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The Fourier series for the signal x(n)=cos√2πn exists.
- a)True
- b)False
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The Fourier series for the signal x(n)=cos√2πn exists.a)Trueb...
Explanation: For ω0=√2π, we have f0=1/√2. Since f0 is not a rational number, the signal is not periodic. Consequently, this signal cannot be expanded in a Fourier series.
Most Upvoted Answer
The Fourier series for the signal x(n)=cos√2πn exists.a)Trueb...
The Fourier series for the signal x(n) = cos(n) can be found by expressing the signal as a sum of sinusoidal components with different frequencies and amplitudes.
The general form of the Fourier series representation for a periodic signal x(n) is:
x(n) = A0 + Σ[Ak * cos(kω0 * n) + Bk * sin(kω0 * n)]
where A0, Ak, and Bk are the coefficients of the Fourier series, ω0 is the fundamental frequency, and k is an integer.
For the signal x(n) = cos(n), we can see that A0 = 0, and the only non-zero coefficient is A1 = 1, since cos(n) can be written as cos(1 * n).
Therefore, the Fourier series for x(n) = cos(n) is:
x(n) = Σ[cos(ω0 * n)]
where ω0 = 1.
The general form of the Fourier series representation for a periodic signal x(n) is:
x(n) = A0 + Σ[Ak * cos(kω0 * n) + Bk * sin(kω0 * n)]
where A0, Ak, and Bk are the coefficients of the Fourier series, ω0 is the fundamental frequency, and k is an integer.
For the signal x(n) = cos(n), we can see that A0 = 0, and the only non-zero coefficient is A1 = 1, since cos(n) can be written as cos(1 * n).
Therefore, the Fourier series for x(n) = cos(n) is:
x(n) = Σ[cos(ω0 * n)]
where ω0 = 1.
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