How to generate 6 cycle or 10 cycle hanning modulated tone burst excitation

I have been trying to generate a hanning window modulated tone burst excitation with a centre frequency of 100 KHz and with 6 or 10 cycles as shown in the picture. Once generated I want to use this as an input excitation to my transducer.

Please let me know how can i acheive this?

Answer 1

[Mathieu NOE]

hello

try this

all the best

fsin = 100e3; % sine freq
Fs = 100*fsin; % sampling freq
dt = 1/Fs;

duration = 0.2e-3; % signal duration in seconds
samples = ceil(duration*Fs);;
t = dt*(0:samples-1);
offset = dt*(samples/2); % to center the gaussian envelope

a = 200*Fs;  % increase or decrease a to change number of periods 
signal = sin(2*pi*fsin*t).*exp(-a*(t-offset).^2);

figure(1),plot(t,signal);grid

[Yaser Arafath Gulam Dhasthagir]

Hi. Thanks for your answer. But what the rationale for the equation in signal and why are you multiplying a with 200*Fs? because i want to change the cycles between 3-10 cycles per signal. The below is the equation used in the paper for the signal i attached above.

Also how can i change the amplitudes of the signal? Currently it goes from 1 to -1.

[Mathieu NOE]

hello again

I you prefer that formula , why not...

here you are

fsin = 100e3; % sine freq
Fs = 100*fsin; % sampling freq (at least 16 x sine freq if you want a clean wave)
dt = 1/Fs;
cycles = 10; % 3 to 10 cycles 
duration = cycles/fsin; % signal duration in seconds
samples = ceil(duration*Fs);
t = dt*(0:samples-1);
offset = dt*(samples/2); % to center the gaussian envelope
angle1 = 2*pi*fsin*t;  % sine wave (with appropriate number of cycles)
angle2 = linspace(0,pi,numel(angle1)); % for envelope generation (bell curve) has always only a half period , whatever the number of cycles
signal = sin(angle1).*(sin(angle2)).^2;

figure(1),plot(t,signal);grid

[Yaser Arafath Gulam Dhasthagir]

Hi,

Thanks for the code. I will try it.

[Mathieu NOE]

hello again

problem solved ?

[Yaser Arafath Gulam Dhasthagir]

Hi mathieu, thanks for following up. I mad few changes to the code. But when i upload the wave to my picoscope 5443B, the frequency is changing (see attached pictures). I want this 10 cycle pulse to repeat once in every 1 second (1 Hz). But it repeats at 10KHz with the correct carrier frequency of 100 KHz. If I change the repeat frequency then the carrier frequency is also changing. If you have any idea on this please feel free to share. I also have a function generator BK precision 4053.

[Mathieu NOE]

hello Yaser

there are a few things that surprises me

the matlab code generates a signal at 100 kHz with a total duration of only 0.1 ms

Fs = 81.92 MHz (so dt = 12.2070 ns)

how can then your scope show a period of 25.34 ms (so f = 39.4633 hz instead of 100 kHz)

there are completely different orders of magnitude in terms of signal frequency and duration....

Also the sampling rates are different : from matlab we have dt = 12.2070 ns , from the picture I can read dt = 176 ns

I personnaly don't have the Picoscope (I use other hardwares) but I wonder if there is a conflict either what the sampling rate we define in the matlab code vs what the picoscope sampling rate actually is

how do you format the waveform to the picoscope ? is it a table with time and signal ? is there a specific format to respect ?

in the matlab code , where does the factor 819.2 comes from ? (Fs = 819.2*fsin;)

fsin = 100e3; % sine freq
Fs = 819.2*fsin; % sampling freq (at least 16 x sine freq if you want a clean wave)
dt = 1/Fs;
cycles = 10; % 3 to 10 cycles 
duration = cycles/fsin; % signal duration in seconds
samples = ceil(duration*Fs);
t = dt*(0:samples-1);
angle1 = 2*pi*fsin*t;  % sine wave (with appropriate number of cycles)
signal = 0.5*sin(angle1).*(1-cos(angle1/cycles));
figure(1),plot(t,signal);grid

[Yaser Arafath Gulam Dhasthagir]

In picoscope, we have the option to generate arbitrary waveforms and it will be generated based on number of points. 8192 is one the standard number of points in the device. So to achieve that number I used it in MATLAB to get 8192 sample points. Then I saved the sample points in a CSV and uploaded to the awg generator in picoscope. Hence 8192 points fill the awg window. Whatever is displayed in awg window will be used to generate signal. Hence I set 1 Hz as my repeating frequency in the awg generator. I think it has something to do with sampling frequency limitations of picoscope. But I'm just confused on how to achieve 1 excitation per second. Weird fact this excitation has been used in many research paper still I'm unable to find a good source on how to generate it. The same csv I tried to import in BK 4053 but it's not accepting the file.

[Mathieu NOE]

ok, still I don't get why the displayed sample rate of 176 ns on the scope

[Yaser Arafath Gulam Dhasthagir]

ok. Then can we do this with the any of the standard sample points. Available points in the picoscope are 1024,2048,4096,8192

I can try creating 999.9 ms of zeros and then 0.1 ms of the waveform (10 cycle 100 KHz sine).

Possible to do this in matlab?

[Mathieu NOE]

if you want to create 1 s of data and you are limited to 8192 points, then your effective max sampling rate is 8192 Hz. How do you want to create a 100 kHz signal in this case ?

[Mathieu NOE]

hello

Problem solved ?