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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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