This measurement column addresses frequently asked questions received by our customer support center and provides answers to those questions.
In FFT analysis (Fourier analysis), a time-domain waveform is segmented to a certain length, and the segmented time-domain waveform is Fourier transformed to obtain a Fourier series. The resulting Fourier series, plotted on the frequency axis, is called a Fourier spectrum.
FFT time length and frequency resolution
The relationship between the length of the time waveform extracted for FFT analysis (FFT time length [seconds]), the frequency resolution [Hz], the frequency range of the FFT analysis, and the number of sample points is as follows:
Number of lines [points] = Number of sample points [points] ÷ 2.56
Frequency resolution [Hz] = Frequency range [Hz] ÷ Number of lines [points]
FFT time length [seconds] = 1 ÷ frequency resolution [Hz]
For a frequency range of 8 kHz and 2048 sample points, the number of lines is 800, the FFT time length is 0.1 seconds, and the frequency resolution is 10 Hz. A time waveform of 0.1 seconds is used for each FFT. Performing an FFT analysis yields spectral values at 10 Hz intervals. Frequency component values at finer intervals than 10 Hz, such as the 995 Hz component or the 1001 Hz component, cannot be obtained.
Spectrum of a signal that does not fit within the frequency resolution
When a sine wave with an amplitude of 1 V and a frequency of 1 kHz is subjected to FFT analysis under the conditions of a frequency range of 8 kHz and 2048 sample points, the value of the 1 kHz component in the resulting spectrum is 1 V.
The spectrum obtained when performing an FFT analysis on a signal with frequencies outside the frequency resolution depends on the window function used.
When performing an FFT analysis using a Hanning window on a sine wave with an amplitude of 1 V and a frequency of 995 Hz, under the conditions of a frequency range of 8 kHz and 2048 sample points, the values of the 990 Hz component and the 1 kHz component are approximately 0.85 V, respectively. The amplitude of a signal at a frequency that does not fall within the frequency resolution (10 Hz) is split into the frequency components immediately before and after that frequency.
Examples of measured spectra of signals that do not fit within the frequency resolution.
Table 1 shows the component values from 980 Hz to 1020 Hz, obtained by performing an FFT analysis on a sine wave with amplitude 1, varying its frequency from 990 Hz to 1010 Hz. The sum of squares is the sum of the squares of these five components. The square root of the sum of squares is also shown. The Hanning window function was used.
Table 1. FFT analysis results of sine waves (990–1010 Hz)
| Sine wave frequency | 980 Hz component | 990 Hz component | 1000 Hz component | 1010 Hz component | 1020 Hz component | Sum of squares | The square root of the sum of squares |
| 990 Hz | 0.500 | 1.000 | 0.500 | 0.000 | 0.000 | 1.500 | 1.225 |
| 991 Hz | 0.430 | 0.994 | 0.571 | 0.018 | 0.004 | 1.500 | 1.225 |
| 992 Hz | 0.352 | 0.974 | 0.652 | 0.047 | 0.010 | 1.500 | 1.225 |
| 993 Hz | 0.289 | 0.945 | 0.719 | 0.079 | 0.015 | 1.499 | 1.225 |
| 994 Hz | 0.222 | 0.898 | 0.792 | 0.124 | 0.021 | 1.499 | 1.224 |
| 995 Hz | 0.170 | 0.849 | 0.849 | 0.170 | 0.024 | 1.499 | 1.224 |
| 996 Hz | 0.124 | 0.792 | 0.898 | 0.222 | 0.026 | 1.500 | 1.225 |
| 997 Hz | 0.079 | 0.719 | 0.944 | 0.289 | 0.026 | 1.500 | 1.225 |
| 998 Hz | 0.047 | 0.652 | 0.974 | 0.352 | 0.022 | 1.500 | 1.225 |
| 999 Hz | 0.018 | 0.571 | 0.994 | 0.430 | 0.013 | 1.500 | 1.225 |
| 1000 Hz | 0.000 | 0.500 | 1.000 | 0.500 | 0.000 | 1.500 | 1.225 |
| 1001 Hz | 0.014 | 0.426 | 0.994 | 0.575 | 0.020 | 1.500 | 1.225 |
| 1002 Hz | 0.022 | 0.354 | 0.975 | 0.650 | 0.046 | 1.500 | 1.225 |
| 1003 Hz | 0.026 | 0.287 | 0.943 | 0.721 | 0.080 | 1.500 | 1.225 |
| 1004 Hz | 0.027 | 0.225 | 0.901 | 0.788 | 0.121 | 1.500 | 1.225 |
| 1005 Hz | 0.024 | 0.170 | 0.849 | 0.849 | 0.170 | 1.499 | 1.225 |
| 1006 Hz | 0.020 | 0.121 | 0.788 | 0.901 | 0.225 | 1.499 | 1.224 |
| 1007 Hz | 0.015 | 0.080 | 0.721 | 0.943 | 0.287 | 1.499 | 1.225 |
| 1008 Hz | 0.010 | 0.046 | 0.650 | 0.975 | 0.354 | 1.500 | 1.225 |
| 1009 Hz | 0.005 | 0.020 | 0.575 | 0.994 | 0.426 | 1.500 | 1.225 |
| 1010 Hz | 0.000 | 0.000 | 0.500 | 1.000 | 0.500 | 1.500 | 1.225 |
Looking at the 1000 Hz component of a 1000 Hz sine wave, its amplitude is 1, correctly representing the amplitude of a sine wave. The 990 Hz and 1010 Hz components are 0.5. This is because the spectrum is spread horizontally due to the Hanning window. The sum of the squares of the 980 Hz to 1020 Hz components is 1.5. This is also due to the horizontal spread caused by using the Hanning window. This value of 1.5 indicates the degree to which the spectrum is spread horizontally and is called the equivalent noise bandwidth.
The 990 Hz and 1000 Hz components of a 995 Hz sine wave show a value of 0.849. This is because the magnitude of the 995 Hz sine wave has been split into the preceding and succeeding frequency components. Looking at the peak value of the spectrum, it will be smaller than the original amplitude. In the worst case, this difference is 0.849 times, or -1.42 dB. The frequency at which the spectrum peaks will be shifted by up to half the frequency resolution, or 5 Hz.
For sine waves with frequencies that are not integer multiples of the frequency resolution (10 Hz), the magnitude of its components is divided into preceding and succeeding components, and is less than 1. However, the sum of the squares of the five preceding and succeeding components is 1.5. Dividing this by the equivalent noise bandwidth (1.5) and then taking the square root gives the precise amplitude of the sine wave (1).
summary
This time, we presented the results of FFT analysis (Fourier analysis) of a sine wave using the Hanning window function. In FFT analysis, spectra are obtained for each frequency resolution, so if the frequency of the sine wave is not an integer multiple of the frequency resolution, its amplitude will be separated into preceding and succeeding frequency components and will be a smaller value. To obtain accurate values, it is necessary to change the settings to a finer frequency resolution.
(Excerpt from the email newsletter issued on October 21, 2020)
