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Quiz 1
Thursday, 3 October, 7:30-9:30pm, room TBD
No recitation next week, no lecture on Thursday 3 Oct.
The quiz is closed book.
No electronic devices (including calculators).
You may use on 8.5x11” sheet of notes (handwritten or printed,
front and back).
Coverage: lectures, recitations, homeworks, and labs up to and in-
cluding 1 Oct.
Practice materials (old exams) are available from the web site.
Last 2 Weeks: Fourier Series
CTFS:
x(t) = x(t+ T ) =∞∑
k=−∞X[k]ej
2πktT
X[k] = 1T
∫ t0+T
t0x(t)e−j
2πktT dt
Last 2 Weeks: Fourier Series
DTFS:
x[n] = x[n+N ] =k+N−1∑k=k0
X[k]ej2πknN
X[k] = X[k +N ] = 1N
n0+N−1∑n=n0
x[n]e−j2πknN
Today: Fourier Transform
Last two weeks: representing periodic signals as sums of sinusoids.
This representation provides insights that are not obvious from other
representations.
However:
• only works for periodic signals
• must know signal’s period before doing the analysis
This is impractical (or impossible) for a large category of signals,
which we still want to be able to analyze using these methods.
Today
Fourier analysis of aperiodic signals: the Fourier transform
Fourier Transform
Consider the following (aperiodic) function of time:
t
x(t)
−S 0 S
1
Can we represent it as a sum of sinusoids?
Toward the Fourier Transform
Let’s start by considering a related signal xp(·), which we create by
summing shifted copies of x(·):
xp(t) =∞∑
m=−∞x(t−mT )
t
xp(t)
−S 0 S−T T
1
Now we can directly find the Fourier Series coefficients of this new
signal (for arbitrarily-chosen T).
However, maybe that’s not really that helpful, since this signal
doesn’t look much like our original signal x(·). How can we fix that?
Toward the Fourier Transform
t
xp(t)
−S 0 S−T T
1
This signal doesn’t really look much like our original. How can we
fix that?
Toward the Fourier Transform
Consequences of T →∞:
The frequency ω0 associated with k = 1 is defined to be 2πT . As we
increase T , ω0 gets smaller, and the spacing between the coefficients
in terms of rad/sec gets smaller and smaller.
For S = 0.5 and for different values of T :
10 5 0 5 10
0.0
0.2
0.4
0.6
0.8
1.0
X p[k
]
T=1
10 5 0 5 10
0.1
0.0
0.1
0.2
0.3
0.4
0.5
X p[k
]
T=2
10 5 0 5 100.05
0.00
0.05
0.10
0.15
0.20
X p[k
]
T=5
10 5 0 5 10
0.02
0.00
0.02
0.04
0.06
0.08
0.10
X p[k
]
T=10
10 5 0 5 10
0.01
0.00
0.01
0.02
0.03
0.04
0.05
X p[k
]
T=20
10 5 0 5 100.005
0.000
0.005
0.010
0.015
0.020
X p[k
]
T=50
Fourier Series to Fourier Transform
Once we have a periodic signal, we can find the FSC:
Xp[k] = 1T
∫Txp(t)e−jkω0tdt
where ω0 = 2πT .
Now we want to think about T → ∞ Let’s replace 1T with ω0
2π , and
explicitly pick a period to integrate over:
Xp[k] = ω02π
∫ T/2
−T/2xp(t)e−jkω0tdt
Fourier Series to Fourier Transform
Now, substitute into the synthesis equation:
xp(t) =∞∑
k=−∞Xp[k]ejkω0t
=∞∑
k=−∞
ω02π
∫ T/2
−T/2xp(t)e−jkω0tdt
ejkω0t
As we take T →∞, a few things happen:
• xp(t)→ x(t)• w0 becomes an infinitesimally small value, ω0 → dω• kω0 becomes a continuum, kω0 → ω (continuous)
• The bounds of integration approach ∞ and ∞ (respectively)
• The outer sum becomes an integral.
x(t) = 12π
∫ ∞−∞
∫ ∞−∞
x(t)e−jωtdtejωtdω
Fourier Series to Fourier Transform
x(t) = 12π
∫ ∞−∞
∫ ∞−∞
x(t)e−jωtdtejωtdω
From here, we’ll define X(ω) such that:
X(ω) =∫ ∞−∞
x(t)e−jωtdt
x(t) = 12π
∫ ∞−∞
X(ω)ejωtdω
X(·) is the Fourier Transform of x(·).
Very similar to the Fourier series, except:
• x(·) need not be periodic
• x(·) can contain all possible frequencies
Example: FT of a Square Pulse
Find the Fourier Transform of a rectangular pulse:
t
x(t)
−1 0 1
1
Check Yourself!
The FT of a rectangular pulse contains almost all frequencies ω.
t
x(t)
−1 0 1
1
−4π −2π 0 2π 4π
2X(ω)
ω
• Why is X(mπ) = 0?
• What is special about those frequencies?
• Why isn’t it zero at ω = 0?
Check Yourself!
A signal and its Fourier transform are shown below:
−2 2
x2(t)
1
t
b
ω1
X2(ω)
ω
Which of the following is true?
1. b = 2 and ω1 = π/22. b = 2 and ω1 = π
3. b = 4 and ω1 = π/24. b = 4 and ω0 = π
5. none of the above
Recap: CT Fourier Relations
Fourier series/transforms express signals by frequency content.
CTFS
x(t) = x(t+ T ) =∞∑
k=−∞X[k]ej
2πktT
X[k] = 1T
∫ t0+T
t0x(t)e−j
2πktT dt
CTFT
x(t) = 12π
∫ ∞−∞
X(ω)ejωtdω
X(ω) =∫ ∞−∞
x(t)e−jωtdt
Recap: CT Fourier Relations
All information in a periodic signal is contained in one period. Infor-
mation in an aperiodic signal is spread across all time.
CTFS
x(t) = x(t+ T ) =∞∑
k=−∞X[k]ej
2πktT
X[k] = 1T
∫ t0+T
t0x(t)e−j
2πktT dt
CTFT
x(t) = 12π
∫ ∞−∞
X(ω)ejωtdω
X(ω) =∫ ∞−∞
x(t)e−jωtdt
Recap: CT Fourier Relations
Harmonic frequencies kω0 are samples of a continuous frequency ω.
CTFS
x(t) = x(t+ T ) =∞∑
k=−∞X[k]ej
2πktT
X[k] = 1T
∫ t0+T
t0x(t)e−j
2πktT dt
CTFT
x(t) = 12π
∫ ∞−∞
X(ω)ejωtdω
X(ω) =∫ ∞−∞
x(t)e−jωtdt
Recap: CT Fourier Relations
Periodic signals can be synthesized from a discrete set of harmonics.
Aperiodic signals generally require all frequencies.
CTFS
x(t) = x(t+ T ) =∞∑
k=−∞X[k]ej
2πktT
X[k] = 1T
∫ t0+T
t0x(t)e−j
2πktT dt
CTFT
x(t) = 12π
∫ ∞−∞
X(ω)ejωtdω
X(ω) =∫ ∞−∞
x(t)e−jωtdt
DT Fourier Transform
We can apply this same idea to DT signals. If x[·] is an arbitrary DT
signal:
X(Ω) =∞∑
n=−∞x[n]e−jΩn
x[n] = 12π
∫2πX(Ω)ejΩndΩ
DT Sinusoids: Aliasing
Because n is an integer, we can only faithfully represent frequencies
in the base band: 0 ≤ Ω ≤ π.
Frequencies outside that range alias to frequencies in that range.
For example, the following graphs are of cosines with Ω = 0.2π, 2.2π,
and 4.2π, respectively:
0 2 4 6 8 10 12 14
1.00
0.75
0.50
0.25
0.00
0.25
0.50
0.75
1.00
0 2 4 6 8 10 12 14
1.00
0.75
0.50
0.25
0.00
0.25
0.50
0.75
1.00
0 2 4 6 8 10 12 14
1.00
0.75
0.50
0.25
0.00
0.25
0.50
0.75
1.00
They all produce exactly the same samples!
Aliasing and DTFT
DT frequencies alias, which has consequences for DTFT analysis.
In particular, where m is an integer:
X(Ω + 2mπ) =∞∑
n=−∞x[n]ej(Ω+2πm)n
=∞∑
n=−∞x[n]ejΩn ej2πmn︸ ︷︷ ︸
=1
=∞∑
n=−∞x[n]ejΩn
= X(Ω)
X(Ω) is periodic in 2π, so all information is contained one period.
Only need to integrate over 2π.
Example: DTFT of Square Pulse
Find the DTFT of a DT rectangular pulse, which is zero outside
the range shown:
0 2-2
Summary
Today, extended Fourier analysis to aperiodic signals by introducing
CTFT and DTFT:
CTFT
x(t) = 12π
∫ ∞−∞
X(ω)ejωtdω X(ω) =∫ ∞−∞
x(t)e−jωtdt
DTFT
x[n] = 12π
∫2πX(Ω)ejΩndΩ X(Ω) =
∞∑n=−∞
x[n]e−jΩn
Next time, properties of Fourier transforms.
