- #1
Jncik
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Homework Statement
I have trouble understanding what the components are
suppose we have this discrete time signal
[tex] x[n] = e^{j \frac{2\pi}{3} n} [/tex]
find the number of harmonic components
Homework Equations
The Attempt at a Solution
in my book it says that we have a set of harmonically related signals, that is signals that have a fundamental frequency that is multiple of [tex]\omega_{0}[/tex] where this omega is the fundamental frequency of x[n] in this case
so the set of this signals is [tex] x_{k}[n] = e^{j k \frac{2\pi}{3} n} [/tex]
now,
for k = 0 we have a constant
for k = +1 and -1 we have the first harmonic component since the frequency is the smae
for k = +2 and -2 we have the second harmonic component
for k = +3 and -3 the third
for k = +4 and -4 the fourth which has frequency equal to the one where k = 0
now I don't understand the questions "how many harmonic components the signal has"
assuming that for k = +4, -4 and k = 0 we have the same frequency we can count this as one
for this reason can I say that we have 4 harmonic components?
the first for k = +1 and -1, the second for k = +2 and -2, the third for k = +3 and -3 and finally the fourth for k = +4 and -4?
last question: would it be the same if I started counting from k = 0, to k = 3,-3?