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The picture above is the block diagram of an echo filter. My question is, how does the delay, and gain block affect the output signal? I think that when the input signal is outputted with a delay from when the user and a smaller amplitude.

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closed as too broad by Marcus Müller, lennon310, A_A, jojek Feb 13 '18 at 11:38

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Can you please edit the question for clarity? The way it stands now, it requires a complete explanation of delay units and their effect on loops which can go on for a while. In the meantime, are you clear with the basics of the type of filter your question is about? $\endgroup$ – A_A Feb 12 '18 at 8:11
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Without the attenuation compensation block the output of your echo filter can be written like:
$$output = input + delayed~ input × gain$$The delay block delays your input signal by a specific number of samples and then the delayed signal is multiplied by the gain ($gain<1$ due to losses in the echo path). So you are basically adding a delayed version of your original signal to the original signal.

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The output is the sum of the input and a scaled and delayed version of the input

$$y[n] = 2 \cdot (x[n] + 0.8 \cdot x[n-4000])$$

The multiplier of 2 seems a little suspicious. Divide by 2 (or multiply by 0.5) would make more sense.

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