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I'm trying to model the effect a radiofrequency pulse will have on a nuclear spin at different detunings. The pulse has a sech lineshape, a pulse area (time integral of the pulse envelope) of $\frac{π}{2}$ and a width of 5MHz centred at 123MHz. This can (I believe) be represented by the equation

$$P(\omega)=\frac{1}{4}\operatorname{sech}\left(\frac{\omega-123}{2}\right)$$

Then I want to calculate the overlap between this pulse and a range of nuclear spins at different detunings from the pulse. The nuclear spins can be represented by a rectangular function in frequency space with a linewidth of 10kHz:

$$R(\omega)= \begin{cases} R_0 & \text{if } \omega_R-\Delta\leq\omega\leq\omega_R+\Delta\\ 0 & \text{otherwise} \end{cases}$$

where $\omega_R$ is the resonant frequency of the pulse and $2\Delta$ is the linewidth of the pulse.

I believe the overlap can be represented by the integral

$$\int_{-\infty}^{\infty}|R(\omega)P(\omega)|^2d\omega$$

but am unsure on a couple of details. I am unclear on whether the value of $R_0$ can be arbitrarily chosen to be $1$, as the only details I have about the nuclei are their frequency and linewidth and I am unclear whether the absolute square I have used is in the correct place in the equation or whether it should be outside the integral (or not included at all).

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  • $\begingroup$ Is $S(\omega)$ same as $P(\omega)$? If not then what is $S(\omega)$ $\endgroup$ – Jitendra Oct 14 '18 at 16:56
  • $\begingroup$ Yes it is, I have edited. $\endgroup$ – JJH Oct 14 '18 at 17:00
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Overlap is generally calculated by convolution. Convolution is a mathematical operation that takes two functions and measures their overlap. To be more specific, it measures the amount of overlap as you slide one function over another. For example, if two functions have zero overlap, the value of their convolution will be equal to zero. If they overlap completely, their convolution will be equal to one. As we slide between these two extreme cases, the convolution will take on values between zero and one.

Overlap will be $$\int_{-\infty}^{\infty}R(\omega)P(\omega)d\omega = \int_{\omega_R-\Delta}^{\omega_R+\Delta}\frac{R_0}{4}\operatorname{sech}\left(\frac{\omega-123}{2}\right)d\omega = \frac{R_0}{2}{tan}^{-1}(sinh(\Delta))$$

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