- #36
Fisher92
- 80
- 0
Yep, I had Pavg = Avg[(Ec^2 sin^2(wc t)] * [1 + avg{m^2 sin^2(wi t)}] but i also tried the integration with yours Pavg = Avg[(Ec^2 cos^2(wc t)] * [1 + avg{m^2 sin^2(wi t)}] directly, with no success.
Fisher92 said:When I pick up from post 16 I get a big mess with wc and wi on the denominator and a string of sin and cos on the numerator? Can't seem to get what you have for my power?
Fisher92 said:Hey RM, sorry for harassing you about this but I have to submit this in 24hrs and I can't for the life of me figure out how you went from the avg formula in post #16 to the power?
Once I have the power, I think the SNR is given by P/(2N_0*f_IF)?. I can put that in dBm pretty easily.
So using the power from your post 24, I get SNR =(Ec^2 * m^2/4 * G_T * C)/(2N_0*f_IF)?
Thanks
Fisher92 said:'E.g. avg[sin2(wt)] = ∫0Tsin2(ωt)dt/T where T = 2π/ω?'
I get pi/w as my result but this can't be right?
http://www.wolframalpha.com/input/?i=integrate+%28sin%28w*t%29%29%5E2+dt+from+0+to+2pi%2Fw
Just never seems to come out - I've been using my nspire, not wolfram but I am getting results with w and pi which shouldn't be in the answer?
You also need to use the general fact that
avg(a + b) = avg(a) + avg(b) and avg(ab) = avg(a) * avg(b). a and b are assumed independent of each other, which is the case since a = sin(wi t) and b = cos(wc t) here.
Expand [1 + m sin(wi t)]^2] out as 1 + 2m sin(wi t) + m^2 sin^2(wi t), take averages as I've shown, then you wind up with the expression for power as Ec^2 * m^2/4.
Fisher92 said:Thanks RM, very close with the power now. I've got:
[tex] \frac{(E_c^2*1/2*(1+m^2/2)*G_T^2)}{R_T} [/tex]
Fisher92 said:Ok, thanks so for the SNR I would essentially have the power above with a (4R) as the denominator of the numerator...and 2fIFNo as the denominator?
Fisher92 said:I have:
[tex] SNR=\frac{\frac{E_c^2*1/4*(m^2+2)*G_T^2}{4R_T} *C))}{2f_{IF} N_0} [/tex]
?
I think that's good.Fisher92 said:[tex]SNR=\frac{\frac{E_c^2 G_T^2}{4R_T}*C}{2f_{IF}*N_0} [/tex]
If the total power amplification through the receiver stages is GR find the output power of the receiver.
As we've established the power at the input is the information or sidebands power
[tex]P_{out}=\frac{E_c^2 G_T^2*G_R^2}{4R_T}*C [/tex]
Assuming the gains are voltage, & ill make a comment on that
e. If the noise power at the output is identical to the noise power at the input of the receiver, find the NF for the receiver
NF = input SNR/output SNR
If the above, power output is correct, this is easy enough - algebra may simplify but my calculator will figure that out
?
Thanks
I'm in time zone GMT-7 (Arizona USA) all year. (No dst).
Fisher92 said:-Gotta love the internet!
For part d, the noise figure, I ended up getting [tex] N_0=\frac{1}{G_R^2} [/tex]
Probably not right (doesn't make much sense I don't think) but its only one part of a question.
Thanks for all your help with these two questions RM!
Fisher92 said:Thanks, will do - the uni has a two week turn around for assignment marking.
Fisher92 said:Central QLD, Australia (3/4 the way up (North is up) on the east coast - the coast is still central for some reason -because everyone in Australia lives on the coast I guess)