Where Does the δ(t) Term Come From in First Order Unit Step Response?

Click For Summary
SUMMARY

The δ(t) term in the first order unit step response equation x' + kx = (r/k + c2)δ(t) represents an impulse function that is crucial for understanding system behavior at t=0. This term evaluates to zero for t≠0, indicating that it only influences the system at the exact moment of the impulse. The confusion arises from the interpretation of δ(t) as it does not evaluate to (r/k + c2) outside of t=0, but rather acts as a mathematical tool to model instantaneous changes in the system.

PREREQUISITES
  • Understanding of differential equations, specifically first-order linear equations.
  • Familiarity with the Dirac delta function and its properties.
  • Basic knowledge of control systems and unit step responses.
  • Experience with mathematical modeling in engineering contexts.
NEXT STEPS
  • Study the properties of the Dirac delta function in detail.
  • Learn about impulse response and its significance in control theory.
  • Explore first-order linear differential equations and their solutions.
  • Investigate the application of Laplace transforms in analyzing system responses.
USEFUL FOR

Students and professionals in engineering, particularly those focused on control systems, signal processing, and mathematical modeling, will benefit from this discussion.

charlie.elvers
Messages
2
Reaction score
0
I think I mostly understand how this works, and it makes intuitive sense. However, I'm a little bit confused by one step in the proving of this.

On the second page of the attached PDF, there is this statement:

x' + kx = (r/k + c2)δ(t) ... (followed by the cases of t<0 and t>0

What comes after δ(t) is easy for me to follow, but I don't know where this δ(t) term comes from. Does this term evaluate to 0? If so, how? I would think it'd evaluate to (r/k + c2) because the (r/k + c2) is independent of t

Thanks.
 

Attachments

Physics news on Phys.org
Sorry folks. I just read the forum rules, and I think this is probably the wrong forum for such a thread.
 

Similar threads

Replies
8
Views
1K
  • · Replies 0 ·
Replies
0
Views
3K
  • · Replies 1 ·
Replies
1
Views
9K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K