### #61 Optical coherence tomography-III 光學相干斷層掃描_中二

We have introduced the TDOCT(time domain OCT) in the last episode and discusses about its difficulty. Today we are going to talk about its good friend -- the FDOCT (Fourier domain OCT).

## FDOCT

From previous episodes we are now clear bout the current intensity we will get from our detector:

Expressed in exponential form to facilitate the Fourier transform. The Fourier transform of exp(i~) would be delta function. We assume the Fourier transform of S(k) takes the form:

The Fourier transform and inverse transform demonstrated here is done by multiplying exp(i2kz) & exp(-i2kz) because the optical path length are always doubled. (Actually the above is done by Fourier cosine transform.)

Before we keep going on, we must briefly explain the concept of convolution. If we have 2 functions a(x) & b(x) and their Fourier transforms are A(k) & B(k), let C(k) = A(k)*B(k), what is the inverse Fourier transform of C(k) expressed in a(x) & b(x)? The answer is the convolution of a(x)&b(x). It is usually expressed as a∗b or a⊗b. However, a∗b is a more common notation for the a⊗b usually means tensor product. a∗b is defined as the following:

Therefore we know the Fourier cosine transform of current would be:

expand the convolution and we will get:

$i_D(z)&space;=&space;\frac{\rho}{4}[\gamma(z)(R_R+R_{S_1}+R_{S_2}+...)]&space;\\&space;+\frac{\rho}{4\pi}[\sum_n&space;\sqrt{R_R&space;R_{S_n}}(\gamma(z+(z_R-z_{S_n}))+\gamma(z-(z_R-z_{S_n})))]\\&space;+\frac{\rho}{4\pi}[\sum_{n\neq&space;m}\sqrt{R_{S_n}&space;R_{S_m}}(\gamma(z+(z_{S_n}-z_{S_m}))+\gamma(z-(z_{S_n}-z_{S_m})))]\\$
How does this guy looks like? It looks just like this:

We have transformed our current originally in k-domain (wave number) into z-domain (position). The peaks could be grouped into 3 parts. In the central zone are the DC terms with largest peak height. The second zone contains numerous small peaks with smallest peak height, and they are the autocorrelation terms. The most peripheral ones are peaks of intermediate height, and they are the cross-correlation terms that we want. The distance between them and their height are determined by the distance between each reflection layers and their reflectivities. It is easily known that the conditions for clear resolution of the cross correlation terms would be:
1. The distance between each reflection layers << The distance difference between the reference optical path length and the optical path length of the first reflection layer. This keep the autocorrelation terms far from the cross-correlation terms.
2. The reflectivity between different layers should be smaller.
3. The reflectivity of the reference optical path should be larger.
4. The coherence of light source should be smaller to keep the peaks sharp enough.
(其實只需要畫一半就可以了，因為另一半是mirror image artifact。)我們可以明顯看到peak分三區，一區是最中央的DC terms，peak的高度最高。第二區是很多細細小小的peak，高度最小，這些是autocorrelation terms。第三區是中等高度的peak，這些是我們要的crosscorrelation terms。為何會呈現這樣的分布，當然就和組織中各個反射層之間的距離，還有各個反射層的反射率與參考光徑的反射率之間的大小比例有關係囉。我們可以很明顯看出來，要能最輕易解析出cross correlation term，最好以下條件成立：
1. 組織中各反射層之間的距離<<參考光徑與第一個反射面之間的距離差，這樣autocorrelation term和crosscorrelation terms的peak才不會混在一起。
2. 各組織層反射率越小越好。
3. 參考光徑的反射率越大越好。
4. 光源的同調性越低越好，每個peak才會夠窄。

It is noteworthy that, because the reference optical path has a fixed length, we could measure the interfered light intensity of different wavelength simply by spectrometer and then transform I_D(k) into i_D(z), and we could derive the position of each reflective layers and their reflectivities. However, if we choose the reference optical path length improperly, the peaks of different zones would mix up because of the mirror image artifact, which would definitely perplex our analysis. So the reference optical path length is of great importance.

We are going to have a simple MATLAB program to simulate the efficacy of FDOCT in the next episode. Stay tuned!