Answer
Question Dec 21 Final exam 2001 question 5 part c finding the peak phase error: Why are we dividing 200 KHz by N=545 to get the step in frequency? Aren.t we stepping by 200 KHz every time we switch between adjacent channels? In other words, shouldn.t the step in frequency be 200 KHz?
Answer The output frequency is being stepped by 200 kHz, but at the phase detector, which is the input, this is equivalent to stepping by 200k/545 at the input, so this is what we use for the analysis. We could of course refer our answer to anywhere in the loop, but since we are interested in the phase error, then doing an analysis with respect to the phase detector makes sense.
Question Dec 21 On page 77 in our books I see the possible colpitts oscillator arrangements. I wanted to check if the common collector was done correctly. I can't re-arrange it into the Op-Amp config listed on the same page.
I think if C1 & C2 are swapped it will work.
I'm assuming the Collector as my negative input, Base as positive and Emitter as output.
Answer This one is a bit unusual in that if you work out the math, it turns out the collector, in spite of being grounded can still be treated as the output, the base is the negative input and the emitter is the positive input. Since this is not at all obvious, I have included this information as part of the question in the 2003 exam
Question Dec 21 For the tuned amplifier question in the 2002 exam, how do we find the gain at th e centre frequency? In the solution to assignment 1 we found the total gain (but it wasn't done for any specific frequency) My guess is that we include the output C and L impedance at fo into RL. Is this his correct?
Also, is the bandwidth of each matching network (ie. both input and output)equal to 2pi*450kHz/0.642 as I've done in part (c)? Is 0.642 a constant, or what for mula can be used to find it, in the case that it isn't given on an exam?
Answer At the centre frequency, the inductance and capacitance resonates out so we use only resistors to do our calculation. And the gain in Assignment 1 was done at the center frequency - that is why we didn't have to be concerned with inductance or capacitance. And that is why the transformer can be used - at the center frequency we have cancelled out the reactive parts so that only resistance is left. I note that in the tuned amplifier question in the 2002 exam, the gain is calculated in nearly the same way as you did for assignment 1 - it is 1/2 x N2/N1 x gm x RL. However, RL as given is made up only of the transistor output resistance, so there should really be a comment about a load resistor or the equivalent Q of the inductor.
As for bandwidth, the factor of 0.642 is the ratio of the -1.5 dB bandwidth to the -3 dB bandwidth. If this were to be required on an exam, it would be provided. However, it isn't all that hard to obtain this value from the simple expression for a tuned circuit.
Question Dec 21 In the time domain, AM and DSBSC represents the same waveform, but in frequency domain, they aren't the same, how does that work? also, how does AM waveform is time domain represent three fequency components and how can we tell each of them just based on the time domain waveform?
Answer To the eye the two waveforms look similar, but if superimposed, they are not exactly the same - there is an example in
http://www.doe.carleton.ca/~cp/telecom/exam99soln.html.
The eye cannot distinguish them that well, but it is no problem with an FFT.
Question Dec 21 When does one use 10 logX as opposed to 20logX when converting from a dB value to numerical value?
Answer If you have voltage gain, it is 20 x log(V2/V1), if it is power gain it is 10 x log(P2/P1). Note that power is related to V squared, so 10 log of a power ratio is in fact exactly the same as 20 log of a voltage ratio.
Question Dec 21 on the 1999 exam Question 6, 1/Tau_1>1/Tau_2. Is that possible? I always thought Tau_1 was larger then Tau2 or does it not matter? Won't that give a different transfer function then the usual?
Answer If this were a lowpass filter with phase lead correction, then definitely I would expect the pole frequency to be lower than the zero frequency - otherwise this really wouldn't be a lowpass filter.
However, for an integrator with phase-lead correction the pole is actually at DC in spite of what tau_1 is. Another way of looking at this is that in such a filter with a simpel s tau_1 on the bottom, we can't distingiush between A and 1/tau_1, so if you don't like 1/tau-1 being so large, you can blame it on A and have a smaller tau_1. Note the main effect of taur_ 1 is to set omega_n, but the equation has a square root of K/tau_1, where K = Kphase x Kvoc x A so again, we can't distinguish between A and 1/tau_1.
Question Dec 21 for the 2001 # 4 that we went through the other day, once we find the f(mod)=50K and f(carrier)=2M, then shouldnt it be 1.95M at one side and 2.05M on the other side, so that the seperation of two sidebands is 100K? the answer on the back of the notes says 1.9M and 2.1M and i couldn't see why
Answer In this case, all we have are the two sidebands, we don't know the carrier. What we know is that the biggest signal, (which is not the carrier but one of the sidebands) is at 2 MHz and the other sideband is 100 kHz away (this is in agreement with a modulating signal of 50 kHz). Since we don't know which sideband we are looking at we gave two possible answers, 1.9 MHz and 2.1 MHz.
Question Dec 21 I have a question about output RMS voltage (Vrms = V/sqrt2 --FORMULA I). From the solution you did on the Monday Exam Review for Final 2002-Q4-b:
Vrms = sqrt[(0.27^2)/2 + (0.27^2)/2]=0.27 V rms
Why is that?
When I did this question, I took the ampltude of the resulting modulating signal, hence (0.54/2) as my V and then sub into the Vrms formula given FORMULA I. Therefore my Vrms is 0.1909 V rms.
Could you please explain to me what happened?
Answer Amplitudes were given as peak voltages. Thus RMS is lower by a factor of sqare root of 2 (multiply peak by 0.707) Then to combine the signals, you add the square of the voltages. And the result is one half of vpeak squared.
Question Dec 21 I'm going though the problems we did on yesterday's review session, from what i copied, you drew T=1/f=1.04719us for the Vc waveform, but shouldn't it be the period of Ic? since Ic is the input waveform and 6*10^6 rad/s is INPUT frequency? therefore period of output waveform which is Vc is T/3 correspondently? Or i just copied that wrong?
Answer You are quite right. 1.04719us is the period of the input signal, thus f = 1/T is the input frequency. And 3/T would be the output frequency.
Question Dec 21 for part (c) why is it that we assume the center frequency to stay the same? or put another way, why doesnt a QL affect the center frequency and only the BW?
Answer The centre frequency is determined by L and C, so if we don't change L and C, but only parallel resistance, frequency won't change.
Question Dec 21 On exam 2003 #1e I get F=2.09 whereas the back of the book gets NF=2.09
Answer Oops, it should be F = 2.09 so NF = 3.20 dB.
Question Dec 21 For PM, AM, FM, TV, etc. Are we just expected to know briefly what you convered in class or more detail where we should be able to = draw out a block diagram of an Armstrong modulator, PLL modulator, etc.
Answer Regarding PM, AM, FM, we did this very briefly, so while we might expect that you will have general knowledge, I din't want to go into a lot of detail of commercial AM or FM. So, don't memorize Foster Seely Discriminators, or ratio detectors, TV block diagrams, etc.
Question Dec 21 For the 2002 exam, Question 2c). The answer says that the single Tuned bandwidth=3D700KHz, and the system = bandwidth was given as 450KHz. But wouldn't each single tuned bandwidth = be 0.642(450KHz)=3D288.9KHz. Where the two narrow single-tuned bandwidth combine to give a broader system bandwidth of 450KHz.
Answer 1. If the two tuned circuits are synchronously tuned (same fo) then where one has - 3dB, the two combined will have -6 dB of gain. Thus, the overall -3 dB bandwidth occurs where each has -1.5 dB of attenuation.
If you stagger tune them (different fo for each), there may be a way of making the bandwidth wider. But the combined transfer function of two ientical but isolated resonators the combined transfer function will have a narrower bandwidth.
Question Dec 21 Can you verify the statement: "We studied 3 types of phase detectors each of which has its own phase error beyond which the PLL will lose lock. The first one is analogue (sinusoidal) and it loses lock at 90 degrees, the secodn type is digital and also loses lock at 90 degress and finally the third type is the 3-state phase detector which loses lock at 360 degrees."
And, if 3-state PD can go plus/minus 360 does that mean it will always attain lock?
Answer It sounds correct, assuming we are currently at the nominal phase, then both the analog and x-or gate style can go plus or minus 90 degrees, whiel the three state can go plus or minus 360 degrees from the nominal.
So, how can a phase shift be more than 360 degrees? For example a frequency step causes a large phase shift that can be more than 360 degrees. Remember the frequency step problem in assignment 2? For a particular damping constant, the predicted maximum phase error normalized to delta omega/omega_n was a constant, e.g., about 0.46 for a damping constant of 0.707. Thus, actual phase error is 0.46 x delta omeag/omega_n. Thus, if delta omega /omega_n is equal to 15, maximum phase error is 6.9 radians which is more than 2 pi. Another example is when the two frequencies are different, there is a constantly growning phase shift between the two signals as they are sliding past each other. Each time they have moved past each other by a complete cycle, they had gone through 360 degrees of phase shift. Question Dec 20 In number 1b from the 2000 final, they ask to find the bandwidth of the tuned output. Now the only way that I can get the answer in the back of the book, is to include the 5k base resistance in parallel with rp and ro. This doesn't seem quite right to me, is there a mistake in that question or is there some trick that I'm missing in this. The answer I get is 106.96kHz using only ro and rp, but the back of the book says 356.65kHz.
Answer Bandwidth is determined as BW = 1/RC where R is the parallel combination of three components:
Question Dec 19 or Q2 (b) Tuned Amplifier on 2003 final,i got Req on sedondary is 3k//10k=2.3K. How come the answer on the notes is 1.154K?
Answer 2.3k is matched to 50 Ohms through the transformer, thus looking into the transformer is another 2.3k resistor. So the equivalent resistance is 2.3k parallel with 2.3k or 1.154k.
Question Dec 18 I guess this is kind of a dumb question because we probably learned it in ELEC 2501 - For the 2002, final 1(a) , I can get the correct value of the image at 728MHz. However, how is the number of decades found?
Answer Actually it is not a dumb question at all because the answer is not that obvious. The important thing to note is that everything is done with reference to the centre frequency fo and the edge of the passband fb. Thus, in linear terms you have to consider the ratio of (f0-fi)/(f0-fb) if this number is 10, it is one decade, if this number is 100 it is 2 decades. For other numbers you take 10 x log [(f0-fi)/(f0-fb)]. With f0 at 915M, fb at 902M and fi at 728M this is 10 log [187/13] = 10 log 13.38 = 1.126 decade, then at -40 dB/decade, there is 45 dB of image attenuation.
So, in ELEC 2501 we certainly never calculated decades for bandpass filters, I'm not sure we even did it for lowpass filters. But we can solve this using an equivalent lowpass filter - such a filter is found by shifting f0 to 0 Hz and setting the corner to be at (f0-fb)=13 MHz, and the image is at f0-fi = 187 MHz and the rolloff is -40 dB/decade. Another equivalent way of looking at this is that -40 dB/decade is a second order rolloff, or vo/vi is inversely proportional to the square of the frequency ratio fi/fb. Thus gain is 1/13.38^2 = 0.005586. In dB this is 20 log 0.005586 = -45 dB.
Question Dec 15 Im looking at the 2000 exam (DEC 7), and for Q 1 A, I am bit confused. I find the center frequency to be 4.5016 MHz, by using omega_0 = 1/sqrt(LxC). And the answer given, is 4.4572 MHz. Should I make use of the Q of a 100 associated with the inductor? Or, is it related to reflecting R_L on the secondary side to the primary?
Any helpful hints would be appreciated
Answer The frequency discrepancy is probably due to not having included the transistor parasitic output capacitance in your calculation. Parasitic capacitance has a habit of reducing the resonant frequency.
Question Dec 14 In regards to question 8 on the 1998 exam. When you solved question 8 on the 1998 exam, after finding solving for tau_1 and tau_2, you wrote down that at 1KHz, LG = K/s.
Then you wrote 10^6*1*A/(2*3.14*1k) = 100, and solved for A. I'm a little confused about this and was wondering where the 100 comes from.
Answer The open loop response was shown as part of the question. At 1 kHz, the open loop gain was shown as 40 dB, which would be a linear voltage gain of 100.
The general equation for loop gain is:
L.G. = Kphase x A x F(s) x Kvco/s = KF(s)/s.
From the shape of the open loop gain we can see that at low frequency, there is a first order rolloff. Since the VCO always has rolloff, the filter cannot be contributing to low frequency rolloff. Hence the loop filter it is some form of lowpass filter with phase lead correction with response given by:
F(s)) = (1+s tau_2)/(1+s tau_1).
At low frequencies up to the first breakpoint, the filter is simply:
F(s)=1,
hence the loop gain is: K/s where K is the product of Kvco, Kphase and A. The magnitude of the loop gain is then:
|L.G.| = (Kvco x Kphase x A)/(2 x pi x f)
And from the plot, this is 100 at 1 kHz. We are told that Kvco is 10^6 rad/sec/V, and that Kphase = 1 V/rad, and we know the frequency so the only unknown is A which we can then solve for.
Question Dec 12 I have been having problems spelling my name correctly. How many marks will name be worth ? (i.e. how much study time should I devote this part)
Answer As for spelling your name, I wish you the best of luck with that. I typically don't get mine right either - that's why there is a backspace key. And in response to your next "Is there a backspace key on the exam?" my answer is: your backspace key will be your pencil eraser, or your pen (use to draw a line through mistakes, or if the mistake is embarrasing scribble over it thoroughly so I can't read it).
Question Dec 12 I have a quick question. I took a look at the formula sheet today and noticed that there is a section on televisions. Since we did not cover this material in class, I was wondering why you left it on the formula sheet. Wouldn't that space be more useful for something that we covered (for example impedance matching)
Answer I try not to change the formula sheet too drastically, so things tend to get left on them. Can you suggest which formulas for impedance matching should be on the formula sheet? In my mind, the impedance matching stuff isn't really formulas as much as concepts - so I'm not sure any of that would have made it to the formula sheet even if there were more room. The transformer equation for impedance is on the page, and for LC matching the main thing is knowing where to start. (See if Re{Zin} is less than or greater than 50 Ohms).
Answer You aren't required to verify this equation for the assignment. This is a somewhat lengthy derivation. You equate the original circuit (C1, C2 and re+Re) with the new parallel circuit (Ceq parallel to Req). Note that Ceq is only approximately equal to C1 in series with C2, (this approximation becomes more exact when re+Re is large). The derivation is shown in the books "Modern Communications Circuits" by Jack Smith and "Radio Frequency Integrated Circuit Design" by Rogers and Plett. Both books are referenced in the course outline.
Question Dec 2 Hello it's 3AM, my assignment is due today, and I'm getting a strange result. I have calculated about 2.7 mA, and this current through a 10k resistor results in a 27V drop. What?
Answer The current of about 2.7 mA you are talking about is DC current. This current will find its way to or from the power supply through the path of least resistance (especially at 3AM). So, this current will not choose to go through a 10k resistor when it has a perfectly good Zero Ohm resistor (the inductor) to go through.
Question Dec 1 In the assignment it asks to perform an open loop frequency domain analysis. Is this a pen and paper exercise involving loop equations, etc?
Answer No, this is strictly a simulation exercise. You break the loop, apply a signal source on one side of the break (appropriately coupled so the DC conditions are maintained), then measure the magnitude and phase on the other side of the break. This information is used to predict if the closed-loop oscillator will oscillate and to predict the frequency of oscillation.
Question Nov 26 Your marking scheme gives marks for finding the new omega_n and eta after changing the filter gain to 1, but nowhere in the lab does it ask to do this....
Answer You are right, I did say so for the divider and thought it was also there for the change of gain. So, you are quite correct. This part is not asked for, so is not expected and it will be removed from the marking scheme.
Question Nov 25 For question 2 b, in order to calculate RL, the peak-to-peak output voltage is required. You have mentioned that the peak-to-peak output voltage will be the same as the peak-to-peak voltage at the collecter. However, the collector has a DC component to it, which is blocked at Cc capacitor before the RL output. Thus, would the Vout be nominal at 5V or 0V with only AC component? I'm assuming this AC component will be the swing by VE as mentioned in the question.
Answer At the collector, there is a DC connection to VCC (through the inductor) thus the DC voltage (average voltage) is at 5V. At the output, there is a DC connection (through the load resistor) to ground, so the Dc or average voltage at the output is at 0V. All of this can be verified by simulation. As for swing, 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. Since the output is coupled with a large capacitor which behaves like an AC short circuit the peak output swing will be the same. Thus, at the output with the average at 0V, the swing would be from -2V to +2V.
Question Nov 25 How come the closed loop frequency is different from open loop > frequency? Shouldn't they be the same?
Answer One reason is that closed loop results in transistors going into saturation, there could be extra phase shifts which can cause the frequency to shift. Note: for similar questions and answers, check out previous year's question lists.
Question Nov 25 if the gain at 0 degrees phase is 2.8, then the phase margin is : > (2.8-1)=1.8 . Is this how you find the margin?
Answer This sounds more like gain margin than phase margin. Gain margin is 2.8, not 1.8. You could reduce the gain by a factor of 2.8 and still have a gain of 1.
Question Nov 19 In class you gave formulas for parallel and series resonant frequencies that didn't agree with previously derived formulas. Did I miss recent amendment to the laws of physics?
Answer Oops again - resonant frequencies should always be omega = 1/square root{LC} or or omega squared = 1/{LC}. I believe I had an equation with both the "squared" and the "square root" in the same equation.
Answer Another oops! Before the printed circuit board came out, we recommended to students to use an extra divide by two to square up the waveform (the output waveform of the divide by 3 is not 50% duty cycle). But this extra divider was not incorporated into the printed circuit board so the text saying "6:1 or 8:1" is not correct, it should be "3:1 or 4:1".
Question Nov 18 Page 142/142 says loop bandwidth is 2000 Hz while at the front of the lab it says natural frequency is 2000 Hz. They aren't the same, are they?
Answer Oops, you are right. That should be natural frequency is 2000 Hz. Then loop bandwdidth can be calculated by one of the formulas given with the result that bandwidth is always bigger than natural frequency.
Answer It turns out that 2.5V is the nominal voltage everywhere - the nominal average phase detector output is 2.5V but can vary from 0V to 5V, the nominal VCO input voltage is also 2.5V, although you have characterized it from 0V to 5V. So everything starts from 2.5V and the ac (or small signal) voltage varies around this nominal value of 2.5V.
Answer First of all, how did you find your equivalent input components? You can do it by theory, or from your simulations. I know it will certainly work out by theory, so if simulations don't work out, I would suggest using theory.
From theory, it should be r_pi (hie) parallel to C_total where C_total is C_pi + Miller x C_mu. In the notes, page 114, C_pi (or C_be) is given as 8 pF, c_mu (or C_cb) is given as 1.6 pF. Typical gain was 15 to 20 so total capacitance would be from 32 to 40 pF. This is safely less than your calculated 59 pF and so the calculation would work out. With my value of 3000 Ohms the required capacitance goes up to about 71 pF.
If in fact you did have Ctotal < Ctransistor, then you would probably have no choice but to add parallel resistors. Thus, for the same bandwidth we could live with a larger Ctotal. We would rather avoid this approach since this will increase the noise, and possibly reduce the gain. So, instead, I suggest check your numbers, and if all else fails, use the theoretical, calculated numbers.
Question Oct 10 Should we include the 10 nF bypass capacitor across R1 in our calculations? Since it is in series with Ctotal and less than it, their equivalent capacitance would be smaller than Ctotal. Therefore BW and fc would be affected.
Answer The 10 nF capacitor is much larger than Ctotal (not much smaller as you say) so provides an ac ground at this node. Thus the bias resistors are shorted out, and the transformer is located between virtual ground and the input of the transistor amplifier.
Question Oct 8 I have a question concerning the calculation of the turns ratio on the input side of my amplifier. At the beginning I don't know the inductor value, so how can I include the parallel resistance of the inductor R_P? As well, I don't know what is the transformed resistance from the source until I know the equivalent resistance on the secondary. So, should I start with leaving out both of these incuctors, then iterating?
Answer At the start, I would suggest ignoring the loss due to finite inductor Q. However, you should include the transformed source resistance which will equal R_in. Thus, in the first round, R_eq is R_in/2. Then calculate C_tot, find L. This is the end of the first round. Now you can determine R_p. Then you should do at least one iteration including R_p in your calculations. Thus in this second round, you are matching Rs to Rin || R_p. so your turns ratio will change. R_eq will change, hence C_tot will change, and L will change. You could keep going, but after shoiwing that you know what you are doing there really isn't much point in continuing. Question Oct 8 Also, are we required to simulate our results or are hand calculation sufficient?
Answer You are not expected to simulate this - just do the hand calculations.