Finding Impulse Response from Block Diagram

Click For Summary
SUMMARY

The discussion centers on deriving the impulse response from a block diagram involving a transfer function represented as $$h(t) = b u(t) + v(t) = b \delta(t) + e^{at}b \mu(t)$$. Participants clarify that the exponential term arises from the Laplace transform of the system's differential equation, specifically $$\dot{V} = abu + ay$$. The correct output is confirmed to be $$y = b\delta(t) + ab e^{at} u(t)$$, with notation distinctions made between the unit step function (u) and the input (μ). The importance of accurately interpreting the block diagram and applying Laplace transforms is emphasized.

PREREQUISITES
  • Understanding of Laplace transforms and their application in control systems.
  • Familiarity with block diagram representations of dynamic systems.
  • Knowledge of impulse and step functions in signal processing.
  • Basic concepts of differential equations as they apply to system dynamics.
NEXT STEPS
  • Study the application of Laplace transforms in deriving system responses.
  • Learn about block diagram reduction techniques in control theory.
  • Explore the differences between impulse functions and step functions in signal analysis.
  • Investigate the use of partial fraction decomposition in inverse Laplace transforms.
USEFUL FOR

Students and professionals in control systems engineering, signal processing, and anyone involved in analyzing dynamic systems using Laplace transforms and block diagrams.

Graphite
Messages
7
Reaction score
0

Homework Statement


Capture.PNG


Where $$v = y - bu$$
It's given that the transfer function is
$$h(t) = b u(t) + v(t) = b \delta(t) + e^{at}b \mu(t)$$

Homework Equations

The Attempt at a Solution


I can't seem to figure out how the impulse response above was found. I understand that the impulse response is y when ##u = \delta(t)##, but I'm not sure where the exponential comes from. One of the steps in the problem is
$$
y(t) = bu(t) + \int_{t_{0}}^{t} e^{a(t-\tau)} abu (\tau) d \tau
$$
where the second term on the right is v. Going by the diagram, shouldn't v be equal to the following?
$$
v = \int_{t_{0}}^{t} u(\tau) ab + v(\tau)a \; d\tau
$$
 
Last edited:
Physics news on Phys.org
Is v' = abu + av or abu - av or what? You need to put + or - signs in front of all your arrows.
The evolution of the exponential term will be apparent if you replace the integrator with a 1/s block and solve for V(s) or Y(s) and then invert back to the time domain.
You wrote "μ" in the exponential term, I think you meant "u".
 
  • Like
Likes   Reactions: Graphite
It's v' = abu + av. The arrows are all positive.
So taking the Laplace transform of v' gives me
$$
\dot{V} = abu + ay \\
s V(s) = ab + a V(s)
$$
Since u is a unit impulse, the transform would just be 1, right? After that,
$$
V(s) = \frac{ab}{s-a} \frac{1}{s}
$$
Using partial fractions, I'm getting
$$
\frac{b}{s-a} - \frac{b}{s} \rightarrow b e^{at} - b
$$
I understand where the exponential is coming from, but I know I'm making a mistake somewhere above since I should be getting ##be^{at} \mu##

I double checked also, and the it is mu that is being used, not u. I think my professor is just using different notation, where u is input and mu is the unit step function.
 
Graphite said:
It's v' = abu + av. The arrows are all positive.
So taking the Laplace transform of v' gives me
$$
\dot{V} = abu + ay \\
No. Look at the diagram.
Further, your "given' output y is wrong. It's y = bδ(t) + ab eat u(t).
But I think you have the right idea, and yes, the laplace transform of δ(t) is 1.

BTW I use u(t) to mean the step function. This is standard notation. I've never seen anyone use μ(t) for that. Use x(t) for the input, not u(t).
 
Last edited:
I'm now seeing where I made my mistake.

Thanks for the help.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 17 ·
Replies
17
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
23
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
6K