**Proposition 1.** *4-WFRFT is used as the basis function, so the M-WFRFT has unitarity.*

**Proof.** The definition of the 4-WFRFT is shown in Equation (1), and Equation (13) can be obtained as

$$\begin{aligned} F^a[f(t)] &= \left( A\_0^a \cdot I + A\_1^a \cdot F + A\_2^a \cdot F^2 + A\_3^a \cdot F^3 \right) f(t) \\ &= \left( I, F, F^2, F^3 \right) \begin{pmatrix} A\_0^a \\ A\_1^a \\ A\_2^a \\ A\_3^a \end{pmatrix} f(t). \end{aligned} \tag{13}$$

From Equations (4) and (13), we obtain

$$F^{a}[f(t)] = \frac{1}{4} \left( I, F, F^{2}, F^{3} \right) \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & -i & -1 & i \\ 1 & -1 & 1 & -1 \\ 1 & i & -1 & -i \end{pmatrix} \begin{pmatrix} B\_{0}^{a} \\ B\_{1}^{a} \\ B\_{2}^{a} \\ B\_{3}^{a} \end{pmatrix} f(t), \tag{14}$$

where *B<sup>α</sup> <sup>k</sup>* <sup>=</sup> exp2*πik<sup>α</sup>* <sup>4</sup> , *k* = 0, 1, 2, 3. Here, let

$$\begin{cases} P\_0 = I + F + F^2 + F^3 \\ P\_1 = I - F \ast i - F^2 + F^3 \ast i \\ P\_2 = I - F + F^2 - F^3 \\ P\_3 = I + F \ast i - F^2 - F^3 \ast i \end{cases} \tag{15}$$

Then, the 4-WFRFT can be re-expressed as

$$T\_{4W}^{\alpha}[f(t)] = \frac{1}{4}(P\_{0\prime}P\_{1\prime}P\_{2\prime}P\_{3})\begin{pmatrix} B\_0^{\alpha} \\ B\_1^{\alpha} \\ B\_2^{\alpha} \\ B\_3^{\alpha} \end{pmatrix} f(t). \tag{16}$$

Thus, the discrete 4-WFRFT can be expressed as

$$T\_{4W}^{\alpha} = \frac{1}{4} (P\_{0\prime}, P\_{1\prime}, P\_{2\prime}, P\_3) \begin{pmatrix} B\_0^{\alpha} \\ B\_1^{\alpha} \\ B\_2^{\alpha} \\ B\_3^{\alpha} \end{pmatrix} . \tag{17}$$

From Equation (10), *Yk* can be expressed as

$$Y\_k = u^{0 \times k} \times I + u^{1 \times k} \times F^{\frac{4}{M}} + \cdots + u^{(M-1) \times k} \times F^{\frac{4(M-1)}{M}};\tag{18}$$
 
$$k = 0, 1, \cdots, M - 1,$$

The 4-WFRFT as the basis function is

$$Y\_k = u^{0 \times k} \times T\_{4M}^0 + u^{1 \times k} \times T\_{4M}^{\frac{4}{M}} + \dots + u^{(M-1) \times k} \times T\_{4M}^{\frac{4(M-1)}{M}}.\tag{19}$$

From Equations (17) and (19), we can obtain

*Yk* = <sup>1</sup> <sup>4</sup> (*P*0, *P*1, *P*2, *P*3) ⎛ ⎜⎜⎜⎜⎜⎜⎝ *<sup>u</sup>*0×*<sup>k</sup>* <sup>×</sup> ⎛ ⎜⎜⎝ *B*0 0 *B*0 1 *B*0 2 *B*0 3 ⎞ ⎟⎟⎠ <sup>+</sup> *<sup>u</sup>*1×*<sup>k</sup>* <sup>×</sup> ⎛ ⎜⎜⎜⎜⎜⎝ *B* 4 *M* 0 *B* 4 *M* 1 *B* 4 *M* 2 *B* 4 *M* 3 ⎞ ⎟⎟⎟⎟⎟⎠ <sup>+</sup> ··· <sup>+</sup> *<sup>u</sup>*(*M*−1)×*<sup>k</sup>* <sup>×</sup> ⎛ ⎜⎜⎜⎜⎜⎜⎝ *B* 4(*M*−1) *M* 0 *B* 4(*M*−1) *M* 1 *B* 4(*M*−1) *M* 2 *B* 4(*M*−1) *M* 3 ⎞ ⎟⎟⎟⎟⎟⎟⎠ ⎞ ⎟⎟⎟⎟⎟⎟⎠ = <sup>1</sup> <sup>4</sup> (*P*0, *P*1, *P*2, *P*3) ⎛ ⎜⎜⎜⎜⎜⎜⎝ *<sup>u</sup>*0×*<sup>k</sup>* <sup>×</sup> *<sup>B</sup>*<sup>0</sup> <sup>0</sup> <sup>+</sup> *<sup>u</sup>*1×*<sup>k</sup>* <sup>×</sup> *<sup>B</sup>* 4 *M* <sup>0</sup> <sup>+</sup> ··· <sup>+</sup> *<sup>u</sup>*(*M*−1)×*<sup>k</sup>* <sup>×</sup> *<sup>B</sup>* 4(*M*−1) *M* 0 *<sup>u</sup>*0×*<sup>k</sup>* <sup>×</sup> *<sup>B</sup>*<sup>0</sup> <sup>1</sup> <sup>+</sup> *<sup>u</sup>*1×*<sup>k</sup>* <sup>×</sup> *<sup>B</sup>* 4 *M* <sup>1</sup> <sup>+</sup> ··· <sup>+</sup> *<sup>u</sup>*(*M*−1)×*<sup>k</sup>* <sup>×</sup> *<sup>B</sup>* 4(*M*−1) *M* 1 *<sup>u</sup>*0×*<sup>k</sup>* <sup>×</sup> *<sup>B</sup>*<sup>0</sup> <sup>2</sup> <sup>+</sup> *<sup>u</sup>*1×*<sup>k</sup>* <sup>×</sup> *<sup>B</sup>* 4 *M* <sup>2</sup> <sup>+</sup> ··· <sup>+</sup> *<sup>u</sup>*(*M*−1)×*<sup>k</sup>* <sup>×</sup> *<sup>B</sup>* 4(*M*−1) *M* 2 *<sup>u</sup>*0×*<sup>k</sup>* <sup>×</sup> *<sup>B</sup>*<sup>0</sup> <sup>3</sup> <sup>+</sup> *<sup>u</sup>*1×*<sup>k</sup>* <sup>×</sup> *<sup>B</sup>* 4 *M* <sup>3</sup> <sup>+</sup> ··· <sup>+</sup> *<sup>u</sup>*(*M*−1)×*<sup>k</sup>* <sup>×</sup> *<sup>B</sup>* 4(*M*−1) *M* 3 ⎞ ⎟⎟⎟⎟⎟⎟⎠ , (20)

where *k* = 0, 1, ··· , *M* − 1 and *u* = exp(−2*πi*/*M*). Therefore, we obtain

$$\begin{split} Y\_{k} &= \frac{1}{4} (\mathbb{P}b\_{1}, \mathbb{P}\_{1}, \mathbb{P}\_{2}, \mathbb{P}\_{3}) \begin{pmatrix} 1 + \exp\left(\frac{-2\pi i 1 (k-1)}{M}\right) + \exp\left(\frac{-2\pi i (2k-1)k}{M}\right) \\ 1 + \exp\left(\frac{-2\pi i (k-1)}{M}\right) + \exp\left(\frac{-2\pi i (2k-1)}{M}\right) + \dots + \exp\left(\frac{-2\pi i (M-1)(k-1)}{M}\right) \\ 1 + \exp\left(\frac{-2\pi i (k-2)}{M}\right) + \exp\left(\frac{-2\pi i (k-2)}{M}\right) + \dots + \exp\left(\frac{-2\pi i (M-1)(k-2)}{M}\right) \\ 1 + \exp\left(\frac{-2\pi i (k-3)}{M}\right) + \exp\left(\frac{-2\pi i (k-3)}{M}\right) + \dots + \exp\left(\frac{-2\pi i (M-1)(k-3)}{M}\right) \\ &= \frac{1}{4} (\mathbb{P}b\_{1}, \mathbb{P}\_{1}, \mathbb{P}\_{2}, \mathbb{P}\_{3}) \begin{pmatrix} \mathbb{S}\_{0}(k) \\ \mathbb{S}\_{1}(k) \\ \mathbb{S}\_{2}(k) \\ \mathbb{S}\_{3}(k) \end{pmatrix}. \end{split} \tag{21}$$

For sequence *S*0(*k*), it can be expressed as

$$S\_0(k) = \frac{a\_1 \left(1 - q^M\right)}{1 - q} = \frac{1 - \exp\left(\frac{-2\pi ik}{M}\right)^M}{1 - \exp\left(\frac{-2\pi ik}{M}\right)}.\tag{22}$$

where *a*<sup>1</sup> = 1. Then, we obtain

$$S\_0(k) = \begin{cases} M, & k \equiv 0 \bmod M \\ 0, & k \not\equiv 0 \bmod M. \end{cases} \tag{23}$$

For sequence *S*1(*k*),

$$S\_1(k) = \frac{1 - \left(e^{-2\pi i (k-1)/M}\right)^M}{1 - e^{-2\pi i (k-1)/M}},\tag{24}$$

we obtain

$$S\_1(k) = \begin{cases} M, & k \equiv 1 \bmod M \\ 0, & k \not\equiv \bmod M. \end{cases} \tag{25}$$

For sequence *S*2(*k*),

$$S\_2(k) = \frac{1 - \left(e^{-2\pi i (k-2)/M}\right)^M}{1 - e^{-2\pi i (k-2)/M}},\tag{26}$$

we obtain

$$S\_2(k) = \begin{cases} M, & k \equiv 2 \bmod M \\ 0, & k \not\equiv 2 \bmod M. \end{cases} \tag{27}$$

For sequence *S*3(*k*),

$$S\_3(k) = \frac{1 - \left(\varepsilon^{-2\pi i(k-3)/M}\right)^M}{1 - \varepsilon^{-2\pi i(k-3)/M}},\tag{28}$$

we obtain

$$S\_3(k) = \begin{cases} M, & k \equiv 3 \bmod M \\ 0, & k \not\equiv 3 \bmod M. \end{cases} \tag{29}$$

Then, Equation (21) can be expressed as

$$Y\_k = \begin{cases} \frac{M}{4} P\_{k\prime} & k = 0, 1, 2, 3 \\ 0, & k = 4, 5, \cdots, M - 1. \end{cases} \tag{30}$$

Therefore, the M-WFRFT Equation (11) is written as

$$\begin{array}{ll} T\_{\text{AMW}}^{\text{ax}} &= \frac{1}{M} (Y\_0, Y\_1, \dots, Y\_{M-1}) \begin{pmatrix} B\_0^{\text{ax}} \\ B\_1^{\text{ax}} \\ \vdots \\ B\_{M-1}^{\text{ax}} \end{pmatrix} \\ &= \frac{1}{4} (P\_0, P\_1, P\_2, P\_3, 0, \dots, 0) \begin{pmatrix} B\_0^{\text{ax}} \\ B\_1^{\text{ax}} \\ \vdots \\ B\_{M-1}^{\text{ax}} \end{pmatrix} \\ &= \frac{1}{4} (P\_0, P\_1, P\_2, P\_3) \begin{pmatrix} B\_0^{\text{ax}} \\ B\_1^{\text{ax}} \\ \vdots \\ B\_2^{\text{ax}} \end{pmatrix} .\end{array} \tag{31}$$

From the expressions, we notice that Equations (17) and (31) are the same, but in fact they are different. The difference is that for Equation (31), *B<sup>α</sup> <sup>k</sup>* <sup>=</sup> exp2*πik<sup>α</sup> <sup>M</sup>* ; *k* = 0, 1, ··· , *M* − 1. However, this does not affect the proof of unitarity. -

**Remark 2.** *In our previous work [1], we proved the unitarity of Equation (17). When the 4-WFRFT is selected as the basis function, the M-WFRFT has unitarity. From Equation (31), we notice that the weighted sum of the M-WFRFT is only four terms.*
