Recent content by Jason-Li

PeroK, I should've simplified it down first like you have e.g. abc... Thanks again!

So my final equation is: ##\frac {1} {2700} + \frac {1} {3930n^2} + 10^{-5}## I need to boil this down, the learning materials has the following working, but I can't seem to get it $$\frac {1} {2700} + \frac {1} {3930n^2} + 10^{-5}$$ $$\frac {3930n^2+2700+2700*3930n^2*10^{-5}}...

Ah I see, I find it strange that you managed to assume hfe of 100 and reach a closer answer than mine when I used the values given with the question (below), not sure how my value is further off than yours? So after finding the gain of the first stage (16.16 in mine and 17.86 for yours) how do...

Hi Tom, Thanks for the reply! Sorry how did you get the first stage IE without ß as you would be unable to use Ib = (Vcc-Vbe)/(Rb-(ß+1)Re) ? The calculation I completed is as follows: I used Av (voltage gain) so used Av=-Rc/(Re+re') where re' = Hie/Hfe so that would be Av=-1000/(47+(5500/370)...

Hi Tom, thanks for the reply. I see that makes sense that it is just further approximation. Out of interest without: -21.27659574 and with -16.16426387. Seems like quite a large difference? I realize the discussion above but just wanted to gain some clarity on the full methodology: So if I...

Hi all, Sorry to resurrect a thread, one thing, why is the formula used Gv=-Rc/(Re+re) ? In the learning materials supplied I can only find ones stating that gain is simply Gv=-Rc/Re ?

Hi thanks for the reply and assurance, do you have any idea why my learning materials is citing the noise figure equation incorrectly?

So pretty confident I understand part (a) however for part (b) I'm not sure if I have carried it out correctly if someone could give me a pointer? (b) 5dBmV Input as a voltage: 5=20Log(V/1mV) V=105/10 V=1.77827941mV Then the noise level is 20dbμV so changing to a voltage: 20=20*Log(V/1μV)...

Edited - I was an idiot and didn't use radians. @.Scott & @DaveE I think I have the correct answer now - thank you both! All of this has helped it click.

Ah of course that makes sense. I can see the first part of the large equation does again equal 0.036 so that would mean that the two other functions would dictate the amplitudes of each wave and in terms of being trapezoidal as the first part of the function has no mention of frequency or which...

Ah so the cosine is indicate how far into the cycle the wave is ranging effectively across the y-axis between -1 & 1? Yeah that is why I was struggling to understand I think because "##A⋅sin(\omega t)## is ##A##" I was seeing this simplified as ##A⋅B## so how could amplitude (A) be effectively...

Scott, this it probably a better explanation and linked materials than all of my learning materials combined thank you, there isn't even mention of duty cycles in the material... So as per that slides the Sin is just a "scaling factor" for the amplitude - is that why I can just ignore the cos...

Due to them both being odd waves? Still not sure how the high state can be 1.8V but the 3rd harmonic is so low at 10.9662222 *10^-6V if I've used the formula correctly

I think the question is looking for a "simple" answer as it is stated in our learning materials (figure 1 is the trapezoidal waveform) :

Thanks for your reply Scott Apologies I am not familiar with it no. Had a quick look and definitely not seen it before I don't think. Could I take the formula x(t) = An*Sin(n*2*π*f*t) And integrate it between say 0 & 1 which would return: Which I would then enter my figures into for f & n -...