Recent content by Graphite

I'm now seeing where I made my mistake. Thanks for the help.

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...

Homework Statement 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 EquationsThe 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...

Doing that, I'm getting y = \frac{1}{2} b c^{3} t^{2} e^{(a-c)t} = be^{at}\delta(t) which is much simpler. I just realized that I forgot to post the DE at the beginning, which is \dot{y} = ay + b \dot{u} When I try taking the derivative of y, I'm getting \dot{y} = abe^{at}\delta(t) +...

Homework Statement y(t) = e^{a(t-t_{0})} y(t_{0}) + \int_{t_{0}}^{t} e^{a(t-\tau)}b \dot{u} (\tau) d\tau u(t) = \delta(t) = \frac{1}{2} c^{3} t^{2} e^{-ct} where c >> |a|, t_0 = 0, and y(0) = 0 Find y(t) and represent the unit impulse, delta, in the solution. The remaining terms should not...

So the current for both the 8 ohm and 7 ohm resistor would be ##i_{1}## and the current through the 2 ohm resistor would be 3 A. I can see how to solve it now. I'm getting v1 = 14 and v2 = 6, which looks correct. Thanks for the help.

Homework Statement Consider the circuit shown. Find the values of v1 and v2. Homework EquationsThe Attempt at a Solution I tried using KVL on the left closed loop and got: $$v_{1} = 30 - 8i$$ I then used the current law between the two loops to get: $$ i_{1} - v_{1} / 7 = v_{2} / 2 - 3$$...