# Tag Info

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It is common in DSP practice to define some convenient center for a filter as being at time 0, even though we cannot build non-causal systems in practice. You see this most when you're designing a symmetrical filter, and you define t = 0 as the filter center, but it happens elsewhere. You do this because it makes the analysis easier, and you justify it by ...

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Actual channels are always causal (like everything else in the physical universe). Actual (discrete-time) channels also sometimes have one tap that is considerably larger than the rest; an example impulse response would be h = [0.1, 1.5, 0.2]. Some authors prefer to define h[0] as the largest tap; in my example, we'd have h[-1] = 0.1, h[0] = 1.5, and h[1] = ...

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For $\mu > 0$, $z^L \in \left ( \sqrt[L]{\mu} \right) e^{j 2\pi n/L} \,\forall\, n \in 1 \cdots L$. I.e., the roots are evenly spaced on a circle $\sqrt[L]{\mu}$ in radius, and there's an $L$ of a lot of them.

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If you are not concerned on the phase and just want to approximate a magnitude response, then your first option should be the frequency sampling method implemented in Matlab/Octave fir2() function. You would provide the frequency grid and corresponding frequency response magnitude at those frequencies. As you have also mentioned, least-squares approach is ...

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This is inspired by the excellent answer by @Richard Lyons (which I upvoted) and the comment by @DSP Novice. It basically combines what they said. If the width of the 5000-amplitude pulse is always the same, then simply do as Richard Lyons suggested: it works fine. If the width varies, then the following scheme could be used: I imported your raw data and ...

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Is the width of your 5000-amplitude pulse always the same? If so, then just discard your signal samples that occur after seven seconds. If the width of your 5000-amplitude pulse varies then try lowpass filtering (experimenting with different lowpass filter bandwidths) your signal to see if the filtered signal contains the information you desire. If the width ...

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HINT: It looks like they shifted the frequency response by half the sampling frequency, i.e., $$H_{HP}\left(e^{j\omega}\right)=H_{LP}\big(e^{j(\omega-\pi)}\big)\tag{1}$$ Frequency shifting corresponds to modulation (multiplication) in the time domain. Now you just have to figure out the modulation sequence that achieves the correct frequency shift.

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Watch after 30mins... above question is addressed. Consider: $$A_{4} (z) = 1 + h_1 z^{-1} + h_2 z^{-2} + h_1 z^{-3} + z^ {-4}$$ $$A_4 (z) = A_3 (z) + k_4 * B_3 (z)$$ $$A_4 (z) = A_3 (z) + k_4 * z^ {-4} (A_3 (z^ {-1}))$$ Try to break $A_4(z)$ into two equal halves:  A_4 (z) = (1 + h_1 z^ {-1} + \frac{h_2}{2} z^ {-2} ) + z^ {-4} (...

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Scanned below is the fred harris' "Rule of Thumb" which included the filter taps as well as his other "rules". This is from DSP World ICSPAT Class Notes DSP World Workshops, Orlando, Florida, November 1-4, 1999. That course along with other similar presentations and courses by fred harris have significantly influenced my views and thinking of signal ...

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