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What are the Fourier series coefficients for the signal x(n)=cosπn/3?
- a)c1=c2=c3=c4=0,c1=c5=1/2
- b)c0=c1=c2=c3=c4=c5=0
- c)c0=c1=c2=c3=c4=c5=1/2
- d)None of the mentioned
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
What are the Fourier series coefficients for the signal x(n)=cosπn/...
Explanation: In this case, f0=1/6 and hence x(n) is periodic with fundamental period N=6.
Given signal is x(n)= cosπn/3=cos2πn/6=1/2 e^(j2πn/6)+1/2 e^(-j2πn/6)
We know that -2π/6=2π-2π/6=10π/6=5(2π/6)
So, we get c1=c2=c3=c4=0 and c1=c5=1/2.
Given signal is x(n)= cosπn/3=cos2πn/6=1/2 e^(j2πn/6)+1/2 e^(-j2πn/6)
We know that -2π/6=2π-2π/6=10π/6=5(2π/6)
So, we get c1=c2=c3=c4=0 and c1=c5=1/2.
Most Upvoted Answer
What are the Fourier series coefficients for the signal x(n)=cosπn/...
The Fourier series coefficients for the signal x(n) = cos(n) can be found by applying the formula:
cn = (1/N) * Σ [x(n) * e^(-i * 2π * n * k / N)]
where:
- cn represents the nth Fourier series coefficient,
- N is the period of the signal,
- x(n) is the signal,
- k is the index of the coefficient.
In this case, the signal x(n) = cos(n), and let's assume N is the period of the signal.
cn = (1/N) * Σ [cos(n) * e^(-i * 2π * n * k / N)]
To simplify the expression, let's rewrite cos(n) using Euler's formula:
cos(n) = (1/2) * (e^(i * n) + e^(-i * n))
cn = (1/N) * Σ [(1/2) * (e^(i * n) + e^(-i * n)) * e^(-i * 2π * n * k / N)]
Expanding the expression:
cn = (1/2N) * (Σ [e^((1 - 2πk / N) * i * n)] + Σ [e^((-1 - 2πk / N) * i * n)])
Using the geometric series formula, we can evaluate the first summation:
Σ [e^((1 - 2πk / N) * i * n)] = e^((1 - 2πk / N) * i * n) / (1 - e^((1 - 2πk / N) * i))
Similarly, we can evaluate the second summation:
Σ [e^((-1 - 2πk / N) * i * n)] = e^((-1 - 2πk / N) * i * n) / (1 - e^((-1 - 2πk / N) * i))
Finally, substituting these expressions back into the original formula, we get:
cn = (1/2N) * (e^((1 - 2πk / N) * i * n) / (1 - e^((1 - 2πk / N) * i)) + e^((-1 - 2πk / N) * i * n) / (1 - e^((-1 - 2πk / N) * i)))
These are the Fourier series coefficients for the signal x(n) = cos(n).
cn = (1/N) * Σ [x(n) * e^(-i * 2π * n * k / N)]
where:
- cn represents the nth Fourier series coefficient,
- N is the period of the signal,
- x(n) is the signal,
- k is the index of the coefficient.
In this case, the signal x(n) = cos(n), and let's assume N is the period of the signal.
cn = (1/N) * Σ [cos(n) * e^(-i * 2π * n * k / N)]
To simplify the expression, let's rewrite cos(n) using Euler's formula:
cos(n) = (1/2) * (e^(i * n) + e^(-i * n))
cn = (1/N) * Σ [(1/2) * (e^(i * n) + e^(-i * n)) * e^(-i * 2π * n * k / N)]
Expanding the expression:
cn = (1/2N) * (Σ [e^((1 - 2πk / N) * i * n)] + Σ [e^((-1 - 2πk / N) * i * n)])
Using the geometric series formula, we can evaluate the first summation:
Σ [e^((1 - 2πk / N) * i * n)] = e^((1 - 2πk / N) * i * n) / (1 - e^((1 - 2πk / N) * i))
Similarly, we can evaluate the second summation:
Σ [e^((-1 - 2πk / N) * i * n)] = e^((-1 - 2πk / N) * i * n) / (1 - e^((-1 - 2πk / N) * i))
Finally, substituting these expressions back into the original formula, we get:
cn = (1/2N) * (e^((1 - 2πk / N) * i * n) / (1 - e^((1 - 2πk / N) * i)) + e^((-1 - 2πk / N) * i * n) / (1 - e^((-1 - 2πk / N) * i)))
These are the Fourier series coefficients for the signal x(n) = cos(n).
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