对于Non-Delaying Integrator
( u − z − 1 v ) 1 1 − z − 1 + e = v v = u + ( 1 − z − 1 ) e ( u − z − 1 v ) 1 − z − 1 1 + e = v v = u + ( 1 − z − 1 ) e
对于Delaying Integrator
( u − v ) z − 1 1 − z − 1 + e = v v = u z − 1 + ( 1 − z − 1 ) e ( u − v ) 1 − z − 1 z − 1 + e = v v = u z − 1 + ( 1 − z − 1 ) e
所以噪声传递函数（这里的ω = 2 π f ω = 2 π f T T
N T F ( z ) = 1 − z − 1 = 1 − e j ω T N T F ( z ) = 1 − z − 1 = 1 − e j ω T
积分上下限为：
2 π / T 2 ⋅ 1 O S R = π O S R ⋅ T 2 2 π / T ⋅ O S R 1 = O S R ⋅ T π
依据噪声的帕萨瓦尔定理
P n o i s e , l p f = 1 2 π ∫ − π O S R ⋅ T π O S R ⋅ T ∣ 1 − e − j ω T ∣ 2 d ω = 1 π ∫ 0 π O S R ⋅ T ( 1 − cos ( ω T ) ) 2 + sin ( ω T ) 2 d ω = 1 π ∫ 0 π O S R ⋅ T 4 sin ( ω T 2 ) 2 d ω P n o i s e , l p f = 2 π 1 ∫ − O S R ⋅ T π O S R ⋅ T π ∣ ∣ ∣ 1 − e − j ω T ∣ ∣ ∣ 2 d ω = π 1 ∫ 0 O S R ⋅ T π ( 1 − cos ( ω T ) ) 2 + sin ( ω T ) 2 d ω = π 1 ∫ 0 O S R ⋅ T π 4 sin ( 2 ω T ) 2 d ω
(*Mathematica 12.3*)
FullSimplify[TrigExpand[TrigReduce[(1 - Cos[2*\[Omega]])^2 + Sin[2*\[Omega]]^2]]]
FullSimplify[TrigExpand[TrigReduce[(1 - Cos[\[Omega]])^2 + Sin[\[Omega]]^2]]]
上述推导，用到了积化和差公式
sin α ⋅ sin β = − 1 2 ( cos ( α + β ) − cos ( α − β ) ) ( sin α 2 ) 2 = − 1 2 ( cos α − 1 ) = 1 2 ( 1 − cos α ) sin α ⋅ sin β ( sin 2 α ) 2 = − 2 1 ( cos ( α + β ) − cos ( α − β ) ) = − 2 1 ( cos α − 1 ) = 2 1 ( 1 − cos α )
接着
P n o i s e , l p f ≈ 1 π ∫ 0 π O S R ⋅ T 4 ( ω T 2 ) 2 d ω = 1 π ω 3 T 2 3 ∣ 0 π OSRT = π 2 3 O S R 3 T P n o i s e , l p f ≈ π 1 ∫ 0 O S R ⋅ T π 4 ( 2 ω T ) 2 d ω = π 1 3 ω 3 T 2 ∣ ∣ ∣ ∣ ∣ 0 OSRT π = 3 O S R 3 T π 2
Understanding SDM 2nd Edition, page 44给出的带内噪声IBN结果是
I B N = V L S B 2 12 ⋅ π 2 3 O S R 3 I B N = 1 2 V L S B 2 ⋅ 3 O S R 3 π 2
进行化简后，可以看出，是归一化处理采样时间T = 1 T = 1
4 ( ω T 2 ) 2 ⇒ T = 1 ω 2 4 ( 2 ω T ) 2 T = 1 ω 2
(*Mathematica 12.3*)
ClearAll["Global`*"]
(*一阶 Delta-Sigma Modulator 的传递函数，噪声的频谱密度*)
NTF = (Abs[1 - E^(-I*\[Omega]*T)])^2
(*NTF在"fS/(2*OSR)"内的积分RMS噪声*)
pnoise = (1/Pi)*Integrate[NTF, {\[Omega], 0, Pi/(OSR*T)}, Assumptions -> OSR > 1]
(*经过近似的一阶积分噪声，与书本上推导一致*)
Series[pnoise, OSR -> Infinity]
SNR O S R + S D M 1 − S N R = 10 10 P s i g n a l P n o i s e π 2 3 O S R 3 T P s i g n a l P n o i s e = 10 10 3 O S R 3 π 2 = 3 × 3.02 × 2 O S R − 5.17 SNR O S R + S D M 1 − S N R = 1 0 log 1 0 P n o i s e P s i g n a l P n o i s e 3 O S R 3 T π 2 P s i g n a l = 1 0 log 1 0 π 2 3 O S R 3 = 3 × 3 . 0 2 × log 2 O S R − 5 . 1 7
所以ENOB的效率相当于1.5 × 2 1 . 5 × log 2
常用的对数公式
a b = c b c a a b c = a b + a c a b c a b + a c a b c = c a b log a b = log c a log c b log a b c = log a b + log a c log a c b log a b + log a c log a b c = c log a b
对于一个信号幅度是M的正弦波，其信号功率为：
P s i g n a l = ( 2 2 V F u l l S c a l e 2 ) 2 = M 2 2 P s i g n a l = ( 2 2 2 V F u l l S c a l e ) 2 = 2 M 2
量化器V L S B = 2 V L S B = 2 1 1
P n o i s e − s h a p p i n g − l p f = V L S B 2 12 π 2 3 O S R 3 = 2 2 12 π 2 3 O S R 3 = π 2 9 O S R 3 P n o i s e − s h a p p i n g − l p f = 1 2 V L S B 2 3 O S R 3 π 2 = 1 2 2 2 3 O S R 3 π 2 = 9 O S R 3 π 2
信噪比为：
P s i g n a l P n o i s e − s h a p i n g = 9 O S R 3 2 π 2 M 2 P n o i s e − s h a p i n g P s i g n a l = 2 π 2 M 2 9 O S R 3
换一个角度看这个问题，M越大，信号的幅值就越大，MOD1可实现的SQNR就越高。其实Fullscale本来是固定的，在understanding SDM这本书里，固定了STEP-SIZE Δ = 2 Δ = 2
似乎在SDM中说量化器，M表示信号的magnitude，同时也表示M-STEP的量化器的M，这里面似乎是有些不严谨，对于bipolar的mid-rise或者mid-tread，量化结果总是M-1。需要根据具体情况分析M所代表的意思。在这里，M表示幅值，而量化器的STEP-SIZE=2。
NTF的冲击响应就是NTF(z)，冲击响应时域为h(n)
N T F ( z ) = 1 − z − 1 N T F ( z ) = ∑ n = 0 ∞ h ( n ) z − n = h ( 0 ) z 0 + h ( 1 ) z − 1 + h ( 2 ) z − 2 + h ( 3 ) z − 3 … N T F ( z ) N T F ( z ) = 1 − z − 1 = n = 0 ∑ ∞ h ( n ) z − n = h ( 0 ) z 0 + h ( 1 ) z − 1 + h ( 2 ) z − 2 + h ( 3 ) z − 3 …
将这个对照到N T F ( z ) N T F ( z )
h ( 0 ) = 1 , h ( 1 ) = − 1 h ( 0 ) = 1 , h ( 1 ) = − 1
对于DC增益，
ω = 0 → z = e j ω T s = 1 → N T F ( z ) = 1 − z − 1 = 0 ω = 0 → z = e j ω T s = 1 → N T F ( z ) = 1 − z − 1 = 0
对于高频增益
ω = ∞ → z = e j ω T s = ∞ → N T F ( ∞ ) = 0 ω = ∞ → z = e j ω T s = ∞ → N T F ( ∞ ) = 0
那么如果没有滤波器的话，总的噪声为(SSB)，频域积分
P n o i s e − s h a p p i n g − t o t a l = 1 π ∫ 0 π P n o i s e ⌈ 1 − e − j Ω ⌉ 2 d Ω = 2 P n o i s e P n o i s e − s h a p p i n g − t o t a l = π 1 ∫ 0 π P n o i s e ⌈ 1 − e − j Ω ⌉ 2 d Ω = 2 P n o i s e
依据Parseval’s theorem，更简单的办法是，时域求和
P n o i s e − s h a p p i n g − t o t a l = P n o i s e 1 π ∫ 0 π ⌈ 1 − e − j Ω ⌉ 2 d Ω = ∑ n = 0 ∞ h 2 ( n ) = 1 2 + ( − 1 ) 2 = 2 P n o i s e P n o i s e − s h a p p i n g − t o t a l = P n o i s e π 1 ∫ 0 π ⌈ 1 − e − j Ω ⌉ 2 d Ω = n = 0 ∑ ∞ h 2 ( n ) = 1 2 + ( − 1 ) 2 = 2 P n o i s e
假设这里积分的信号的是噪声，那DC处积分噪声是无穷大的。事实显然不是如此，因为运放的速度与增益是有限的。这里推导在有限增益A下的积分器传递函数：
Delaying Integrator
( V o ( n ) − A − V o ( n ) ) + ( 0 − V i ( n ) ) = ( V o ( n + 1 ) − A − V o ( n + 1 ) ) + ( V o ( n + 1 ) − A ) V o V i = z − 1 ( A + 2 A ) − ( 1 + A A ) z − 1 V o V i = ( A A + 2 ) z − 1 1 − ( A + 1 A + 2 ) z − 1 ( − A V o ( n ) − V o ( n ) ) + ( 0 − V i ( n ) ) V i V o V i V o = ( − A V o ( n + 1 ) − V o ( n + 1 ) ) + ( − A V o ( n + 1 ) ) = ( A A + 2 ) − ( A 1 + A ) z − 1 z − 1 = 1 − ( A + 2 A + 1 ) z − 1 ( A + 2 A ) z − 1
Non-Delaying Integrator
( V o ( n ) − A − V o ( n ) ) + ( 0 − 0 ) = ( V o ( n + 1 ) − A − V o ( n + 1 ) ) + ( V o ( n + 1 ) − A − V i ( n + 1 ) ) V o V i = − A A + 2 1 − ( A + 1 A + 2 ) z − 1 ( − A V o ( n ) − V o ( n ) ) + ( 0 − 0 ) V i V o = ( − A V o ( n + 1 ) − V o ( n + 1 ) ) + ( − A V o ( n + 1 ) − V i ( n + 1 ) ) = 1 − ( A + 2 A + 1 ) z − 1 − A + 2 A
暂时认为传递函数变成了以下的形式，这里p = A / p = A /
T F ( D e l a y i n g ) = p z − 1 1 − p z − 1 T F ( N o n − D e l a y i n g ) = p 1 − p z − 1 T F ( D e l a y i n g ) T F ( N o n − D e l a y i n g ) = 1 − p z − 1 p z − 1 = 1 − p z − 1 p
代入到MOD1的传递函数中，发现NTF的形式比较简洁
( 0 − V ) p z − 1 1 − p z − 1 + E = V N T F = 1 − p z − 1 ( 0 − V ) 1 − p z − 1 p z − 1 + E = V N T F = 1 − p z − 1
继续代入，计算噪声
P n o i s e , l p f , F i n i t e G a i n = 1 2 π ∫ − π O S R ⋅ T π O S R ⋅ T ∣ 1 − A A + 1 e − j ω T ∣ 2 d ω P n o i s e , l p f , F i n i t e G a i n = 2 π 1 ∫ − O S R ⋅ T π O S R ⋅ T π ∣ ∣ ∣ ∣ ∣ 1 − A + 1 A e − j ω T ∣ ∣ ∣ ∣ ∣ 2 d ω
根据Understanding Delta-Sigma Data Converters, 2nd Edition, p50的描述，传递函数简化后的结果是，N T F ( e j ω ) = A − 2 + ω 2 N T F ( e j ω ) = A − 2 + ω 2
∣ 1 − A A + 1 e − j ω T ∣ 2 = ∣ 1 − p ( cos ( ω T ) − j s i n ( ω T ) ) ∣ 2 = ( 1 − p c o s ( ω T ) ) 2 + 2 ( ω T ) ≈ ( 1 − A A + 1 ) 2 + 2 ( ω T ) ≈ A − 2 + ( ω T ) 2 ∣ ∣ ∣ ∣ ∣ 1 − A + 1 A e − j ω T ∣ ∣ ∣ ∣ ∣ 2 = ∣ ∣ ∣ ∣ ∣ ∣ 1 − p ( cos ( ω T ) − j s i n ( ω T ) ) ∣ ∣ ∣ ∣ ∣ ∣ 2 = ( 1 − p c o s ( ω T ) ) 2 + sin 2 ( ω T ) ≈ ( 1 − A + 1 A ) 2 + sin 2 ( ω T ) ≈ A − 2 + ( ω T ) 2
这里我们直接求积分，假设A =OSR，求解后发现，SNR仅仅提高了1.15dB，也就是书中说的不到1.2dB。
P n o i s e , l p f , G a i n = O S R = 3 + π 2 3 O S R 3 T P n o i s e , l p f = π 2 3 O S R 3 T 10 log ( P n o i s e , l p f , G a i n = O S R P n o i s e , l p f ) = 10 log ( 3 + π 2 π 2 ) = 1.15 d B P n o i s e , l p f , G a i n = O S R P n o i s e , l p f 1 0 log ( P n o i s e , l p f P n o i s e , l p f , G a i n = O S R ) = 3 O S R 3 T 3 + π 2 = 3 O S R 3 T π 2 = 1 0 log ( π 2 3 + π 2 ) = 1 . 1 5 d B
(*Mathematica 12.3*)
ClearAll["Global`*"]
(*一阶 Delta-Sigma Modulator 的传递函数，噪声的频谱密度*)
A = p*OSR;
NTF = (Abs[1 - A/(A + 1)*E^(-I*\[Omega]*T)])^2
(*NTF在"fS/(2*OSR)"内的积分RMS噪声*)
pnoise = (1/Pi)*
Integrate[NTF, {\[Omega], 0, Pi/(OSR*T)}, Assumptions -> {OSR > 1, A > 1, p > 0}]
(*经过近似的一阶积分噪声*)
Series[pnoise, OSR -> Infinity]
直观理解的话，不考虑积分时间，不考虑噪声，量化误差是一个DC值，refer到输入时，应该能被前级抑制，但是前级的增益不是无穷大，所以量化误差不能被无限抑制，所以对于单级积分器，其量化误差 VSTEP/GAIN就是DeadZone，此时SDM无法区分该输入信号与0的区别。
也可以这样理解，存在一个输入信号，无论积分多久，输出总是积不满一个量化STEP，这样量化结果永远不会因为输入信号产生变化。
假设输入u=0， y[0]=0，对于单bit量化器，输出就是v=+1, −1, +1, −1…，这些反馈通过有限增益积分器后，得到输出y，可以表示为
y [ n ] = p z − 1 1 − p z − 1 ∣ z = e j π = − 1 cos ( π n ) = − p 1 + p cos ( π n ) y [ n ] = 1 − p z − 1 p z − 1 ∣ ∣ ∣ ∣ ∣ z = e j π = − 1 cos ( π n ) = − 1 + p p cos ( π n )
这里为了表示 + 1, − 1的反馈信号，用cos ( π n ) cos ( π n ) ω = 2 π / T v ω = 2 π / T v t = n T s t = n T s
cos ( π n ) = cos ( 2 π T v n T s ) cos ( π n ) = cos ( T v 2 π n T s )
当信号的v v T v T v T s T s v v
cos ( ω t ) = cos ( 2 π 2 T s n T s ) ∣ 归一化 T s = 1 = cos ( π n ) cos ( ω t ) = cos ( 2 T s 2 π n T s ) ∣ ∣ ∣ ∣ ∣ 归 一 化 T s = 1 = cos ( π n )
该信号只有一个频率，信号和这个系统卷积后，只会留下ω = ω v ω = ω v
z = e j ω T s ∣ ω = ω v = π T s = e j π z = e j ω T s ∣ ∣ ∣ ω = ω v = T s π = e j π
此时，在输出端额外增加的y o s y o s v v
∣ y o s ∣ < p 1 + p ∣ y o s ∣ < 1 + p p
输出结果-p/(1+p)是用积分器的TF计算得出的，是在[+1,-1...]这个信号的长期激励下的周期结果，这里代入的z = e j ω T s z = e j ω T s
由于我们的积分器增益有限，增加一个y o s y o s z − 1 = 1 z − 1 = 1
∣ u o s ∣ = ∣ y o s ∣ p ( 1 − p ) = 1 − p 1 + p = 1 A + 1 2 A + 1 A + 1 ≈ 1 2 A ∣ u o s ∣ = ( 1 − p ) p ∣ y o s ∣ = 1 + p 1 − p = A + 1 2 A + 1 A + 1 1 ≈ 2 A 1
如果增益无穷大的话，∣ u o s ∣ ∣ u o s ∣
p = A ( A + 1 ) ⟺ A = 1 ( 1 − p ) p = ( A + 1 ) A ⟺ A = ( 1 − p ) 1
DC inputs with rational value (a/b) lead to cyclic patterns of output codes, this can be problematic if period is very long.
1 100 = 101 × ( + 1 ) + 99 × ( − 1 ) 200 1 0 0 1 = 2 0 0 1 0 1 × ( + 1 ) + 9 9 × ( − 1 )
如果去计算这个cyclic pattern的话，那么有
N D = p − m p + m D N = p + m p − m
both D and N are odd m = D − N 2 , p = D + N 2 , p e r i o d = D either D and N is even m = D − N , p = D + N , p e r i o d = 2 D both D and N are odd either D and N is even m = 2 D − N , p = 2 D + N , p e r i o d = D m = D − N , p = D + N , p e r i o d = 2 D
假设采样间隔是
T s = 1 F s T s = F s 1
那么这个cyclic pattern的周期是
D F s or 2 D F s F s D or F s 2 D
Limit Cycle，当OSR越高，宽度越窄，高度越高。当增益无穷高时，Limit Cycle影响不大，因为只有准确的rational输入时才能体现，这个非常窄（暂未找到书中的位置）。
Understand SDM 2nd Edtion, page 54
Comparing Figure (a) with Figure (b), we see that as OSR is increased, the widths and absolute heights of the anomalous regions decrease, but relative to the mean noise level the noise spikes are higher. Candy and Benjamin [1] showed that the central noise peaks have a height − 20 log ( 2 ⋅ O S R ) d B − 2 0 log ( 2 ⋅ O S R ) d B O S R − 1 O S R − 1
Candy-1981-The-Structure-of-Quantization-Noise-from-Sigma-Delta-Modulation.pdf
Friedman1988-The-Structure-of-the-Limit-Cycles-in-Sigma-Delta-Modulation.pdf
