where *k* = 0, 1, ··· , *M* − 1.Therefore, we can obtain

$$\begin{aligned} \mathbf{V}\_{k} &= V \begin{pmatrix} u^{0,k} \times D^{0} + u^{1,k} \times \mathbf{1}^{k,0} + \cdots + u^{(M-1),k} \times D^{(M-1),0} \end{pmatrix} \mathbf{V}^{0} \\ &- V \begin{pmatrix} u^{0,k} \lambda\_{1}^{0} + u^{1,k} \lambda\_{1}^{1,0} + \cdots + u^{(M-1),k} \lambda\_{1}^{1,0} + u^{(M-1),k} \end{pmatrix} \mathbf{V}^{0} \\ &- V \begin{pmatrix} 0 & u^{0,k} \lambda\_{2}^{0} + u^{1,k} \lambda\_{2}^{1,0} + \cdots + u^{(M-1),k} \lambda\_{2}^{1,0} + \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & u^{(M-1),k} \end{pmatrix} \mathbf{V}^{0} \\ &- V \begin{pmatrix} Q\_{1}(k) & 0 & \cdots & 0 \\ 0 & Q\_{2}(k) & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & Q\_{n}(k) \end{pmatrix} \mathbf{V}^{0} \end{aligned} \tag{36}$$

Here, let

$$\begin{cases} Q\_1(k) = u^{0 \times k} \lambda\_1^0 + u^{1 \times k} \lambda\_1^{4/M} + \dots + u^{(M-1) \times k} \lambda\_1^{4(M-1)/M} \\ Q\_2(k) = u^{0 \times k} \lambda\_2^0 + u^{1 \times k} \lambda\_2^{4/M} + \dots + u^{(M-1) \times k} \lambda\_2^{4(M-1)/M} \\ \vdots \\ Q\_n(k) = u^{0 \times k} \lambda\_n^0 + u^{1 \times k} \lambda\_n^{4/M} + \dots + u^{(M-1) \times k} \lambda\_n^{4(M-1)/M} \end{cases} \tag{37}$$

The multiplicities of the DFT eigenvalues [26,27] are shown in Table 1. Therefore,

where is 
$$\begin{aligned} \lambda\_r &= \{1, i, -1, -i\} \\ &= \left\{ e^{4n\pi i/2}, e^{(4n+1)\pi i/2}, e^{(4n+2)\pi i/2}, e^{(4n+3)\pi i/2} \right\} \\ &= \left\{ e^{2n\pi i} e^{0\pi i/2}, e^{2n\pi i} e^{\pi i/2}, e^{2n\pi i} e^{2\pi i/2}, e^{2n\pi i} e^{3\pi i/2} \right\} \\ &= \left\{ e^{0\pi i/2}, e^{\pi i i/2}, e^{2\pi i i/2}, e^{3\pi i i/2} \right\}. \end{aligned} \tag{38}$$

**Table 1.** Multiplicities of the DFT eigenvalues.


For Equation (37), *Qr*(*k*),*r* = 1, 2, ··· , *n* can be expressed as

$$Q\_r(k) = \mu^{0 \times k} \lambda\_r^0 + \mu^{1 \times k} \lambda\_r^{4/M} + \dots + \mu^{(M-1) \times k} \lambda\_r^{4(M-1)/M}.\tag{39}$$

When the eigenvalues *<sup>λ</sup><sup>r</sup>* <sup>=</sup> *<sup>e</sup>*0*πi*/2 <sup>=</sup> 1 and *<sup>u</sup>* <sup>=</sup> *<sup>e</sup>*−2*πi*/*M*, *<sup>Q</sup>*(1) *<sup>r</sup>* (*k*) can be expressed using Equation (39), as

$$\begin{split} Q\_r^{(1)}(k) &= u^{0 \times k} \lambda\_r^0 + u^{1 \times k} \lambda\_r^{4/M} + \dots + u^{(M-1) \times k} \lambda\_r^{4(M-1)/M} \\ &= 1 + e^{-2\pi i \mathbf{1}(k-0)/M} + e^{-2\pi i 2(k-0)/M} + \dots + e^{-2\pi i (M-1)(k-0)/M} \\ &= \frac{1 - \left(e^{-2\pi i (k-0)/M}\right)^M}{1 - e^{-2\pi i (k-0)/M}}. \end{split} \tag{40}$$

Therefore, we obtain

$$Q\_r^{(1)}(k) = \begin{cases} 0, & k \equiv 0 \bmod M \\ M, & k \not\equiv 0 \bmod M. \end{cases} \tag{41}$$

When the eigenvalue *<sup>λ</sup><sup>r</sup>* <sup>=</sup> *<sup>e</sup>πi*/2 <sup>=</sup> *<sup>i</sup>*, *<sup>Q</sup>*(*i*) *<sup>r</sup>* (*k*) can be expressed using Equation (39), as

$$\begin{split} Q\_r^{(\vec{1})}(k) &= u^{0 \times k} \lambda\_r^0 + u^{1 \times k} \lambda\_r^{4/M} + \cdots + u^{(M-1) \times k} \lambda\_r^{4(M-1)/M} \\ &= 1 + e^{-2\pi i \mathbf{1} (k-1)/M} + e^{-2\pi i \mathbf{2} (k-1)/M} + \cdots + e^{-2\pi i (M-1)(k-1)/M} \\ &= \frac{1 - \left( e^{-2\pi i (k-1)/M} \right)^M}{1 - e^{-2\pi i (k-1)/M}}. \end{split} \tag{42}$$

Therefore, there is

$$Q\_r^{(i)}(k) = \begin{cases} 0, & k \equiv 1 \bmod M \\ M, & k \not\equiv 1 \bmod M. \end{cases} \tag{43}$$

When the eigenvalue *<sup>λ</sup><sup>r</sup>* <sup>=</sup> *<sup>e</sup>*2*πi*/2 <sup>=</sup> <sup>−</sup>1, *<sup>Q</sup>*(−1) *<sup>r</sup>* (*k*) can be expressed using Equation (39), as

$$\begin{split} Q\_r^{(-1)}(k) &= u^{0 \times k} \lambda\_r^0 + u^{1 \times k} \lambda\_r^{4/M} + \dots + u^{(M-1) \times k} \lambda\_r^{4(M-1)/M} \\ &= 1 + e^{-2\pi i 1 (k-2)/M} + e^{-2\pi i 2 (k-2)/M} + \dots + e^{-2\pi i (M-1)(k-2)/M} \\ &= \frac{1 - \left(e^{-2\pi i (k-2)/M}\right)^M}{1 - e^{-2\pi i (k-2)/M}}. \end{split} \tag{44}$$

Then, we can obtain

$$Q\_r^{(-1)}(k) = \begin{cases} 0, & k \equiv 2 \bmod M \\ M, & k \not\equiv 2 \bmod M. \end{cases} \tag{45}$$

When the eigenvalue *<sup>λ</sup><sup>r</sup>* <sup>=</sup> *<sup>e</sup>*3*πi*/2 <sup>=</sup> <sup>−</sup>*i*, *<sup>Q</sup>*(−*i*) *<sup>r</sup>* (*k*) can be expressed using Equation (39), as

$$\begin{split} Q\_r^{(-i)}(k) &= u^{0 \times k} \lambda\_r^0 + u^{1 \times k} \lambda\_r^{4/M} + \dots + u^{(M-1) \times k} \lambda\_r^{4(M-1)/M} \\ &= 1 + e^{-2\pi i 1 (k-3)/M} + e^{-2\pi i 2 (k-3)/M} + \dots + e^{-2\pi i (M-1)(k-3)/M} \\ &= \frac{1 - \left( e^{-2\pi i (k-3)/M} \right)^M}{1 - e^{-2\pi i (k-3)/M}}. \end{split} \tag{46}$$

Therefore, there is

$$Q\_r^{(-i)}(k) = \begin{cases} 0, & k \equiv 3 \bmod M \\ M, & k \not\equiv 3 \bmod M. \end{cases} \tag{47}$$

Using Equations (41), (43), (45) and (47), we can formulate Equation (36) as

$$\mathcal{Y}\_k = \begin{cases} \mathcal{Y}\_{k'} & k = 0, 1, 2, 3 \\ 0, & k = 4, 5, \cdots, M - 1. \end{cases} \tag{48}$$

In this way, the M-WFRFT of Equation (11) can also be expressed as Equation (49).

$$\begin{aligned} \left(T\_{\rm MV}^{a}\right)\_{\rm MV} &= \frac{1}{M} (Y\_0, Y\_1, \dots, Y\_{M-1}) \begin{pmatrix} B\_0^a \\ B\_1^a \\ \vdots \\ B\_{M-1}^a \end{pmatrix} \\ &= \frac{1}{M} (Y\_0, Y\_1, Y\_2, Y\_3, 0, \dots, 0) \begin{pmatrix} B\_0^a \\ B\_1^a \\ \vdots \\ B\_M^a \end{pmatrix} \\ &= \frac{1}{M} (Y\_0, Y\_1, Y\_2, Y\_3) \begin{pmatrix} B\_0^a \\ B\_1^a \\ \vdots \\ B\_2^a \\ B\_3^a \end{pmatrix} .\end{aligned} \tag{49}$$

where *B<sup>α</sup> <sup>k</sup>* <sup>=</sup> exp2*πik<sup>α</sup> <sup>M</sup>* ; *k* = 0, 1, ··· , *M* − 1.

The effective weighted sum of the M-WFRFT based on the fractional-order matrix is also four terms. In order to prove its unitarity, we denote

$$(T\_{MW}^{a})^{H} = \frac{1}{M} \left( \mathbf{Y}\_{0}^{H} \mathbf{B}\_{0}^{-a} + \mathbf{Y}\_{1}^{H} \mathbf{B}\_{1}^{-a} + \mathbf{Y}\_{2}^{H} \mathbf{B}\_{2}^{-a} + \mathbf{Y}\_{3}^{H} \mathbf{B}\_{3}^{-a} \right), \tag{50}$$

Therefore, there is

$$\left(T\_{\rm MW}^{a} \left(T\_{\rm MW}^{a}\right)\right)^{H} = \frac{1}{M^{2}} \left(\sum\_{k=0}^{3} \sum\_{l=0}^{3} \mathbb{Y}\_{k} \mathbb{Y}\_{l}^{H} B\_{k}^{a} B\_{l}^{-a}\right). \tag{51}$$

From Equation (36), we can obtain

$${}^{H}Y\_{k}Y\_{l}^{H} = V \left( \begin{array}{cccc} Q\_{1}(k) & 0 & \cdots & 0 \\ 0 & Q\_{2}(k) & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & Q\_{\pi}(k) \end{array} \right) V^{H} \left[ V \left( \begin{array}{cccc} Q\_{1}(l) & 0 & \cdots & 0 \\ 0 & Q\_{2}(l) & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & Q\_{\pi}(l) \end{array} \right) V^{H} \right] . \tag{52}$$

The eigenvector *V* of the DFT can be defined as a real symmetric matrix [27–29]; and through Equations (41), (43), (45) and (47), we know that the value of *Qr*(*k*) is 0 or *M* (*M* is an integer greater than 4). Therefore, *Y<sup>H</sup> <sup>l</sup>* = *Yl*. Then, Equation (52) can be expressed as

$$\mathbf{Y}\_{k}\mathbf{Y}\_{l}^{H} = \mathbf{Y}\_{k}\mathbf{Y}\_{l} = V \begin{pmatrix} Q\_{1}(k)Q\_{1}(l) & 0 & \cdots & 0\\ 0 & Q\_{2}(k)Q\_{2}(l) & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & Q\_{n}(k)Q\_{n}(l) \end{pmatrix} V^{H} \,,\tag{53}$$

and

$$Q\_r(k)Q\_r(l) = \begin{cases} M^2, & k=l\\ 0, & k \neq l. \end{cases} \tag{54}$$

Therefore, we can obtain

$$\mathcal{Y}\_{\mathbf{k}}\mathcal{Y}\_{\mathbf{l}}^{H} = \begin{cases} M\mathcal{Y}\_{\mathbf{k}\prime} & k=l\\ 0, & k \neq l. \end{cases} \tag{55}$$

Then, the result of Equation (51) is

$$\begin{aligned} \left(T\_{MW}^{\mathbf{u}} \left(T\_{MW}^{\mathbf{x}}\right)^{H}\right)^{H} &= \frac{1}{M^{2}} \left(\sum\_{k=0}^{3} \sum\_{l=0}^{3} Y\_{k} Y\_{l}^{H} B\_{k}^{\mathbf{x}} B\_{l}^{-\mathbf{a}}\right) \\ &= \frac{1}{M} (Y\_{0} + Y\_{1} + Y\_{2} + Y\_{3}) \\ &= \frac{1}{M} \left( V \begin{pmatrix} M & 0 & \cdots & 0 \\ 0 & M & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & M \end{pmatrix} V^{H} \right) \\ &= I. \end{aligned} \tag{56}$$


**Remark 3.** *With the help of theoretical analysis, we can confirm that the M-WFRFT based on the fractional-order matrix has unitarity. However, we find that the theoretical analysis deviates from the previous numerical simulation [1], which we will discuss further in Section 4.*

*3.3. Eigendecomposition-Type FRFT as the Basis Function*

**Proposition 3.** *Eigendecomposition-type FRFT is used as the basis function, so the M-WFRFT has unitarity.*

**Proof.** In [2], Zhu et al. proposed the M-WFRFT and stated that the basis function is the FRFT, as shown in Equation (57).

$$F^{a}[f(t)] = \int\_{-\infty}^{\infty} K\_{a}(u, t) f(t) dt,\tag{57}$$

where the transform kernel is given by

$$K\_{\mathfrak{a}}(u,t) = \begin{cases} A\_{\mathfrak{a}} e^{i\frac{u^2+t^2}{2}\cot\phi - iut\csc\phi} & \mathfrak{a} \neq k\pi \\\ \delta(u-t) & \mathfrak{a} = 2k\pi \\\ \delta(u+t) & \mathfrak{a} = (2k+1)\pi \end{cases},\tag{58}$$

where *<sup>φ</sup>* = *απ*/2 is interpreted as a rotation angle in the phase plane and *<sup>A</sup><sup>α</sup>* <sup>=</sup> <sup>3</sup>(<sup>1</sup> <sup>−</sup> *<sup>i</sup>* cot *<sup>α</sup>*)/2*π*.

As we know, Equation (57) is a continuous FRFT, and a discrete FRFT is used for numerical simulation. At present, the discrete definition [29] closest to the continuous FRFT is

$$F^{\mathfrak{a}}(m,n) = \sum\_{k=0}^{N-1} v\_k(m) e^{-i\frac{\mathfrak{a}}{2}k\mathfrak{a}} v\_k(n),\tag{59}$$

where *vk*(*n*) is an arbitrary orthonormal eigenvector set of the *N* × *N* DFT matrix. Equation (59) can be written as

$$F^a = V D^a V^H,\tag{60}$$

where *<sup>V</sup>* <sup>=</sup> (*v*0, *<sup>v</sup>*1, ··· , *vN*−1), *vk* is the *<sup>k</sup>*th-order DFT Hermite eigenvector, and *<sup>D</sup><sup>α</sup>* is a diagonal matrix, defined as

$$D^{\mathfrak{a}} = \text{diag}\left(1, e^{-i\frac{\pi}{2}a}, \cdot, \cdot, e^{-i\frac{\pi}{2}(N-2)a}, e^{-i\frac{\pi}{2}(N-1)a}\right), \text{ when } N \text{ is odd},\tag{61}$$

and

$$D^{\mathfrak{a}} = \text{diag}\left(1, e^{-i\frac{\mathfrak{a}}{2}a}, \cdot, \cdot, e^{-i\frac{\mathfrak{a}}{2}(N-2)a}, e^{-i\frac{\mathfrak{a}}{2}(N)a}\right), \text{ when } N \text{ is even.}\tag{62}$$

We only prove that *N* is odd (when *N* is even, the proof process is the same). Therefore, there is

$$D^{\mathfrak{a}} = \text{diag}\left(\left(1\right)^{\mathfrak{a}}, \left(-i\right)^{\mathfrak{a}}, \left(-1\right)^{\mathfrak{a}}, \left(i\right)^{\mathfrak{a}}, \left(1\right)^{\mathfrak{a}}, \left(-i\right)^{\mathfrak{a}}, \left(-1\right)^{\mathfrak{a}}, \left(i\right)^{\mathfrak{a}}, \dots, \dots, \left(1\right or - 1\right)^{\mathfrak{a}}\right). \tag{63}$$

Then, Equation (10) can be written as

$$F\_k \quad \implies \mathfrak{a}^{\otimes \times \mathbb{A}} \times \colon F^{\otimes} \twoheadrightarrow \mathfrak{a}^{\otimes \times \mathbb{A}} \times \colon F \stackrel{\mathbb{A}}{\dashrightarrow} \dashv \dashv \colon \mathfrak{a}^{\otimes (\mathcal{M}-1)} \times \mathbb{A} \times F \stackrel{\underline{a}(\mathfrak{a}\mathfrak{b}-1)}{\dashrightarrow}$$

$$\mathbf{u}^{\text{flat}} = \mathbf{u}^{\text{flat}} V \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & (-i)^0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & (1 \, \text{or} \, -1)^0 \end{pmatrix} \mathbf{V}^{\text{flat}} + \mathbf{u}^{\text{flat}} V \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & (-i)^\mathbf{\bar{n}} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & (1 \, \text{or} \, -1)^\mathbf{\bar{n}} \end{pmatrix} \mathbf{V}^{\text{flat}} + \cdots + \mathbf{u}^{\text{flat}} V^{\text{flat}} V \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & (-i)^\frac{\mathbf{\bar{n}}(\mathbf{I} - 1)}{\mathbf{I}} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & (1 \, \text{or} \, -1)^\mathbf{\bar{n}} \end{pmatrix} \mathbf{V}^{\text{flat}}.\tag{64}$$

We can further obtain Equation (65) as

$$Y\_k = V \begin{pmatrix} Q^{(1)}(k) & 0 & \cdots & 0 \\ 0 & Q^{(-i)}(k) & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & Q^{(1 or -1)}(k) \end{pmatrix} V^H. \tag{65}$$

The diagonal matrix of Equation (65) can be expressed as

$$\text{diag}\left(\mathcal{Q}^{(1)}(k), \mathcal{Q}^{(-i)}(k), \mathcal{Q}^{(-1)}(k), \mathcal{Q}^{(i)}(k), \mathcal{Q}^{(1)}(k), \mathcal{Q}^{(-i)}(k), \dots, \dots, \mathcal{Q}^{(1 \ or r - 1)}(k)\right). \tag{66}$$

Then, *Q*(1)(*k*) is the same as Equation (40), *Q*(−*<sup>i</sup>*)(*k*) is the same as Equation (46), *Q*(−<sup>1</sup>)(*k*) is the same as Equation (44), and *Q*(*i*)(*k*) is the same as Equation (42). Thus, *Yk* can be obtained as

$$Y\_k = \begin{cases} Y\_{k\prime} & k = 0, 1, 2, 3 \\ 0, & k = 4, 5, \cdots \text{ } \text{ } \text{M}-1. \end{cases} \tag{67}$$

All the following proofs are the same as Section 3.2. In other words, the M-WFRFT has unitarity. -

**Remark 4.** *From Equation (67), it is not difficult to find that there are only four weighted terms of the M-WFRFT based on the eigendecomposition-type FRFT.*
