# Why does fft energy leakage was solved

if we take a sinus with Fd=100KHz with fs=1000KHz sampling rate We do an FFT with 256 members we get energy leakage. Why choosing Fd=113.28Khz solves this problem? Thanks

## 1 Answer

See my answer here on how to work with the units in question:

DFT exercise in the book Understanding digital signal processing 3 Ed

Then apply the fact that leakage does not occur when the DFT frame covers a whole number of cycles of your tone signal.

$$Fs = 1000000 \frac{Samples}{Second}$$

$$Fd = 113280 \frac{Cycles}{Second}$$

$$N = 256 \frac{Samples}{Frame}$$

$$\frac{Fd}{Fs} \cdot N = \frac{\frac{Cycles}{Second}}{\frac{Samples}{Second}} \cdot \frac{Samples}{Frame} = \frac{Cycles}{Frame}$$

$$\frac{113280}{1000000} \cdot 256 = 28.99968$$

Which is pretty darn close to a whole number.

• we have a range of 1000KHz which is divided by 256,so each bin is 3.90625KHz. (113.28Khz/256)*1000K – rocko445 Jul 20 '20 at 16:49