Chapter 1 : A Calibration Technique for Phase Noise Canceling Fractional-N Phase-
B. Calibration Loop Performance
VI. Conclusions
In practice, component mismatches could result in errors in the calibration technique. With mismatches in the component values of approximately 5%, the phase noise at the output of the PLL remains well behaved as shown in Figure 11, demon-strating the robustness of the calibration technique. It can be shown that component mismatches simply result in a replica of the CP and DAC charges appearing at the in-tegrator input. Since these charges have zero dc content due to the PLL loop, this re-sults in noise on the calibration signal, not misalignment. This demonstrates the vi-ability of this calibration technique for phase-noise canceling delta-sigma fractional-N PLL.
A
PPENDIXA
Consider the continuous-time calibration loop model shown in Figure 8b. The impulse response, g(t), from the input of the filter to w(t) can be derived as
{
/ 1}
( ) m ( ) 1 t RC
int DAC
g t g u t e
C I
= ⋅ − − . (19)
The worst-case variation of (19) during a DAC pulse event occurs when the DAC gain is zero. Using (6) and (19), the deviation of w(t) over a duration of TDAC during the nth DAC and CP event can be expressed as
{
( ) / 1 / 1}
0
( ) ( DAC) m n CP[ ] kT TDAC RC kT RC
int DAC k
w nT w nT T g Q n e e
C I
− + −
=
− + =
∑
− (20)Recalling from (3) that |QCP[n]| ≤ ICPTVCO, the right side of (20) can be bounded by
{
1}
1{
1}
1 11
1
( 1) /
/ / /
/ 0
/ /
1 1 1
1 1
1
DAC DAC
DAC
n T RC
T RC n kT RC T RC
CP VCO m CP VCO m
T RC
int DAC k int DAC
T RC
CP VCO m
T RC int DAC
I T g I T g e
e e e
C I C I e
I T g e
C I e
− − − − +
= −
−
−
− = − −
−
< −
−
∑
(21)In order to ensure that w(t) does not change over the duration of a DAC pulse, it is necessary to keep (21) small. This leads to
1
1
/ /
1 1
TDAC RC CP VCO m
T RC int DAC
I T g e
C I e
−
−
− < ∆
− (22)
where ∆ is the specified level of gain calibration required for w(t).
A
PPENDIXB
This appendix provides the claims necessary to prove the convergence of the calibration loop. Consider the signal-processing model shown in Figure 8c. With c[n]
given by (5), Claim 1 proves that the charge supplied by the CP and DAC are ergodic, i.e. C0 exists, and Claim 2 provides an intermediate claim to prove the stability of the calibration loop.
Definition: Equant[n] is an integer-valued sequence representing the quantizer noise from the ∆Σ modulator such that
[ ] [ ] 2
quant
quant N
E n
e n ,
where 2N is the quantizer step size of the ∆Σ modulator. This definition is provided to ensure congruency with the theorems outlined in [14].
Claim 1: Suppose that the correlation signals given by (5) are generated in conjunc-tion with a 2nd order ∆Σ modulator designed to satisfy the conditions of Theorem 3 in [14] with a quantizer step size of 2N. Then C0 is given by
2 0
2 1
2
N
CP VCO N
C =I ⋅T ⋅ − − .
Proof: Consider the case where c[n]=sgn(equant[n−1])= sgn(Equant[n−1]). Then from (3),
( )
( )
0 [ ] [ ] [ ] [ 1] sgn [ 1] [ 2]
2
CP VCO
CP CP N quant quant quant
C +C n =Q n c n⋅ = I T ⋅ E n− − E n− ⋅E n− .(23)
From (14), C0 is defined as the sample average of (23). First consider the sam-ple average of |Equant[n]|. Theorem 1 in [14] proves that Equant[n] asymptotically ap-proaches a uniform random distribution as n→∞ given by
1 1
( ) 2 , N 2N 1 2N
P uU = − − − + ≤ ≤u − . (24)
And therefore,
lim E quant[ ] 2N 2
n E n E u −
→∞ = = . (25)
For any postive number m, define
[ ] 2N 2
n quant
X = E n m+ − − , (26)
and Lemma 3 from [14] shows that E[Xn]→0 is sufficient to prove that the sample av-erage of |Equant[n]| converges in probability to 2N-2, or in other words
1 2
1 2
lim [ ]
2
n m N
quant N
n k m
e k
n
+ − −
→∞ =
∑
= . (27)Now consider the second term in (23). Theorem 2 in [14] also proves that the joint pmf of any two samples of the quantizer error converges in distribution to a jointly uniform random variable given by
2 1 1
, ( , ) 2 N, 2N 1 , 2N
PU V u v = − − − + ≤u v≤ − , (28)
and therefore
( ) [ ]
lim E sgn quant[ ] quant[ 1] E sgn( ) 1
n E n E n u v
→∞ ⋅ − = ⋅ = . (29)
For any positive number m, let
( )
sgn [ ] [ 1] 1
n quant quant
X = E n m E+ n m+ − − , (30)
and Lemma 3 from [14] shows that E[Xn]→0 is sufficient to prove that the sample av-erage of sgn(Equant[n])⋅Equant[n−1] converges in probability to 1. Therefore due to the linearity of the sample mean operator,
( )
{ }
21 1 2 1
lim [ ] sgn [ ] [ 1]
2
n m N
quant quant quant N
n k m
e k e k e k
n
+ − −
→∞ =
− ⋅ − = −
∑
, (31)which proves Claim 1 for c[n]=sgn(equant[n]), and the convergence is irrespective of the initial start index, m. Now consider the second correlation signal given by
( ) ( )
sgn [ ] sgn [ 1]
[ ] [ ]
2
quant quant
e n e n
c n − − s n
= + ,
where s[n] is by definition a zero-mean ergodic sequence. For this correlation signal, (23) can be expressed as
( )
{
( ) ( ) }
0
[ ] 1 [ ] [ 1] sgn [ ] [ 1]
2 2
sgn [ 1] [ ] 2 [ ] [ ] [ 1]
CP VCO
CP N quant quant quant quant
quant quant quant quant
I T
C C n E n E n E n E n
E n E n s n E n E n
+ = ⋅ ⋅ + − − ⋅ −
− − ⋅ + ⋅ ⋅ − −
. (32)
The first 4 terms have already been shown to converge to (31), and since s[n] is a zero-mean ergodic sequence uncorrelated from the quantization error, C0 is given by
2 0
2 1
2
N
CP VCO N
C =I ⋅T ⋅ − −
Therefore Claim 1 is proven for both correlation signals.
For the following stability claim, consider a system of non-linear difference equations defined by
1 2
[ ] [ ] [ ] [ ]
x n =u n −w n u n⋅ (33)
[ ] [ ] [ ]
w n =x n h n∗ (34)
where u1[n] and u2[n] are bounded inputs, and h[n] is a filter with the following char-acteristics
( 1)
[ ] 0 for 0, [1] 0, and [ ] [ 1] n
h n = n≤ h > h n −h n− =Ke− − α (35)
where K and α are positive constants. Without loss of generality, the system is as-sumed to start at n = 0 such that u1[n] = u2[n] = 0 for n < 0.
Claim 2: Suppose that
[1] 2[ ] 1
h ⋅u n < (36)
Then the system of equations defined by (33) and (34) are BIBO stable if for any posi-tive integer m,
1 2
lim 1 N m [ ] 0
N n m
u n U N
+ −
→∞ =
= >
∑
(37)Proof: First, consider the product given by
(
2)
1
1 [1] [ ]
n
l k
h u l
= +
∏
− (38)It follows from (36) and the fact that |1 − x| ≤ e−x that
(
2)
[1] 12[ ]1
1 [1] [ ]
n
l k
n h u l
l k
h u l e− = +
= +
− ≤ ∑
∏
(39)(37) implies that for any index m, and positive number ε, there exists an Nε,m such that for all N > Nε,m,
( )
1 2[ ]
N m
n m
u n U Nε
+ −
=
− <
∑
(40)For all m and ε, Nε,m is finite. Therefore, for some ε < U, define
{ }
, 0' max m
N N ε m∞
= = , (41)
and for n − k > N’, the right side of (39) is bounded by
( ) 1( 2 ) ( )( )
[1] [ ]
[1] [1]
n
l k
h u l U
h n k U h n k U
e e = + e ε
− −
− − ∑ − − −
≤ (42)
Therefore, there exists positive numbers D, and β such that for all n and k,
(
2)
( )1
1 [1] [ ]
n n k
l k
h u l De− − β
= +
− ≤
∏
, (43)and therefore (38) is bounded by an exponentially decreasing function. The non-linear difference equations given by (33), (34) and (35) can be expressed as
( )
( ) ( )
( )
1 2
0
2
1 1 2
0 2
2 0
[ ] [ ] [ ] [ ] [ ]
[1] [ 1] [ ] [ 1 ] [ ] [ 1] 1 [1] [ 1]
[ ] [ 1 ] [ ] [ ].
n
k
n
k n
k
w n h n k u k w k u k
h u n h n k h n k u k w n h u n
h n k h n k w k u k
=
−
=
−
=
= − −
= − + − − − − + − − −
− − − − −
∑
∑
∑
(44)
Recursively substituting (44) into itself to eliminate w[k], k = 0, …, n − 1 yields
( )
( ) ( ) ( )
1 2
1 1
0 0
1
2 2 2
2 1
[ ] [1] [ 1] [ ] [ 1] [ ]
1 [1] [ ] [ ] [ ] [ 1] 1 [1] [ ] .
n n k
k l
k k k
m
m o m
w n h u n k h n l h n l u l
h u n m u n k h m h m h u n k o
− − −
= =
−
=
= =
= − − + − − − −
⋅ − − − − − − ⋅ − − +
∑ ∑
∏ ∑ ∏
(45)The above equation consists of four separate terms. For a bounded input, there exists a B such that u1[n] and u2[n] are bounded by B in magnitude. From (43), the first term can be bounded by
( )
1 1
1 2
0 1 0
[1] [ 1] 1 [1] [ ] [1]
1 [1]
[1] 1 1
k
n n
k
k m k
n
h u n k h u n m h B De
e h B D h B D
e e
β
β
β β
− −
−
= = =
−
− −
− − − − <
− ⋅
= ⋅ <
− −
∑ ∏ ∑
(46)
Using (35) the second term can be expressed and bounded by
( ) ( )
( )
1 2
1 2
0 0 1
1 2 1
1 ( 1)
0 0 0 0
[ ] [ 1] [ ] k 1 [1] [ ]
n n k
k l m
n n k n n k
n l k k k l
k l k l
h n l h n l u l h u n m
B Ke α De β KBD e β e α e α
− − −
= = =
− − − − −
− − − − − − + −
= = = =
− − − − − −
≤ ⋅ = ⋅
∑ ∑ ∏
∑ ∑ ∑ ∑
. (47)
The expression on the right side of (47) can be further reduced resulting in
( )( )
( 1) ( )
1 ( 1)
( )
0
( )
1 1
1 1 1
1 1
n k n
n k k
k
e e e
B De Ke KBD
e e e
KBDe
e e
α α α β
β α
α α α β
α
α α β
− − + − − +
− − − +
− − − +
=
−
− − +
− −
⋅ ≤
− − −
≤ − −
∑
(48)
The third term from (45) can be expressed and bounded by
( )
1( )
1
1 2 2
0 2
2 1 ( 1) ( )
0 2
[1] [ 1] [ ] [ ] [ 1] 1 [1] [ ]
[1]
k
n k
k m o m
n k
m k m
k m
h u n k u n k h m h m h u n k o
h B Ke α De β
− −
= = =
− − − − −
= =
− − ⋅ − − − ⋅ − − +
≤ ⋅
∑ ∑ ∏
∑∑
(49)
The expression on the right side of (49) can be bounded by
( 1)( )
1 1
2 ( 1) 2 2( )
( )
0 2 0
2 2( ) ( )
( )
( )
2 2( )
( )
[1] [1] 1
1
1 1 1
[1] 1 1 1
1 1
[1] 1 1 1
n k n k
k m m k
k m k
n n
h B KD e e e h B KDe e e e
e
e e
h B KDe e e
e e e
h B KDe e e
e e e
β α β α α β β α β
α β
β α
α α β α β
α β β α
α α β α β
α β β α
− − −
− −
− − − − − −
= = = − −
− −
− − −
− − − −
− − −
− − − −
⋅ = −
−
− −
= − − − −
≤ −
− − −
∑ ∑ ∑
(50)
Finally, the fourth term in (45) can be expressed and bounded by
( )
( ) ( )
1 2
1 2
0 0
1
2 2
1 2
2 ( 1) ( 1) ( )
0 0 2
1 2
2 ( 1) ( )
0 0 2
[ ] [ 1] [ ] [ ]
[ ] [ 1] 1 [1] [ ]
n n k
k l
k k
m o m
n n k k
n l m k m
k l m
n n k k
k k l m
k l m
h n l h n l u l u n k
h m h m h u n k o
B e Ke De
KDB e e e e e
α α β
α β α α β α
− − −
= =
−
= =
− − −
− − − − − − −
= = =
− − −
− − + − −
= = =
− − − − ⋅ −
⋅ − − ⋅ − − +
≤ ⋅
= ⋅
∑ ∑
∑ ∏
∑ ∑ ∑
∑ ∑ ∑
(51)
Solving the right side of (51) yields
( ) ( )
( )( ) ( )
( )( )
( 1) 2( ) ( 1)( )
2 1 ( 1)
( )
0
2 2( ) 1
( ) ( 1)( )
( )
0
2 2( ) 1 1
( ) ( )
( )
0 0
1 1
1 1
1 1 1
1 1
n k k
n k k
k
n k k
k
n n
k k
k k
e e e
KDB e e e
e e
KDB e
e e
e e
KDB e
e e e
e e
α β α β α
α β α
α β α
β α β α β α
α β α
β α β α β α β α β α
α β α
− − − − − − −
− − − +
− − −
=
− −
− + − − −
− − −
=
− − −
− + − − + + −
− − −
= =
− −
⋅ − −
≤ ⋅ −
− −
⋅
= − − −
∑
∑
∑ ∑
( )( )
( )( )( )( )
2 2( ) ( ) 2
( ) 2
( )
2 2( )
( ) ( ) 2
1 1
1 1
1 1
1 1 1 1
n n
KDB e e e
e e
e e
KDB e
e e e e
β α β α β
β α β
α β α
β α
α β α β α β
− − + −
− + −
− − −
−
− − − − + −
⋅ − −
= − − − −
≤ ⋅
− − − −
(52)
which proves the BIBO stability of (33) and (34).
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