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No problem! And yes, a load resistor would change the overall gain of the amplifier circuit. You basically take Rout and put it in parallel to RL. I have another example here where I include it in the gain calculation: th-cam.com/video/V8VCzWV8Kgo/w-d-xo.html

So if you have y(0) = 2 and y'(0) = 1 for example, you take the final solution ygen, and plug t = 0 and have the LHS = 2. After doing so, you'll get C1 = 2. Then, take the derivative for ygen', and plug t = 0 and set the LHS = 1. After doing so, you'll get C2 = 3. Now, using C1 = 2 & C2 = 3, you can plug those constants into ygen = 2e^-t + 3te^-t + 1/2 * t^2 * e^-t

Given that there's a DC blocking capacitor in between RL and Rc, under DC conditions, the capacitor is open. This means that current flows straight from 20V through Rc and into the collector. So you are correct in your assumption. Due to this, the load resistor has no effect on the collector current Ic, which we found to be 4.125 mA.

I wouldn't consider it common enough of a base to put it on this list, but you're more than welcome to include it!

Np bb

Could you elaborate on the identity that you mentioned where cosx = sinx? Do you mean via phase shift? Because otherwise there's no situation that I could think of in which they're directly equivalent. Honestly I don't think substitution is needed, since the power reduction (half angle identity) is the most straightforward way to approach this problem. As a side note, the integral of cos(ax) = 1/a * sin(ax). Since the coefficient was already = 2 in 2cos(2x), the integral would be 2/2 * sin(2x), reducing to sin(2x).

I can see why you're confused. The schematic symbol for a MOSFET has a disconnected line for the gate (G). On the other hand, for a JFET, the line for the gate (G) is connected directly to the Drain (D) and Source (S). I've attached a link explaining more: www.electricaltechnology.org/2021/04/difference-jfet-mosfet.html By Ig, do you mean gate current? If so, Ig = 0 is because the gate terminal is isolated from the substrate by a dielectric medium. This is of course under ideal conditions. Realistically, there's always a minuscule amount of leakage current going through. But we're talking about pA to nA range, so we could usually neglect that. Does that make sense?

Starting from the general solution Yh = C1*e^(-3t) + C2*t*e^(-3t), we have an initial value problem (IVP) given the conditions y(0) = 1, and y'(0) = 7. Start with simply plugging in y(0) = 1 into Yh. It becomes 1 = C1*e^(-3*0) + C2*(0)*e^(-3*0) since y = 1, t = 0. Simplifying, 1 = C1 + 0, then 1 = C1. We can conclude from the first condition that C1 = 1. Now, to plug in y'(0) = 7, we have to take the derivative of Yh. Yh' becomes -3*C1*e^(-3t) + C2*e^(-3t) + (C2*t)(-3*e^(-3t)) using chain rule on the second term. Plugging in y'(0) = 7 with y' = 7, t = 0 -> 7 = -3*C1*e^(-3*0) + C2*e^(-3*0) - 3*C2*0*e^(-3*0). Simplifying and plugging in C1 = 1 from the previous equation, 7 = -3*1 + C2. Then C2 = 7 + 3 = 10. Therefore C2 = 10. Plugging in C1 = 1 and C2 = 10 into the original Yh equation -> Yh = e^(-3t) + 10t*e^(-3t)

Yes, but that requires advanced techniques. This is the most straightforward way to evaluate this problem in my opinion.

Yes my past 8 videos or so are variation of parameters. Laplace will be my next series!

Thank you

jGRASP

@@JJHanna thankyou

You're qualifying assumption is correct. Since Beta is zero (no imaginary component) the sine term cancels out. And Alpha = 1 (real component), the Ce^rt term remains. However, in this particular case, there are repeated roots. Therefore the general form must be C1e^(r1*t) + C2 * t * e^(r2*t). It would only ever be C1e^r*t if there was *one* *real* *root.* But again, we have two repeated ones. Does that make sense?

I separated each stage so that I can take each external voltage gain into account during the final AC analysis gain calculation starting at 24:20.

ДДСҰ?

Derivative of -cost = sint. 1 is a constant so it turns to zero.

You're forgetting the 1/2 coefficient but true!

@@JJHannathanks

You could move C around like that if you want. If it's multiplied by any constant it will always stay as C. I just left it in the log so that there's only 1 term as an answer.

For this specific problem, the x and y terms can't be completely separated from each other such that x is on one side being multiplied by dy/dx and y on the other. So no, this is not a separable equation.

I understand what you mean. The property where you can add 1 to the exponential numerator and put it over the denominator when integrating only applies to simple polynomial functions. Every other expression requires a special sort of identity or trick as I've shown in the video. So no, you can't just say cos^5(x)/5 as the answer.

@@JJHanna AHH that explains why I was confused.You could do so if it was cos^2thetha tho right?

only if it's cos(constant * x). Then you could use u-sub to make it = 1/constant * sin(constant * x). No matter what, you can't linearly add powers to the trig terms.

@@JJHanna AHH thnx man you earned a sub

@@sharequsman596 thanks so much! If you have any other question let me know.

Yup!

Bro needs to get off TikTok with that 0 attention span

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