Fourier integrals and the
sampling theorem
Anna-Karin Tornberg
Mathematical Models, Analysis and Simulation
Fall semester, 2011
Fourier Integrals
Read: Strang, Section 4.3.
Fourier series are convenient to describe periodic functions
(or functions with support on a finite interval).
What to do if the function is not periodic?
- Consider f : R → C. The Fourier transform fˆ = F(f ) is given by
� ∞
fˆ(k) =
f (x)e −ikx dx, k ∈ R.
−∞
Here, f should be in L1 (R).
- Inverse transformation:
1
f (x) =
2π
�
∞
fˆ(k)e ikx dk
−∞
Note: Often, you will se a more symmetric version by using a
different scaling.
- Theorem of Plancherel: f ∈ L2 (R) ⇔ fˆ ∈ L2 (R) and
� ∞
� ∞
1
|f (x)|2 dx =
|fˆ(k)|2 dk.
2π
−∞
−∞
Fourier Integrals: The Key Rules
�
df
/dx = ik fˆ(k)
�
f (x
− d) = e −ikd fˆ(k)
Examples of Fourier transforms:
f (x) = sin(ωx)
f (x) = cos(ωx)
f (x) = δ(x)
�
1,
f (x) =
0,
� x�
f (ξ)dξ = fˆ(k)/(ik)
−∞
icx f = fˆ(k − c)
e�
fˆ(k) = −iπ(δ(k − ω) − δ(k + ω))
fˆ(k) = π(δ(k − ω) + δ(k + ω))
fˆ(k) = 1 for all k ∈ R.
if −L ≤ x ≤ L,
if |x| > L.
sin kL
fˆ(k) = 2
= 2Lsinc(kL),
k
where sinc(t) = sin(t)/t is the Sinus cardinalis function.
Sampling of functions - The Nyquist sampling rate
How many samples do we need to identify sin ωt or cos ωt?
Assume equidistant sampling with step size T .
Definition: Nyquist sampling rate: T = π/ω.
The Nyquist sampling rate, is exactly 2 equidistant samples over a full
period of the function. (period is 2π/ω).
For a given sampling rate T , frequencies higher than the Nyquist
frequency ωN = π/T cannot be detected. A higher frequency harmonic is
mapped to a lower frequency one. This effect is called aliasing.
Figure from Strang.
Band-limited functions
– When the continuous signal is a combination of many frequencies ω,
the largest frequency ωmax sets the Nyquist sampling rate at
T = π/ωmax .
– No aliasing at any faster rate. (Do not set sampling rate equal to
the Nyquist sampling rate. Consider sin ωT to see why.)
– A function f (x) is called band-limited if there is a finite W such that
its Fourier integral transform fˆ(k) = 0 for all |k| ≥ W .
The Shannon-Nyquist sampling theorem states that such a function f (x)
can be recovered from the discrete samples with sampling frequency
T = π/W .
(Note that relating to above, W = ωmax + ε, ε > 0. )
The Sampling Theorem
Theorem: (Shannon-Nyquist) Assume that f is band-limited by W , i.e.,
fˆ(k) = 0 for all |k| ≥ W . Let T = π/W be the Nyquist rate. Then it
holds
∞
�
f (x) =
f (nT )sinc(π(x/T − n))
n=−∞
where sinc(t) = sin(t)/t is the Sinus cardinalis function.
Note: The sinc function is band-limited:
�
1, if −π ≤ k ≤ π,
�
sinc(k)
=
0, elsewhere.
Also note: This interpolation procedure is not used in practice. Slow
decay, summing an infinite number of terms. Approximations to the sinc
function are used, introducing interpolation errors.
The Sampling Theorem: Proof
Assume for simplicity W = π.
By the inverse Fourier transform,
� ∞
� π
1
1
f (x) =
fˆ(k)e ikx dk =
fˆ(k)e ikx dk.
2π −∞
2π −π
Define
f˜(k) =
�
fˆ(k),
periodic continuation,
if −π < k < π,
if |k| ≥ π.
f˜ can be represented as a Fourier series in k-space,
∞
�
˜
f (k) =
fˆ˜n e ink ,
n=−∞
where
˜n = 1
fˆ
2π
�
π
1
f˜(k)e −ink dk =
2π
−π
Hence, for −π < k < π,
fˆ(k) = f˜(k) =
�
∞
�
π
fˆ(k)e −ink dk = f (−n).
−π
f (−n)e ink .
n=−∞
The Sampling Theorem: Proof, contd.
Therefore,
1
f (x) =
2π
�
π
fˆ(k)e ikx dk
−π
�
� π � �
∞
1
=
f (−n)e ink e ikx dk
2π −π n=−∞
� π
∞
�
1
=
f (−n)
e ik(x+n) dk
2π −π
n=−∞
=
=
∞
�
n=−∞
∞
�
n=−∞
f (−n)
f (n)
sin π(x + n)
π(x + n)
sin π(x − n)
π(x − n)
If W different from π, rescale x by π/W and k by W /π to complete the
proof.