# Finding Impulse Response from Block Diagram

1. Apr 26, 2015

### Graphite

1. The problem statement, all variables and given/known data

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

3. 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: Apr 26, 2015
2. Apr 26, 2015

### rude man

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".

3. Apr 26, 2015

### 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.

4. Apr 26, 2015

### rude man

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: Apr 26, 2015
5. Apr 26, 2015

### Graphite

I'm now seeing where I made my mistake.

Thanks for the help.