Discrete Time Convolution Part 1-Digital Signal Processing-Assignment, Exercises of Digital Signal Processing

This assignment wa given by Prof. Bhuvanesh Sankuratri at Baddi University of Emerging Sciences and Technologies for Digital Signal Processing course. Its main points are: Discrete, Time, Convolution, Impulse, Response, Sequences, Start, Variable, System

Typology: Exercises

2011/2012

Uploaded on 07/14/2012

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Discrete-Time Convolution:
1. Find the impulse response for each of the following discrete-time systems:
a) y[n] + 0.2y[n-1] = x[n]-x[n-1]
b) y[n] + 1.2y[n-1] = 2x[n-1]
c) y[n] = 0.24(x[n]+x[n-1]+x[n-2]+x[n-3])
d) y[n] = x[n] + 0.5x[n-1] + x[n-2]
2. Perform the following convolutions, x[n]*v[n]
a) x[n] = u[n] - u[n-4], v[n] = 0.5nu[n]
b) x[n] = [1 4 8 2]; v[n] = [0 1 2 3 4] (the sequences both start at n=0)
c) x[n] = u[n], v[n] = 2(0.8)nu[n]
d) xn un[] [ ]=−1, vn un
n
[] (.) []=205
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Discrete-Time Convolution:

  1. Find the impulse response for each of the following discrete-time systems: a) y[n] + 0.2y[n-1] = x[n]-x[n-1] b) y[n] + 1.2y[n-1] = 2x[n-1] c) y[n] = 0.24(x[n]+x[n-1]+x[n-2]+x[n-3]) d) y[n] = x[n] + 0.5x[n-1] + x[n-2]
  2. Perform the following convolutions, x[n]*v[n] a) x[n] = u[n] - u[n-4], v[n] = 0.5 nu[n] b) x[n] = [1 4 8 2]; v[n] = [0 1 2 3 4] (the sequences both start at n=0) c) x[n] = u[n], v[n] = 2(0.8) nu[n]

d) x n[ ] = u n[ − 1 ] , v n[ ] = 2 0 5(. ) nu n[ ]

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