sigpy.mri.samp.spiral

[ZhengguoTan]

Dear sigpy developers,

I may have a question regarding the implementation of the spiral function in sigpy.mri.samp. I try a toy example:

from sigpy.mri import samp
import numpy as np

fov = 200E-3
N = 160
f_sampling = 0.1
R = 6
ninterleaves = 1
alpha = 4
gm = 0.054  # T/m
sm = 150    # T/m/s

gamma = 42.577478E6  # Hz/T

k = samp.spiral(fov, N, f_sampling, R, ninterleaves, alpha, gm, sm)

dt = 4.12E-7  # s

grad = (k[1:, :] - k[:-1, :]) / dt / gamma

print(k.shape, grad.shape)

assert np.all(abs(grad) < gm)

Here, dt is the time increment that I got from https://github.com/mikgroup/sigpy/blob/main/sigpy/mri/samp.py#L312. With dt, I then convert the k to gradients grad. In principle, the gradients should not exceed the max gradient amplitude gm. However, the assert does not pass.

Do I miss anything in the spiral trajectory computation function? Any suggestion would be appreciated.

Best Regards,
Zhengguo

[wgrissom]

Don’t you need to use 1 / f_sampling as your dt, when calculating the gradient waveform?

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On Sun, Apr 6, 2025 at 1:21 PM Zhengguo Tan ***@***.***> wrote: Dear sigpy developers, I may have a question regarding the implementation of the spiral function in sigpy.mri.samp. I try a toy example: `from sigpy.mri import samp import numpy as np fov = 200E-3 N = 160 f_sampling = 0.1 R = 6 ninterleaves = 1 alpha = 4 gm = 0.054 # T/m sm = 150 # T/m/s gamma = 42.577478E6 # Hz/T k = samp.spiral(fov, N, f_sampling, R, ninterleaves, alpha, gm, sm) dt = 4.12E-7 # s grad = (k[1:, :] - k[:-1, :]) / dt / gamma print(k.shape, grad.shape) assert np.all(abs(grad) < gm)` Here, dt is the time increment that I got from https://github.com/mikgroup/sigpy/blob/main/sigpy/mri/samp.py#L312. With dt, I then convert the k to gradients grad. In principle, the gradients should not exceed the max gradient amplitude gm. However, the assert does not pass. Do I miss anything in the spiral trajectory computation function? Any suggestion would be appreciated. Best Regards, Zhengguo — Reply to this email directly, view it on GitHub <#146>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AAICTWRLCJNGLCXOQYNQM3T2YFPA3AVCNFSM6AAAAAB2R5JWEKVHI2DSMVQWIX3LMV43ASLTON2WKOZSHE3TKMBYGQYDQOA> . You are receiving this because you are subscribed to this thread.Message ID: ***@***.***> [image: ZhengguoTan]*ZhengguoTan* created an issue (mikgroup/sigpy#146) <#146> Dear sigpy developers, I may have a question regarding the implementation of the spiral function in sigpy.mri.samp. I try a toy example: `from sigpy.mri import samp import numpy as np fov = 200E-3 N = 160 f_sampling = 0.1 R = 6 ninterleaves = 1 alpha = 4 gm = 0.054 # T/m sm = 150 # T/m/s gamma = 42.577478E6 # Hz/T k = samp.spiral(fov, N, f_sampling, R, ninterleaves, alpha, gm, sm) dt = 4.12E-7 # s grad = (k[1:, :] - k[:-1, :]) / dt / gamma print(k.shape, grad.shape) assert np.all(abs(grad) < gm)` Here, dt is the time increment that I got from https://github.com/mikgroup/sigpy/blob/main/sigpy/mri/samp.py#L312. With dt, I then convert the k to gradients grad. In principle, the gradients should not exceed the max gradient amplitude gm. However, the assert does not pass. Do I miss anything in the spiral trajectory computation function? Any suggestion would be appreciated. Best Regards, Zhengguo — Reply to this email directly, view it on GitHub <#146>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AAICTWRLCJNGLCXOQYNQM3T2YFPA3AVCNFSM6AAAAAB2R5JWEKVHI2DSMVQWIX3LMV43ASLTON2WKOZSHE3TKMBYGQYDQOA> . You are receiving this because you are subscribed to this thread.Message ID: ***@***.***>

[ZhengguoTan]

Thanks @wgrissom for the quick reply!

Perhaps I misunderstand the logic here.

In the code:
https://github.com/mikgroup/sigpy/blob/main/sigpy/mri/samp.py#L312

You the have time array t, which, to my understanding, is the sampling time for the spiral trajectory. Therefore, the "raster time" is the time interval in the array t.

I don't understand why and how we should use 1 / f_sampling to convert the trajectory to gradients.

Look forward to hearing from you.

Best Regards,
Zhengguo

[wgrissom]

Oh sorry, I misunderstood the f_sampling definition to be sampling frequency of the waveforms themselves. Yes, you do want to use the raster time of the time vector. Is that equal to the ~4us dt you are using? Best, Will

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On Apr 6, 2025, at 1:45 PM, Zhengguo Tan ***@***.***> wrote: ZhengguoTan left a comment (mikgroup/sigpy#146) Thanks @wgrissom for the quick reply! Perhaps I misunderstand the logic here. In the code: https://github.com/mikgroup/sigpy/blob/main/sigpy/mri/samp.py#L312 You the have time array t, which, to my understanding, is the sampling time for the spiral trajectory. Therefore, the "raster time" is the time interval in the array t. I don't understand why and how we should use 1 / f_sampling to convert the trajectory to gradients. Look forward to hearing from you. Best Regards, Zhengguo — Reply to this email directly, view it on GitHub, or unsubscribe. You are receiving this because you were mentioned. <https://github.com/wgrissom> <https://github.com/mikgroup/sigpy/blob/main/sigpy/mri/samp.py#L312> <#146 (comment)> <https://github.com/notifications/unsubscribe-auth/AAICTWU5NORNQK2CX52SFED2YFR4BAVCNFSM6AAAAAB2R5JWEKVHI2DSMVQWIX3LMV43OSLTON2WKQ3PNVWWK3TUHMZDOOBRGUZDSNJZHE> ZhengguoTan left a comment (mikgroup/sigpy#146) <#146 (comment)> Thanks @wgrissom <https://github.com/wgrissom> for the quick reply! Perhaps I misunderstand the logic here. In the code: https://github.com/mikgroup/sigpy/blob/main/sigpy/mri/samp.py#L312 You the have time array t, which, to my understanding, is the sampling time for the spiral trajectory. Therefore, the "raster time" is the time interval in the array t. I don't understand why and how we should use 1 / f_sampling to convert the trajectory to gradients. Look forward to hearing from you. Best Regards, Zhengguo — Reply to this email directly, view it on GitHub <#146 (comment)>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AAICTWU5NORNQK2CX52SFED2YFR4BAVCNFSM6AAAAAB2R5JWEKVHI2DSMVQWIX3LMV43OSLTON2WKQ3PNVWWK3TUHMZDOOBRGUZDSNJZHE>. You are receiving this because you were mentioned.

[ZhengguoTan]

Hi Will,

Thanks! I try to output the time array t in sigpy.mri.samp.spiral:

from sigpy.mri import samp
import numpy as np

fov = 200E-3
N = 160
f_sampling = 0.5
R = 6
ninterleaves = 1
alpha = 4
gm = 0.054  # T/m
sm = 150    # T/m/s

gamma = 42.577478E6  # Hz/T

k, t = samp.spiral(fov, N, f_sampling, R, ninterleaves, alpha, gm, sm)

dt = t[1] - t[0]
# dt = 4.12E-7  # s

grad = (k[1:, :] - k[:-1, :]) / dt / gamma

print(dt, k.shape, grad.shape)

print(np.max(abs(grad)))
assert np.all(abs(grad) < gm)

The raster time is about 0.412 us. In the end, the converted gradient grad exceeds the max gradient amplitude gm.

I am not sure whether the conversion from k to grad is problematic, or how I should use the spiral function...

[wgrissom]

It does look like kt is in units of radians/second/meter, but your gamma is in Hz/T, could that be it?

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On Apr 6, 2025, at 2:35 PM, Zhengguo Tan ***@***.***> wrote: ZhengguoTan left a comment (mikgroup/sigpy#146) Hi Will, Thanks! I try to output the time array t in sigpy.mri.samp.spiral: from sigpy.mri import samp import numpy as np fov = 200E-3 N = 160 f_sampling = 0.5 R = 6 ninterleaves = 1 alpha = 4 gm = 0.054 # T/m sm = 150 # T/m/s gamma = 42.577478E6 # Hz/T k, t = samp.spiral(fov, N, f_sampling, R, ninterleaves, alpha, gm, sm) dt = t[1] - t[0] # dt = 4.12E-7 # s grad = (k[1:, :] - k[:-1, :]) / dt / gamma print(dt, k.shape, grad.shape) print(np.max(abs(grad))) assert np.all(abs(grad) < gm) The raster time is about 0.412 us. In the end, the converted gradient grad exceeds the max gradient amplitude gm. I am not sure whether the conversion from k to grad is problematic, or how I should use the spiral function... — Reply to this email directly, view it on GitHub, or unsubscribe. You are receiving this because you were mentioned. <#146 (comment)> <https://github.com/notifications/unsubscribe-auth/AAICTWTLJRIXYSZ6563ROKT2YFXXPAVCNFSM6AAAAAB2R5JWEKVHI2DSMVQWIX3LMV43OSLTON2WKQ3PNVWWK3TUHMZDOOBRGU2TCMZUGY> ZhengguoTan left a comment (mikgroup/sigpy#146) <#146 (comment)> Hi Will, Thanks! I try to output the time array t in sigpy.mri.samp.spiral: from sigpy.mri import samp import numpy as np fov = 200E-3 N = 160 f_sampling = 0.5 R = 6 ninterleaves = 1 alpha = 4 gm = 0.054 # T/m sm = 150 # T/m/s gamma = 42.577478E6 # Hz/T k, t = samp.spiral(fov, N, f_sampling, R, ninterleaves, alpha, gm, sm) dt = t[1] - t[0] # dt = 4.12E-7 # s grad = (k[1:, :] - k[:-1, :]) / dt / gamma print(dt, k.shape, grad.shape) print(np.max(abs(grad))) assert np.all(abs(grad) < gm) The raster time is about 0.412 us. In the end, the converted gradient grad exceeds the max gradient amplitude gm. I am not sure whether the conversion from k to grad is problematic, or how I should use the spiral function... — Reply to this email directly, view it on GitHub <#146 (comment)>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AAICTWTLJRIXYSZ6563ROKT2YFXXPAVCNFSM6AAAAAB2R5JWEKVHI2DSMVQWIX3LMV43OSLTON2WKQ3PNVWWK3TUHMZDOOBRGU2TCMZUGY>. You are receiving this because you were mentioned.

[ZhengguoTan]

I am curious how you see kt contains radians? If so, we will need to divide the grad by 2pi. Is that right?

[wgrissom]

Because it says that in the comment on line 313. Can’t say I’ve checked the math though ;). Yes, that means you would also need to divide k by 2pi, to get g.

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On Sun, Apr 6, 2025 at 6:08 PM Zhengguo Tan ***@***.***> wrote: I am curious how you see kt contains radians? If so, we will need to divide the grad by 2pi. Is that right? — Reply to this email directly, view it on GitHub <#146 (comment)>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AAICTWTSXBXTXN4CGRIMD632YGQUBAVCNFSM6AAAAAB2R5JWEKVHI2DSMVQWIX3LMV43OSLTON2WKQ3PNVWWK3TUHMZDOOBRGY4TKNJZHA> . You are receiving this because you were mentioned.Message ID: ***@***.***> [image: ZhengguoTan]*ZhengguoTan* left a comment (mikgroup/sigpy#146) <#146 (comment)> I am curious how you see kt contains radians? If so, we will need to divide the grad by 2pi. Is that right? — Reply to this email directly, view it on GitHub <#146 (comment)>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AAICTWTSXBXTXN4CGRIMD632YGQUBAVCNFSM6AAAAAB2R5JWEKVHI2DSMVQWIX3LMV43OSLTON2WKQ3PNVWWK3TUHMZDOOBRGY4TKNJZHA> . You are receiving this because you were mentioned.Message ID: ***@***.***>

[ZhengguoTan]

Dear Will,

I tested the function sigpy.mri.samp.spiral on the scanner - the spiral trajectory runs! However, the readout duration is internally computed in the function as Tend. In the user point of view, I may want to have a user-definable Tend. Some user may want to have a long spiral, whereas others may want to have a short spiral. Would there be any chance to modify this function? Any instruction is welcome.

Best Regards,
Zhengguo