Check My Find Convolution of n+1, 0<=n<=2

  • Thread starter Thread starter angel23
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on finding the convolution of the discrete-time signals defined as x[n] = n + 1 for 0 ≤ n ≤ 2 and x[n] = 0 otherwise, with h[n] = a^n u[n]. The derived solution for the convolution is y[n] = a^n + 2a^(n-1) + 3a^(n-2). However, a critical point raised is the oversight regarding the impact of the step function u[n] on the solution, which must be accounted for to ensure accuracy.

PREREQUISITES
  • Understanding of discrete-time signals and systems
  • Familiarity with convolution operations in signal processing
  • Knowledge of the unit step function u[n]
  • Basic algebraic manipulation of exponential functions
NEXT STEPS
  • Study the properties of the convolution operation in discrete-time systems
  • Learn about the effects of the unit step function on signal behavior
  • Explore examples of convolution involving exponential signals
  • Investigate the implications of boundary conditions in signal processing
USEFUL FOR

Students and professionals in signal processing, electrical engineering, and applied mathematics who are working with discrete-time systems and convolution operations.

angel23
Messages
21
Reaction score
0
find convolution of.


n+1 0<=n<=2
x[n]=

0 otherwise

h[n]= a^n u[n]

solution:
y[n]= a^n + 2a ^n-1 + 3 a^n-2


is it right?
 
Physics news on Phys.org
Looks to be, except for the part where you seem to ignore what the step function u[n] does to the solution.
 
Last edited by a moderator:

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
5
Views
2K
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K