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C. A. Bouman: Digital Image Processing - January 8, 2018 1
1-D Rate Conversion
• Decimation
– Reduce the sampling rate of a discrete-time signal.
– Low sampling rate reduces storage and computation
requirements.
• Interpolation
– Increase the sampling rate of a discrete-time signal.
– Higher sampling rate preserves fidelity.
C. A. Bouman: Digital Image Processing - January 8, 2018 2
1-D Periodic Subsampling
• Time domain subsampling of x(n) with period D
y(n) = x(Dn)
Dx(n) y(n)
• Frequency domain representation
Y (ejω) =1
D
D−1∑
k=0
X(
ej(ω−2πk)/D)
• Problem: Frequencies above π/D will alias.
• Example when D = 2
aliasedsignal
aliasedsignal
ωπ−π −π/2 π/2
• Solution: Remove frequencies above π/D.
C. A. Bouman: Digital Image Processing - January 8, 2018 3
Decimation System
ωj D y(n)x(n) H(e )
• Apply the filter H(ejω) to remove high frequencies
– For |ω| < π
H(ejω) = rect
(
Dω
2π
)
– For all ω
H(ejω) =
∞∑
k=−∞
rect
(
Dω − k2π
2π
)
= prect2π/D(ω)
– Impulse response
h(n) =1
Dsinc(n/D)
• Frequency domain representation
Y (ejω) =1
D
D−1∑
k=0
H(
ej(ω−2πk)/D)
X(
ej(ω−2πk)/D)
C. A. Bouman: Digital Image Processing - January 8, 2018 4
Graphical View of Decimation for D = 2
• Spectral content of signal
X(
ejω)
−π −π/2 3π/2 2πππ/2 5π/2ω
1
• Spectral content of filtered signal
H(
ejω)
X(
ejω)
−π −π/2 3π/2 2πππ/2 5π/2ω
1
• Spectral content of decimated signal
Y (ejω) =1
D
D−1∑
k=0
H(
ej(ω−2πk)/D)
X(
ej(ω−2πk)/D)
π 2π 4πω
1/D
−2π −π 3π 5π
C. A. Bouman: Digital Image Processing - January 8, 2018 5
Decimation for Images
• Extension to decimation of images is direct
• Apply 2-D Filter
f (i, j) = h(i, j) ∗ x(i, j)
• Subsample result
y(i, j) = f (Di,Dj)
• Ideal choice of filter is
h(m,n) =1
D2sinc(m/D)sinc(n/D)
• Problems:
– Filter has infinite extent.
– Filter is not strictly positive.
C. A. Bouman: Digital Image Processing - January 8, 2018 6
Alternative Filters for Image Decimation
• Direct subsampling
h(m,n) = δ(m,n)
– Advantages/Disadvantages:
∗ Low computation
∗ Excessive aliasing
• Block averaging
h(m,n) = δ(m,n) + δ(m + 1, n)
+δ(m,n + 1) + δ(m + 1, n + 1)
– Advantages/Disadvantages:
∗ Low computation
∗ Some aliasing
• Sinc function
h(m,n) =1
D2sinc(m/D, n/D)
– Advantages/Disadvantages:
∗ Optimal, if signal is band limited...
∗ High computation
C. A. Bouman: Digital Image Processing - January 8, 2018 7
Decimation Filters
05
1015
2025
3035
0
10
20
30
400
0.2
0.4
0.6
0.8
1
2−D Filter Impulse Response used for Block Averaging
Block averaging filter
05
1015
2025
3035
0
10
20
30
40−0.01
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
2−D Sinc Filter used for Ideal Subsampling
Sinc filter
C. A. Bouman: Digital Image Processing - January 8, 2018 8
Original Image
• Full resolution
C. A. Bouman: Digital Image Processing - January 8, 2018 9
Image Decimation by 4 using Subsampling
• Severe aliasing
C. A. Bouman: Digital Image Processing - January 8, 2018 10
Image Decimation by 8 using Subsampling
• More severe aliasing
C. A. Bouman: Digital Image Processing - January 8, 2018 11
Image Decimation by 4 using Block Averaging
• Sharp, but with some aliasing
C. A. Bouman: Digital Image Processing - January 8, 2018 12
Image Decimation by 4 using Sinc Filter
• Theoretically optimal, but not necessarily the best visual
quality
C. A. Bouman: Digital Image Processing - January 8, 2018 13
1-D Up-Sampling
• Up-sampling by L
y(n) =
{
x(n/L) if n = KL for some K0 otherwise
or the alternative form
y(n) =
∞∑
m=−∞
x(m)δ(n−mL)
• Example for L = 3
0 5 10 15 20−1
0
1Discrete time signal x(n)
0 10 20 30 40 50 60−1
0
1x(n) up−sampled by 3
C. A. Bouman: Digital Image Processing - January 8, 2018 14
Up-Sampling in the Frequency Domain
• Up-sampling by L
y(n) =
∞∑
m=−∞
x(m)δ(n−mL)
• In the frequency domain
Y (ejω) = X(ejωL)
• Example for L = 3
−4 −2 0 2 40
5
10
15DTFT of x(n)
−4 −2 0 2 40
5
10
15DTFT of y(n)
C. A. Bouman: Digital Image Processing - January 8, 2018 15
Interpolation in the Frequency Domain
ωjH(e ) y(n)x(n) L
• Interpolating filter has the form
Y (ejω) = H(ejω)X(ejωL)
• Example for L = 3
−4 −2 0 2 40
5
10
15DTFT of y(n)
−4 −2 0 2 40
20
40Interpolated Signal in Frequency
C. A. Bouman: Digital Image Processing - January 8, 2018 16
Interpolating Filter
ωjH(e ) y(n)x(n) L
• In the frequency domain
H(ejω) = Lprect2π/L(ω)
• In the time domain
h(n) = sinc(n
L
)
0 10 20 30 40 50 60−0.5
0
0.5
1Interpolating filter in Time
−4 −2 0 2 40
1
2
3Interpolating filter in Frequency
C. A. Bouman: Digital Image Processing - January 8, 2018 17
Interpolation in the Time Domain
ωjH(e ) y(n)x(n) L
• In the frequency domain
Y (ejω) = X(ejωL)
• Example for L = 3
0 5 10 15 20−1
0
1Discrete time signal x(n)
0 10 20 30 40 50 60−1
0
1Interpolated Signal in Time
C. A. Bouman: Digital Image Processing - January 8, 2018 18
2-D Interpolation
H(e ,e ) y(n)jυ jµx(n) (L,L)
• Up-sampling
y(m,n) =
∞∑
k=−∞
∞∑
l=−∞
x(k, l)δ(m− kL, n− lL)
• In the frequency domain
H(ejµ, ejν) = prect2π/L(µ)prect2π/L(ν)
h(m,n) = sinc (m/L, n/L)
• Problems: sinc function impulse response
– Infinite support; infinite computation
– Negative sidelobes; ringing artifacts at edges
C. A. Bouman: Digital Image Processing - January 8, 2018 19
Pixel Replication
H(e ,e ) y(n)jυ jµx(n) (L,L)
• Impulse response of filter
h(m,n) =
{
1 for 0 ≤ m ≤ L− 1 and 0 ≤ n ≤ L− 10 otherwise
• Replicates each pixel L2 times
C. A. Bouman: Digital Image Processing - January 8, 2018 20
Bilinear Interpolation
H(e ,e ) y(n)jυ jµx(n) (L,L)
• Impulse response of filter
h(m,n) = Λ(m/L)Λ(n/L)
• Results in linear interpolation of intermediate pixels