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Is this correct? No. This only works if filter and signal have the same sample rate. Either up-sample the signal or down-sample the impulse response, depending on what you want your output sample rate to be. It might be the easiest to down-sample the impulse response to 44.1kHz Resampling from 48kHz to 44.1kHz is fairly awkward and involves a fair bit of ...

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The approach to frequency shift a signal is best described with complex signals (the easiest way I find to explain SSB as well). We can frequency translate a real waveform x(t) by converting it to an analytic signal with a Hilbert Transform (which for narrow band signals simply means using a quadrature splitter), and then multiplying this resulting complex ...

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even though everything important has been said I wanted to add some code and more visual keys on why the following formulas require positve and negative frequencies to make clear that negative frequencies are important for canceling out the counterparts in the inverse dft. So lets see for $\sin z = \frac{1}{2\mathrm{i}} \left(\mathrm{e}^{\mathrm{i}z} - \... 0 Assuming that you know the set of fundamental frequencies, you should be able to calculate the THD for either signal (original or encoded) as $$\frac{\sqrt{v_2^2 + v_3^2 + \ldots + v_n^2}}{v_1}$$ where$v_1$is the signal amplitude at the fundamental frequency and$v_2, \ldots, v_n$are the signal amplitudes at the harmonics of interest for that fundamental ... 2 To visualize the frequencies of discrete time signals beyond the sampling rate, simply insert$M-1$zeros in between each sample and scale the signal by$M$. This will extend the frequency axis by$M$where$M\$ is any positive integer. What you will see is the periodicity in the frequency domain as given for discrete time signals.

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