New answers tagged dft
6
Start with what we know:
The DFT works perfectly for a single cycle waveform input. For a single cycle waveform, all harmonics have an integer number of cycles (by the definition of “harmonic”).
“Spectral leakage” occurs when the input is not a single cycle waveform. As such, at least one frequency component is not represented by an integer number of cycles....
2
IMHO, leakage is a poor term for it. Best representation is a much more descriptive though wordy description.
Imagine a stereo with one of those displays with the bars that correspond to frequency. They usually bounce around a lot, but they do give a representative view of the frequency content.
Now, let's take the stereo into the lab and feed it a sweep ...
3
Imagine that a continuous signal has a frequency component of amplitude $A$ at exactly 20 Hz (you can imagine it is alone, a single perfect wave).
With a FFT, or any discrete Fourier transform, you could hope that, if you have acquired the signal correctly (above Nyquist), the discrete spectrum will give you a clear peak with amplitude $A$ at 20 Hz. This is ...
1
staying faithful to the subject "removing window effect in freq domain via convolution" (despite the OP perhaps wanted to achieve something else or something similar), I feel to add my comment having personal experience with this specific topic.
Often I have the necessity to remove a Hann window in frequency domain, working in a STFT framework which uses ...
0
I came back to this and tried deriving the discrete version which helped make things more sense:
Somehow $f_k t_n = f(n, k, N)$
$ f_k = \frac{f_s}{N}k$ and $ t_n = \frac{T}{N}n $
$f_s = \frac{N}{T}$
So
$f_k t_n = \frac{f_s}{N}k\frac{T}{N}n = \frac{N}{TN}k\frac{T}{N}n = \frac{kn}{N}$
Done!
0
Two conjugate poles:
$(z-(0.51+0.7i))(z-(0.51-0.7i)) = z^{2} - 1.02z +0.76$
Two conjugate zeros:
$(z-(0.57+0.78i))(z-(0.57-0.78i)) = z^{2} - 1.14z +0.94$
1
The idea here is to build the problem in its Matrix Form.
We have the filter $ h $ which is represented by the matrix $ H $ to represent Linear Convolution operation:
$$ H = \begin{bmatrix}
{h}_{1} & 0 & 0 & \ldots & & 0 \\
{h}_{2} & {h}_{1} & 0 & \ldots & & 0 \\
\vdots & & \ddots & & & \...
0
Check LMS in Wikipedia, it is an iterative way, works for any size.
0
The area under an impulse function is one. So if you multiply a continuous time waveform with an impulse, the result will be a weighted impulse with the area equal to whatever the waveform was at that moment in time.
1
Assume $x[n]$ and $y[n]$ represent two jointly WSS (wide-sense stationary) discrete-time random processes. Further assume that they have zero means, without losing the generality.
If $x[n]$ and $y[n]$ are independent, then they are also uncorrelated, and since they have zero means, they are also orthogonal; i.e.
$$ E\{ x[n]y^*[n-k]\} = 0 \tag{1}$$
Let $$...
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