## 2 System Model and Problem Analysis
### Joint Effect of Group Delay Distortion and Nonlinear Distortion  
According to Figure 1, the proposed method is expected to focus on baseband signals after the RF front-end and DDC. In order to see how a non-constant group delay in the passband influences the signals with nonlinearity, a simplified grey-box model containing the LNA and band-pass filter (BPF) in series is built in Figure 2. When passing through the receive chain, signals are distorted. Even though all kinds of modules with non-ideal linear characteristics in the RF front-end could bring in nonlinearity and non-constant group delay, we should pay attention to the main influence and make the following assumptions that the main source where nonlinearities are generated is at the LNA stage, and non-constant group delay distortion in the passband is due to the non-ideal phase-frequency response of the BPF. Let \(x(n)\) be the chosen signals with a fixed center frequency and bandwidth. After the LNA, \(y(n)\) are signals containing nonlinearity. \(z(n)\) means the signals passing through the BPF are distorted both by nonlinear products and non-constant group delay distortion. If no BPF exists, our previous  
Figure 1: Broadband reconfigurable direct RF front-end. Figure 2: Simplified model. method [27] presents a blind identification criterion of forming a nonlinear behavior model to compensate for nonlinear interference:  
\[s_{y}(n)=y(n)-f\{y(n)\} \tag{1}\]  
where \(s_{y}(n)\) are the compensated signals and \(f\{\cdot\}\) indicates the nonlinear behavior model, such as the Volterra model and Hammerstein model. Generally speaking, a constant group delay in the passband of RF filter circuits is desirable. However, it is difficult to achieve in broadband systems, or even in narrowband communications, when tight filtering is required [15]. Since the hypothesis is that the BPF only brings the non-constant group delay to \(y(n)\), it can be regarded as a linear filter adding nonlinear phase increments to each frequency component in the passband. From the perspective of the frequency domain, to each frequency spectrum:  
\[Z(k)=Y(k)\cdot H(k)=Y(k)\cdot A(k)e^{i\varphi(k)} \tag{2}\]  
where \(Y(k)\) and \(Z(k)\) are the frequency spectra of \(y(n)\) and \(z(n)\), respectively. Their frequency value is represented as \(k\) (\(k=0,1,2,...,N-1\)). \(N\) is the number of points carrying out the discrete Fourier transform (DFT). \(H(k)\) represents the frequency response function of the BPF whose amplitude is \(A(k)\) and phase is \(\varphi(k)\). Then, the Inverse Discrete Fourier Transform (IDFT) of \(Z(k)\) is expressed as:  
\[z(r)=\frac{1}{N}\sum_{k=0}^{N-1}Z(k)e^{i\frac{2\pi}{N}kr} \tag{3}\]  
and the DFT of \(y(n)\) is:  
\[Y(k)=\sum_{n=0}^{N-1}y(n)e^{-j\frac{2\pi}{N}kn} \tag{4}\]  
Substituting Equations (2) and (4) into Equation (3):  
\[ \begin{array}{ll}z(r)&=\frac{1}{N}\sum\limits_{k=0}^{N-1}Y(k)A(k)e^{i\varphi( k)}e^{j\frac{2\pi}{N}kr}\\ &=\frac{1}{N}\sum\limits_{k=0}^{N-1}A(k)e^{i\varphi}(k)\sum\limits_{n=0}^{N-1} y(n)e^{-j\frac{2\pi}{N}kn}e^{j\frac{2\pi}{N}kr}\\ &=\frac{1}{N}\sum\limits_{n=0}^{N-1}y(n)\sum\limits_{k=0}^{N-1}A(k)e^{i[\frac {2\pi}{N}k(r-n)+\varphi(k)]}\end{array}  \tag{5}\]  
In Reference [34], the phase characteristic of the BPF is modeled by a \(P\)th order polynomial function and, without loss of generality, we obtain:  
\[\varphi(k)=\frac{2\pi}{N}\sum\limits_{p=1}^{P}c_{p}k^{p} \tag{6}\]  
Next, we employ FIR to simulate which has the same amplitude response and phase response as the BPF in Figure 2. Since the phase response of the BPF is nonlinear in the passband, there exists no linear relationship between \(\varphi(k)\) and \(k\). Meanwhile, considering that the time domain response coefficients of FIR belong to the real number field, to meet to the symmetry condition, we suppose that \(A(k)\) has even symmetry and \(\varphi(k)\) has odd symmetry relative to \(k=\frac{N}{2}\) (\(N\) is an even number), i.e., \(A(k)=A(N-k)\), \(\varphi(k)=-\varphi(N-k)\), and \(A(\frac{N}{2})=0\), \(\varphi(\frac{N}{2})=0\).