rho = x .* (x < pi) + (x - 2*pi) .* (x > pi); rho(N/2+1)=0
k = [ 0:N/2-1, -N/2:-1 ]; k(1) = 1;
phi = ifft(- fft(rho) ./ k.^2);
18.336 Problem Set 1
Due Tuesday, 21 February 2006.
Problem 1:
(The k(1)=1 command is a hack to avoid divid-
Trigonometric interpo-
ing zero by zero when we divide by
lation polynomials
(a)
f (xn ) = fn at N points
2π
n
, n = 0, · · · , N − 1, and you construct
N
the trigonometric interpolation polynomial
φ00N (x)
You are given a function
ation of
xn =
mand twice to phi and scaling by
f (x) =
N
−1
X
φ00N (x)
(b)
e
e
`k = 1,
term with
showed for
i(k+`k N )x
ck
∆N
if you replace any
for any integer
`k
(we
but other values follow by
N = 10
many possible trigonometric interpolations with
(N −1)/2
ikx
ck e
on a
Consider the Cooley-Tukey decimation-in-time
(DIT) radix-2 FFT algorithm applied to compute the DFT of length
N
−1
X
+
k=0
(b)
from
Problem 3: Fast Fourier Transforms
`Rk 's that min2π
1
0
2
imizes the mean-square slope 2π 0 |f (x)| dx
(for arbitrary fn ) is the symmetric polynomial:
f (x) =
∆N for a sequence of N values
N = 10000 and plot ∆N vs. N
to
to be odd for simplicity. Show
that the unique choice of integer
X
φ2N (x), and compute
dence (a straight line). What is the exponent?
fn .
N
points to get
log-log scale. You should get a power-law depen-
the same ck coecients passing through the same
Assume
2N
v
u
N
u1 X
2π
2π
|φN ( n) − φ2N ( n)|2
≡t
N n=0
N
N
Compute
induction). This means that there are actually
(a)
that
Estimate the error by repeating the calcu-
via the DFT. We showed in class that one would
ikx
1/∆x2 )
the root-mean-square dierence
k=0
get the same coecient
(applying Matlab's di com-
≈ ρ(x).
lation for
ck eikx
k 2 .)
Show by a simple nite-dierence evalu-
fn
As we showed
in class, this recursively decomposes the prob-
i(k−N )x
ck e
lem into two DFTs of length
k=(N +1)/2
Suppose that the
N = 2m .
N/2
of the even-
and odd-indexed inputs, respectively.
(a)
values are real num-
Assume that the inputs are complex num-
to be
bers, and that a complex addition takes 2 real
Show that the
additions and a general complex multiplication
above symmetric polynomial satises this (i.e.
takes 4 real multiplications and 2 real additions,
bers. We would like the interpolated
real as well for all
real
f (x)
x ∈ [0, 2π).
for arbitrary real
polynomial with real
fn ).
f (x)
and that all twiddle factors (cosine and sine val-
Is it the unique
ues) are precomputed. The exact count of real
f (x)?
adds+multiplies is of the form # · N log2 N +
O(N ). Show what # is. (Count multiplications
by ±1 and ±i as free, or equivalently subtraction
Problem 2: Solving Poisson's equation
counts as addition.)
(b)
Here, you will use Matlab to explore the solution
of Poisson's equation
boundary conditions
Suppose that the inputs are purely real,
in which case the outputs
00
φ (x) = ρ(x) with periodic
on x ∈ [0, 2π) via spectral
try
cN −k = c∗k .
ck
have the symme-
Show that, by applying radix-2
methods and FFTs. In particular, the following
Cooley-Tukey directly to this data and tossing
φ(x) given a sawtooth
function ρ(x) = x for 0 ≤ x < π and ρ(x) =
x − 2π for π < x < 2π using N = 100 points to
get an approximate solution φN (x):
out the redundant computations, we can com-
Matlab code solves for
pute the DFT in
(#/2) · N log2 N + O(N )
adds+multiplies.
(Numerical Recipes fans may
real
know that you can compute a real-input DFT
for even
N=100;
x = linspace(0,2*pi, N+1); x = x(1:end-1);
N
by embedding it into a complex-input
DFT of length
1
N/2.
Don't use this trick here.)