Question Dec 14
Answer
Question Dec 14 Referring to year 2000 PLL question why is one of the Tau ....which is Tau 2 at 1000Hz. I don't see any 3db or corner frequency there. in fact how would u calculate tau 2?
Answer Actually, I believe you may have tau 1 and tau 2 turned around, but the question is still valid for tau 1.
For an integrator, the transfer function is 1/(s tau) but there is no break point. It is just a straight line at -20 dB/ decade. The tau can be pulled out and lumped with K, so it really has no meaning. However, we can still calculate a value because the gain changes as you change tau.
See the example below where I have shown two lines, one with tau = 2 and one with tau = 4. For gain = 1/2s, magnitude is 1/2w gain goes to unity at w = 0.5. For For gain = 1/4s, magnitude is 1/4w and gain goes to unity at w = 0.25.
.
| .
Gain| .
|. .
| . .
| . . 1/2s
| . .
| . .
| 1/4s . .
| . .
-------------+-----+----
0.25 . 0.5 . omega
. .
. .
This happens in the question, where an integrator is used with phase lead correction. This has a break point defined by the zero, but no break point for the pole since it is an integrator. In solving the circuit, tau2 is easy to find as it is represented by the break point and we solved for tau2=15.914u. For tau1 we solved for the gain at high frequency where the gain is Atau2/tau1 = 0.1. We note that the gain and time constant A/tau1 cannot be separated so I arbitrarly set A =1 then 1/tau1= 2 pi x 1000.
Question Dec 14 I'm having some problems with 1999 #7 that you solved today in class. From my understanding, the graph represents the same time domain response for a DSBC and AM wave. For AM: Tc comes from 6us/12 (or Fc=2M) however, today in class you solved Tmod as 12us or Fmod=83kHz. However the exam solutions on your webpages shows sidebands at 2.167M and 1.833M, which corresponds to the Tmod of 6us. Which is correct? As for DSBC, are the frequency components solved the exam same way as AM, or did the value happen to be the same for this.
Answer For AM, the modulating freqeuncy is at 1/6usec or 167kHz. Thus sidebands are at 2M +- 167 kHz. For DSBSC Tmod is 12 usec so fmod is 83 kHz. Thus sidebands are at fc +- 83 kHz, but fc is suppressed so all we have left are sidebands separated by 2x83 kHz.
So, the following are the possiblities. Note for DSBSC there are two possible answers. For all of these answers it is not possible to distinguish between them by eye, however, an FFT of the waveform would be able to tell you which one is the correct one.
^
AM |
|
^ | ^
| | |
-------------------------------------------------
1.83 2 2.167
DSBSC + Filter ^ . - - suppressed carrier
| .
| .
| . ^
| v |
-------------------------x-----------------------
2 2.167
DSBSC + Filter ^
|. - - - suppressed carrier
. |
^ . |
| v |
------------------x------------------------------
1.83 2
Question Dec 13 We sort of touched on this in class today but I think we danced around the answer. In part "a" we are to determine if the loop filter is a LPF with phase lead or an integrator with phase lead and finite DC gain....how can we tell the difference when the transfer functions are the same for each? The only difference is that the integrator has DC gain of R3/R1 and the LPF has DC gain of 1...but what if R3/R1 is 1? I tried to do the question by determining the gain of the VCO at low frequencies and seeing if it was the same as the loop gain. Then I'd know that the filter must have gain of 1. For a frequency of 300 Hz, I got the gain of VCO is 55 dB and the graph says 50 dB for the loop. I started to think this is close enough but that's really off by over 300% right? Thus, the gain of the filter must be more than 1 at 300 Hz. I would therefore say we must have an integrator, but the answer says no, a LPF and an A=0.628? This even further confuses me. The difference of 5 dB = 1.68 gain. Therefore A should = 1.68 if anything...no?...and if we have a LPF with DC gain=1 and a VCO with gain = 50 dB, where the heck does A come from? In short....how do we tell from the graph if the loop filter is an integrator or a LPF? Please help!
Later: Actually, after a second look....I make a little progress. If the gain of the VCO is 55 dB or 530 at 300 Hz, and the gain of the loop (picture) says 55 dB or 316, then the gain of the filter needs to be 316/530 = 0.6. Therefore I see how A=0.68 or about 0.6? BUT....why couldn't we take this to mean that we have an integrator with phase lead and DC gain of R3/R1 = 0.68?
Answer An integrator with finite gain is really an active lowpass filter. What I called a lowpass filter is the resistor capacitor passive one. So, I should maybe have called it that, an active low pass filter with phase lead correction.
So, for a filter gain of 0.68, we could use a passive or an active filter, and either one would have been acceptable. However, if the gain were bigger than one, you have no choice but to use an active filter or an extra gain block somewhere.
Question Dec 13 I've been working on Class C amplifier question (Q4) from 2000 and i'm getting different values.
Idc = 5.58mA I1 = 10.0mA RL= 113.15 ohms
is there any chance the answers on the web may be wrong? or am i way off base?
(yes, i divided the conduction angle by 2, and yes, i'm working in rads)
Answer I can only get your answer if I make the following two mistakes:
1. theta = 30 (wrong) 2. Ip = 100 mA, (wrong)
Without knowing your Theta and Ip, I can't otherwise see what you did wrong. For theta, see the next question.
Question Dec 13 In question 4 of 1998 exam, I am not getting the same values as what you posted. I was wondering what value of theta you got and how you found it ?
Answer Theta = 60 degrees since it conducts for one third of the time, that is 360/3 = 120 degrees. Theta is half of the conduction angle.
Question Dec 12 Exam 1998, Question 7. In the solutions it says efficiency max = 70.1. I believe that answer is correct for just efficiency and my calculated efficiency max is 89.7 %
Answer You are correct.
Question Dec 12 Exam 1997, Question 1. I get part a) answers but b) and c) don't make any sense to me. My n2/n1 = 7.75, L = 47.75 uH, Cextra = 484.6 pF. Wasn't Rx = 100 ohms and Ry = 6 kohms. n2/n1 = sqrt = (6000/100). How did you get those answers?
Answer You have forgotten about the Q of the inductor. This places a resistor Rp in parallel with the 6k resistor. However, you can't calculate Rp until you know L. You can't find L until you've calculated the total capacitance Ctot. You can't find Ctot until you know the equivalent resistance Req which includes Rp. Thus we have come full circle.
Thus we have a problem which needs iteration. You have done the first round correctly. Now using your value for L, predict Rp which allows you to calculate a new turns ratio, a new Rp, new Ctot, new L, and so on.
Question Dec 12 In general when you want us to calculate BW to we solve it in Hz or rad/sec? Also for turns ratio do you want it in n1/n2 or n2/n1?
Answer If it is 1/RC then it is in radians/second. If you have calculated it as 1/(2pi x RC) then it is in Herz. Where you will lose marks is if you calculate 1/RC then call it Hz.
A similar logic holds for n1 or n2. If n2/n1 = 10 then n1/n2 = 0.1 and both would be correct. Since turns ratio is determined by impedance ratio where the higher impedance side has more turns (higher voltage, lower current equals higher impedance). If you end up with an answer where the higher impedance side has fewer turns, you've obviously got it backwards and you will lose marks.
Question Dec 12 With respect to sections 6-8 on the course outline (AM, FM, TV): how in-depth are we expected to know these sections since we did not spend much time on them (I don't think we touched the AM section)?
Answer I left out the AM section because we had effectively covered most of it in other section. For example, at the beginning of the course while discussing the image problem, I used the AM example from this section. The AM waveforms were discussed in the mixer section and the mixer lab. We didn't talk about AM receivers, but it is often simply a diode-capacitor peak detector, so everyone know that anyway.
As for FM, we did it fairly quickly (30 minutes at most). However, we did cover more FM when we explored the response of a PLL to an FM input. Thus if there were any FM type questions they would either be very superficial or related to one of the other sections like FM in a PLL. I would suggest that you don't study the FM section any more than what was done in the class.
There won't be any quesitons about TV since we only talked about this for about 5 minutes.