Finding Impulse Response from Block Diagram

AI Thread Summary
The discussion centers on deriving the impulse response from a block diagram involving a transfer function. The impulse response is identified as y when u equals the delta function, but confusion arises regarding the origin of the exponential term in the response. Participants clarify that the Laplace transform of the system leads to an output that includes both the exponential and a unit step function, with notation differences noted between u and μ. The correct formulation of the output y is confirmed as y = bδ(t) + ab e^{at} u(t). Overall, the conversation highlights the importance of consistent notation and understanding the relationship between the input and output in control systems.
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Homework Statement


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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?
$$
v = \int_{t_{0}}^{t} u(\tau) ab + v(\tau)a \; d\tau
$$
 
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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".
 
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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.
 

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