Recent content by JeeebeZ

I figured it out, Those are not in dB, even though I thought they were. So I was converting everything from dB or nothing from dB so the integral ends up being -exp(gamma/gamma_bar) evaluated from a to b. -exp(-52.98/ ( 10^(18/10) )) + exp(-10.6/ ( 10^(18/10) )) = 0.413505 -exp(-222.53/ (...

Homework Statement 2. Homework Equations The Attempt at a Solution I can't figure out how to get those numbers, I think I have to convert from dB when actually calculating it but I'm unable to get those ones. I have PQPSK = 0.502368 P16QAM = 0.0526855 P64QAM = 0.000004 Pno = 0.4450

yes that's exactly what I'm looking to do

Sorry, the "1 2 2 2 2 ..." bit was basically was how I wanted each part of the summation to equate to, so essentially it would be "1 + 2 + 2 + 2 ..." "2(∑an) - a0" is the exact same as "a0 + 2(∑an)" with the exception that n goes from 0->N for the first one and 1->N for the second one...

Homework Statement I need a summation where the answer is 1 2 2 2 2 2 2 2 Homework Equations a(0) + sum(2*a(1) + 2*a(2) +2*a(3)) The Attempt at a Solution I unfortunately have no idea where to start, basically it is taking a symmetrical function from 0 to N-1. where the function...

Homework Statement 2. The attempt at a solution So, I know that you can test a Diode with a ohmmeter. Based off of the R you get Very large reversed biased and fairly low forward biased. But I don't know why it would show these values. The Large one, revered biased. Kinda makes...

I've never used latex before I just quoted yours and edited it as needed but I'll keep it in mind for the future. Thx

Because |Z|^2 = z * zbar we have $$|H(e^{j\omega})|^2=(\frac{1-(1.25)e^{-j\omega}}{1-(0.8)e^{-j\omega}})(\frac{1-(1.25)e^{j\omega}}{1-(0.8)e^{j\omega}})$$ substituting $$x=e^{j\omega}$$ $$|H(e^{j\omega})|^2=(\frac{1-(1.25)x^-1}{1-(0.8)x^-1})(\frac{1-(1.25)x}{1-(0.8)x})$$...

$$Magnitude = \sqrt{Re(z)^2 + Im(z)^2}$$

Homework Statement H(e^jw) = (1-1.25e^(-jw))/(1-0.8e^(-jw)) Prove |H(e^jw)|^2 = G^2, and what is G Find Magnitude & Phase Homework Equations H(e^jw) = (e^(jw)-1.25)/(e^(jw)-0.8) H(e^jw) = 1 - (0.45e^(-jw))/(1-0.8e^(-jw)) H(e^jw) = 1 - 0.45/(e^(jw)-0.8) The Attempt at a...

the link is linked to my account, its forbidden i guess\

32.62 gain, 3162 R That way you have 1V over the 100ohm = 10mW and 32.62(100000/(3162+100000)=31.62V -> 31.62^2 / 100000 = 9.998mW

h... and I got it to work. https://www.circuitlab.com/circuit/283w8y/screenshot/1024x768/

It is really easy to do with 2 difference circuits. But using one circuit I don't see how changing one resistor could allow it to get 10mW just on the one resistor that has been changed. In order for the 100k ohm, it needs ~31V, while the 100 ohm only needs 1V.

Homework Statement see image Homework Equations The Attempt at a Solution I can make 2 different circuits to solve this. However, Using just op amps and resistors I am not sure how to make a single circuit that could do this