Recent content by PenDraconis

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Homework Statement Tell whether following systems are i. linear and ii. time-invariant: $$y(t) = \int_{-\infty}^t x(\tau)d\tau$$ $$y(t) = \int_{0}^t x(\tau)d\tau$$ $$y(t) = \int_{t-1}^{t+1} x(\tau)d\tau$$Homework Equations N/AThe Attempt at a Solution I'm a little thrown off by the integrals...

Thank you! I understood my error basically as soon as I wrote out the question despite struggling for a while. For anyone else with a similar problem looking at this, basically there's only charge in the "shell" between b and a. So to find the charge density you'd have to do the equation...

Homework Statement An insulator in the shape of a spherical shell is shown in cross-section above. The insulator is defined by an inner radius a = 4 cm and an outer radius b = 6 cm and carries a total charge of Q = + 9 μC(You may assume that the charge is distributed uniformly throughout the...

Yup! You're right. Those are the impedances I gave. The resistor and inductor are in series and they're both in parallel with the capacitor.

Thank you! I was just confused as to why this point was glossed over whenever I mentioned it in class, I wasn't able to ever get a straight answer. The second part of the circuit I seem to understand, basically I have a RLC circuit and I must used phasor analysis to find the capacitance given R...

Homework Statement I have ##V = 25cos(1000t)## and ##I = Acos(1000t)## for a circuit, I'd like to make those into phasors.Homework Equations ##V = Acos(\omega t+\theta)## ##\leftrightarrow## ##V = Ae^{j\theta}##The Attempt at a Solution It seems like this is relatively simple, but that seems...

Hah! It was so easy but for some reason I was battling myself the whole time thinking it was much more complicated. Thank you for the assistance Simon and Rude Man.

Using that information we have: ##4e^{j\theta}+4e^{j(-\theta)}## Expanding that out we have: ##4(\cos(\theta) + j\sin(\theta)) + 4(\cos(-\theta) + j\sin(-\theta))## We know that ##\cos(\theta)+\cos(-\theta)## is ##2\cos(\theta)## due to trigonometric identities. We also know that the...

So, basically ##j\theta=jt##, thus ##\theta=t##? So then we'd have ##4e^{j\theta}+4e^{-j\theta}##? How do I account for the negative then? If that's the case (and ##Mcos(ωt+\theta)## is interchangeable with ##Me^{j\theta}##) wouldn't the end result be: ##4cos(ωt+\theta) +...

Homework Statement "Put each of the following into the form Acos(ωt+θ)..." (a.) 4ejt+4e-jt Homework Equations Euler's Identity: ejθ = cos(θ)+jsin(θ) Phasor Analysis(?): Mcos(ωt+θ) ←→ Mejθ j = ej π/2 Trignometric Identities The Attempt at a Solution I attempted to use phasor analysis to...