東京大学 情報理工学系研究科 電子情報学専攻 2024年8月実施 専門 第5問 Author adj-matrix , 祭音Myyura
Description Answer the following questions about discrete-time signal processing.
(1) Describe briefly the functionality and role of an anti-aliasing filter.
(2) For a linear time-invariant and causal discrete-time signal processing system L L L
H ( z ) = α z 2 α β z 2 − α β z + β − γ , H(z) = \frac{\alpha z^2}{\alpha\beta z^2 - \alpha\beta z + \beta - \gamma}, H ( z ) = α β z 2 − α β z + β − γ α z 2 , where z z z α , β , γ \alpha, \beta, \gamma α , β , γ α , β > 0 \alpha, \beta > 0 α , β > 0 x [ n ] x[n] x [ n ] y [ n ] y[n] y [ n ] L L L n n n
(3) In (2), find the range of γ \gamma γ L L L α = 4 \alpha = 4 α = 4 β = 1 \beta = 1 β = 1
(4) In (2), when α = 4 \alpha = 4 α = 4 β = γ = 1 2 \beta = \gamma = \frac{1}{2} β = γ = 2 1 ∣ H ( e j Ω ) ∣ |H(e^{j\Omega})| ∣ H ( e j Ω ) ∣ ∠ H ( e j Ω ) \angle H(e^{j\Omega}) ∠ H ( e j Ω ) L L L j j j Ω \Omega Ω
(5) In (2), when α = 4 , β = 1 \alpha = 4, \beta = 1 α = 4 , β = 1 γ = 1 9 \gamma = \frac{1}{9} γ = 9 1 h [ n ] h[n] h [ n ] L L L u [ n ] u[n] u [ n ]
u [ n ] = { 1 ( n ≥ 0 ) 0 ( n < 0 ) u[n] = \begin{cases} 1 & (n \ge 0) \\ 0 & (n < 0) \end{cases} u [ n ] = { 1 0 ( n ≥ 0 ) ( n < 0 ) 题目描述
回答下列离散时间信号处理问题。
(1) 简要说明抗混叠滤波器的功能与作用。
(2) 对线性、时不变、因果的离散时间信号处理系统 L L L
H ( z ) = α z 2 α β z 2 − α β z + β − γ , H(z)=\frac{\alpha z^2}
{\alpha\beta z^2-\alpha\beta z+\beta-\gamma}, H ( z ) = α β z 2 − α β z + β − γ α z 2 , 其中 z z z α , β , γ \alpha,\beta,\gamma α , β , γ α , β > 0 \alpha,\beta>0 α , β > 0 n n n x [ n ] x[n] x [ n ] y [ n ] y[n] y [ n ]
(3) 在 (2) 中取 α = 4 , β = 1 \alpha=4,\beta=1 α = 4 , β = 1 L L L γ \gamma γ
(4) 在 (2) 中取 α = 4 \alpha=4 α = 4 β = γ = 1 2 \beta=\gamma=\frac12 β = γ = 2 1 ∣ H ( e j Ω ) ∣ |H(e^{j\Omega})| ∣ H ( e j Ω ) ∣ ∠ H ( e j Ω ) \angle H(e^{j\Omega}) ∠ H ( e j Ω ) j j j Ω \Omega Ω
(5) 在 (2) 中取 α = 4 , β = 1 , γ = 1 9 \alpha=4,\beta=1,\gamma=\frac19 α = 4 , β = 1 , γ = 9 1
u [ n ] = { 1 , n ≥ 0 , 0 , n < 0 u[n]=
\begin{cases}
1,&n\ge0,\\
0,&n<0
\end{cases} u [ n ] = { 1 , 0 , n ≥ 0 , n < 0 求系统 L L L h [ n ] h[n] h [ n ]
Kai (1)
An anti-aliasing filter is a filter used before a signal sampler to restrict the bandwidth of a signal to satisfy the Nyquist-Shannon sampling theorem over the band of interest.
(2)
Given H ( z ) = Y ( z ) X ( z ) = α z 2 α β z 2 − α β z + β − γ H(z) = \frac{Y(z)}{X(z)} = \frac{\alpha z^2}{\alpha\beta z^2 - \alpha\beta z + \beta - \gamma} H ( z ) = X ( z ) Y ( z ) = α β z 2 − α β z + β − γ α z 2
i.e., Y ( z ) ( α β − α β z − 1 + ( β − γ ) z − 2 ) = X ( z ) α Y(z)(\alpha\beta - \alpha\beta z^{-1} + (\beta - \gamma)z^{-2}) = X(z)\alpha Y ( z ) ( α β − α β z − 1 + ( β − γ ) z − 2 ) = X ( z ) α
Since y [ n − k ] ↔ z − k Y ( z ) y[n-k] \leftrightarrow z^{-k}Y(z) y [ n − k ] ↔ z − k Y ( z ) α β y [ n ] − α β y [ n − 1 ] + ( β − γ ) y [ n − 2 ] = α x [ n ] \alpha\beta y[n] - \alpha\beta y[n-1] + (\beta - \gamma) y[n-2] = \alpha x[n] α β y [ n ] − α β y [ n − 1 ] + ( β − γ ) y [ n − 2 ] = αx [ n ]
Divide by α β \alpha\beta α β y [ n ] − y [ n − 1 ] + β − γ α β y [ n − 2 ] = 1 β x [ n ] y[n] - y[n-1] + \frac{\beta - \gamma}{\alpha\beta} y[n-2] = \frac{1}{\beta} x[n] y [ n ] − y [ n − 1 ] + α β β − γ y [ n − 2 ] = β 1 x [ n ]
(3)
Substitute parameters into the denominator characteristic equation: 4 z 2 − 4 z + ( 1 − γ ) = 0 4z^2 - 4z + (1 - \gamma) = 0 4 z 2 − 4 z + ( 1 − γ ) = 0
Roots: z = 4 ± 16 − 16 ( 1 − γ ) 8 = 1 ± γ 2 z = \frac{4 \pm \sqrt{16 - 16(1 - \gamma)}}{8} = \frac{1 \pm \sqrt{\gamma}}{2} z = 8 4 ± 16 − 16 ( 1 − γ ) = 2 1 ± γ
For stability, poles must be inside the unit circle: ∣ z ∣ < 1 |z| < 1 ∣ z ∣ < 1 ∣ 1 ± γ 2 ∣ < 1 ⟹ ∣ 1 ± γ ∣ < 2 \left| \frac{1 \pm \sqrt{\gamma}}{2} \right| < 1 \implies |1 \pm \sqrt{\gamma}| < 2 2 1 ± γ < 1 ⟹ ∣1 ± γ ∣ < 2
If γ ≥ 0 \gamma \ge 0 γ ≥ 0 − 3 < ± γ < 1 ⟹ 0 ≤ γ < 1 -3 < \pm \sqrt{\gamma} < 1 \implies 0 \le \gamma < 1 − 3 < ± γ < 1 ⟹ 0 ≤ γ < 1
If γ < 0 \gamma < 0 γ < 0 1 + ( − γ ) < 4 ⟹ − 3 < γ < 0 1 + (-\gamma) < 4 \implies -3 < \gamma < 0 1 + ( − γ ) < 4 ⟹ − 3 < γ < 0
i.e., − 3 < γ < 1 -3 < \gamma < 1 − 3 < γ < 1
(4)
Given α = 4 , β = γ = 1 2 \alpha = 4, \beta = \gamma = \frac{1}{2} α = 4 , β = γ = 2 1
H ( z ) = 4 z 2 2 z 2 − 2 z = 2 z z − 1 H(z) = \frac{4z^2}{2z^2 - 2z} = \frac{2z}{z - 1} H ( z ) = 2 z 2 − 2 z 4 z 2 = z − 1 2 z Since z = e j Ω z = e^{j\Omega} z = e j Ω H ( e j Ω ) = 2 e j Ω e j Ω − 1 = 2 e j Ω e j Ω 2 ( e j Ω 2 − e − j Ω 2 ) = e j Ω 2 j sin Ω 2 H(e^{j\Omega}) = \frac{2e^{j\Omega}}{e^{j\Omega} - 1} = \frac{2e^{j\Omega}}{e^{j\frac{\Omega}{2}}(e^{j\frac{\Omega}{2}} - e^{-j\frac{\Omega}{2}})} = \frac{e^{j\frac{\Omega}{2}}}{j \sin\frac{\Omega}{2}} H ( e j Ω ) = e j Ω − 1 2 e j Ω = e j 2 Ω ( e j 2 Ω − e − j 2 Ω ) 2 e j Ω = j s i n 2 Ω e j 2 Ω
Therefore, ∣ H ( e j Ω ) ∣ = ∣ e j Ω 2 j sin Ω 2 ∣ = 1 ∣ sin Ω 2 ∣ = { 1 sin Ω 2 if sin Ω 2 > 0 − 1 sin Ω 2 if sin Ω 2 < 0 |H(e^{j\Omega})| = \left| \frac{e^{j\frac{\Omega}{2}}}{j \sin\frac{\Omega}{2}} \right| = \frac{1}{|\sin\frac{\Omega}{2}|} = \begin{cases} \frac{1}{\sin\frac{\Omega}{2}} & \text{if } \sin\frac{\Omega}{2} > 0 \\ -\frac{1}{\sin\frac{\Omega}{2}} & \text{if } \sin\frac{\Omega}{2} < 0 \end{cases} ∣ H ( e j Ω ) ∣ = j s i n 2 Ω e j 2 Ω = ∣ s i n 2 Ω ∣ 1 = ⎩ ⎨ ⎧ s i n 2 Ω 1 − s i n 2 Ω 1 if sin 2 Ω > 0 if sin 2 Ω < 0
Similarly, ∠ H ( e j Ω ) = { Ω 2 − π 2 if sin Ω 2 > 0 Ω 2 + π 2 if sin Ω 2 < 0 ( m o d 2 π ) \angle H(e^{j\Omega}) = \begin{cases} \frac{\Omega}{2} - \frac{\pi}{2} & \text{if } \sin\frac{\Omega}{2} > 0 \\ \frac{\Omega}{2} + \frac{\pi}{2} & \text{if } \sin\frac{\Omega}{2} < 0 \end{cases}\pmod{2\pi} ∠ H ( e j Ω ) = { 2 Ω − 2 π 2 Ω + 2 π if sin 2 Ω > 0 if sin 2 Ω < 0 ( mod 2 π )
In summary (phases are modulo 2 π 2\pi 2 π
If sin Ω 2 > 0 \sin\frac{\Omega}{2} > 0 sin 2 Ω > 0 ∣ H ( e j Ω ) ∣ = 1 sin Ω 2 , ∠ H ( e j Ω ) = Ω 2 − π 2 |H(e^{j\Omega})| = \frac{1}{\sin\frac{\Omega}{2}}, \quad \angle H(e^{j\Omega}) = \frac{\Omega}{2} - \frac{\pi}{2} ∣ H ( e j Ω ) ∣ = s i n 2 Ω 1 , ∠ H ( e j Ω ) = 2 Ω − 2 π
If sin Ω 2 < 0 \sin\frac{\Omega}{2} < 0 sin 2 Ω < 0 ∣ H ( e j Ω ) ∣ = − 1 sin Ω 2 , ∠ H ( e j Ω ) = Ω 2 + π 2 |H(e^{j\Omega})| = -\frac{1}{\sin\frac{\Omega}{2}}, \quad \angle H(e^{j\Omega}) = \frac{\Omega}{2} + \frac{\pi}{2} ∣ H ( e j Ω ) ∣ = − s i n 2 Ω 1 , ∠ H ( e j Ω ) = 2 Ω + 2 π
At Ω = 2 π k ( k ∈ Z ) \Omega=2\pi k\ (k\in\mathbb Z) Ω = 2 πk ( k ∈ Z ) h [ n ] = 2 u [ n ] h[n]=2u[n] h [ n ] = 2 u [ n ] ∣ z ∣ > 1 |z|>1 ∣ z ∣ > 1 H ( z ) H(z) H ( z )
(5)
Given α = 4 , β = 1 , γ = 1 9 \alpha = 4, \beta = 1, \gamma = \frac{1}{9} α = 4 , β = 1 , γ = 9 1
H ( z ) = 4 z 2 4 z 2 − 4 z + 8 9 = 1 1 − z − 1 + 2 9 z − 2 = 1 ( 1 − 2 3 z − 1 ) ( 1 − 1 3 z − 1 ) H(z) = \frac{4z^2}{4z^2 - 4z + \frac{8}{9}} = \frac{1}{1 - z^{-1} + \frac{2}{9}z^{-2}} = \frac{1}{(1 - \frac{2}{3}z^{-1})(1 - \frac{1}{3}z^{-1})} H ( z ) = 4 z 2 − 4 z + 9 8 4 z 2 = 1 − z − 1 + 9 2 z − 2 1 = ( 1 − 3 2 z − 1 ) ( 1 − 3 1 z − 1 ) 1 i.e.,
H ( z ) = 6 3 − 2 z − 1 + − 3 3 − z − 1 = 2 1 − ( 2 3 ) z − 1 − 1 1 − ( 1 3 ) z − 1 H(z) = \frac{6}{3 - 2z^{-1}} + \frac{-3}{3 - z^{-1}} = \frac{2}{1 - (\frac{2}{3})z^{-1}} - \frac{1}{1 - (\frac{1}{3})z^{-1}} H ( z ) = 3 − 2 z − 1 6 + 3 − z − 1 − 3 = 1 − ( 3 2 ) z − 1 2 − 1 − ( 3 1 ) z − 1 1 Inverse Z-transform using u [ n ] u[n] u [ n ]
h [ n ] = ( 2 ( 2 3 ) n − ( 1 3 ) n ) u [ n ] h[n] = \left( 2\left(\frac{2}{3}\right)^n - \left(\frac{1}{3}\right)^n \right) u[n] h [ n ] = ( 2 ( 3 2 ) n − ( 3 1 ) n ) u [ n ]