Given the following Fourier series coefficients for a voltage:
av = 10 V, ak = (12/k)* cos (k * pi/2), and bk = (12/k) * ( 1 – sin(k*pi/2))
Find the trigonometric form of the Fourier series coefficients A1 through A3 and theta1 through theta3.
If the period T is 104.7 ms, write a expression for the voltage using the trigonometric coefficients (equation 16.20).
What is 1) the average voltage and 2) the frequency for which there is no voltage contribution (in Hz or rad/sec)?
If the voltage is applied to a 30 ohm resistor, what is an estimate of the average power delivered to the resistor based on up to the first three harmonics. Don’t forget to include the dc component.
The correct answer and explanation is:
To solve this problem, we need to break it into steps:
Step 1: Given Fourier coefficients
- av=10 Va_v = 10 \, \text{V} (DC component).
- ak=12kcos(kπ/2)a_k = \frac{12}{k} \cos(k \pi / 2) (cosine coefficients for k≥1k \geq 1).
- bk=12k(1−sin(kπ/2))b_k = \frac{12}{k} (1 – \sin(k \pi / 2)) (sine coefficients for k≥1k \geq 1).
Step 2: Trigonometric form of the Fourier coefficients
The trigonometric form involves amplitude and phase: Ak=ak2+bk2,θk=tan−1(bkak)A_k = \sqrt{a_k^2 + b_k^2}, \quad \theta_k = \tan^{-1}\left(\frac{b_k}{a_k}\right)
For k=1,2,3k = 1, 2, 3:
- a1=121cos(π/2)=0a_1 = \frac{12}{1} \cos(\pi/2) = 0, b1=121(1−sin(π/2))=12(1−1)=0b_1 = \frac{12}{1}(1 – \sin(\pi/2)) = 12(1 – 1) = 0, so A1=0A_1 = 0.
- a2=122cos(2π/2)=122cos(π)=−6a_2 = \frac{12}{2} \cos(2\pi/2) = \frac{12}{2} \cos(\pi) = -6, b2=122(1−sin(2π/2))=122(1−0)=6b_2 = \frac{12}{2}(1 – \sin(2\pi/2)) = \frac{12}{2}(1 – 0) = 6, so A2=(−6)2+62=62A_2 = \sqrt{(-6)^2 + 6^2} = 6\sqrt{2} and θ2=tan−1(6/(−6))=3π/4\theta_2 = \tan^{-1}(6/(-6)) = 3\pi/4.
- a3=123cos(3π/2)=0a_3 = \frac{12}{3} \cos(3\pi/2) = 0, b3=123(1−sin(3π/2))=4(1+1)=8b_3 = \frac{12}{3}(1 – \sin(3\pi/2)) = 4(1 + 1) = 8, so A3=8A_3 = 8 and θ3=tan−1(8/0)=π/2\theta_3 = \tan^{-1}(8/0) = \pi/2.
Step 3: Voltage expression using AkA_k and θk\theta_k
The Fourier series in trigonometric form is: v(t)=av+∑k=1∞Akcos(kω0t+θk)v(t) = a_v + \sum_{k=1}^{\infty} A_k \cos(k \omega_0 t + \theta_k)
For the first three harmonics: v(t)=10+62cos(2ω0t+3π/4)+8cos(3ω0t+π/2)v(t) = 10 + 6\sqrt{2} \cos(2\omega_0 t + 3\pi/4) + 8 \cos(3\omega_0 t + \pi/2)
Here, ω0=2πT\omega_0 = \frac{2\pi}{T} and T=104.7 msT = 104.7 \, \text{ms}, so ω0=2π0.1047≈60 rad/s\omega_0 = \frac{2\pi}{0.1047} \approx 60 \, \text{rad/s}.
Step 4: Average voltage and frequency with no contribution
- Average voltage: The average voltage is the DC component, av=10 Va_v = 10 \, \text{V}.
- Frequency with no contribution: Harmonics with zero coefficients (Ak=0A_k = 0) have no contribution. For k=1k = 1, A1=0A_1 = 0, so the first harmonic f1=ω0/(2π)≈60/(2π)≈9.55 Hzf_1 = \omega_0 / (2\pi) \approx 60 / (2\pi) \approx 9.55 \, \text{Hz} has no contribution.
Step 5: Average power delivered to a 30-ohm resistor
Power is calculated using Parseval’s theorem: P=1R(av22+∑k=1NAk22)P = \frac{1}{R} \left( \frac{a_v^2}{2} + \sum_{k=1}^N \frac{A_k^2}{2} \right)
Including up to three harmonics: P=130(1022+(62)22+822)P = \frac{1}{30} \left( \frac{10^2}{2} + \frac{(6\sqrt{2})^2}{2} + \frac{8^2}{2} \right) P=130(50+36+32)=11830≈3.93 W.P = \frac{1}{30} \left( 50 + 36 + 32 \right) = \frac{118}{30} \approx 3.93 \, \text{W}.
Explanation
The voltage v(t)v(t) is represented using Fourier series coefficients, including a DC component ava_v, cosine terms aka_k, and sine terms bkb_k. These coefficients describe the amplitude and phase of harmonics at frequencies kω0k\omega_0, where ω0=2π/T\omega_0 = 2\pi/T. Using the provided coefficients, the trigonometric form is derived, where Ak=ak2+bk2A_k = \sqrt{a_k^2 + b_k^2} gives the amplitude, and θk=tan−1(bk/ak)\theta_k = \tan^{-1}(b_k/a_k) gives the phase for each harmonic. This form facilitates writing the voltage expression clearly.
The DC component, av=10 Va_v = 10 \, \text{V}, represents the average voltage. Additionally, harmonics with zero coefficients contribute no voltage; in this case, the first harmonic (k=1k = 1) has A1=0A_1 = 0, so its frequency (9.55 Hz9.55 \, \text{Hz}) is identified as one with no contribution. The power delivered to a 30-ohm resistor is estimated using Parseval’s theorem, summing contributions from the DC component and the first three harmonics. The total power is calculated to be approximately 3.93 W3.93 \, \text{W}, demonstrating the voltage’s effect when applied to a resistive load. This approach combines frequency domain analysis and circuit theory to estimate power accurately.