I'm reading about a linearly polarized field (in the context of NMR). The field is given by

$$ {\bf H_{lin}}=2H_1({\bf i}\cos(\omega_zt)).$$

This can be created by having a pulse field plus its mirror image; i.e.

$${\bf H_1}=H_1({\bf i}\cos(\omega_zt)+{\bf j}\sin(\omega_zt)).$$

The mirror image field in the lab frame is obviously the same as ${\bf H_1}$ expect $\omega_z \rightarrow -\omega_z.$

Now, my question is what is the expression for the counter-rotating "mirror image" field in the rotating frame? I'm having some trouble working through this.


1 Answer 1


I'm not really sure about your notation, so I'm going to assume that $\mathbf{i}$ = $\vec{x}$ and $\mathbf{j}$ = $\vec{y}$ in the lab frame (I think I've seen this in physics textbooks before). Assuming that's the case, the transform from the lab frame to the rotating frame is: $$\vec{x} = \vec{x}'\cos{\omega_0t} - \vec{y}'\sin{\omega_{0}t}$$ $$\vec{y} = \vec{x}'\sin{\omega_0t} + \vec{y}'\cos{\omega_{0}t}$$

So if you take your equation:

$$\mathbf{H_{1}} = H_1\left(\vec{x}\cos{\omega_{z}t} + \vec{y}\sin{\omega_{z}t}\right),$$

and convert to the rotating frame: $$\mathbf{H_{1}} = H_{1}\left[\cos{\omega_{z}t}\left(\vec{x}'\cos{\omega_0t} - \vec{y}'\sin{\omega_{0}t}\right) + \sin{\omega_{z}t}\left(\vec{x}'\sin{\omega_0t} + \vec{y}'\cos{\omega_{0}t}\right)\right]$$ $$ \begin{gathered} \mathbf{H_{1}} & = & H_{1}\left[\vec{x}'\left(\cos{\omega_0 t}\cdot\cos{\omega_{z}t} + \sin{\omega_{0}t}\sin{\omega_{z}t}\right) + \\ \quad \vec{y}'\left(\cos{\omega_{0}t}\sin\omega_{z}t - \cos{\omega_{z}t}\sin{\omega_{z}t}\right)\right] \end{gathered} $$

$$\mathbf{H_{1}} = H_{1}\left[\vec{x}'\cos{(\omega_{z}-\omega_{0})t} + \vec{y}'\sin{(\omega_{z}-\omega_{0})t}\right]$$

If you define $\Delta\omega = \omega_{z}-\omega_{0}$, then you have: $$\mathbf{H_{1}} = H_{1}\left[\vec{x}'\cos{\Delta{\omega}t} + \vec{y}'\sin{\Delta\omega t}\right]$$

So if $\omega_{z}\rightarrow-\omega_{z}$, then the precession frequency is $-\left(\omega_{z}+\omega_{0}\right)$, which is $\Delta\omega-2\omega_{z}$, so you have:

$$\mathbf{H_{1}} = H_{1}\left[\vec{x}'\cos{(\Delta\omega - 2\omega_{z})t} + \vec{y}'\sin{(\Delta\omega - 2\omega_{z}) t}\right]$$

Which is to be expected, because you're mirroring the frequency about 0, not about $\omega_{0}$, so you pick up one factor of $\omega_{z}$ by going from $\omega_{z} \rightarrow 0$ and another factor of $\omega_{z}$ going from $0 \rightarrow -\omega_{z}$. Again, I'm a bit unclear on your notation, so if you were intending that $\omega_{z}$ is the frequency of precession of the rotating frame (what I'm calling $\omega_{0}$), then $\Delta\omega = 0$ and your signal will precess at $-2\omega_{0}$. You can see why from this diagram (the trace in blue is the on-resonance signal, the trace in green is the $-\omega_{0}$ signal):

Frequency diagram, full acquisition

The rotating frame is on resonance with $\omega_{0}$, and the signal is at $-\omega_{0}$, so in the rotating frame, the apparent frequency will double. One thing to note is that with some (maybe most?) NMR instruments, you're not actually going to be acquiring 2 channels of signal, so you're not actually getting a full quadrature signal. This is equivalent to looking at only the real channel of the signal. If you look at the Fourier transform of a real (or imaginary) channel only signal, you'll see that it's mirrored about 0, with negative and positive frequency components showing up on the spectrum:

Negative and positive frequency with real channel acquisitions only.

So if $\omega_{z} = \omega_{0}$ and you acquire only one signal channel, then mirroring the field will have no effect in the rotating field.

If anyone wants it, here is the code I used to generate the above figures.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.