EE518 Digital Signal Processing
Autumn 2001
University of Washington
Dept. of Electrical Engineering
Lecture 3: DTFT, Convergence and Symmerty Properties of DTFT
Mon., Oct 8, 2001
Prof: J. Bilmes <[email protected]>
TA: Mingzhou Song <[email protected]>
(Review of Previous Lecture)
H(e jω ) =
∞
∑
h[k]e− jωk
k=−∞
is the frequency response of a system.
• ammount of attenuation for an input consisting of a complex exponential e jωn (sinusoid)
• can think of as a response for each frequency.
Important: since many signals can be represented as
∞
∑
x[n] =
ak e− jωk n
k=−∞
Ex: H(e jω ) for moving average system. The impulse response of the moving average system is
h[n] =
1
(u[n + M1 ] − u[n − M2 − 1])
M1 + M2 + 1
The frequency response is
H(e jω ) =
M2
1
∑ e− jωn
M1 + M2 + 1 n=−M
1
N2
1
e jωM1 − e− jω(M2 +1)
=
M1 + M2 + 1
1 − e− jω
sin ω 1 2 2
1
M1 + M2 + 1
sin ω2
M +M +1
=
e− jω
∵
∑
k=N1
aN1 − aN2 +1
ak =
1−a
!
M2 −M1
2
Question: Assume M1 = 0 and M2 = 4. For what type of signals (think in time domain) will the output be zero?
1. x[n] = 0
2. x[n] = e jω0 n+φ0
ω0 = 2π/5
3. x[n] = e jω1 n+φ1
ω1 = 4π/5
Note where frequency response is zero, at ω =
of five samples) is shown in Figure 3.1.
2π 4π
5 , 5 .
Time domain interpretation when M1 = 0 and M2 = 4 (average
Magnitude response |H(e jω )| is shown in Fig 3.2.
3-1
3-2
ω = 2π/5
1
0.5
0.5
0
0
−0.5
−0.5
−1
0
0.2
0.4
0.6
0.8
−1
1
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
1
2
ω = 4π/5
1
3
0
4
0
0.5
1
1.5
2
0
1
2
3
4
Figure 3.1: Sampling a sinusoid at ω = 2π
5 cancells out so the frequency response at ω =
regardless of the phase of the input signal. It is the same for ω = 4π
5
.
2π
5
is zero. This is true
1
0.9
0.8
0.7
|H(ejω)|
0.6
0.5
0.4
0.3
0.2
0.1
0
−4
−3
−2
−1
0
ω
1
2
3
4
4π
Figure 3.2: Magnitude response |H(e jω )| of moving average system (M1 = 0, M2 = 4). Zeros are located at ω = 2π
5 , 5
5
2.5
as expected. The periods correspond to the sampling rates of 2π samples/radian and 2π samples/radian, respectively,
or 5 samples/cycle and 2.5 samples/cycle.
3-3
3.1
Representations of Sequences by Discrete-Time Fourier Transform
x[n] =
1
2π
Z π
X(e jω )e jωn dω
Inverse DTFT or synthesis equation
(3.1)
−π
− jωk n .
Note: This is like a continuous version of x[n] = ∑∞
k=−∞ ak e
∞
X(e jω ) ,
∑
x[n]e− jωn
DTFT or analysis equation
(3.2)
n=−∞
We will call DTFT FT in this course.
Note: x[n] is infinite superposition of weighted complex sinusoids of the form
1
X(e jω )e jωn
2π
Definition 3.1 (Magnitude Spectrum and Phase Spectrum). |X(e jω )| is the magnitude spectrum. ]X(e jω ) is the
phase spectrum, but it is not unique.
The ARG() function has a range from −π to π. The arg() function has a range from 0 to 2π. They can be used to find
the phase spectrum of X(e jω ), i.e., ARG(X(e jω )) and arg(X(e jω ))
Note:
H(e jω ) ,
∞
∑
h[n]e− jωn
n=−∞
so FT of impulse response for LTI system is the frequency response. This means
h[n] =
1
2π
Z π
H(e jω )e jωn dω
−π
Note:
Question: On a computer, do we compute |X(e jω )| from a given x[n] stored on a computer?
Answer: No. This requires infinite sequence x[n] and produces continuous frequency variable ω. We can do this
analytically for many signals as we will see in this course, but we cannot do it for real world signals. Instead on
computers (MATLAB), we compute DFT (& FFT), which we will see later in this course.
Comparison:
∞
X(e jω ) ,
Z
X( jΩ) ,
∑
x[n]e− jωn
Discrete-Time Fourier Transform
(3.3)
Continuous-Time Fourier Transform
(3.4)
n=−∞
∞
x(t)e− jΩt dt
−∞
Similar conventions:
• discrete time frequency variable ω
• continuous time frequency variable Ω
3.2
Convergence and Wide-Sense Fourier Transform
Note: X(e jω ) is an infinite sum, hence may not converge. We need |X(e jω )| < ∞ ∀ω.
|X(e jω )| = |
∞
∑
n=−∞
x[n]e− jωn | ≤
∞
∑
n=−∞
|x[n]||e− jωn | ≤
∞
∑
n=−∞
|x[n]| < ∞
(3.5)
3-4
The last inequality is called absolute summability, which is a sufficient condition for the existence of X(e jω ). Substitute
jω
H(e jω ) of some system for X(e jω ). Since ∑∞
n=−∞ |h[n]| < ∞ is the condition for stability for a LTI system, H(e ) must
exist for stable systems.
Ex: Signals with existing FT’s.
x[n] = δ[n − nd ]
x[n] = δ[n] − δ[n − 1]
x[n] =
(backward difference)
1
(u[n + M1 ] − u[n − M2 − 1])
M1 + M2 + 1
and any FIR with finite values.
x[n] = an u[n]
∞
X(e jω ) =
∑
∞
n=−∞
1
∑ (ae− jω )n = 1 − ae− jω
an u[n]e− jωn =
if |ae− jω | = |a| < 1
n=0
The existence of FT requires
∞
1
∑ |a|n = 1 − |a| < ∞
n=0
which is the absolute summability as a sufficient condition.
Definition 3.2 (Uniform Convergence). Given some X(e jω ), and let
XM (e jω ) =
M
∑
x[n]e− jωn
n=−M
then FT of x[n] exists if ∀ε > 0 and ∀ω, ∃M0 such that
|X(e jω ) − XM (e jω )| < ε
(3.6)
whenever M > M0 .
But, some sequences are not absolutely summable, so they don’t achieve uniform convergence to some X(e jω ). Some
sequence are square summable. For those that achieve mean-square convergence, we can give them FT’s, too.
Definition 3.3 (Mean Square Convergence). Given some X(e jω ), and let
XM (e jω ) =
M
∑
x[n]e− jωn
n=−M
we say XM (e jω ) converges in mean-square sense if
Z π
lim
|X(e jω ) − XM (e jω )|2 dω = 0
(3.7)
FT {x[n]} = X(e jω )
(3.8)
M→∞ −π
In these cases, we say that
So we might have it that
|X(e jω ) − XM (e jω )| 6→ 0
as
M→∞
but the total energy of the difference between the two does, i.e., could be an infinite number of points that don’t
converge, but they have zero-measure.
Important Example:
jω
H(e ) =
Question: What kind of filter is this?
1
0
|ω| ≤ ωc
ωc < |ω| ≤ π
3-5
It is an ideal lowpass (LP) filter with the cutoff frequency at ωc .
The impulse response is
hl p [n] =
1
2π
Z π
H(e jω )e jωn dω =
−π
1
2π
Z ωc
e jωn dω =
−ωc
Note sinc function is
sinc(n) =
sin ωc n
πn
−∞ < n < ∞
sin πn
πn
The zeros of hl p [n] are at ωc n = kπ. Whether we see any zeros depends on the value of ωc .
Note: hl p [n] is not absolutely summable. Why?
1
Theorem 3.4. ∑∞
n=−∞ nρ converges if ρ > 1, diverges if ρ ≤ 1.
By the above theorem, hl p [n] is square summable, i.e.,
∞
sin ωc n − jωn
e
πn
n=−∞
∑
diverges, but
∞
∑
1
0
sin ωc n
∑ πn e− jωn = FT
n=−M
n=−∞
sin ωc n
πn
2
e− jωn
converges.
Let
y[n] =
−M ≤ n ≤ M
else
Consider
HM (e jω ) =
M
Z
sin ωc n
1 ωc sin[(2M + 1)(ω − θ)/2]
y[n] =
dθ
πn
2π −ωc
sin[(ω − θ)/2]
Different versions of HM (e jω ) are plotted in the Fig 3.3 when M takes different values.
So HM (e jω ) does not converge uniformly to H(e jω ), i.e., the magnitude of the difference does not converge to 0 as
M → ∞. But since hl p [n] is square summable,
Z π
lim
M→∞ −π
|Hl p (e jω ) − HM (e jω )|2 dω = 0
so, if places where values of the two functions are different have infinitely small spectral extent (zero-measure),
Hl p (e jω ) is ok as an FT. This is important for filter design (Chapter 7).
Definition 3.5 (Impulse Function in Frequency Domain). δ(ω) is defined as
1. δ(ω) = 0 w 6= 0
2. f (ω)δ(ω) = f (0)δ(ω)
3.
R∞
−∞ δ(ω)dω
=1
4. δ(ω) ∗ X(e jω ) = X(e jω )
Ex: Other important signals that are not summable but we want FT.
Consider
X(e jω ) =
∞
∑
r=−∞
2πδ(ω − ω0 + 2πr)
3-6
M=1
1.2
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
−0.2
−4
−2
0
2
−0.2
−4
4
M=8
1.2
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
−2
0
−2
2
−0.2
−4
4
0
2
4
2
4
M=20
1.2
1
−0.2
−4
M=4
1.2
−2
0
Figure 3.3: HM (e jω ). The oscillation at ω = ωc is called the Gibbs phenomenon.
Then
1
x[n] =
2π
Z π
X(e jω )e jωn dω = e jω0 n
−π
When ω0 = 0, we get x[n] = 1, a constant, and only the dc component exists.
Note: x[n] is not absolutely or square summable, but we still find it useful to define its FT as such.
3.3
Let
Some Symmetry Properties of Fourier Transform
F
x[n] ←→ X(e jω )
then
F
x∗ [n] ←→ X ∗ (e jω )
1
F
Re(x[n]) ←→ Xe (e jω ) = [X(e jω ) + X ∗ (e− jω )]
2
(3.9)
conjugate symmetry pair
Note:
x[n] = xe [n] + xo [n]
1
xe [n] = (x[n] + x∗ [−n]) = xe∗ [−n] conjugate symmetry or even if x[n] is real
2
1
xo [n] = (x[n] − x∗ [−n]) = −xo∗ [−n] conjugate anti-symmetry or odd if x[n] is real
2
(3.10)
3-7
Similarly
X(e jω ) = Xe (e jω ) + Xo (e jω )
1
Xe (e jω ) = (X(e jω ) + X ∗ (e− jω )) = Xe∗ (e− jω )
2
1
jω
Xo (e ) = (X(e jω ) − X ∗ (e− jω )) = −Xo∗ (e− jω )
2
Symmetry Properties of FT for Real Signals
For any real x[n]
X(e jω ) = X ∗ (e− jω )
X(e jω ) = XR (e jω ) + jXI (e jω )
i.e., FT is conjugate symmetry for real signals. So, real part is even,
XR (e jω ) = XR (e− jω )
and imaginary part is odd,
XI (e jω ) = −XI (e− jω )
Also,
X(e jω ) = |X(e jω )|]X(e jω )
So the magnitude is even
|X(e jω )| = |X(e− jω )|
and the phase is odd
]X(e jω ) = ]X(e− jω )
Why? Becuase of conjugate symmetry of X(e jω )
F
xe [n] ←→ XR (e jω )
So, a signal that is real and even in time has its FT real and even in frequency.
Other Properties
F
Re{x[n]} ←→ Xe (e jω )
F
Im{x[n]} ←→ Xo (e jω )
F
xe [n] ←→ XR (e jω )
F
jω
xo [n] ←→ jXI (e )
Other properties are given on O&S Table 2.1.
(3.11)
(3.12)
(3.13)
(3.14)