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Question Dec 14 2001 Exam, question 3a...asks us to find the input and output frequencies for a multiplier of 2. ic is on for 10us and off for 40 us, meaning that the input must have a frequency of 1/50us = 20kHz, and therefore the output is 40kHz. The solution to this question gives input of 40kHz. Can you tell me if I'm mi ssing out on something. I'm looking at the diagram on page 32 of the course not es and I think i'm right.
Answer It looks like you are right. It should have said output frequency is 40 kHz and input frequency is 20 kHz. The transistor on and off time is always controlled by the input signal. Looks like I was in a hurry as I also swapped numbering for Question 3 and 4.
Question Dec 7 I have been reviewing the course notes for the exam and I noticed in class that we did not cover some of the material at the end of the notes. I am wondering up to what page exactly I should study up to? I was thinking page 100 but I wasn't sure.
Answer I believe we covered all sections except TV so through page 100 is correct. However, we went lightly on the historical FM detectors, (e.g., Ratio Detector or Foster Seely Discriminator) so I would not ask for such on the exam.
Question Dec 6 For Question 1 of last year's exam, the transmitting frequency is 2.45GHz (Q1a). I would have expected the image frequency (Q1b) to be 2.45 + 2.65 (LO) = 5.1GHz, but you have the answer as 2.45 - 0.2 = 2.25GHz. I would have done that for a receiver but not for the transmitter in this case. Am I wrong? Or is there a mistake here?
Answer I don't have the full solution in front of me to see why I put down that answer, but a quick look indicates you are correct.
Question Nov 28 in part 2.b, I calculted the pk-pk voltage to be about 8V (from about 1 to 9V), which results in an rms value of about 5.6V, and thus RL= ((5.6V)^2)/ 2mW or about 17k, way off from the expected 5 k value. did i go wrong anywhere in my calculation?
Answer Yes, if peak to peak swing is about 8V, then peak voltage is only 4 V. RMS value is 1/\sqrt{2} of peak value, not peak to peak value (remember ELEC 2501 way back there somewhere?). So, your rms voltage is high by a factor of 2, when you square it, your answer is high by a factor of 4.
Question Nov 26 My question is related to problem 2b on the assignment. I understand that the output voltage swing is required to obtain the peak-to-peak voltage and thus the rms voltage for use in the Po = (v_rms)^2/R_L equation. What is confusing me is that in one of your answers to a previous year question stated that "if at the collector the swing goes down to 3V, that is a -2V swing from the average value of 5V, the positive swing would be expected to be +2V up to 7V. This is 5V plus and minus 2V or 4V peak to peak."
However, on the assignment it is written that the collector voltage is nominally at Vcc with a peak downward swing to about Ve. My calculated Ve is approximately 1V, which means the swing would be from from 5V down to ~1V and the peak to peak is then 8V. Does this mean I've made an error in my operating point calculations? Could it be that my voltage divider calculation for Vb is to blame? Any help is appreciated.
Answer My answer was a general hypothetical answer starting with the important word "if". In your case, knowing what your particular your ve is equal to, i.e., about 1V you can make this sentence more definite, "since the collector voltage goes down to 1V ..." Later in your simulation, you can and should check your prediction.
Answer First of all, I should have said natural frequency is 2000 Hz, not that loop bandwidth is 2000 Hz. In the lab manual, there is a hint there saying there isn't a direct relationship. Loop bandwidth is an ac phenomenon - it means that if you frequency modulate the input signal, the loop bandwidth will determine how much of the input signal bandwidth around the carrier is transferred to the output. However, in contrast, what you have done is to sweep the input frequency slowly in order to check out the lock range. There is no modulated component, and any changes you made were done very slowly, that is, at low frequency, so loop bandwidth does not get involved.
Question Nov 20 My slope for phase detector 3 seems to be high by a factor of 2. What could be the problem?
Answer One possibility is incorrect measurement of phase - since you use narrow pulses for this phase detector, you are directly using the outputs of the Johnson Counter, so the ref and var signal have not experienced the divide-by-two, hence the frequency is 400 kHz, not 200 kHz as it is for the other phase detectors. You can not use the listed phases (18 degrees, 36 degrees, etc) but instead phases are now 36 degrees, 72 degrees, etc. Of course, if you measure phase by measuring time shift relative to measured period, all should still be correct.