Lecture 4: Frequency domain
representation, DTFT, IDTFT, DFT, IDFT
1
Instructor:
Dr. Gleb V. Tcheslavski
Contact: gleb@ee.lamar.edu
Office Hours: Room 2030
Class web site:
http://ee.lamar.edu/gleb/dsp/ind
ex.htm
ELEN 5346/4304 DSP and Filter Design
Fall 2008
2
Some history
Jean Baptiste Joseph Fourier was born
in France in 1768. He attended the Ecole Royale
Militaire and in 1790 became a teacher there.
Fourier continued his studies at the Ecole
Normale in Paris, having as his teachers
Lagrange, Laplace, and Monge. Later on, he,
together with Monge and Malus, joined Napoleon
as scientific advisors to his expedition to Egypt
where Fourier established the Cairo Institute.
In 1822 Fourier has published his most
famous work: The Analytical Theory of Heat.
Fourier showed how the conduction of heat in
solid bodies may be analyzed in terms of infinite
mathematical series now called by his name, the
Fourier series.
ELEN 5346/4304 DSP and Filter Design
Fall 2008
3
Frequency domain representation
Let xn = A cos(ω0 n + ϕ ) − a (co)sinusoidal input
(4.3.1)
frequency
A
A
A
A
xn = e j (ω0n+ϕ) + e− j (ω0n+ϕ) = e jϕe jω0n + e− jϕe− jω0n
2
2
2
2
(4.3.2)
complex exponent of ω0
complex exponent of -ω0
A sinusoidal signal is represented by TWO complex exponents of
opposite frequencies in the frequency domain.
ELEN 5346/4304 DSP and Filter Design
Fall 2008
4
Frequency domain representation (cont)
⎛
⎞
Let xn = e jω0 n ⇒ yn = ∑ hk e jω0 ( n − k ) = ⎜ ∑ hk e− jω0 k ⎟ e jω0 n H e jω0 xn
k
⎝ k
⎠
jω
(4.4.1)
The frequency response : H ( e jω ) = ∑ hk e− jωk = H ( e jω ) e
k
(4.4.2)
(
)
( )
j H e
phase
magnitude
Iff it exists!
*
⎛
⎞
For a real system : H ( e jω ) = ∑ hn e − jωn = ∑ hn*e − jω n = ⎜ ∑ hn e − jωn ⎟ = H * ( e − jω )
n
n
⎝ n
⎠
⇒ H ( e jω ) e
( )
j H e jω
= H ( e− jω ) e
⎧ H ( e jω ) = H ( e − jω )
⎪
⇒ ⎨
jω
− jω
⎪⎩ H ( e ) = − H ( e )
H ( e jω ) has a period of 2π
ELEN 5346/4304 DSP and Filter Design
Fall 2008
(
− j H e− jω
(4.4.3)
)
(4.4.4)
5
Frequency domain representation (cont 2)
10
Magnitude (dB)
5
For an arbitrary
real LTI system:
0
-5
-10
Symmetric with
respect to π
-15
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Normalized Frequency (×π rad/sample)
1.6
1.8
2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Normalized Frequency (×π rad/sample)
1.6
1.8
2
Phase (degrees)
100
Anti-symmetric
with respect to π
50
0
-50
-100
ELEN 5346/4304 DSP and Filter Design
Fall 2008
6
Frequency domain representation (cont 3)
Combining (4.3.2) and (4.4.4) – back to our sinusoid!
(
)
(
)
A jϕ
A
e H e jω0 e jω0 n + e − jϕ H e− jω0 e− jω0 n = {real system}
2
2
jω0
j
H
e
− j H ( e jω0 ) − jω n
A jϕ
A
(
)
e jω0 n + e− jϕ H e jω0 e
e 0
= e H e jω0 e
2
2
⎛ j ⎡ω0 n +ϕ + H ( e jω0 ) ⎤⎦⎥ − j ⎡⎣⎢ω0 n +ϕ + H ( e jω0 ) ⎤⎦⎥ ⎞
A
= H e jω0 ⎜ e ⎣⎢
+e
⎟
2
⎝
⎠
Stady-state response: yn =
(
(
LTI filtering:
)
(
)
)
(
)
(
(
yn = A H e jω0 cos ω0 n + ϕ + H e jω0
))
due to the
same as from the
input
the input input
change due to the system
phase change due to the system
Via design, we manipulate H(ejω), therefore, hn, and, finally, manipulate the
coefficients in the Linear Constant Coefficient Difference Equation (LCCDE)
ELEN 5346/4304 DSP and Filter Design
Fall 2008
(4.6.1)
(4.6.2)
(4.6.3)
(4.6.4)
7
Frequency domain representation (cont 4)
LCCDE:
xn = e jω0 n , n ≥ 0
⎫
− jω n
n
⎬ ytot ,n = ∑ ci λi + ke 0
initial conditions ≠ 0 ⎭
i
ytot ,n → ke− jω0 n
for large enough n:
yn = ∑ hk e
jω0 ( n − k )
k
(4.7.1)
(4.7.2)
⎡ n
⎤
un − k = ⎢ ∑ hk e − jω0 k ⎥ e jω0 n = H (e jω0 )e jω0 n
k =0
⎣
⎦
(4.7.3)
n →∞
H (e jω ) ≤ ∑ hne− jωn = ∑ hn < ∞ − if the system is BIBO, the frequency response exists
n
(4.7.4)
n
ELEN 5346/4304 DSP and Filter Design
Fall 2008
8
Frequency domain representation (cont 5)
for an LTI:
xn = e jω0 nun ⇒ yn = ytr ,n + yss ,n
(4.8.1)
We don’t need systems of order higher than 2: can always make cascades.
yss ,n = H (e jω0 )e jω0 n − exists only for LTI , BIBO systems
for a real, LTI, BIBO system:
xn = A cos(ω0 n + ϕ ) → yss ,n = A H (e jω0 ) cos(ω0 n + ϕ + H (e jω0 )
effects of filtering
We cannot observe ANY frequency components in the output that are not
present in the input (in steady state). We may see less when H (e jωr ) = 0
ELEN 5346/4304 DSP and Filter Design
Fall 2008
(4.8.2)
9
Frequency domain representation (cont 6)
In continuous time:
xt = st + nt ; {we want} yt = c ⋅ st −τ − distortion − less transmission
signal
noise
∞
Y ( Ω) =
∫ ye
t
− j Ωt
const
∞
dt =
−∞
∫ c⋅s
t −τ
(4.9.1)
delay
e− jΩ (t −τ ) d (t − τ )e− jΩτ = ce− jΩτ S (Ω) = H (Ω) S (Ω)
(4.9.2)
−∞
We need a constant magnitude and linear phase for the frequencies of interest.
Ideal filters:
LPF
H (e jω0 )
∠H (e jω0 )
ω
HPF
BPF
H (e jω0 )
H (e jω0 )
π
ω
BSF
ω
π
π
H (e jω0 )
ω
π
ω
π
Ideal filters are non-realizable!
ELEN 5346/4304 DSP and Filter Design
Fall 2008
10
CTFT and ICTFT
∞
CTFT:
X (Ω ) =
∫ x(t )e
− jωt
dt
(4.10.1)
−∞
ICTFT:
x(t ) =
1
2π
∞
∫ X (Ω )e
−∞
Examples:
ELEN 5346/4304 DSP and Filter Design
Fall 2008
jωt
dΩ
(4.10.2)
11
DTFT
X ( e jω ) =
X re ( e
∞
∑xe
← a complex function, 2π periodic, continuous on ω
− jω n
n
n =−∞
if exists
jω
) = X ( e ) cos ( X ( e ) ) ,
jω
jω
X im ( e jω ) = X ( e jω ) sin
( X (e ) )
jω
X ( e jω ) = X re 2 ( e jω ) + X im 2 ( e jω )
(4.11.1)
(4.11.2)
2
(4.11.3)
What’s about convergence???
1. Absolute convergence:
< ∞, ⇒ X ( e jω ) converges absolutely or uniformly
∑x
if xn ∈ l1 , i.e.
n
(4.11.4)
n
Let X k (e jω ) =
k
∑xe
n =− k
− jω n
n
⇒ lim X ( e jω ) − X k ( e jω ) = 0
k →∞ (4.11.5)
abs . error
ELEN 5346/4304 DSP and Filter Design
Fall 2008
12
DTFT (cont)
Absolutely summable sequences always have finite energy. However,
finite energy sequences are not necessary absolutely summable.
⎫
⎪
2
a
⎬ are not l1. In fact , they are not l2 : ∑ xn < ∞
n
A cos(ω0 n + ϕ ) ⎪⎭
must be
un
n
2. Mean-square convergence:
π
lim
k →∞
ω
ω
∫π X ( e ) − X ( e ) = 0
j
j
(4.12.1)
k
−
energy of the error
The total energy of the error must approach zero, not an error itself!
ELEN 5346/4304 DSP and Filter Design
Fall 2008
13
IDTFT
Let p = 2π is a period , ⇒ X ( e jω ) =
∞
∑xe
n =−∞
1
1
jω
jω n
IDTFT: xn = ∫ X (e )e d ω =
2π
pp
Combining (4.11.1) and (4.12.2)
⎛ 1
xl ⎜
∑
l =−∞
⎝ 2π
∞
−j
2π
ωn
p
(4.13.1)
n
π
∫π X (e
jω
)e jω n dω
(4.13.2)
−
xn =
1
2π
π
⎛ ∞
⎞
xl e − jωl ⎟ e jωn dω
∫−π ⎜⎝ l∑
=−∞
⎠
⎞ ∞
sin [π (n − l )] ∞
jω ( n −l )
ω
e
d
=
x
xlδ n −l = xn
⎟
∑ l π (n − l ) = ∑
∫
−∞
−π
⎠ l =−∞
(4.13.3)
π
(4.13.4)
X ( e jω ) shows where xn “lives” in the frequency domain.
ELEN 5346/4304 DSP and Filter Design
Fall 2008
14
Back to ideal filters
jω
1 H (e )
0
Ideal LPF:
0
ωc π
ω
2π
⎪⎧1, 0 ≤ ω ≤ ωc
H LP ( e jω ) = ⎨
⎪⎩0, ωc < ω < π
(4.14.1)
Using IDTFT:
hLP ,n =
1
2π
ωc
∫
e jω n d ω =
−ωc
1 ⎛ e jωc n e − jωc n ⎞ sin ωc n
−
, −∞ < n < ∞
⎜
⎟=
2π ⎝ jn
jn ⎠
πn
(4.14.2)
1. The response in (4.14.2) is not absolutely summable, therefore, the filter is
not BIBO stable!
2. The response in (4.14.2) is not causal and is of an infinite length.
As a result, the filter in (4.14.1) is not realizable.
Similar derivations show that none of the ideal filters in slide 9 is realizable.
ELEN 5346/4304 DSP and Filter Design
Fall 2008
15
DTFT properties
Linearity :
ag n + bxn ↔ aG ( e jω ) + bX ( e jω )
(4.15.1)
Shift in time :
g n − n0 ↔ e − jω n0 G ( e jω )
(4.15.2)
Shift in frequency :
e− jω0 n g n ↔ G e j (ω −ω0 )
Differenciation :
n ⋅ gn ↔ j
Linear convolution :
g n ∗ xn
DTFT
DTFT
DTFT
(4.15.3)
dG ( e jω )
(4.15.4)
dω
↔ G ( e jω ) ⋅ X ( e jω )
∞
∑
n =−∞
(4.15.5)
DTFT
DTFT
ELEN 5346/4304 DSP and Filter Design
)
DTFT
Periodic convolution : g n ⋅ wn ↔
Parseval ' s theorem :
(
1
2π
g n ⋅ w*n =
π
ω
∫π G ( e )W ( e
j
j (ω −θ )
) dθ
(4.15.6)
−
1
2π
π continuous, periodic functions
*
jω
jω
∫ G ( e )W ( e ) dθ
(4.15.6)
−π
Fall 2008
16
DTFTs of commonly used sequences
ELEN 5346/4304 DSP and Filter Design
Fall 2008
17
DTFT examples
Let xn = e jω0 n ↔ X ( e jω ) = 2π ∑ δ (ω − ω0 + 2π k )
(4.17.1)
ω0 = 0 ⇒ xn = 1 ↔ X ( e jω ) = 2π ∑ δ (ω + 2π k )
(4.17.2)
DTFT
k
DTFT
k
un − un −1 = δ n ↔ U ( e jω ) ⋅ (1 − e − jω ) = 1 ⇒ U ( e jω ) =
DTFT
1
+ c 2π ∑ δ (ω + l 2π )
1 − e − jω
l
½ of DC value of un
1
+ π ∑ δ (ω + l 2π )
1 − e − jω
l
un ↔
DTFT
(4.17.3)
(4.17.4)
*
⎛
⎞
∑n hn*e− jωn = ⎜⎝ ∑n hn e jωn ⎟⎠ = H * ( e jω )
π
1
1
∑n hn g n e− jωn = ∑n 2π ∫ H ( e jθ ) e jθ n g n e− jωn dθ = 2π
−π
=
1
2π
π
θ
∫ H (e ) G (e
j
j ( ω −θ )
−π
ELEN 5346/4304 DSP and Filter Design
(4.17.5)
π
θ
∫π H ( e ) ∑ g e
j
− j (ω −θ ) n
n
−
dθ
n
) dθ
(4.17.6)
Fall 2008
18
DTFT examples (cont)
We can re-work the Parseval’s theorem (4.15.6) as follows:
Energy : En = ∑ hn 2 =
n
1
2π
π
π
θ
θ
H ( e θ ) dθ
∫π H ( e ) H ( e ) dθ = 2π ∫π j
*
j
−
1
j
−
2
(4.18.1)
Sh ( e jω )
energy density (spectrum)
Autocorrelation function:
rgg ,l = ∑ g n g n −l * = ∑ g n g − (l − n )* = g l ∗ g − l * ↔ G ( e jω ) G * ( e jω ) ≡ S g ( e jω )
n
ELEN 5346/4304 DSP and Filter Design
n
DTFT
Fall 2008
(4.18.2)
19
DTFT examples (cont 2)
One obvious problem with DTFT is that we can never compute it since xn
needs to be known everywhere! which is impossible!
Therefore, DTFT is not practical to compute.
Often, a finite dimension LTI system is described by LCCDE:
N
∑a y
i =0
i n−i
M
= ∑bm xn−m ↔
DTFT
m=0
N
∑a e
i =0
M
Y ( e ) ∑b e
⇒
=
jω
X ( e jω )
m=0
N
Y ( e jω ) = ∑bme− jωm X ( e jω )
M
− jωi
i
− jωm
m
∑aie− jωi
(4.19.1)
m=0
= H (e
jω
)
practical (finite
dimensions)
(4.19.2)
i =0
Prediction of steady-state behavior of LCCDE
ELEN 5346/4304 DSP and Filter Design
Fall 2008
How to measure frequency response
of an actual (unknown) filter?
20
1. Perform two I/O experiments:
(
(
) (
) (
( ))
( ))
⎧1) xn ( c ) = A cos(ω0 n + ϕ ) → yn ( c ) = A H e jω0 cos ω0 n + ϕ + ∠H e jω0 + ytr ,n ( c )
⎪
LTI
⎨
jω
jω
(s)
(s)
(s)
⎪2) xn = A sin(ω0 n + ϕ ) → yn = A H e 0 sin ω0 n + ϕ + ∠H e 0 + ytr ,n
LTI
⎩
(4.20.1)
(4.20.2)
2. Analyze these measurements and form:
⎧ xn = xn ( c ) + jxn ( s ) = Ae j (ω0 n +ϕ )
⎪
⎨
j ⎡⎢ω0 n +ϕ +∠H ( e jω0 ) ⎤⎥
jω
(c)
(s)
⎦
+ ytr ,n
⎪⎩ y n = yn + jyn = A H e 0 e ⎣
j∠H ( e jω0 )
y
+ ztr ,n
Finally ⇒ n = H e jω0 e
xn
(
(
)
)
That’s a good way to measure/estimate a frequency response for every ω.
ELEN 5346/4304 DSP and Filter Design
Fall 2008
(4.20.3)
(4.20.4)
(4.20.5)
21
DFT and IDFT
Consider an N-sequence xn (at most N non-zero values for 0 ≤ n ≤ N-1)
N −1
X ( e jω ) = ∑ xn e − jω n = ∑ xn e − jω n = X ( e jω )
n =0
n
N −1
X k = ∑ xn e
DFT:
−j
n=0
2π
kn
N
ωk =
2π
k
N
= X k , k = 0,1,..., N − 1
(4.21.1)
uniformly spaced frequency samples
, k = 0,1,..., N − 1
(4.21.2)
Finite sum! Therefore, it’s computable.
Using notation: WN = e
−j
2π
N
(4.21.3)
N −1
X k = ∑ xnWN kn
(4.21.1) can be rewritten as:
1
N
⇒ IDFT : xn =
(4.21.4)
n =0
N −1
∑X W
k
n =0
− kn
N
,
n = 0,1,..., N − 1
(4.21.5)
Btw, DFT is a sampled version of DTFT.
ELEN 5346/4304 DSP and Filter Design
Fall 2008
22
DFT and IDFT (cont)
Let us verify (4.21.5). We multiply both sides by WN l ⋅n
N −1
∑x W
n =0
⇒
n
N
l ⋅n
N −1
⎛1
= ∑⎜
n =0 ⎝ N
N −1
∑ xnWN l⋅n =
n =0
∑W
n =0
⇒
− kn
N
N −1
N −1
k =0
n =0
⎞ l ⋅n 1
⎟ WN = N
⎠
N −1 N −1
∑∑ X W
n =0 k =0
k
− ( k −l ) n
N
∑ X k ∑WN −( k −l ) n
− ( k −l ) n
N
N −1
∑x W
n =0
ELEN 5346/4304 DSP and Filter Design
k
k =0
1
N
N −1
Since
N −1
∑X W
n
N
⎧ N , for k − l = rN , r is Integer
=⎨
⎩0, otherwise
l ⋅n
= Xl
(4.22.1)
(4.22.2)
(4.22.3)
(4.22.4)
Fall 2008
23
FFT
X = DN x − DFT
x = DN −1 X − IDFT
In the matrix form:
where:
(4.23.1)
(4.23.2)
x = [ x0 x1 ... xN −1 ] − time domain signal
(4.23.3)
X = [ X 0 X 1 ... X N −1 ] − N DFT samples
(4.23.4)
1
1
⎡1
⎢1 W 1
2
W
N
N
⎢
2
DN = ⎢1 WN
WN 4
⎢
#
#
⎢#
⎢⎣1 WN N −1 WN 2⋅( N −1)
(4.23.5)
T
T
⎤
⎥
⎥
2⋅( N −1)
⎥ − DFT matrix
" WN
⎥
%
#
⎥
( N −1)⋅( N −1) ⎥
" WN
⎦
"
1
"
WN N −1
1
1
⎡1
⎢1 W −1
−2
W
N
N
1 ⎢
−1
−2
−4
DN = ⎢1 WN
WN
N⎢
#
#
⎢#
⎢⎣1 WN − ( N −1) WN −2⋅( N −1)
"
1
⎤
WN − ( N −1) ⎥⎥
"
" WN −2⋅( N −1) ⎥ − IDFT matrix
⎥
%
#
⎥
− ( N −1)⋅( N −1) ⎥
−1
" WN
⎦
DN
(4.23.6)
=
This is actually FFT…
ELEN 5346/4304 DSP and Filter Design
1
DN *
N
(4.23.7)
Fall 2008
24
Relation between DTFT and DFT
1. Sampling of DTFT
Let xn ↔ X ( e jω )
DTFT
Sample X ( e
jω
) at ω
k
= 2π
{ ( ) }=Y
k
, k = 0,1,...N − 1 ⇒ N frequency samples X e jωk
N
k
− N − point DFT
Yk ↔ yn , n = 0,1,..., N − 1
IDFT
⎛ j 2π k ⎞ ∞
Yk = X e jωk = X ⎜ e N ⎟ = ∑ xlWN kl
⎝
⎠ l =−∞
N −1
N −1 ∞
∞
1
1
⎡1
yn = ∑ YkWN − kn = ∑ ∑ xlWN klWN − kn = ∑ xl ⎢
N k =0
N k =0 l =−∞
l =−∞
⎣N
(
)
Considering (4.22.3) ⇒ yn =
∞
∑x
m =−∞
n + mN
(4.24.1)
N −1
∑W
k =0
, n = 0,1,..., N − 1
N
− k ( n −l )
⎤
⎥
⎦
(4.24.2)
(4.24.3)
yn is an infinite sum of shifted replicas of xn. Iff xn is a length M sequence (M ≤ N)
than yn = xn. Otherwise, time-domain aliasing ⇒ xn cannot be recovered!
ELEN 5346/4304 DSP and Filter Design
Fall 2008
25
Relation between DTFT and DFT (cont)
2. DTFT from DFT by Interpolation
↔
Let xn be a length N sequence: xn
N − point DFT
Xk
Let us try to recover DTFT from DFT (its sampled version).
N −1
X ( e jω ) = ∑ xn e − jω n =
n =0
N −1 − j ⎛ ω − 2π k ⎞ n
⎜
⎟
N⎠
⎝
Since ∑ e
n =0
1
N
N −1
⎡ N −1
∑ ⎢⎣∑ X W
n =0
k =0
k
− kn
N
1 − e − j ( ω N − 2π k ) e − j ( ω N − 2 π k ) / 2
=
=
⋅
1 − e − j ( ω − 2π k / N ) e − j ( ω − 2π k ) / 2 N
n
⎤ − jωn 1 N −1 N −1 j 2π N − jωn
=
e
X
e
e
∑
∑
k
⎥
N k =0 n =0
⎦
(4.25.1)
⎛ ω N − 2π k ⎞
⎛ ω N − 2π k ⎞
sin ⎜
⎟ sin ⎜
⎟ − j ⎜⎛ ω − 2π k ⋅ N −1 ⎟⎞
2
2
⎝
⎠=
⎝
⎠ ⋅ e ⎝ N 2 ⎠ (4.25.2)
⎛ ω N − 2π k ⎞
⎛ ω N − 2π k ⎞
sin ⎜
sin
⎟
⎜
⎟
2N
2N
⎝
⎠
⎝
⎠
⎛ ω N − 2π k ⎞
⎛ ωN ⎞
sin ⎜
N −1
⎟ − j ⎛⎜ ω − 2π k ⋅ N −1 ⎞⎟ 1 N −1 sin ⎜
⎟
2
⎝
⎠
⎝ 2 ⎠ e− jω 2
N
2
⎠
(4.25.3)
Xk
⋅e ⎝
= ∑ Xk
∑
N k =0
⎛ ω N − 2π k ⎞
⎛1⎞
k =0
sin ⎜
sin
⎟
⎜ ⎟
2N
⎝
⎠
⎝2⎠
j
ω
It’s possible to determine DTFT X(e ) from its uniformly sampled version uniquely!
1
X (e ) =
N
jω
N −1
ELEN 5346/4304 DSP and Filter Design
Fall 2008
26
Relation between DTFT and DFT (cont 2)
3. Numerical computation of DTFT from DFT
Let xn is a length N sequence:
xn ↔ X ( e jω )
DTFT
defined by N uniformly spaced samples
We wish to evaluate
X ( e jω ) at more dense frequency scale.
ωk = 2π k / M , k = 0,1,..., M − 1, where M N
(
)
N −1
N −1
n =0
n=0
X e jωk = ∑ xn e − jωk n = ∑ xn e
⎧ xn ,
⎩0,
Define: xl , n = ⎨
(
− j 2π k
n
M
(4.26.1)
0 ≤ n ≤ N −1
zero-padding
N ≤ n ≤ M −1
)
M −1
⇒ X e jωk = ∑ xl ,n e
− j 2π k
n =0
n
M
M −1
= ∑ xl ,nWM kn
n =0
No change in information, no change in DTFT… just a better “plot resolution”.
ELEN 5346/4304 DSP and Filter Design
Fall 2008
(4.26.2)
(4.26.3)
27
A note on WN
Recall that: WN = e
Nth
WN is also called an
−j
2π
N
(4.27.1)
WN = e
root of unity, since
−j
2π
N
≡1
(4.27.2)
Im
Im
2π
N
Re
Re
2π
n
N
ELEN 5346/4304 DSP and Filter Design
Fall 2008
28
DFT properties
1. Circular shift
xn is a length N sequence defined for n = 0,1,…N-1.
An arbitrary shift applied to xn will knock it out of the 0…N-1 range.
Therefore, a circular shift that always keeps the shifted sequence in
the range 0…N-1 is defined using a modulo operation:
xc ,n = x( n − n0 ) modulo N = x n − n0
N
n0 ≤ n ≤ N − 1
⎧⎪ xn − n0 ,
=⎨
⎪⎩ xN − n0 + n , 0 ≤ n ≤ n0
x n − n0
xn
x n −1 = x n +5
6
ELEN 5346/4304 DSP and Filter Design
6
x n−4 = x n+ 2
6
1
N
6
Fall 2008
=
N
1
N
N −1
∑X W
k =0
N −1
∑W
k =0
N
kn0
(4.28.1)
X kWN
k
− k ( n − n0 )
=
↔ WN
kn0
N
(4.28.2)
− kn
DFT
Xk
29
DFT properties (cont)
2. Circular convolution
A linear convolution for two length N sequences xn and gn has a length 2N-1:
N −1
yL ,n = ∑ g m xn − m , 0 ≤ n ≤ 2 N − 2
(4.29.1)
m=0
A circular convolution is a length-N sequence defined as:
N −1
yC ,n = ∑ g m x n − m ≡ g n
N
m=0
N −1
∑g
m=0
N −1
m
x n−m = ∑ g m
N
m =0
N
xn ≡ xn
1 N −1
1
X kWN − k ( n − m ) =
∑
N k = 0 N
x n−m
N
gn
(4.29.2)
N −1 N −1
∑∑ g W
k =0 m=0
l
N
km
X kWN − kn
N
1 N −1
= ∑ Gk X kWN − kn ↔ Gk X k
DFT
N k =0
Procedure: take two sequences of the same length (zero-pad if needed),
DFT of them, multiply, IDFT: a circular convolution.
ELEN 5346/4304 DSP and Filter Design
(4.29.3)
Fall 2008
30
DFT properties (cont 2)
Example:
X ( e jω ) = ∑ xn e − jω n
(4.30.1)
n
Take N frequency samples of (4.30.1) and then IDFT:
xn =
1
N
N −1
∑ X kWN − kn =
k =0
= ∑ xl
l
1
N
N −1
∑∑ xl e
k =0
−j
l
2π
kl
N
2π
−j
⎧
⎫
WN − kn = ⎨WN = e N ⎬
⎩
⎭
N −1
1
WN − k ( n −l ) = ∑ xn + rN
∑
N
k =0
r
∑δ n−l +rN (4.22.3)
(4.30.2)
aliased version of xn
r
The results of circular convolution differ from the linear convolution “on the edges”
– caused by aliasing.
To avoid aliasing, we need to use zero-padding…
ELEN 5346/4304 DSP and Filter Design
Fall 2008
31
Linear filtering via DFT
Often, we need to process long data sequences; therefore, the input must
be segmented to fixed-size blocks prior LTI filtering. Successive blocks
are processed one at a time and the output blocks are fitted together…
We can do it by FFT: IFFT{FFT{x}FFT{h}}…
Problem: DFT implies circular convolution – aliasing!
yn = hn ∗ xn
(4.31.1)
Assuming that hn is an M-sequence, we form an N-sequence:
⎧x
xm ,n = ⎨ n
⎩0
mL ≤ n ≤ mL + L − 1
(4.31.2)
otherwise
N >> M; L >> M; N = L + M - 1 and is a power of 2
ELEN 5346/4304 DSP and Filter Design
Fall 2008
32
Linear filtering via DFT (cont)
Next, we compute N-point DFTs of xm,n and hn, and form
Ym,k = H k X m,k , k = 0,1,...N − 1
Ym ,k → ym ,n
IDFT
(4.32.1)
- no aliasing!
Since each data block was terminated
with M -1 zeros, the last M -1 samples
from each block must be overlapped
and added to first M – 1 samples of
the succeeded block.
An Overlap-Add method.
ELEN 5346/4304 DSP and Filter Design
Fall 2008
33
Linear filtering via DFT (cont 2)
Alternatively:
Each input data block contains M -1
samples from the previous block followed
by L new data samples; multiply the DFT
of the filter’s impulse response and the
DFT of the input block, take IDFT.
Keep only the last L data samples from
each output block.
An Overlap-Save method.
ELEN 5346/4304 DSP and Filter Design
Fall 2008
34
DFT properties: General
from Mitra’s
book
Btw, g[n] = gn
ELEN 5346/4304 DSP and Filter Design
Fall 2008
35
DFT properties: Symmetry
from Mitra’s
book
xn is a real
sequence
ELEN 5346/4304 DSP and Filter Design
Fall 2008
36
DFT properties: Symmetry (cont)
from Mitra’s
book
xn is a complex
sequence
ELEN 5346/4304 DSP and Filter Design
Fall 2008
37
N-point DFTs of 2 real sequences
via a single N-point DFT
Let gn and hn are two length N real sequences.
Form
xn = gn + jhn ↔ Xk
(4.37.1)
{
}
1
X k + X * −k
N
2
1
Hk =
X k − X * −k
N
2j
Gk =
{
Note that
X * −k
N
= X * N −k
(4.37.2)
}
(4.37.3)
(4.37.4)
N
ELEN 5346/4304 DSP and Filter Design
Fall 2008
38
Summary
Algorithm
Time
Frequency
CTFT
Continuous
Continuous
DTFT
Discrete
Continuous
DFT
Discrete
Discrete
ELEN 5346/4304 DSP and Filter Design
Fall 2008