## FFT example

This example is aimed at demonstrating the effect of bandwidth on the measurement of some standard waveforms. The waveforms considered are

1. A single frequency (narrow band),

The single frequency and the ringwave have the same unit maximum amplitude. The sample rate used is 1.024 MHz. This is suggested in IEC Standard 61000 4-30. This gives a 2 kHz resolution (or measurement bandwidth) for 512 samples. The program initialises at that resolution. The single frequency wave considered has a unit amplitude and frequency of 64 kHz.

The 0.5 μs-100 kHz ring wave is used to represent inductively induced lightning surges in circuits and is often used to test surge protection devices it has a rise time of 0.5 μs and a ringing frequency of 100 kHz and is defined by:

$V\left(t\right)=AV_p\left(1-e^\frac{-t}{\tau_1}\right)e^\frac{-t}{\tau_2}\cos{\left(\omega t\right)}$

where $$\tau_1 = 0.533~\mu s$$, $$\tau_2 = 9.788~\mu s$$, $$\omega = 2\pi \cdot 10^5$$ rads and $$A =1.59$$.

The white noise has a normal distribution with zero mean and a standard devation of 0.1.

The slider enables the bandwidth to be adjusted by $$2^n$$ kHz where $$n$$ is an integer. This is to make the FFT efficient and to avoid the complication of spectral leakage. Bandwidth changes are achieved by changing the time window of the data as illustrated in the time domain plot.

Notice that each signal responds differently to changes in bandwidth. The narrow band signal amplitude is independent of bandwidth in the frequency domain. The broadband ringwave amplitude is proportional to bandwidth in the frequency domain and the white noise amplitude varies according to the square root of the bandwidth.