Find an echo filter 'h'. 'x' is a vector containing the original recording.

'h * x' is the convolution of the original signal 'x' and the filter 'h' and it contains the original recording plus an echo of the original recording delayed by 0.25 seconds with half the amplitude. Build filter 'h' using Julia (Julia code is not necessary. Just want to understand the logic and thinking behind this). What does 'h' look like?

sampling rate = 44100 per sec.

Original recording is 10 seconds long. Hence the total is 441000 samples.

(Hint) - The result vector 'y' of the convolution 'h * x' satisfies the equation of the form

y_i= x_i+ 0.5x_(i-k)

where x_j= 0 for j outside the range 1,...,441000, and k = number of samples in 0.25 seconds.