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the impulse-response and frequency-response of a signal Only systems have impulse responses and frequency responses, signals don't. I assume that what you mean here. A System describes the relation ship between its input signal and its output signal. For an LTI system, that relationship can be captured through either the transfer function or the impulse ...


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Is this because of poles? Yes. A steep drop or rise in a filter's frequency-domain response can only be achieved with a filter that has a long memory. In a FIR filter, this long memory can only come from, well, being long. In IIR filter, this long memory can come from having poles that are close to the unit circle.


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Your assumption why IIR filters can have steeper transitions from passbands to stopbands compared to FIR filters of the same order is correct: IIR filter have poles away from the origin of the complex plane, and poles inside the unit circle close to zeros on the unit circle cause the corresponding frequency response to change rapidly with frequency. The FIR ...


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This is meant as a stepping stone up to Dan's answer. The units for frequency at the sample level are radians per sample. You've got: $$ 400 \frac{cycles}{second} \cdot 2 \pi \frac{radians}{cycle} / 8000 \frac{samples}{second} = \frac{\pi}{10} \frac{radians}{sample} $$ That is your target $\omega_t$. Sound is a real valued signal, so you can model it like ...


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Bottom Line: $$A <1$$ $$B =2\cos(\omega_n)$$ Where $\omega_n$ is the normalized angular frequency of the desired notch location (in this case for the OP with a sampling rate of 8KHz and notch at 400 Hz this would be $\omega_n = 2\pi400/8000 = \pi/10$), resulting in $B \approx 1.902$ and $A$ is the frequency notch bandwidth parameter; the closer $A$ is to ...


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You just need to evaluate the transfer function on the unit circle at the frequency of interest: $$H(e^{j\omega_0})=\frac{\displaystyle\sum_{k=0}^Nb_ke^{-jk\omega_0}}{\displaystyle\sum_{k=0}^Na_ke^{-jk\omega_0}}\tag{1}$$ and take the magnitude. For the special values $\omega_0=0$ and $\omega_0=\pi$, Eq. $(1)$ simplifies to $$H(e^{j0})=\frac{\displaystyle\...


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