The definition of echo time (TE) in MRI is the time difference between a 90-degree RF signal and the echo peak, which feels like an intrinsic property of the proton.
How can we even control how long does it take for the echo peak to happen?
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The echo peak is not an intrinsic property of the proton. The echo time describes the magnetic environment, which can be manipulated. Specifically, there are two types of echoes in MR.
Gradient echoes form whenever $\int G(t) \ dt =0$ where $G$ is the gradient waveform.
Spin echoes form whenever $t=2\tau$ where $\tau$ is the time between excitation and refocusing.
So we can control both when a gradient echo occurs or when a spin echo occurs.
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$\begingroup$ This is what confuse me, using gradient echo for example, the gradient echo, when the integration is 0 is not something one can easily pick/choose, also it is not really true, as one typically only do half of negative phase for gradient to re-phase. And duration of gradient is computed based on flip angle which is based of the instinct property of proton $\endgroup$– yupbankCommented Apr 5, 2023 at 2:24
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$\begingroup$ It is, in fact, very easy to pick and choose the time when the integral of the gradient is 0. We literally time that down to the microsecond on a routine basis, and we have been doing so for decades. I am not sure why you think that is difficult, but it is not. $\endgroup$– DaleCommented Apr 5, 2023 at 3:32
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$\begingroup$ Ok, the peak of Gradient echoes happens only when the integral of the gradient is 0 at the readout dimension. Both the phase encoding and slice selection gradient is not necessarily 0. And the peak is only theoretical, meaning we need an ADC on all the time to capture the symmetric shape. But in reality it won't be possible, since the RF signal we send and FID we receive would convolute $\endgroup$– yupbankCommented Apr 5, 2023 at 11:59
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$\begingroup$ @yupbank yes, of course it is convolved. That isn’t a problem, it is the desired behavior. Look, comments here are not intended for this sort of back and forth discussion. If you want to have a discussion please go to a discussion oriented site like physicsforums.com where I am also active as Dale $\endgroup$– DaleCommented Apr 5, 2023 at 16:49