Featured Post

#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).
在上集我們介紹了TDOCT(time domain OCT),也跟大家講了他的困難點在哪裡。所以今天我們就要來介紹他的好朋友FDOCT(Fourier domain OCT)囉!


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:
欸欸欸等一下我們上次不是已經把它寫成cosine的形式了嗎?為什麼還要轉回來呢??因為我們今天要來做Fourier transform阿,exp(i~)的傅立葉轉換是delta function,所以遠比cosine函數來的方便。所以我們廢話不多說,就要對電流來做傅立葉轉換了。我們假設S(k)的傅立葉轉換為:
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.)
這裡的傅立葉轉換比較特別(或者說是奇怪,因為參考資料裡面的數學彼此不一致,所以小編以下寫的是彼此consistent的結果,但和參考資料的寫法不同。),是乘上exp(i2kz) & exp(-i2kz),因為干涉儀裡的光徑都必須乘上2倍。(而且嚴格來說,上面那個算式,必須是用Fourier cosine transform才會得到那個結果)。

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:
再繼續下去之前,我們必須先簡單說明convolution(摺積)的觀念。事情是這樣子,如果我們有兩個函數a(x) & b(x),他們的傅立葉轉換分別是A(k) & B(k),則如果我們令C(k) = A(k)*B(k),那麼C(k)的反傅立葉轉換c(x)如何以a(x) & b(x)表示呢?答案就是a(x) & b(x)的convolution,數學上以a∗b或a⊗b表示。通常是a∗b比較常用,a⊗b通常是被拿來表示tensor product。a∗b的定義如下,至於為什麼會是這樣子,讀者可以自己試試看證明。

Therefore we know the Fourier cosine transform of current would be:
所以我們就可以知道,對電流做Fourier cosine transform之後我們就得到:
expand the convolution and we will get:

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.
值得注意的是,因為參考路徑的長度是固定的,我們只要用分光儀,能測得不同波長的干涉光強度,再把I_D(k)用傅立葉轉換到i_D(z),就可以求出反射層的位置和他的反射率。不過如果參考路徑的長度沒有選好,那麼由於有mirror image artifact,到時候在DC terms某側的peak就會混合了真實的peak和mirror image,就會讓分析非常麻煩,所以參考路徑的長度也是要選取適當的大小的。

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