# Tag Info

## New answers tagged digital-filters

0

In the first formulation, $\Omega_c$ is the $3$ $\textrm{dB}$ cut-off frequency, because for $\Omega=\Omega_c$ you obtain $|H(\Omega_c)|^2=\frac12$. In the other formulation, the attenuation at the pass band edge $\Omega_p$ is determined by the constant $\epsilon$. For $\Omega=\Omega_p$ you have $|H(\Omega_p)|=\frac{1}{1+\epsilon^2}$. I.e., by choosing the ...

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My personal point of view: what matters much more than the distribution is the second-order statistics of the channel. In particular things like delay spread / coherence bandwidth and Doppler spread / coherence time. These parameters tell you a thing or two that you must take into account when estimating your channels: The coherence bandwidth (roughly equal ...

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Cleanliness of the eye pattern sample point is also used for symbol synchronization for NDA timing recovery. My communications book essentially ignored this, I don't know why. But carrier and symbol synchronization seems to be completely ignored in many newer digital communications texts.

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When considering a transceiver system, if we do a certain filtering operation on the signal, will the BER of the system be affected? Yes it does. The whole of equalization concept in digital communication is to mitigate the effect of transmit signal getting filtered by channel (which includes right from the Transmit chain components, the communication ...

4

It is the objective of the receiver to make the best estimate for each symbol as to what was transmitted. This is often done by ultimately determining a decision time in each sample (through timing recovery) on the waveform after it has been processed by the receiver (equalization and matched filtering) in which to sample the waveform and make a decision as ...

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Help design my system would be far more detail than what we typically provide here, but I can give you some suggestions that may help or at least lead to a more concise question. I do not get correct BER, which means my signal is getting aliased as I receive through the DDC. Not getting the correct BER does not necessarily conclude this. In your ...

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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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If you're familiar with LaPlace transforms, you can see the Z transform by analogy. The unit circle is equivalent to the jw axis, with zero frequency at 1+j0 and the Nyquist rate at -1+j0.

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BIBO stability of LTI systems implies that their impulse response is absolutely summable, that is, $$\sum_{n=-\infty}^{+\infty}|h(n)| < +\infty$$ That exact same relationship is a sufficient condition for the Fourier Transform of the impulse response - the so-called Frequency Response - to converge. Convergence of the ...

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We get the Fourier Transform of a signal at the unit circle. If the ROC does not include the unit circle, that means that the Fourier transform does not converge which means that the system is unstable. Also, please read bores signal processing basics website and Alan Oppenheim. Its explained really well there.

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