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Because they made a mistake. The very first equation on the right-hand side is wrong. It should be $$\textrm{DTFT}\big\{x[2n+1]\big\}=\sum_{n=-\infty}^{\infty}x[2n+1]e^{-jn\omega}=\sum_{n\textrm{ odd}}x[n]e^{-j(n-1)\omega/2}\tag{1}$$ Using the trick they suggested, $(1)$ can be written as $$\textrm{DTFT}\big\{x[2n+1]\big\}=\frac{e^{j\omega /2}}{2}\left[\...


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So, assuming you already know that the two streams you're multiplying point-wise are aligned, and you want the output of your system to be the correlation coefficient of zero shift for the length of the CIC-built averager, this is the way to go. So, the output of your CIC will be the sum of all point-wise products of the last N samples flying through. Now, ...


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The IDTFT of $X(e^{j\omega})=1$ is indeed $$x[n]=\frac{\sin(n\pi)}{n\pi}\tag{1}$$ Now, what happens for indices $n\neq 0$? As it turns out, you can safely rewrite $(1)$ as $$x[n]=\delta[n]\tag{2}$$ where $\delta[n]$ is the discrete-time unit impulse. (HINT: think about where the zeros of $\sin(x)$ are).


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In order to filter a discrete-time signal with a linear and time-invariant (LTI) filter, you need to implement a difference equation: $$y[n]=-a_1y[n-1]-a_2y[n-2]-\ldots -a_Ny[n-N]+\\+b_0x[n]+b_1x[n-1]+\ldots+b_Nx[n-N]\tag{1}$$ where $x[n]$ is the input sequence, $y[n]$ is the output sequence, and $a_i$ and $b_i$ are the filter coefficients, which determine ...


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What you need to do is find a digital filter $H(z)$ that behaves similar to the $H(s)$ of your Butterworth filter. That is, you need to read about IIR implementation. There are several methods described in signal processing books about this subject. I recommend you to check for example "Essentials of Digital Signal Processing" from Lathi and Green. Two of ...


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An order-200 IIR filter? You're playing with fire you know that? High-order IIR filter are prone to instability and/or misleading behavior even if they are stable. If I understand correctly, you designed a 50-Hz notch filter, then a 100-Hz notch filter and so on? Do you really need to filter out every harmonic? Do you know how high your harmonics are? ...


3

Your two transfer functions are in parallel, i.e they simply add up. So your feedback transfer function is simply $G(z) = H_1(z)+H_2(z)$. You want to makes sure that the magnitude of $G(z)$ is smaller than one. and the overall closed loop transfer function is $$H(z) = \frac{Y(z)}{X(z)} = \frac{1}{1+H_1(z)+H_2(z)}$$


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Most likely your filter is unstable, i.e some poles on or outside the unit circle. You can test this by executing p = roots(a); display(abs(p) >= 1); If you see any "1" in the result, you have bad poles. This happens primarily through lack of numerical precision. To find the poles from the transfer function you have to calculate the roots of the ...


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I'm pretty sure that your filter is unstable. As mentioned in a comment, it would help if you provided the coefficients in an easily readable form to your question. You can try abs(roots(a)) to compute the magnitude of the filter's poles. If any of these values is larger than $1$ then the filter is unstable, and that would be the explanation why your output ...


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First part for an answer. There are several unknowns in you question. The fist is of course: why do you really want to bring the signal to the same level? and the second: what is this "same level"? Assuming that the baseline is the lower levels, do those simulated signals approximately match your expectations?


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The Python function scipy.signal.filtfilt increases the order of the filter by a factor of 2 because you are doing forward-backward filtering. Traditional filtering techniques employ forward filtering, which generates a phase. To negate the phase, backward filtering is applied to the output of the forward filter. For example, if your original filter (...


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The sharper the filter is in the frequency domain, the longer the impulse response will be. This typically leads to "time blur" or "ringing" in the time domain. In addition, a zero phase filter is non-causal, so you get "pre-ringing" and any sharp onsets or transients in the time domain get degraded. The long impulse response also leads to a long "...


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