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Example 6 Express the signal x(t)x(t) as Fourier series. Express the Fourier seri...
May 9, 2024
Solution by Steps
step 1
To find the Fourier series coefficients of the given function x(t)=1+sin⁡(ω0t)+2cos⁡(ω0t)+cos⁡(2ω0t+π4)x(t) = 1 + \sin(\omega_0 t) + 2 \cos(\omega_0 t) + \cos(2 \omega_0 t + \frac{\pi}{4}), we will use the Fourier series formula for periodic functions
step 2
The Fourier series of a function x(t)x(t) is given by a0+∑n=1∞[ancos⁡(nω0t)+bnsin⁡(nω0t)]a_0 + \sum_{n=1}^{\infty} [a_n \cos(n \omega_0 t) + b_n \sin(n \omega_0 t)], where a0a_0 is the average value of the function, ana_n are the cosine coefficients, and bnb_n are the sine coefficients
step 3
The coefficient a0a_0 is the average value of the function over one period, which is a0=1a_0 = 1 since the constant term in x(t)x(t) is 1
step 4
The coefficient a1a_1 corresponds to the cosine term with the fundamental frequency ω0\omega_0, which is a1=2a_1 = 2 due to the term 2cos⁡(ω0t)2 \cos(\omega_0 t)
step 5
The coefficient b1b_1 corresponds to the sine term with the fundamental frequency ω0\omega_0, which is b1=1b_1 = 1 due to the term sin⁡(ω0t)\sin(\omega_0 t)
step 6
The term cos⁡(2ω0t+π4)\cos(2 \omega_0 t + \frac{\pi}{4}) can be expanded using the cosine of a sum formula: cos⁡(A+B)=cos⁡(A)cos⁡(B)−sin⁡(A)sin⁡(B)\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)
step 7
Applying the formula from step 6, we get cos⁡(2ω0t+π4)=cos⁡(2ω0t)cos⁡(π4)−sin⁡(2ω0t)sin⁡(π4)\cos(2 \omega_0 t + \frac{\pi}{4}) = \cos(2 \omega_0 t)\cos(\frac{\pi}{4}) - \sin(2 \omega_0 t)\sin(\frac{\pi}{4})
step 8
Since cos⁡(π4)=sin⁡(π4)=22\cos(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}, the term becomes 22cos⁡(2ω0t)−22sin⁡(2ω0t)\frac{\sqrt{2}}{2} \cos(2 \omega_0 t) - \frac{\sqrt{2}}{2} \sin(2 \omega_0 t)
step 9
The coefficient a2a_2 is the amplitude of the cosine term with frequency 2ω02 \omega_0, which is a2=22a_2 = \frac{\sqrt{2}}{2}
step 10
The coefficient b2b_2 is the amplitude of the sine term with frequency 2ω02 \omega_0, which is b2=−22b_2 = -\frac{\sqrt{2}}{2}
step 11
The Fourier series for x(t)x(t) is therefore x(t)=1+2cos⁡(ω0t)+sin⁡(ω0t)+22cos⁡(2ω0t)−22sin⁡(2ω0t)x(t) = 1 + 2 \cos(\omega_0 t) + \sin(\omega_0 t) + \frac{\sqrt{2}}{2} \cos(2 \omega_0 t) - \frac{\sqrt{2}}{2} \sin(2 \omega_0 t)
step 12
To express the coefficients in polar form, we convert each rectangular coefficient pair (an,bn)(a_n, b_n) to polar form using rn=an2+bn2r_n = \sqrt{a_n^2 + b_n^2} and θn=arctan⁡(bnan)\theta_n = \arctan(\frac{b_n}{a_n})
step 13
For n=1n=1, we have r1=22+12=5r_1 = \sqrt{2^2 + 1^2} = \sqrt{5} and θ1=arctan⁡(12)\theta_1 = \arctan(\frac{1}{2})
step 14
For n=2n=2, we have r2=(22)2+(−22)2=12r_2 = \sqrt{(\frac{\sqrt{2}}{2})^2 + (-\frac{\sqrt{2}}{2})^2} = \sqrt{\frac{1}{2}} and θ2=arctan⁡(−2222)=arctan⁡(−1)=−π4\theta_2 = \arctan(\frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}) = \arctan(-1) = -\frac{\pi}{4}
Answer
The Fourier series for x(t)x(t) is x(t)=1+2cos⁡(ω0t)+sin⁡(ω0t)+22cos⁡(2ω0t)−22sin⁡(2ω0t)x(t) = 1 + 2 \cos(\omega_0 t) + \sin(\omega_0 t) + \frac{\sqrt{2}}{2} \cos(2 \omega_0 t) - \frac{\sqrt{2}}{2} \sin(2 \omega_0 t). The coefficients in polar form are r1=5r_1 = \sqrt{5}, θ1=arctan⁡(12)\theta_1 = \arctan(\frac{1}{2}), r2=12r_2 = \sqrt{\frac{1}{2}}, and θ2=−π4\theta_2 = -\frac{\pi}{4}.
Key Concept
Fourier Series Coefficients
Explanation
The Fourier series coefficients for a function can be found by identifying the amplitudes of the sine and cosine terms at different frequencies. These coefficients can be expressed in rectangular form as (an,bn)(a_n, b_n) and converted to polar form with magnitude rnr_n and phase θn\theta_n.
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