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6.003: Signal Processing
Discrete-Time Fourier Transform
• Definition
• Examples
• Properties
• Relations between Fourier series and transforms (DT and CT)
Announcements:
• No PSet 4 – practice quiz questions have been posted instead.
• Office Hours today (7-9pm) and Sunday (4-6pm)
− practice quiz problems and general questions
• Solutions to practice quiz questions will be posted Sunday at 6pm
• Quiz 1: Tuesday, March 3, 2-4pm in 32-141.
− Coverage up to and including all of week 4.
− Closed book except for one page of notes (8.5”x11” both sides).
− No electronic devices. (No headphones, cellphones, calculators, ...)
February 27, 2020
From Periodic to Aperiodic
Last time: representing arbitrary (aperiodic) CT signals as sums of sinu-
soidal components using the continuous-time Fourier transform.
Today: generalize the Fourier Transform idea to discrete-time signals.
Fourier Representions of Aperiodic Signals
How can we represent an aperiodic signal as a sum of sinusoids?
n
x[n]
−2 0 2Strategy: make a periodic version of x[n] by summing shifted copies:
xp[n] =∞∑
m=−∞x[n−mN ]
n
xp[n]
−2 0 2−N N
Since xp[n] is periodic, it has a Fourier series (which depends on N).
Find Fourier series coefficients Xp[k] and take the limit of Xp[k] as N →∞.
As N →∞, xp[n]→ x[n], and Fourier series will approach Fourier transform.
Fourier Representions of Aperiodic Signals
Example.
n
x[n]
−2 0 2Strategy: make a periodic version of x[n] by summing shifted copies:
xp[n] =∞∑
m=−∞x[n−mN ]
n
xp[n]
−2 0 2−N N
Calculate the Fourier series coefficients Xp[k]:
Xp[k] = 1N
∑n=〈N〉
xp[n]e−j2πNkn = 1
N+ 2N
cos 2πkN
+ 2N
cos 4πkN
Fourier Representions of Aperiodic Signals
Calculate the Fourier series coefficients Xp[k]:
Xp[k] = 1N
∑n=〈N〉
xp[n]e−j2πNkn = 1
N+ 2N
cos 2πkN
+ 2N
cos 4πkN
Plot the resulting Fourier coefficients for N=8.
k
Xp[k]
What happens if you double the period N?
Fourier Representions of Aperiodic Signals
Calculate the Fourier series coefficients Xp[k]:
Xp[k] = 1N
∑n=〈N〉
xp[n]e−j2πNkn = 1
N+ 2N
cos 2πkN
+ 2N
cos 4πkN
Plot the resulting Fourier coefficients for N=8.
k
Xp[k]
What happens if you double the period N?
• The amplitude will shrink by a factor of 2.
• There will be twice as many samples per period of the cosine functions.
k
Xp[k]
k
Xp[k]
The red samples are at new intermediate frequencies.
Fourier Representions of Aperiodic Signals
Define a new function X(Ω) = NXp[k] where Ω = 2πk/N .
NXp[k] = 1+2 cos 2πkN
+2 cos 4πkN
= 1+2 cos(Ω)+2 cos(2Ω) = X(Ω)
N=8:
Ω = 2πkN
NXp[k] = X(Ω)
N=16:
Ω = 2πkN
N=32:
Ω = 2πkN
The discrete function NXp[k] is a sampled version of the function X(Ω).
Fourier Representions of Aperiodic Signals
We can reconstruct x[n] from X(Ω) using Riemann sums.
xp[n] =∑k=〈N〉
Xp[k]e j2πNkn= 1
2π∑k=〈N〉
NXp[k]ej2πNkn(2πN
)x[n] = lim
N→∞xp[n] = lim
N→∞
12π
∑k=〈N〉
NXp[k]ej2πNkn(2πN
)= 1
2π
∫2πX(Ω)ejΩndΩ
N=8:
Ω = 2πkN
NXp[k] = X(Ω)
−π π−2π 2π
2πN
N=16:
Ω = 2πkN
−π π−2π 2π
2πN
N=32:
Ω = 2πkN
−π π−2π 2π
2πN
Fourier Transform relation: x[n] ft⇐⇒ X(Ω)
Fourier Series and Fourier Transform
Fourier series and transforms are similar:
both represent signals by their frequency content.
Discrete-Time Fourier Series
X[k] = X[k+N ] = 1N
∑n=〈N〉
x[n]e−jkΩon analysis equation
x[n] = x[n+N ] =∑k=〈N〉
X[k]e jkΩon synthesis equation
where Ωo = 2πN
Discrete-Time Fourier Transform
X(Ω)= X(Ω+2π) =∞∑
n=−∞x[n]e−jΩn analysis equation
x[n] = 12π
∫2πX(Ω)ejΩn dΩ synthesis equation
Fourier Series and Fourier Transform
All of the information in a periodic signal is contained in one period.
The information in an aperiodic signal is spread across time.
Discrete-Time Fourier Series
X[k] = X[k+N ] = 1N
∑n=〈N〉
x[n]e−jkΩon analysis equation
x[n] = x[n+N ] =∑k=〈N〉
X[k]e jkΩon synthesis equation
where Ωo = 2πN
Discrete-Time Fourier Transform
X(Ω)= X(Ω+2π) =∞∑
n=−∞x[n]e−jΩn analysis equation
x[n] = 12π
∫2πX(Ω)ejΩn dΩ synthesis equation
Fourier Series and Fourier Transform
Periodic signals can be synthesized from a discrete set of k harmonics.
Aperiodic signals generally require a continuous set of frequencies Ω.
Discrete-Time Fourier Series
X[k] = X[k+N ] = 1N
∑n=〈N〉
x[n]e−jkΩon analysis equation
x[n] = x[n+N ] =∑k=〈N〉
X[k]e jkΩon synthesis equation
where Ωo = 2πN
Discrete-Time Fourier Transform
X(Ω)= X(Ω+2π) =∞∑
n=−∞x[n]e−jΩn analysis equation
x[n] = 12π
∫2πX(Ω)ejΩn dΩ synthesis equation
Fourier Series and Fourier Transform
Harmonic frequencies kΩo are samples of continuous frequency Ω.
Discrete-Time Fourier Series
X[k] = X[k+N ] = 1N
∑n=〈N〉
x[n]e−jkΩon analysis equation
x[n] = x[n+N ] =∑k=〈N〉
X[k]e jkΩon synthesis equation
where Ωo = 2πN
Discrete-Time Fourier Transform
X(Ω)= X(Ω+2π) =∞∑
n=−∞x[n]e−jΩn analysis equation
x[n] = 12π
∫2πX(Ω)ejΩn dΩ synthesis equation
CT and DT Fourier Transforms
DT frequencies alias because adding 2π to Ω does not change e−jΩn.
Continuous-Time Fourier Transform
X(ω)=∫ ∞−∞
x(t)e−jωtdt analysis equation
x(t) = 12π
∫ ∞−∞
X(ω)ejωtdω synthesis equation
where Ωo = 2πN
Discrete-Time Fourier Transform
X(Ω)= X(Ω+2π) =∞∑
n=−∞x[n]e−jΩn analysis equation
x[n] = 12π
∫2πX(Ω)ejΩn dΩ synthesis equation
CT and DT Fourier Transforms
DT frequencies alias because adding 2π to Ω does not change e−jΩn.
Because of aliasing, we need only integrate dΩ over a 2π interval.
Continuous-Time Fourier Transform
X(ω)=∫ ∞−∞
x(t)e−jωtdt analysis equation
x(t) = 12π
∫ ∞−∞
X(ω)ejωtdω synthesis equation
where Ωo = 2πN
Discrete-Time Fourier Transform
X(Ω)= X(Ω+2π) =∞∑
n=−∞x[n]e−jΩn analysis equation
x[n] = 12π
∫2πX(Ω)ejΩn dΩ synthesis equation
Examples of Fourier Transforms
Find the Fourier Transform (FT) of a rectangular pulse of width 2S+1:
pS [n] =
1 −S ≤ N ≤ S0 otherwise
n
p2[n]
−2 0 2
n
p5[n]
−5 0 5
PS(Ω) =∞∑
n=−∞pS [n]e−jΩn =
S∑n=−S
e−jΩn
= ejΩS + ejΩ(S−1) + · · ·+ 1 + · · ·+ e−jΩ(S−1) + e−jΩS
= 1 + 2 cos(Ω) + 2 cos(2Ω) + · · ·+ 2 cos(SΩ)
We would like to reduce this expression to a closed form to better identify
trends across S.
Working with Sums
Closed form sums of exponential sequences.
A =N−1∑n=0
αn
If the series has finite length (here N terms), it will converge for finite α.
A = 1 + α + α2 + · · · + αN−1
αA = α + α2 + · · · + αN−1 + αN
A− αA = 1 − αN
A =
1− αN
1− α if α 6= 1
N if α = 1
If the series has infinite length, it will converge if |α| < 1.
∞∑n=0
αn = limN→∞
N−1∑n=0
αn = limN→∞
1− αN
1− α = 11− α if |α| < 1
Examples of Fourier Transforms
Find the Fourier Transform (FT) of a rectangular pulse of width 2S+1:
pS [n] =
1 −S ≤ N ≤ S0 otherwise
n
p2[n]
−2 0 2
PS(Ω) =S∑
n=−Se−jΩn =
(ejΩS
) 2S∑n=0
e−jΩn =(ejΩ(S+ 1
2 )
ejΩ/2
)1− e−jΩ(2S+1)
1− e−jΩ
=(ejΩ(S+ 1
2 ) − e−jΩ(S+ 12 )
ejΩ/2 − e−jΩ/2
)=
sin(Ω(S+1
2))
sin (Ω/2)
Ω
P2(Ω)
0 2π−2π
5
Examples of Fourier Transforms
Compare Fourier transforms of pulses with different widths.
n
p2[n]
−2 2
1
Ω
P2(Ω)
2π−2π 2π5
5
n
p4[n]
−4 4
1
Ω
P4(Ω)
2π−2π 2π9
9
n
p6[n]
−6 6
1
Ω
P6(Ω)
2π−2π 2π13
13
As the function widens in n the Fourier transform narrows in Ω.
Moments
Similar to CT, the value of X(Ω) at Ω = 0 is the sum of x[n] over time t.
X(0) =∞∑
n=−∞x[n]e−jΩn =
∞∑n=−∞
x[n]
n
p2[n]
−2 2
1
Ω
P2(Ω)
2π−2π 2π5
5”area” = 5
Moments
The value of x[0] is 12π times the integral of X(Ω) over Ω = [−π, π].
x[0] = 12π
∫2πX(Ω) e jΩndΩ = 1
2π
∫2πX(Ω) dΩ
n
p2[n]
−2 2
1
Ω
P2(Ω)
2ππ-π−2π 2π5
5
Ω
P2(Ω)
2ππ-π−2π 2π5
5equal areas
area
2π = 1++
−−
Notice that the form of this relation is similar to the one for CT at the end
of last lecture (and repeated in the next slide).
Moments
The value of x(0) is the integral of X(ω) divided by 2π.
x(0) = 12π
∫ ∞−∞
X(ω) e jωtdω = 12π
∫ ∞−∞
X(ω) dω
−1 1
x1(t)
1
t++
−− ++ −−
2
π
X1(ω) = 2 sinωω
ω
area
2π = 1
2
πω
equal areas !
Stretching Time
Stretching time compresses frequency and increases amplitude
(preserving area).
−1 1
x1(t)
1
t
2
π
X1(ω) = 2 sinωω
ω
−2 2
1
t
4
πω
Very similar in CT and DT.
Compressing Time to the Limit
Alternatively, we could compress time while keeping area = 1.
−12
12
x(t)
1
t
1
2πω
X(ω) = sinω/2ω/2
ω
−14
14
2
t
1
2πω
In the limit, the pulse has zero width but area 1!
We represent this limit with the delta function: δ(t).
1
t
1
ω
Math With Impulses
Although physically unrealizeable, the impulse (a.k.a. Dirac delta) function
is useful as a mathematically tractable approximation to a very brief signal.
Example 1: Find the Fourier transform of a unit impulse function.
X(ω) =∫ ∞−∞
δ(t)e−jωtdt
Since δ(t) is zero except near t=0, only values of e−jωt near t=0 are impor-
tant. Because e−jωt is a smooth function of t, e−jωt can be replaced by
e−jω0:
X(ω) =∫ ∞−∞
δ(t)e−jω0dt = 1
This matches our previous result which was based explicitly on a limit.
Here the limit is implicit.
Math With Impulses
Although physically unrealizeable, the impulse function is extremely useful
as a mathematically tractable approximation to a very brief signal.
Example 2: Find the function whose Fourier transform is a unit impulse.
x(t) = 12π
∫ ∞−∞
δ(ω)ejωtdω = 12π
∫ ∞−∞
δ(ω)ej0tdω = 12π
1 ctft⇐⇒ 2πδ(ω)
Notice the similarity to the previous result:
δ(t) ctft⇐⇒ 1
These relations are duals of each other.
• A constant in time consists of a single frequency at ω = 0.
• An impulse in time contains components at all frequencies.
Math With Impulses
Delta functions are similarly useful in discrete-time transforms.
Let
X(Ω) =∞∑
m=−∞δ(Ω− 2πm)
where the sum results because DT Fourier Transforms are periodic in 2π.
Then
x[n] = 12π
∫2πX(Ω)ejΩndΩ = 1
2π
∫2πδ(Ω)ejΩndΩ = 1
2π
∫2πδ(Ω) dΩ = 1
2π
Thus if x[n] = 1 for all n, the transform is a delta function in frequency.
1 dtft⇐⇒
∞∑m=−∞
2πδ(Ω− 2πm)
We previously showed a similar result for CT:
1 ctft⇐⇒ 2πδ(ω)
Math With Impulses
Using delta functions to calculate the transforms of complex exponentials.
Let
X(ω) = δ(ω−ωo)then
x(t) = 12π
∫ ∞−∞
X(ω)ejωtdω = 12π
∫ ∞−∞
δ(ω−ωo)ejωtdω
The delta function is zero except when ω = ωo. Therefore we can substitute
ejωot for ejωt without changing the result of the integration.
x(t) = 12π
∫ ∞−∞
δ(ω−ωo)ejωotdω = 12πe
jωot
∫ ∞−∞
δ(ω−ωo)dω = 12πe
jωot
Thus the transform of a complex exponential is a delta at that frequency.
ejωot ctft⇐⇒ 2πδ(ω−ωo)
The analgous expresion for DT holds using analogous reasoning.
ejΩon dtft⇐⇒
∞∑m=−∞
2πδ(Ω−Ωo−2πm)
Relations Between Fourier Series and Fourier Transforms
Continuous Time:
ej2πkT
t ctft⇐⇒ 2πδ(ω−2πk
T
)x(t) = x(t+T ) ctfs⇐⇒ X[k]
x(t) = x(t+T ) =∞∑
k=−∞X[k]ej
2πTkt ctft⇐⇒
∞∑k=−∞
2πX[k]δ(ω−2π
Tk)
Discrete Time:
ej2πkN
n dtft⇐⇒
∞∑m=−∞
2πδ(
Ω−2πkN−2πm
)x[n] = x[n+N ] dtfs⇐⇒ X[k]
x[n] = x[n+N ] =∑k=〈N〉
X[k]ej2πNkn dtft⇐⇒
∑k=〈N〉
∞∑m=−∞
2πX[k]δ(
Ω−2πNk−2πm
)
Relation between Fourier Transform and Fourier Series
Each term in the Fourier series is replaced by an impulse in ω.
· · ·· · ·
x(t) =∞∑
k=−∞xp(t− kT )
t0 T
· · ·· · ·
X[k]
k
· · ·· · ·
X(ω) =∞∑
k=−∞2πX[k] δ(ω − k2π
T)
ω0 2π
T
Summary
Discrete-Time Fourier Transform
• Definition
• Examples
• Properties
• Relations between Fourier series and Fourier transforms (DT and CT)
Quiz 1: Tuesday, March 3, 2-4pm in 32-141.
• Coverage up to and including all of week 4.
• Closed book except for one page of notes (8.5”x11” both sides).
• No electronic devices. (No headphones, cellphones, calculators, ...)