Example of Decimation and Interpolation
Sample the signal
x( t) = 5sin(2000t ) cos(20,000t )
at 80 kHz to form a discrete-time signal x[n ] , take every fourth sample of x[n ] to form
xs[n ] and decimate xs[n ] to form xd [n ] . Then upsample xd [n ] by a factor of four to
form xi[ n] and compare it to x[n ] .
( ) (
) (
) (
) (
) (
) (
)
X f = j5 / 2 f + 1000 f 1000 1 / 2 f 10000 + f + 10000 ( ) (
) (
) (
) (
) (
) (
) (
)
X f = j5 / 4 f 9000 + f + 11000 f 11000 f + 9000 or in the form
( ) (
) (
) (
)
X j = j5 / 2 18000 + + 22000 22000 + 18000 .
()
The signal power of x t is the average of its squared magnitude.
0.0005
Px =
(
) (
1
5sin 2000 t cos 20,000 t
0.001 0.0005
0.0005
= 2000 25
(
)
(
)
2
dt
)
sin 2 2000 t cos 2 20,000 t dt
0
0.0005
Px = 50000
0
1 1
1 1
cos 4000 t + cos 40000 t dt
2 2
2 2
(
)
(
)
0.0005
0.0005 1
1
dt cos 4000 t dt
4
0 4
0
=0
Px = 50000 0.0005
0.0005
1
1
+
cos 40000 t dt cos 4000 t cos 4000 t dt 0 4
4
0 =0
=0
(
(
)
)
(
) (
)
Px = 50000 0.0005 / 4 = 6.25
The signal power is also the integral of its power spectral density. The power spectral
density is
( ) (
) (
) (
) (
) (
)
G x f = 25 / 16 f 9000 + f + 11000 f 11000 f + 9000 .
This integral over all frequency is just the sum of the impulse strengths which is 25/4 or
6.25.
The sampled signal is
(
) (
)
(
) (
x n = 5sin 2000 nTs cos 20,000 nTs = 5sin n / 40 cos n / 4
)
and its Fourier transform is
5 1
1 1 1
1 X F = j 1 F + 1 F 1 F + 1 F + 2 80 80 2 8
8 ( )
( ) (t nT ) and the (periodic convolution) operation is defined by
where T t =
n= ()
t0 +T
( ) x ( ) y (t ) d = x ( ) y (t ) d
x t y t =
T
=t0
where t0 is arbitrary and T is any period that is common to both x and y. Also
()
()
() () ()
()
x t y t = x ap t y t = x t y ap t
()
x t
where x ap t = 0
()
, t0 t < t0 + T
, otherwise
()
y t
and y ap t = 0
()
, t0 t < t0 + T
, otherwise
where t0 is arbitrary. Therefore
5 1
1 1
1
X F = j 1 F + 1 F F + F + 4 80 80 8
8 ( )
5 9
11 11 9 X F = j 1 F + 1 F + 1 F 1 F + 4 80 80 80 80 or, in the form
5 18 22 22 18 +
X e j = j 2 +
+
.
2 2 2 2 80 80 80 80 ( )
( )
The signal power is
(
Px = 1 / 80
)
(
79
) (
5sin n / 40 cos n / 4
n=0
(
Px = 25 / 80
79
n=0
) (
2
= 25 / 80
) sin ( n / 40) cos ( n / 4)
79
2
2
n=0
) 12 12 cos 20n 12 12 cos 2n 79
79
79
79
n
n
n
n = 25 / 80 1 / 4 1 + cos + cos + cos cos 20 n=0
2 n=0
20 20 n=0
n=0 =0
=0
=0
(
) (
)
(
) (
)
Px = 25 / 80 1 / 4 80 = 25 / 4
The signal power is also the summation of the power spectral density over one period.
( )
Gx F =
25 9
11 11 9 1 F + 1 F + 1 F 1 F + .
16 80 80 80 80 The summation over one period (1) is 25/4.
Proakis would write these in his notation as
X f = j 5 f 9 k + f + 11 k + f 11 k + f + 9 k 4 k = 80 k = 80 k = 80 k = 80 or
18 22 k
+
+
k
80 k = 80 5 k = X f = j
2 22 18 + k + +
k 80 k = 80 k = ( )
( )
(
) (
x s n = x n 4 n = 5sin n / 40 cos n / 4
) n mN m= 5 9
11 11 9 1
X s F = j 1 F + 1 F + 1 F 1 F + 1/ 4 F
4 80 80 80 80 4
( )
( )
9
11 1
1 F + 1 F + F + F 80 80 4
5 Xs F = j
16 11 9
1
3
1 F 1 F + + F + F 80 80 2
4 ( )
( )
where the period T common to both signals is 1.
9
1 9
1 9
3 9 1 F + 1 F + 1 F + 1 F 80 4 80 2 80 4 80 11 1 11 1 11 3 11 +1 F + + 1 F + + 1 F + + 1 F + 80 4 80 2 80 4 80 5 Xs F = j 16 11 1 11 1 11 3 11 F 1 F 1 F 1 F 1 80 4 80 2 80 4 80 9
1 9
1 9
3 9 1 F + 80 1 F 4 + 80 1 F 2 + 80 1 F 4 + 80 ( )
9
29 49 69 1 F + 1 F + 1 F + 1 F 80 80 80 80 11 9
29 49 +1 F + + 1 F + 1 F + 1 F 80 80 80 80 5 Xs F = j 16 11 31 51 71 F 1 F 1 F 1 F 1 80 80 80 80 9
11 31 51 1 F + 80 1 F 80 1 F 80 1 F 80 ( )
Combining equivalent periodic impulses (using the periodicity of the periodic impulse),
9
29 49 69 1 F + 1 F + 1 F + 1 F 80 80 80 80 5 Xs F = j 8
11 31 51 71 1 F 1 F 1 F 1 F 80 80 80 80 ( )
(
)
The signal power is Px = 25 / 64 8 = 25 / 8 .
s
18 58 98 138 + 2 + 2 + 2 2 80 80 80 80 5 = j
4 22 62 102 142 2 2 2 2 80 80 80 80 ( )
X s e j
x d n = x s 4n Xd
F 9
F 29 F 49 F 69 1 + 1 + 1 + 1 4 80 4 80 4 80 5 4 80 F = Xs F / 4 = j 8
F 11 F 31 F 51 F 71 1 1 1 1 4 80 4 80 4 80 4 80 ( )
(
Xd
)
36 116 196 276 + 4 F + 4 F 4 F + 4 F 80 80 80 80 5 F = j
2
44 124 204 284 4 F 4 F F
F
4
4
80 80 80 80 ( )
5 36 44 X d F = j 1 F 1 F 2 80 80 ( )
(
)
The signal power is Px = 25 / 4 2 = 25 / 2 .
d
or
72 88 X d e j = j5 2 F
2 80 80 ( )
To recover the original x n from x d n , replace the zeros to form x s n and then
lowpass filter x s n .
9
29 49 69 1 F + 1 F + 1 F + 1 F 80 80 80 80 5 Xs F = j 8
11 31 51 71 1 F 1 F 1 F 1 F 80 80 80 80 ( )
If we lowpass filter this with lowpass filter with a transfer function
1 , 0 < F < Fc
H F =
0 , otherwise
( )
where 11 / 80 < Fc < 69 / 80 , we get
5 9
69 11 71 X i F = X s F H F = j 1 F + 1 F 1 F 1 F 8 80 80 80 80 ( )
( ) ( )
or
5 9
9
11 11 X i F = j 1 F 1 F + + 1 F + 1 F 8 80 80 80 80 ( )
The inverse Fourier transform of this is
(
)
(
)
(
)
x i n = 5 / 2 sin 11 n / 40 sin 9 n / 40 .
(
) (
)
The original discrete-time signal is x n = 5sin n / 40 cos n / 4 . Using a
trigonometric identityt this can be written as
(
)
(
)
x n = 5 sin 11 n / 40 sin 9 n / 40 .
The two signals are the same except for a factor of 2. The signal power of x i n is 25/16
which is 1/4 of the signal power of x n . This loss of signal power can be compensated
for by using a lowpass filter with a gain of 2 instead of a gain of one.
|X(e jΩ)|
x[n]
5π
2
5
96
n
-5
-2π
xs[n]
2π
Ω
|Xs (e jΩ)|
5π
4
5
96
n
-5
-2π
xd[n]
|Xd(e jΩ)|
2π
Ω
2π
Ω
5π
5
n
96
-5
-2π