help with complex function
Robert Adams
I have a complex function f(w)
f(w) = 1 + exp(i*a*w) + exp(i*b*w)
where a and b are coefficients and i = sqrt(-1)
For a given value of w (= w0), I am interested in finding the values
of a and b that cause f(w) to become 0. Of course there are infinitely
many combinations of a and b that can satisfy this condition, but I am
interested soltions where, for w != w0, the complex magnitude of f(w)
is as large as possible and does not contain any values that are once
again close to zero. In other words, if I were to plot |h(w)| versus
w, there would be a single well-defined "dip" where the function goes
to zero, and no "dips" anywhere else.
As you can no doubt tell from this posting, I am an engineer and not a
mathematician, and therefore I ask you to have some patience regarding
my imprecise use of terminology!
Bob Adams
jonnie
"Robert Adams" <robert...@analog.com> wrote in message
news:6d7c67cc-640e-4d1c...@a32g2000yqm.googlegroups.com...
restated for a given w find a and b such that
exp(i*a*w) + exp(i*b*w) = -1
trivial for any electrical engineer
goooooooogle for it
Greg Neill
Convert your formula to one using sines and cosines
( using e^iq = cos(q) + i*sin(q) ). Group the real
and imaginary terms.
The function will be zero if and only if the real and
imaginary components are both zero at the same time,
so you have two equations in two unknowns (a and b):
1 + cos(a*w) + cos(b*w) = 0
sin(a*w) + sin(b*w) = 0
Use the properties of the sine function to determine the
relationship between a and b from the second equation:
a = -b or a = b + pi
Substitute for a or b into the first equation and solve
for the remaining unknown. You may find the trig identity
for the sum of cosines comes in handy.
Plot the results to see if you have what you want.
Since your function is periodic, I think you'll find that
the response pattern is going to be repetitive no matter
what you do with constants a and b. The question is, is
w band limited in some fashion?
Robert Adams
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I was hoping that by the correct choice of a and b the function would
NOT be periodic; this is the hard part!
Bob
Greg Neill
> I was hoping that by the correct choice of a and b the function would
> NOT be periodic; this is the hard part!
Hard indeed! Impossible I'd say.
Robert Adams
Well, don't we end up with a sum of two periodic terms with differing
periods? If the two periods were irrationally related to one another,
then I don't see how it's purely periodic.
Bob
Greg Neill
If you examine the magnitude of the function, |f(w)|, it
expands as:
sqrt(3 + 2*cos(a*w) + 2*cos(b*w) + 2*cos(a*w)*cos(b*w) +
2*sin(a*w)*sin(b*w))
combining the product terms using the identity:
cos(A-B) = cos(A)*cos(B) + sin(A)*sin(B)
= sqrt(3 + 2*cos(a*w) + 2*cos(b*w) + 2*cos((a-b)*w))
So you end up with up to three frequencies in the magnitude
response, a*w, b*w, and (a-b)*w. I say up to three frequencies
because you could always choose a=0, or b=0, or a=b.
Such a function will have a period that is the least common
multiple of its component periods.
James Burns
Suppose
f(w) = 1 + exp(i*a*w) + exp(i*b*w)
with a/b irrational.
Then you are right that f(w) would not be periodic,
in that case.
However, we are looking for pairs (a,b) such that
f(w0) = 0. Let us call f(w) = F(w;a,b).
If F(w0; a, b) = 0, then
F(w0; a+2pi/w0, b) = F(w0; a, b+2pi/w0) = 0,
so your solutions will be periodic in a and b.
It also happens that none of the solutions (a,b)
have a/b irrational, so any function f allowing
f(w) = 0 will be periodic in w also, although
that might have turned out differently with a
slightly different problem.
Jim Burns
Robert Adams
Thanks Jim, this is beginning to make sense; could you explain why no
solutions exist with a/b irrantional?
Bob
Lynne Vickson
If we let u = a*w and c = b/a we have f = 1 + exp(i*u) + exp(i*c*u).
For given u, we choose c to make f = 0. The imaginary part gives c =
-1 + 2*n*pi/u, n = 0,1,2, ... (or -1,-2,...); substituting this into
the real part gives 1 + 2*cos(u) = 0 for any n, hence u = arcos(-1/2)
= 2*pi/3 and c = -1 + 3*n.
R.G. Vickson
James Burns
Sure. You need to go back to the solution technique
suggested by Greg upthread.
You get something that looks like
i) sin((A+B)/2)*cos((A-B)/2) = 0
ii) cos((A+B)/2)*cos((A-B)/2) = -1/2
In i, either the sine or the cosine must be zero.
If the sine, (A+B)/2 is an integer multiple of pi.
If the cosine, it's a half-integer multiple.
In ii, we see the cos((A-B)/2) again, and it can't be zero.
So, (A+B)/2 is an integer multiple of pi.
That means cos((A-B)/2) = -1/2 or +1/2. Either way,
(A-B)/2 is an integer multiple of pi/3.
All together, that means A and B are both rational
multiples of pi, and similarly for a and b.
Jim Burns
Robert Adams
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Brilliant; thanks to all for the help!
Bob