The homework from last time was, "What is the combined power of sine waves with amplitudes of 2V and 3V?" The answer is, of course, not 12.5 (= 5 x 5/2) by simply thinking 2 + 3 = 5. We need to add them in terms of power, so the power is...
2 x 2 / 2 ＋ 3 x 3 / 2 ＝ 6.5 （V2）
Therefore, the rms value is 2.55 (V).
(More precisely, it refers to two sine waves with different frequencies.)
In power spectroscopy, complex time-domain signals are decomposed into frequency components, and the power of each component is determined. But what exactly is the frequency resolution capability? This column is about the fundamentals of digital signal processing, so the following explanation will focus on the Discrete Fourier Transform (DFT), which was explained in the third installment. This is the processing of discrete digital time-series signals obtained by A/D conversion of analog time-domain signals.
To perform a DFT calculation, we first extract a finite time window (also called the sampling time) of 1 second from an infinite time signal. This results in a fundamental period of 1 second and a fundamental frequency of 1 Hz. What would the results be if we separately sampled four sine wave signals at 0.5 Hz, 1 Hz, 1.5 Hz, and 2 Hz within this time window and performed a DFT?
As you can intuitively see, 1 Hz and 2 Hz are integer multiples of the fundamental frequency, so line spectra appear at the first and second points.
So, what happens in the case of 0.5 Hz (period of 2 seconds)?
Since only half a period has been sampled, it is impossible to determine whether the signal is 0.5 Hz or not. In this example, the minimum frequency resolution is 1 Hz.
Generally, if the time window length is T seconds, the frequency resolution Δf (Hz) is:
Δf ＝ 1/T
This calculates the frequency for each Δf. In this example, it's 1 Hz, 2 Hz, 3 Hz, ...
So, since 1.5 Hz is outside the resolution range, does that mean no spectral power is generated at all? Actually, that's not the case.
DFT spectral analysis analyzes the signal width Δf. The main power at 1.5 Hz separates into 1 Hz and 2 Hz components. (Strictly speaking, a small amount leaks out above 3 Hz as well.) Similarly, the power at 0.5 Hz separates into a DC component with an infinite period (i.e., a DC component with a frequency of 0) and a 1 Hz component.
When calculating the spectrum using DFT, the frequency resolution is Δf, which is the reciprocal of the time window length. The spectrum then appears with a width of Δf, at intervals of Δf. In this example, with a width of 1 Hz, the spectra appear at 1 Hz, 2 Hz, and 3 Hz. If the time window length is 2 seconds, the spectra appear with a width of 0.5 Hz, at 0.5 Hz, 1 Hz, 1.5 Hz, 2 Hz, and so on.
Next time, we will explain the relationship between the sampling frequency and analysis frequency of time-domain signals.
(Excerpt from the email newsletter issued on February 24, 2003)
