微电子学院
The theory and application of digital signal processing system 本科(大三)
Lecture 6 DSP system: Sampling II
Yang Li yangli@https://www.doczj.com/doc/887390703.html, li@i j d School of Microeletronics Shanghai ao o g U e s y S a g a Jiao Tong Univerisity Spring 2010
2010.春
1
Outline
Downsampling Upsampling (with zero insertion) Upsampling vs Direct upsampling Sampling id titi S li identities Filter implementation with down/up p p sampling Fractional sampling
2
Sampling Rate Conversion
In I applications sampling rate can be changed li ti li t b h d by an integer factor.
Downsampling (also called undersampling or decimation): Reducing the sampling rate by an integer factor. -- Requires fewer registers, i.e., reduces p q computation requirement. Upsampling (also called oversampling or interpolation): Increasing the sampling rate by an integer factor (zero-valued samples are inserted.) -(zero valued oversampling improves ADC resolution by interpolation.
Sampling rate also can be changed by a noninteger number (by cascading up/down sampling.)
3
Downsampling & Upsampling
Normal Sampling
xT [n ]
0 1 2 3 4 5 6 7 8 9 10
n
Downsampling
↓ x 2 [n ]
T ' = MT
(M ≥ 1) Upsampling U li T'= T L (L ≥ 1)
0
n 0 1
↑ x 2 [n ]
2
3
4
5
2 4 6 8 10 12 14 16 18 20
n
4
Downsampling
5
Downsampling Notation
xT [n ]
n
↓ xT ' [n ]
0 1 2 3 4 5
6 7
8 9 10
0
1
2
3
4
5
n
Sampling period = T
Sampling period = MT
M ≥ 1 is an integer
xT [n ]
Sampling period T
xT [n ] = xc (nT )
M
↓ xM [n ] = xMT [n ]
Sampling period T = MT T’
↓ xM [n ] = xT [nM ] = xMT [n ] = xc (nMT )
↓ ? xM [n ] is the sequence sampled from xc(t) with the sampling period T’ = MT MT.
6
Downsampling without Aliasing
xT [n ]
Sampling period T
M
↓ xM [n ] = xMT [n ]
Sampling period T’ = MT
↓ xM [n ] = xT [nM ] = xMT [n ] = xc (nMT )
Suppose X c ( jω ) = 0 for ω ≥ ωN
|H|
If
2π 2π = ≥ ωN = 2ωB T ' MT
ωB
ωN
2π MT
ω
then downsampling by a factor of M will not cause aliasing.
7
Downsampling in Freq Domain
Recall the sampling identity R ll th li id tit (with sampling period = T)
1 XT (e j Ω ) = T
+∞
? ? Ω 2π k ? ? Xc ? j ? ? ? T ?? ? ?T ? k =?∞
∑
+∞
Now for sampling period = MT, we have
Ω = Tω
Ω ' = MT ω = M Ω
n = kM + m
X MT (e
jΩ '
1 )= MT
? ? Ω ' 2π n ? ? ? Xc ? j ? ?? ? ? MT MT ? ? n =?∞
∑
1 M ?1 ? 1 +∞ ? ? Ω ' 2π ( m + kM ) ? ? ? X j? = ? ? ? ?? M m =0 ?T k = ∞ c ? ? MT MT ? ?? ? =?∞
∑
∑
1 M ?1 ? 1 = ? M m =0 ?T
∑
? ? Ω '? 2 π m 2 π k ? ? ? Xc ? j ? ? ? ? T ? ?? ? MT ? ? n =?∞
∑
+∞
↓ xM [n ] = xT [nM ] = xMT [n ] = xc (nMT )
X MT e j Ω '
(
)
1 M ?1 = X M m =0 T
∑
? j ( Ω '?2π m ) ? M ?e ? ? ? ? ?
8
Downsampling Identity
xT [n ]
Sampling period T
M
↓ xM [n ] = xMT [n ]
Sampling period T = MT T’
The Downsampling Identity:
' '
Ω ' = MT ω = M Ω
M ?1
? j ( Ω ? m 2π ) ? 1 ↓ jΩ jΩ X M (e ) = X MT (e ) = ∑0 X T ? e M ? ? ? M m= ? ?
Relation between the sequence sampled with period T and the sequence downsampled with period MT in the frequency domain.
9
Illustration of Decimation
1
X c ( jω )
X MT e j Ω '
ω
(
)
?ω B
ωB
1 M ?1 = X M m =0 T
∑
? j ( Ω '?2π m ) ? M ?e ? ? ? ? ?
Look at this exponent carefully!
1 T
XT ( e j Ω )
Let Ω =
π
2π
Ω '? 2π m M
e
j ( Ω '?2π m ) M
for any m fixed
?2 π
?π
ΩB
Ω = ωT
= e jΩ
X MT e
(
jΩ '
)
1 M ?1 = XT e j Ω M m =0
∑
(
)
Ω ' = M Ω + 2π m
period M2π period 2π
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X MT e j Ω '
1
X c ( jω )
(
)
? j ( Ω '?2π m ) ? 1 = ∑ XT ? e M ? ? ? M m =0 ? ?
M ?1
To avoid aliasing we must have
?ω B
1 T
ωB
XT ( e j Ω )
ω
ωs ≥ MωN = 2MωB
T ωs ≥ MT ωN = 2MT ωB 2π ≥ M Ω N = 2M Ω B
?2 π
?π
ΩB
π
2π
Ω = ωT
Ω ' = M Ω + 2π m
p period M2π period 2π
1 T
XT ( e j Ω )
T ' = MT
2π
?2 π
?π
π M ΩB
M ΩB ≤ π
M2π
Ω ' = ωT ' = M Ω
for m = 0
Add M shifted copies then divide by M
1 T
XT ( e j Ω )
Add M shifted copies then divided by M
for m = 0
π M ΩB
2π
M2π
XT ( e
jΩ
?2π
?π
1 T
Ω ' = ωT ' = M Ω
)
for m = 1
π M ΩB
2π M2π
?2π
?π
Ω ' = ωT ' = M Ω
1 T
XT ( e j Ω )
for m = M-1
M 1 2π?1 ∑ XT e j Ω M m =0
?2 π
?π
π M ΩB
(
)
X MT e
M2π
Ω ' = ωT ' = M Ω
1 MT
X MT (e
jΩ '
(
jΩ '
)
)
T ' = MT
2π
1 M ?1 X = M m =0 T
∑
? j ( Ω '?2π m ) ? M ?e ? ? ? ? ?
?2π
?π
π M ΩB
Ω ' = ωT ' = M Ω
Spectrum after decimation by M
Downsampling Spectrum
1 T
XT ( e j Ω )
Reference spectrum
?2 π
?π
ΩB
π
2π
Ω = ωT
do ownsampling
1 MT
Downsampled spectrum
X MT (e j Ω ' )
T ' = MT
Ω ' = ωT ' = M Ω
π M ΩB
2π
?2 π
?π
Ω ' = MΩ
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D.S. D S Frequency axis mapping
1 T
XT ( e j Ω )
Reference spectrum
ΩB
π
2π
downs sampling axis c change
?2π
?π
1 MT
Ω = ωT
X MT (e j Ω ' )
M=2
Downsampled spectrum
T ' = MT
?2 π
?π
1 T
π M ΩB
XT ( e j Ω )
2π
Ω ' = MΩ
Ω ' = ωT ' = M Ω Ω = ωT
π
2π
?2π
?π
ΩB
Ω = ωT
ωs
2
ωs
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Downsampling might cause aliasing D li i ht li i
Normal sampling Sampling rate too low
0 2π fB
1 fs = > 2fB T
2π T
ω
fs ' =
0
aliasing
1 < 2fB MT
2π MT
ω
15
Prefiltering
In practice the original signal must be bandlimited to fs’/2 before downsampling.
low-pass filter
signal downsampled
M
fs ' 2
16
Where to place the Downsampler? Wh t l th D l ?
X(z)
M
U(z)
H(z)
Y(z)
Downsampling before filtering
Y (e j Ω ) = H (e j Ω )U (e j Ω )
X(z)
H(zM)
V(z) ( )
M
Downsampling after filtering Y(z)
V (e j Ω ) = H (e j Ω M ) X (e j Ω )
The two downsampling schemes are equivalent !
17
Equivalent Downsampler Placement
X(z)
M
U(z)
H(z)
Y(z)
U (e
jΩ
j ( Ω ?2 π m ) ? 1 M ?1 ? )= X ?e M ? ? M m =0 ? ? ?
∑
X(z)
H(zM)
V(z)
M
D.S. j ( Ω ?2 π m ) ? 1 M ?1 ? Y (e j Ω ) = V ?e M ? ? ? M m =0 ? Y(z) ?
∑
V (e j Ω ) = H (e j Ω M ) X (e j Ω )
Use the downsampling Identity:
? j ( Ω ?2 π m ) ? 1 M ?1 H e j ( Ω ?2 π m ) X ? e M = ? ? ? M m =0 ? ?
∑ (
)
Y e
( )
jΩ
j ( Ω ?2 π m ) ? 1 M ?1 ? V ?e M = ? ? ? M m =0 ? ?
j ( Ω ?2 π m ) ? 1 M ?1 ? = H (e j Ω ) X ?e M ? ? M m =0 ? ? ?
∑
∑
= H (e j Ω )U (e j Ω )
D.S.
18
Filter Implementation with D S D.S.
Filter can be restructured by moving the downsampler from output to input.
x[k]
w0 w1 w2 w3 w4 w5
+
+
+
+
+
y’[k]
2
y[k]
= (w 0 x [k ] + w 2 x [k ? 2] + w 4 x [k ? 4] ) + (w1x (k ? 1) + w 3 x (k ? 3) + w 5 x (k ? 5) )
y '[k ] = w 0 x [k ] + w1x [k ? 1] + w 2 x [k ? 2] + w 3 x [k ? 3] + w 4 x [k ? 4] + w 5 x [k ? 5]
19
Filter Implementation with D S D.S.
Filter reimplemented in parallel:
x[k]
w0 w2 w4
+
+
y1’[k]
2
+
w1 w3 w5
y[k]
+
+
y2’[k]
2
y1 '(k ) = w 0 x (k ) + w 2 x (k ? 2) + w 4 x ( k ? 4)
y 2 '(k ) = w1x (k ? 1) + w 3 x (k ? 3) + w 5 x (k ? 5)
20