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All realizable causal IIR (Infinite Impulse Response) systems essentially have feedback. In discrete time systems this is given by poles that are not at the origin. (If the poles are inside the unit circle, we further know the system is stable). Given a generic transfer function as follows with zeros in the numerator ($z_1, z_2 \ldots$) and poles in the ...


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Because the definition of "feedback" is contextual, you can't. If you're talking about physical systems, any transfer function that's not of the form $\frac{A}{s^n}$ has feedback, because the only way to make a non-zero pole is with feedback. So in theory, non-zero poles means there's feedback. By this same theory, any zeros that aren't at ...


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It depends, if you have a graphical representation of your system, you can see that the system has feedback or feedforward elements. Or you can look at the difference equation or transfer function. For example a discrete integrator $y[n] = x[n] - y[n-1]$ has feedback since the previous output is fed back to the current output. On the other hand, a ...


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I'm not gonna give you a complete answer but I can help you. Your open-loop transfer function has 2 poles at $ -1 ±j \sqrt(3) $ Good news, they are stable. However, they are not damped and they are slow. Your strategy is this 1 - Damp the poles, this will reduce the overshoot. 2 - Try to increase the poles frequency. This will reduce the settling time. 3 - ...


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Answering my own question, my S domain filter has a better representation: $$H(s)=\frac{s^2+\mu\omega_bs+\omega_o^2}{s^2+\omega_bs+\omega_o^2}$$ mu is the unitless attenuation of the signal at the bandstop frequency. Wb is the bandwidth in r/s at the -3db point Wo is the bandstop frequency in r/s. The comparison between my existing Z domain filter and the ...


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$$ \begin{align*} \frac{Y(z)}{X(z)}&=g\frac{1+a_1}{1+a_1z^{-1}} &\text{...given transfer function}\\ Y(z)(1+a_1z^{-1})&=X(z)g(1+a_1) &\text{...via cross multiplication}\\ Y(z)+a_1z^{-1}Y(z)&=g(1+a_1)X(z) &\text{...via distribution}\\ y[n]+a_1y[n-1]&=g(1+a_1)x[n] &\text{...via inverse z-transform}\\ y[n]&=g(1+a_1)x[n]-...


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