## 2 Conventional Edge-Based MTF Measurement Methods
### MTF Measurement Process of Edge Method  
Over-sparse ESF obtained from absolute vertical edge images will cause MTF spectrum aliasing. The problem can be solved by rotating the vertical edge target by \(\theta\) angle relative to the pixel column of the imaging device. It is assumed that the sampling interval of the detector in both horizontal and vertical directions is \(p\). If all the sampling points are projected on the coordinate axis perpendicular to the edge, the sampling interval in the horizontal and vertical directions become \(p\)-\(\cos\)\(\theta\) and \(p\)-\(\sin\)\(\theta\), respectively. For the slanted edge method, the ESF sampling interval is \(\varepsilon\) as follows (The schematic diagram of \(\varepsilon\) is given in document [5]):  
\[\varepsilon=\left\{\begin{aligned} p\sin\theta&(0< \theta<\pi/4)\\ p\cos\theta&(\pi/4<\theta<\pi/2)\end{aligned}\right. \tag{6}\]  
Theoretically, by adjusting the edge angle \(\theta\), an oversampled \(ESF(x)\) with an arbitrarily small sampling interval can be obtained. After one-dimensional comb function sampling, the expression of ESF can be obtained as follows: (The following equation is simplified on the basis of Equation (4), in which the heaviside step function is briefly referred to as \(step(x)\), the noise term \(n(x)\) is omitted in order to ease the reading of equation, and the comb sampling interval \(p_{x}\) in the horizontal direction is replaced by \(\varepsilon\) in Equation (6).)  
\[\begin{aligned} ESF(x)&=[step(x)\otimes h(x)]\cdot comb (x/\varepsilon)\\ &=[step(x)\otimes h(x)]\cdot\left[\varepsilon\sum\limits_{n=- \infty}^{\infty}\delta(x-n\cdot\varepsilon)\right]\end{aligned} \tag{7}\]  
The oversampled \(LSF(x)\) can be obtained by computing finite differences of \(ESF(x)\):  
\[\begin{aligned} LSF(x)&=\frac{d}{dx}ESF(x)=ESF(x) \otimes\left[\frac{\delta(x+\varepsilon)-\delta(x-\varepsilon)}{2\varepsilon }\right]\\ &=[step(x)\otimes h(x)]\cdot\left[\varepsilon\sum\limits_{n=- \infty}^{\infty}\delta(x-n\cdot\varepsilon)\right]\otimes\left[\frac{\delta( x+\varepsilon)-\delta(x-\varepsilon)}{2\varepsilon}\right]\end{aligned} \tag{8}\]  
Applying FFT, modulus, and normalization to Equation (8):  
\[\begin{aligned} MTF_{mean}(f)&=C\cdot|\text{FFT}[ LSF(x)]|\\ &=C\cdot\left|\left(\frac{2\cdot i\cdot\sin(2\pi\varepsilon f)}{2 \varepsilon}\right)\cdot\left[\frac{1}{2}\cdot\left(\delta(f)-\frac{i}{\pi f }\right)\cdot h(f)\cdot\sum\limits_{n=-\infty}^{\infty}\delta(f-\frac{n}{ \varepsilon})\right]\right|\end{aligned} \tag{9}\]  
where \(MTF_{mean}(f)\) is the measured MTF, \(C\) is a normalization factor, and FFT (\(\cdot\)) is an FFT operator. According to the Shannon sampling theorem, the sampling interval needs to satisfy the following condition (\(\varepsilon\): Actual projection sampling interval; \(p\): Detector sampling interval; \(\frac{\varepsilon}{p}\): Sampling rate factor):  
\[\frac{1}{\varepsilon}>2\cdot f_{\varepsilon}\Leftrightarrow\varepsilon<\frac{ 1}{2\cdot f_{\varepsilon}}=\frac{1}{2\cdot(1/2p)}=p\Leftrightarrow\frac{ \varepsilon}{p}<1 \tag{10}\]where \(f_{c}\) is the Nyquist frequency of the imaging system. No spectrum aliasing occurs in \(MTF_{meas}(f)\) when the actual projection and the detector sampling interval satisfy Equation (10). Thus, Equation (9) can be simplified as follows:  
\[MTF_{meas}(f)=C\cdot\left|\frac{\sin(2\pi ef)}{2\pi ef}\right|\cdot\left|h(f) \right|=C\cdot\left|\frac{\sin(2\pi ef)}{2\pi ef}\right|\cdot MTF_{rea}(f) \tag{11}\]  
where \(MTF_{rea}(f)\) is real MTF of the imaging system. Therefore, \(MTF_{rea}(f)\) is:  
\[MTF_{rea}(f)=\frac{MTF_{meas}(f)}{C\cdot\left|\frac{\sin(2\pi ef)}{2\pi ef} \right|} \tag{12}\]  
The impact introduced by the differential operator on real MTF of the imaging system can be corrected by use of Equation (12).