**Unformatted text preview: **MEC702 Wk 12 Lect 2 The Fourier Transform of the Convolution
The convolution f ∗g of function ˆ f g and is dened by ˆ ∞ ∞ f (x − p)g(p) dp. f (p)g(x − p) dp = h(x) = (f ∗ g)(x) = −∞ −∞ Theorem. (Convolution Theorem) Suppose that f (x) and g(x) are piecewise
continuous, bounded, and absolutely integrable on the x-axis. Then √ F {f ∗ g} = 2πF {f } · F {g} and the inverse Fourier transform of the convolution is given by ˆ ∞ fˆ(w)ˆ
g (w)eiwx dw (f ∗ g)(x) =
−∞ Discrete Fourier Transform
In the study of Fourier Series, Fourier Integral and Fourier transforms, we assume that a function f (x) that we analyse is continuous or piecewise continuous over some interval.
Very often, a function f (x) is given only in terms of nitely many points. Discrete Fourier Transform (DFT) deals with applying Fourier analysis on these nitely many points.
The main application of such a discrete Fourier analysis concerns large
amounts of equally spaced data, as they occur in telecommunication, time series
analysis, and various simulation problems. In these situations, dealing with sampled values rather than with functions, we can replace the Fourier transform
by the DFT. Let f (x) surements be periodic, for simplicity of period
of the function regularly spaced f (x) 2π . nodes xk = 1 2πk
N N mea0 ≤ x ≤ 2π at We assume that are taken over the interval for k = 0, 1, · · · , N −1. f (x) is being sampled at these points.
complex trigonometric polynomial of the form We also say that Our aim is to nd a q(x) = N
−1
X cn einxk n=0
that interpolates f (x) the symbol fk denoting at the nodes. That is, q(xk ) = f (xk ), written out, with f (xk ), fk = f (xk ) = q(xk ) = N
−1
X cn einxk . n=0
The coecient formula cn is N −1
1 X
fk e−inxk
N cn = k=0 1
N to the other side, hence the
of the given signal We take the factor (DFT) Discrete Fourier Transform f0 f1 f~ = . .. fN −1
is the vector ˆ f0 fˆ1 fˆn = . .. fˆN −1 with fˆn = N cn = N
−1
X fk e−inxk k=0
This is the frequency spectrum of the signal.
In vector notation, we write DFT by the linear system fˆ = FN f~
where the N ×N Fourier matrix FN = [enk ] has the entries enk = e−inxk = e−2πink/N = wnk
with w = e−2πi/N
where n, k = 0, · · · , N − 1.
2 Example. DFT. Sample of N = 4 Values.
Let N =4 measurements (sample values) be given. Then w = e−2πi/N = e−πi/2 = −i
and thus enk = wnk = (−i)nk .
Then F4 e(0)(0) e(0)(1) e(0)(2)
e(1)(0) e(1)(1) e(1)(2)
= e(2)(0) e(2)(1) e(2)(2)
e(3)(0) e(3)(1) e(3)(2) 0 w w0 w0 w0
w0 w1 w2 w3 = w0 w2 w4 w6 w0 w3 w6 w9 1 1
1
1
1 −i −1 i = 1 −1 1 −1 .
1 i −1 −i e(0)(3)
e(1)(3) e(2)(3) e(3)(3) Given a sample signal 0
1 f~ = 4
9
its DFT is 1
1
fˆ = F4 f~ = 1
1 1
−i
−1
i 14
0
1 i 1 = −4 + 8i .
−1 4 −6 −4 − 8i
9
−i 1
−1
1
−1 Remark. We can recreate the original signal again through f~ = FN−1 fˆ
The inverse of the Fourier matrix FN FN−1 =
where FN is the complex conjugate of FN = is given by 1
FN
N FN which itself is given by 1 nk
w
N 3 Fast Fourier Transform
In the more practice sense, DFT incolves alot of calculations and computer
power. For N = 1000 sample points, the straightforward calculations would involve millions of operations.
the so-called However, this diculty can be overcome by Fast Fourier Transform (FFT), for which codes are readily available (eg, in Maple and Matlab).
The FFT is a computational method for the DFT that needs lesser computa- 1
log2 N . It makes the DFT a practical tool for large N. For in1000
stance, take N = 1000, those operations are reduced by a factor
log2 1000 ≈ 100.
N
The breakdown is possible and produces two problems of size M =
2 . Then tions by a factor of w2 = (e−2πi/N )2 = e−4πi/(2M ) = e−2πi/(M ) = wM .
The given signal vector f0 f1 f~ = . .. fN −1 is split into two vectos with M components each, namely, f~even and f~odd For f0 f2 = . .. fN −2 f~even and f~odd , f1 f3 = . . .. fN −1 we determine the DFTs fˆeven = FM f~even
and fˆodd = FM f~odd
M × M matrix FM and ˆ ˆ feven,0
fodd,1 fˆeven,2 fˆodd,3 ˆ
fˆeven = f
= even
.
.
.
. .
.
ˆ
ˆ
feven,N −2
fodd,N −1 involving the same 4 From these vectors, we obtain the components of the DFT of the given vector f~ by the formulas
fˆn = fˆeven,n + (wn ) fˆodd,n , n = 0, · · · , M − 1
fˆn+M = fˆeven,n − (wn ) fˆodd,n , n = 0, · · · , M − 1. Example. FFT. Sample of N = 4 Values from DFT.
Let N = 4, then M = 2. Then w = −i, hence wM = e−2πi/2 = e−πi = −1.
So fˆeven
fˆodd
Since
1 1
f0
f + f2
= F2 f~even =
= 0
1 −1 f2
f0 − f2
1 1
f1
f + f3
= F2 f~odd =
= 1
.
1 −1 f3
f1 − f3 f0 = 0, f1 = 1, f2 = 4 and f3 = 9 from the previous example, then using the formulas fˆ0 = fˆ1 =
fˆeven,0 + w0 fˆodd,0 = (f0 + f2 ) + (1)(f1 + f3 ) = 14
fˆeven,1 + w1 fˆodd,1 = (f0 − f2 ) + (−i)(f1 + f3 ) = −4 + 8i and fˆ0+2
fˆ1+2
fˆeven,0 − w0 fˆodd,0 = (f0 + f2 ) − (1)(f1 + f3 ) = −6
=
fˆeven,1 + w1 fˆodd,1 = (f0 − f2 ) − (−i)(f1 − f3 ) = −4 − 8i. = 5 ...

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- Winter '16
- Alveen Chand
- Fourier Series