New answers tagged finite-impulse-response
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hint only, since this is homework and you could get a lot out of this!
The mathematical operation a filter performs between filter impulse response and signal is called __________(1). When we process both filter and signal with an FFT (which is just a fast implementation of __________(2)), then that operation on the originals is equivalent to __________(3) ...
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However, {f~n} has a time shift comparing to the origin signal, and their amplitudes also have a deviation.
Any causal system has a non-zero group delay. That's to be expected from systems theory!
So, yes, you've got a shift. And, unless your ${h_n}$ is a linear-phase filter, that delay isn't even constant for different frequencies.
This has nothing ...
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The impulse response of a linear phase FIR filter is odd or even symmetrical. This has as a consequence that if $z_0$ is a zero, $1/z_0^*$ must also be a zero. If in addition to the linear phase property you also have real-valued coefficients, complex zeros always occur in complex conjugate pairs. So if a real-valued linear phase filter has a complex root $...
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The canonical way to remove an echo (it can be either electrical echo, or acoustic echo) from a signal is by using adaptive filters. Such an adaptive filter should be able to hold a model of an echo path in the form of an impulse response of this path. Having this model, we can use an original signal to create a copy of its echo, so it can be then subtracted ...
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Down-sampling is pretty straight forward in theory but difficult in practice. IN theory you just need to low-pass filter with a cutoff below half of the new sample rate and then you can just throw away the extra samples.
In practice, the choice of filter involves a lot of trade offs that are highly dependent on the specific requirements of your application, ...
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There are several problems with your approach:
The expression $\displaystyle\frac{1+z^{-1}}{2}$ is the transfer function $H(z)$, NOT its magnitude $|H(z)|$.
Consequently, the expression $\displaystyle \frac{1+e^{-j\Omega}}{2}$ is the complex frequency response $H(e^{j\Omega})$, not its magnitude.
So now you have to compute the squared magnitude of $H(e^{j\...
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