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
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...
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
$$x=e^{j\omega}$$
$$|H(e^{j\omega})|^2=(\frac{1-(1.25)x^-1}{1-(0.8)x^-1})(\frac{1-(1.25)x}{1-(0.8)x})$$...
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