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Math  /  Calculus

QuestionExample: Find the trigonometric Fowrier series representation) of the signal below:

Studdy Solution

STEP 1

1. The function x(t) x(t) is periodic with period T=4 T = 4 .
2. The function has a piecewise definition within one period: x(t)=2 x(t) = 2 for −1≤t<1 -1 \le t < 1 and x(t)=−1 x(t) = -1 for 1≤t<3 1 \le t < 3 .
3. The Fourier series representation of x(t) x(t) will be in the trigonometric form: x(t) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos\left( \frac{2 \pi n t}{T} \right) + b_n \sin\left( \frac{2 \pi n t}{T} \right) \right) \]
4. The coefficients \( a_0 \), \( a_n \), and \( b_n \) are calculated using the integral formulas: a_0 = \frac{1}{T} \int_{0}^{T} x(t) \, dt \] a_n = \frac{2}{T} \int_{0}^{T} x(t) \cos\left( \frac{2 \pi n t}{T} \right) \, dt \] b_n = \frac{2}{T} \int_{0}^{T} x(t) \sin\left( \frac{2 \pi n t}{T} \right) \, dt \]

STEP 2

1. Determine the period T T .
2. Calculate the DC component a0 a_0 .
3. Compute the cosine coefficients an a_n .
4. Compute the sine coefficients bn b_n .
5. Combine the coefficients to form the Fourier series representation.

STEP 3

Determine the period T T of the function.
Given that the function repeats every 4 units, we have: T=4 T = 4

STEP 4

Calculate the DC component a0 a_0 .
a0=1T∫04x(t) dt a_0 = \frac{1}{T} \int_{0}^{4} x(t) \, dt
Since x(t) x(t) is piecewise: a0=14(∫−112 dt+∫13−1 dt) a_0 = \frac{1}{4} \left( \int_{-1}^{1} 2 \, dt + \int_{1}^{3} -1 \, dt \right)

STEP 5

Evaluate the integrals for a0 a_0 .
∫−112 dt=2×(1−(−1))=4 \int_{-1}^{1} 2 \, dt = 2 \times (1 - (-1)) = 4 ∫13−1 dt=−1×(3−1)=−2 \int_{1}^{3} -1 \, dt = -1 \times (3 - 1) = -2 a0=14(4+(−2))=14×2=12 a_0 = \frac{1}{4} (4 + (-2)) = \frac{1}{4} \times 2 = \frac{1}{2}

STEP 6

Calculate the cosine coefficients an a_n .
an=24(∫−112cos⁡(2πnt4) dt+∫13−1cos⁡(2πnt4) dt) a_n = \frac{2}{4} \left( \int_{-1}^{1} 2 \cos\left( \frac{2 \pi n t}{4} \right) \, dt + \int_{1}^{3} -1 \cos\left( \frac{2 \pi n t}{4} \right) \, dt \right) an=12(∫−112cos⁡(πnt2) dt+∫13−1cos⁡(πnt2) dt) a_n = \frac{1}{2} \left( \int_{-1}^{1} 2 \cos\left( \frac{\pi n t}{2} \right) \, dt + \int_{1}^{3} -1 \cos\left( \frac{\pi n t}{2} \right) \, dt \right)

STEP 7

Evaluate the integrals for an a_n .
For the first integral: ∫−112cos⁡(πnt2) dt=2[2πnsin⁡(πnt2)]−11=4πn(sin⁡(πn2)−sin⁡(−πn2)) \int_{-1}^{1} 2 \cos\left( \frac{\pi n t}{2} \right) \, dt = 2 \left[ \frac{2}{\pi n} \sin\left( \frac{\pi n t}{2} \right) \right]_{-1}^{1} = \frac{4}{\pi n} \left( \sin\left( \frac{\pi n}{2} \right) - \sin\left( -\frac{\pi n}{2} \right) \right) Since sin⁡(−θ)=−sin⁡(θ) \sin( -\theta ) = -\sin( \theta ) : =4πn(sin⁡(πn2)+sin⁡(πn2))=8πnsin⁡(πn2) = \frac{4}{\pi n} \left( \sin\left( \frac{\pi n}{2} \right) + \sin\left( \frac{\pi n}{2} \right) \right) = \frac{8}{\pi n} \sin\left( \frac{\pi n}{2} \right)
For the second integral: ∫13−cos⁡(πnt2) dt=−[2πnsin⁡(πnt2)]13=−2πn(sin⁡(3πn2)−sin⁡(πn2)) \int_{1}^{3} - \cos\left( \frac{\pi n t}{2} \right) \, dt = - \left[ \frac{2}{\pi n} \sin\left( \frac{\pi n t}{2} \right) \right]_{1}^{3} = - \frac{2}{\pi n} \left( \sin\left( \frac{3 \pi n}{2} \right) - \sin\left( \frac{\pi n}{2} \right) \right) Since sin⁡(3πn2)=−sin⁡(πn2) \sin \left( \frac{3 \pi n}{2} \right) = - \sin \left( \frac{\pi n}{2} \right) for odd n n and 0 0 for even n n : =−2πn(−sin⁡(πn2)−sin⁡(πn2))=4πnsin⁡(πn2) = - \frac{2}{\pi n} \left( - \sin \left( \frac{\pi n}{2} \right) - \sin \left( \frac{\pi n}{2} \right) \right) = \frac{4}{\pi n} \sin\left( \frac{\pi n}{2} \right)
Combine the results: an=12(8πnsin⁡(πn2)+4πnsin⁡(πn2))=12(12πnsin⁡(πn2))=6πnsin⁡(πn2) a_n = \frac{1}{2} \left( \frac{8}{\pi n} \sin\left( \frac{\pi n}{2} \right) + \frac{4}{\pi n} \sin\left( \frac{\pi n}{2} \right) \right) = \frac{1}{2} \left( \frac{12}{\pi n} \sin\left( \frac{\pi n}{2} \right) \right) = \frac{6}{\pi n} \sin\left( \frac{\pi n}{2} \right)

STEP 8

Calculate the sine coefficients bn b_n .
bn=24(∫−112sin⁡(2πnt4) dt+∫13−1sin⁡(2πnt4) dt) b_n = \frac{2}{4} \left( \int_{-1}^{1} 2 \sin\left( \frac{2 \pi n t}{4} \right) \, dt + \int_{1}^{3} -1 \sin\left( \frac{2 \pi n t}{4} \right) \, dt \right) bn=12(∫−112sin⁡(πnt2) dt+∫13−1sin⁡(πnt2) dt) b_n = \frac{1}{2} \left( \int_{-1}^{1} 2 \sin\left( \frac{\pi n t}{2} \right) \, dt + \int_{1}^{3} -1 \sin\left( \frac{\pi n t}{2} \right) \, dt \right)

STEP 9

Evaluate the integrals for bn b_n .
For the first integral: ∫−112sin⁡(πnt2) dt \int_{-1}^{1} 2 \sin\left( \frac{\pi n t}{2} \right) \, dt By symmetry and the properties of sine (an odd function), the integral over symmetric limits around 0 is zero.
For the second integral: ∫13−sin⁡(πnt2) dt \int_{1}^{3} - \sin\left( \frac{\pi n t}{2} \right) \, dt Again, by symmetry and the properties of sine, the integral over symmetric limits around 2 is zero.
Hence, bn=0 b_n = 0 for all n n .

STEP 10

Combine the coefficients to form the Fourier series representation.
Given the coefficients: a0=12 a_0 = \frac{1}{2} an=6πnsin⁡(πn2) a_n = \frac{6}{\pi n} \sin\left( \frac{\pi n}{2} \right) bn=0 b_n = 0
The Fourier series representation is: x(t)=12+∑n=1∞(6πnsin⁡(πn2)cos⁡(πnt2)) x(t) = \frac{1}{2} + \sum_{n=1}^{\infty} \left( \frac{6}{\pi n} \sin\left( \frac{\pi n}{2} \right) \cos\left( \frac{\pi n t}{2} \right) \right)

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