http://www.k-wave.org/forum/topic/signal-emission-frequency Support for the k-Wave MATLAB toolbox en-US Tue, 12 May 2026 23:08:54 +0000 http://bbpress.org/?v=1.0.2 q http://www.k-wave.org/forum/search.php http://www.k-wave.org/forum/topic/signal-emission-frequency#post-7962 Fri, 27 Nov 2020 13:41:22 +0000 Bradley Treeby 7962@http://www.k-wave.org/forum/ <p>Hi Florian,</p> <p>Thanks for your patience - it took me a while to find time to take a look.</p> <p>The frequency shift is expected behaviour when you have a mass source as used in k-Wave. If you look in the k-Wave manual, in Eq. (2.9) you can see that the mass source term has a time derivative when it appears in the wave equation. If you consider a point source in free-space, the solution to the wave equation can then be written as the convolution between this source term and the Green's function.</p> <p>In 1D, because the Green's function is a unit step, the convolution will integrate out the time derivative, so the output will look exactly as the input. In 3D, because the Green's function is a delta function, the time derivative will remain after the convolution. In 2D, you'll get something in-between.</p> <p>If you have a time derivative, the frequency content will change (d/dt is equivalent to multiplying by i*omega in the frequency domain).</p> <p>I've written a <a href="http://www.k-wave.org/downloads/forum_freq_shift.m">simple example</a> which shows these three behaviours.</p> <p>Hope that helps,</p> <p>Brad. </p> http://www.k-wave.org/forum/topic/signal-emission-frequency#post-7848 Wed, 23 Sep 2020 16:02:58 +0000 flomomo01 7848@http://www.k-wave.org/forum/ <p>Hi,</p> <p>This is how I compute the FFT.</p> <pre><code>figure; plot(sensor_data(1,:)); [f,Y_fft]=fft_amp(sensor_data(1,:),kgrid.dt); Y=abs(fftshift(fft(sensor_data(1,:)))); fs=1/kgrid.dt; N=length(sensor_data(1,:)); fax_bins=[0: N-1]; N2=ceil(length(sensor_data(1,:))/2); fax_Hz=(fax_bins-N2)*fs/N; plot(fax_Hz,Y);</code></pre> <p>Thank you again for your help </p> http://www.k-wave.org/forum/topic/signal-emission-frequency#post-7846 Tue, 22 Sep 2020 17:46:53 +0000 Bradley Treeby 7846@http://www.k-wave.org/forum/ <p>Thanks - can you edit to also include the part of the code where you calculate the frequency? </p> http://www.k-wave.org/forum/topic/signal-emission-frequency#post-7845 Tue, 22 Sep 2020 16:37:46 +0000 flomomo01 7845@http://www.k-wave.org/forum/ <p>Hi Mr.Treeby,</p> <p>Thank you for your help. You can find below the example I worked with.</p> <pre><code>clearvars; % ========================================================================= % SIMULATION % ========================================================================= % create the computational grid Nx = 128; % number of grid points in the x (row) direction Ny = 128; % number of grid points in the y (column) direction dx = 0.1e-3; % grid point spacing in the x direction [m] dy = 0.1e-3; % grid point spacing in the y direction [m] kgrid = kWaveGrid(Nx, dx, Ny, dy); % define the properties of the propagation medium medium.sound_speed = 1500 * ones(Nx, Ny); % [m/s] medium.sound_speed(1:Nx/2, :) = 1800; % [m/s] medium.density = 1000 * ones(Nx, Ny); % [kg/m^3] medium.density(:, Ny/4:Ny) = 1200; % [kg/m^3] % create initial pressure distribution using makeDisc disc_magnitude = 5; % [Pa] disc_x_pos = 50; % [grid points] disc_y_pos = 50; % [grid points] disc_radius = 8; % [grid points] disc_1 = disc_magnitude * makeDisc(Nx, Ny, disc_x_pos, disc_y_pos, disc_radius); disc_magnitude = 3; % [Pa] disc_x_pos = 80; % [grid points] disc_y_pos = 60; % [grid points] disc_radius = 5; % [grid points] disc_2 = disc_magnitude * makeDisc(Nx, Ny, disc_x_pos, disc_y_pos, disc_radius); % source.p0 = disc_1 + disc_2; % define a centered circular sensor sensor_radius = 4e-3; % [m] num_sensor_points = 50; sensor.mask = makeCartCircle(sensor_radius, num_sensor_points); kgrid.Nt=725; kgrid.dt=1.667e-8; sampling_freq = 1/(1*kgrid.dt); % [Hz] tone_burst_freq = 750e3; % [Hz] tone_burst_cycles = 3; source.p = toneBurst(sampling_freq, tone_burst_freq, tone_burst_cycles); source.p=source.p/max(abs(source.p)); source.p_mask=cart2grid(kgrid,sensor.mask(:,1)); % run the simulation with optional inputs for plotting the simulation % layout in addition to removing the PML from the display sensor_data = kspaceFirstOrder2D(kgrid, medium, source, sensor, ... &#39;PlotLayout&#39;, false, &#39;PlotPML&#39;, false); figure; plot(sensor_data(1,:));</code></pre> http://www.k-wave.org/forum/topic/signal-emission-frequency#post-7843 Tue, 22 Sep 2020 13:50:00 +0000 Bradley Treeby 7843@http://www.k-wave.org/forum/ <p>Hi Florian,</p> <p>Can you post a complete working example? I'll take a look. It may be related to spectral leakage in your FFT if you haven't set your sampling to be an integer number of points-per-period.</p> <p>Brad. </p> http://www.k-wave.org/forum/topic/signal-emission-frequency#post-7838 Mon, 21 Sep 2020 22:20:07 +0000 flomomo01 7838@http://www.k-wave.org/forum/ <p>Hi,</p> <p>I am currently working on ultrasound simulations. I have started to work with k-wave and I have an issue regarding signal emission frequency.</p> <p>In your example you provided named "example_ivp_heterogeneous_medium.m", I tried to emit a 750kHz toneburst signal by adding these following lines before running the simulation and commenting the line when you define the uniform pressure:</p> <pre><code>kgrid.Nt=725; kgrid.dt=1.667e-8; sampling_freq = 1/(1*kgrid.dt); % [Hz] tone_burst_freq = 750e3; % [Hz] tone_burst_cycles = 3; source.p = toneBurst(sampling_freq, tone_burst_freq, tone_burst_cycles); source.p=source.p/max(abs(source.p)); source.p_mask=cart2grid(kgrid,sensor.mask(:,1));</code></pre> <p>But when I tried to check the signal emission frequency by doing a FFT and plot the amplitude of the frequency components, I do not have 750kHz as expected but around 814kHz.</p> <p>Is it a phenomenon I should expect with the simulations ?</p> <p>Kind regards,<br /> Florian M. </p>