Quick convolution integral checking

Click For Summary
SUMMARY

The discussion focuses on evaluating the convolution integral for a linear system with the impulse response defined as g(t) = 3τ² - 4τ + 7 for τ > 0. The user attempts to find the output for the input f(t) = t for t ≥ 0 using the convolution integral formula. The computed result is t⁴/4 - 2t³/3 + 7t²/2, which differs from the official answer that includes a positive sign in the middle term. The user expresses confusion over this discrepancy, despite using Wolfram Alpha for verification.

PREREQUISITES
  • Understanding of convolution integrals in linear systems
  • Familiarity with impulse response functions
  • Knowledge of definite integrals and their evaluation
  • Basic proficiency in using computational tools like Wolfram Alpha
NEXT STEPS
  • Review the properties of convolution in linear time-invariant systems
  • Study the evaluation techniques for definite integrals
  • Learn about the implications of impulse response on system output
  • Explore common pitfalls in convolution calculations and how to avoid them
USEFUL FOR

Students studying signal processing, engineers working with linear systems, and anyone involved in evaluating convolution integrals for system analysis.

dfx
Messages
59
Reaction score
1

Homework Statement



Consider a linear system with the impulse response:

g(t) = 3x^2 - 4x + 7 for t>0 and 0 otherwise.

Find the output for the input f(t) = t for t \geq 0 and f(t) = 0 for t<0.

Homework Equations



\[ \int_{-\infty}^t f(t - \tau)g(\tau)\,d\tau\]


The Attempt at a Solution



\[ \int_0^t f(t - \tau)g(\tau)\,d\tau\]

\[ \int_0^t (t - \tau)(3\tau^2 - 4\tau + 7\,d\tau\)]

and the answer I keep getting is

\frac{t^4}{4} - \frac{2t^3}{3} + \frac{7t^2}{2}

whereas the official given answer has the sign in the middle term as a plus: +\frac{2t^3}{3}

I've even tried wolfram and I think I'm correct:

http://img58.imageshack.us/img58/8637/mspzk2.gif (obviously with different variables - x instead of tau, but still evaluted between t and 0).

If anyone could clear up the correct answer that would be much appreciated.
 
Last edited by a moderator:
Physics news on Phys.org
...anyone?
 
Even I got the same answer as you. So I guess not much of a help.
 

Similar threads

Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 11 ·
Replies
11
Views
7K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K