# Confused about how Echo filter diagram works [closed]

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.

## 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.

• 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? – A_A Feb 12 '18 at 8:11

$$output = input + delayed~ input × gain$$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.
$$y[n] = 2 \cdot (x[n] + 0.8 \cdot x[n-4000])$$