)

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