Recent content by MrBondx

Thanks for the pointer. I guess i didn't subtract the lower limit from the the upper limit [325.2692 / (2π x 529)] x [ intergral with upper limit] - [ intergral with lower limit] 31.83 x (2.8345 - 0.3071) = 80.45 Thats it, isn't it? Thanx gneill

Thanks gneill Here it is: [V2 / (2πR)] x [ ½ (θ - (½ sin 2(θ)))] [325.2692 / (2π x 529)] x [ ½ (5π/3 - (½ sin 2(5π/3)))] 31.83 x [ ½ (5.236 -( - 0.433))] 31.83 x 2.8345 90.2

Hi guys I'm getting same as this but if I multiply with 31.83 that I am getting from (3.0086 x 10^-4) * 325.32^2 My answer is 89.9. Just wondering why my answer is off. Please help Thanks

Hi just wondering how you got to have Z1 = 26.26 + j96.52 I did Z1 = VS / IS I'm getting Z1 = 26.26 - j644.75 What am I doing wrong?

I hope it is correct, will send it for marking. Thanks for your help, much appreciated.

Converting Celsius to Kelvin 25degrees = 298.15 K Plugging numbers into equation Rn = (38.15 x 10^-6)^2 / (4 x (1.38 x 10^-23) x 298.15 x 1000000) = 88433.17 Rn = 88.4 kOhms

:smile: Input voltage resolution 18bits = 2^18 steps = 262144 Voltage range / steps = (10 / 262144) Vn = 38.14697 x 10^-6 V Is that correct?

Thanks, Yea my mistake Vn = sqrt(4kTBR) Re-arranging R = (Vn^2 / 4kTB)

Homework Statement A simplified model of ADC noise refers the noise to a noisy source resistance Rn while assuming the rest of the signal path to be noiseless. Figure 3 represents a particular 18-bit ADC that has a 10 V input voltage range. The ADC has a bandwidth of 1 MHz. Calculate the...

Ta

Thanx, Yea I had done that but my answer is coming out as negative value. Is that correct?

Is it correct to then say Reflected = Rsecondary + (Rprimary / n^2) Where n = (N1 / N2) Which gives me RS = (0.007 - (0.010 / n^2) Problem is I've failed to lose n^2 so I'm left with numerals only

Got this question sorted, thanks anyway.

For (a) isn't V supposed to become negative when it goes to the other side of the equal sign. Or what am I missing?

just corrected an error. Now giving me 7.5 + j7.5