傅里叶级数分析：周期信号展开与基波周期、功率等特性求解

Let's break down this periodic signal analysis problem thoroughly, step by step. I'll cover finding the fundamental period, calculating average power using Parseval's theorem, and outlining how to plot the discrete amplitude and phase spectra.

First, let's find the period of each individual component in (s(t)=2\sin (1000\pi t)+0.5\sin (500\pi t)+\cos (250\pi t)):

• For (2\sin(1000\pi t)): Angular frequency (\omega_1 = 1000\pi) rad/s, so period (T_1 = \frac{2\pi}{\omega_1} = \frac{1}{500} = 0.002) seconds.
• For (0.5\sin(500\pi t)): Angular frequency (\omega_2 = 500\pi) rad/s, so period (T_2 = \frac{2\pi}{\omega_2} = \frac{1}{250} = 0.004) seconds.
• For (\cos(250\pi t)): Angular frequency (\omega_3 = 250\pi) rad/s, so period (T_3 = \frac{2\pi}{\omega_3} = \frac{1}{125} = 0.008) seconds.

The fundamental period (T_0) is the least common multiple (LCM) of these individual periods. Converting to fractions with a common denominator: (T_1 = \frac{1}{500}), (T_2 = \frac{2}{500}), (T_3 = \frac{4}{500}). The LCM of the numerators (1, 2, 4) is 4, so (T_0 = \frac{4}{500} = \frac{1}{125} = 0.008) seconds.

Alternatively, using frequencies: the fundamental frequency (f_0) is the greatest common divisor (GCD) of the component frequencies ((f_1=500) Hz, (f_2=250) Hz, (f_3=125) Hz). The GCD is 125 Hz, so (T_0 = \frac{1}{f_0} = 0.008) seconds.

We're using the trigonometric Fourier series form:
$$s(t)=a_0+\sum_{n=1}^{{\infty}a_n\cos(n\omega_0t)+\sum_{n=1}}{\infty}b_n\sin(n\omega_0t)$$
where (\omega_0 = \frac{2\pi}{T_0} = 250\pi) rad/s (the fundamental angular frequency).

Matching each component to the series:

• (a_0 = 0) (no DC component in (s(t)))
• For (n=1): (\cos(1\cdot\omega_0 t) = \cos(250\pi t)), so (a_1 = 1), (b_1 = 0)
• For (n=2): (\sin(2\cdot\omega_0 t) = \sin(500\pi t)), so (a_2 = 0), (b_2 = 0.5)
• For (n=4): (\sin(4\cdot\omega_0 t) = \sin(1000\pi t)), so (a_4 = 0), (b_4 = 2)
• All other (n): (a_n = 0), (b_n = 0)

Parseval's theorem states that the average power of a periodic signal equals the sum of the average powers of its individual frequency components. Mathematically:
$$P = \frac{1}{T_0}\int_{T_0}|s(t)|^2dt = a_0^2 + \frac{1}{2}\sum_{n=1}^{\infty}(a_n2 + b_n^2)$$

Plugging in our coefficients:

• (a_0^2 = 0)
• (n=1): (\frac{1}{2}(1^2 + 0^2) = 0.5)
• (n=2): (\frac{1}{2}(0^2 + 0.5^2) = 0.125)
• (n=4): (\frac{1}{2}(0^2 + 2^2) = 2)

Summing these up gives the total average power:
$$P = 0 + 0.5 + 0.125 + 2 = 2.625$$

We can verify this by calculating the power of each sinusoidal component directly (average power of (A\sin(\omega t)) or (A\cos(\omega t)) is (\frac{A^2}{2})):

• (2\sin(1000\pi t)): (\frac{2^2}{2} = 2)
• (0.5\sin(500\pi t)): (\frac{0.5^2}{2} = 0.125)
• (\cos(250\pi t)): (\frac{1^2}{2} = 0.5)
  Total power is the same: (2 + 0.125 + 0.5 = 2.625).

Amplitude Spectrum #

The amplitude of each harmonic (n) is given by (c_n = \sqrt{a_n^2 + b_n^2}):

• (n=1): (c_1 = \sqrt{1^2 + 0^2} = 1)
• (n=2): (c_2 = \sqrt{0^2 + 0.5^2} = 0.5)
• (n=4): (c_4 = \sqrt{0^2 + 2^2} = 2)
• All other (n): (c_n = 0)

To plot this, use a discrete plot where the x-axis is either the harmonic number (n) or the angular frequency (n\omega_0), and the y-axis is (c_n). You'll have vertical lines at (n=1,2,4) with heights 1, 0.5, 2 respectively.

Phase Spectrum #

The phase (\phi_n) for each harmonic is derived from the trigonometric identity (a_n\cos(n\omega_0t) + b_n\sin(n\omega_0t) = c_n\cos(n\omega_0t + \phi_n)):

• (n=1): The component is (\cos(250\pi t)), which is (1\cdot\cos(250\pi t + 0)), so (\phi_1 = 0) radians (or 0°)
• (n=2): The component is (0.5\sin(500\pi t) = 0.5\cos(500\pi t - \frac{\pi}{2})), so (\phi_2 = -\frac{\pi}{2}) radians (or -90°)
• (n=4): The component is (2\sin(1000\pi t) = 2\cos(1000\pi t - \frac{\pi}{2})), so (\phi_4 = -\frac{\pi}{2}) radians (or -90°)
• All other (n): Phase is undefined (or can be set to 0 since amplitude is 0)

内容的提问来源于stack exchange，提问作者user300045