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I have a doubt in the gauspuls command in matlab, i entered a fbr(fractional bandwidth of 1.5,ie. bandwidth= 0.75Mhz ) but as i plot the signal in the frequency spectrum using fft. i dont see the signal being broad enough. Why is that. the code is shown below: the below code also has general parameters as follows: dt comes out to be 0.0088us.

%% General parameters %%%

dx=0.05; % mm
Vmax=4.0; % mm/us
alpha=0.99;
dt=alpha*dx/(sqrt(2)*Vmax);% µs
f0=0.5; % central frequency, MHz
t0=1.5; % pulse center time
bndwdth=1.5; % pulse -6dB bandwidth
duration=2*t0; % signal length

timebase=(0:round(duration/dt)-1)'*dt;
[signalI,signalQ,signalE]=gauspuls(timebase-t0,f0,bndwdth);
signal=signalQ;
signal=signal/max(signal);

figure(2)
plot(timebase,signal,'.-')
title('source signal')
xlabel('time (µs)')


fs=1e6;
figure(3)
n=length(signal);
f = fs*(0:n)/n; 
Y=fft(signal);
Y=Y';
plot(f,abs(Y)) 
title('Single-Sided Amplitude Spectrum of y(t)')
xlabel('Frequency (Hz)')
ylabel('|Y(f)|')

Shown below are the plots: enter image description here

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  • $\begingroup$ If the center frequency is 0.5 MHz, how can the bandwidth be 0.75 MHz? Then it would have to extend below DC? IMO the fractional bandwidth needs to be less than one for things to make sense. $\endgroup$ – Florian Apr 26 at 8:55
  • $\begingroup$ Even if thats correct, the signal should be broader right? If I reduce the fractional bandwidth to 0.4, i still get a broader signal than when the Fractional bandwidth is 1.5. $\endgroup$ – Kunal Khosla Apr 26 at 10:54
  • $\begingroup$ @Florian, should I just assume that the information is there and go ahead even if the magnitude is less? $\endgroup$ – Kunal Khosla Apr 26 at 11:42
  • $\begingroup$ The way you show the spectrum, it is very hard to see anything. I recommend zooming to the range of interest around fc, then changing the bandwidth parameter and seeing what it does. Btw have a look inside gauspuls.m if you like, it's only a few lines really. $\endgroup$ – Florian Apr 26 at 12:37
  • $\begingroup$ thanks for your feedback @Florian $\endgroup$ – Kunal Khosla Apr 26 at 19:16

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