Answer
Question Dec 6 Exam 2004, Q4a -How can we make out ciruit more linear is it beacuse of having the Re resistance?
Answer Yes, RE is what makes it linear. The input voltage is not only across BE junctions of a transistor, but for a large input, most of it is across the resistor RE.
Question Dec 6 Exam 2000 - Q5..did we do analysis for large signal?
Answer Yes, our large-signal analysis was done through time domain Spice simulations of the closed loop oscillator. This is large signal because the output voltage was large, in assignment 3 about 9V peak to peak, and the signal was limited by hitting the nonlinearity (saturation) of the transistor. This is to be compared to our small-signal, analysis done by Spice open-loop ac simulations.
Question Dec 6 EXM 2002 Q3c- How do we calculate the value of C after we have found out L because I dont get 170pF by using a center frequecy of 18 X 106.
Answer Oops, that's a typo. The answer should be 138.9 pF.
Question Dec 6 I cant remember how to solve the noise figure question from 2003...Its question 1(e).
Answer that was considered one of the hard questions on that exam. It is not directly in the notes, but I did it in class. Here is a summary:
The noise factor is F = N_o,total/N_o,src
where
N_o,src = N_src x G1 x G2.
N_o,total = N_o,src+N_add1 x G2 + N_add2
where N_add1 and N_add2 refer to noise being added at the output of
their respective blocks
Thus,
F = (N_src x G1 x G2 + N_add1 x G2 + N_add2)/(N_src x G1 x G2)
and this is equal to
F = 1 + N_add1/(N_src x G1) + N_add2/(N_src x G1 x G2)
But individually, the noise figure for each block can be found as:
F1 = 1 + N_add1/(N_src x G1) this is the same as the first two terms.
F2 = 1 + N_add2/(N_src x G2) and this is similar to the last term, so
the last term can be written as (F2 - 1)/G1
Thus,
F = F1 + (F2-1)/G1
Note, all gains, F, Noise, etc are all power terms, so to change from
linear to dB, we have to take 10 log, not 20 log.
G1 = 20 dB or 100 linear, G2 = 6 dB or 4 linear,
F1 = 3 dB or 2 linear, F2 = 10 dB or 10 linear.
Thus
F = 2 + (10-1)/100 = 2 + 9/100 = 2.09. In dB this is 3.20 dB.
Question Dec 6 In Exam 2000, Q6 part d, we use the OL gain to find frequeccy at unity gain.Why do we need to do that and how is it we got that OL(open loop) gain euqation?
Answer The open loop magnitude and phase are used to predict stability. If the loop has a gain of 1 and phase of 180 degrees, in the closed loop, negative feedback would become positive feedback and oscillation would occur. Thus, we test the open loop gain at the unity gain point and determine how far the phase is from 180 degrees - this is the phase margin and if it is equal to or bigger than 45 degrees, we say the loop will be stable.
The loop is made up of phase detector with gain Kphase, loop filter with transfer function A_0 F(s) and VCO with gain of Kvco/s. Kphase and Kvco are stated in the problem and the diagram shows F(s) thus we have all the information to find the loop gain. (We are not told what A_0 is so we simply assumed it was 1).
Question Dec 6 Exam 2002 Q1a - could you please explan how we came to the asnwer given in the book - I'm not able to figure it out.
Answer This is a little bit beyond the scope of what we did this year, so such a calculation will not be on the exam this year, but I can give you a quick explanation - actually, this one is not that quick - still interested? First, what is the worst case? For low-side LO with f_IF at 100 MHz f_LO = f_RF-100. The image is another 100 MHz lower than the LO, so f_image = f_RF - 200. The worst case is when f_RF is at 928 MHz since that means the image frequency is closer to the filter edge. So, f_image is 928-200 = 728 MHz. A fourth order BPF is like a second order HPF so we have 40 dB/decade rollof, but we need to figure out how many decades we have. We do this by considering the centre of the filter is at zero, then the band edge is at 13 MHz, and then we compare the offset to 13 MHz. Thus, 130 MHz offset would be exactly one decade. However, the offset here is 902-728=174 MHz. Clearly a bit more than a decade, but how much? We find it by taking the log of the ratio (so that a factor of 10 would give 1 decade, a factor of 100 would give 2 decades). Here it is log(174/13) = 1.127 decades. Thus attenuation is 40 x 1.127 = 45.1 dB.
Question Dec 6 To clarify about the exam. You said it there will definitely be problems on the 3 main topics of tuned amplifiers, PLLs > and oscillators and some minor problems on mixers, C amplifiers, noise, etc. And the question difficulty would they be around the same as the assignments and past exams? Would they be pretty much straighforward or a little bit tricky?
Answer I can't say anything specific about this year's exams, except that my intention every year is to have a mixture of straight forward questions and parts of questions that require a bit of extra thought.
Question Dec 6 For year 2004, Q5, part c) I do not know how to get the settling time t=100usec to 160usec and the max error=0.0465 radians. For part d) I got output error is 0.1, but your answer is 0.628, I think if I use 0.1 *2pi, then I can get your answer, but I can't see why I would need a factor of 2pi?
Answer You typically would only get one answer, but I gave the range from 10% settling to 2% settling (that is the answres are for omega_n t_settling = 5 to omega_n t_settling = 8) For part d, you have Delta omega/omega n - The question gives Delta f instead of Delta omega, and since we have omega_n in radians we have to convert both to the same units.
Question Dec 6 For year 2004, Q4, part b) how do you see the 1MHz and 0.9MHz frequency from the waveform?
Answer I see 10 cycles per 10usec (or 1 MHz) for the unfiltered signal, then from the envelope, I see a period of 10 usec, but that is twice the modulating signal of 20usec (so the modulating frequency is 50 kHz). Recall that DSBSC has output frequencies at omega_c plus and minus omega_m so the two outputs are separated by twice the modulating frequency, and that is 100kHz.
Question Dec 6 We all believe the unit is tau_1 is seconds, so the unit of 1/tau_1 should be Hz, but why does it show up as "rad/sec" in your answer?
Answer From ELEC 2501 RC is a time constant with units of seconds, 1/RC is always the frequency in radians per second. Herz has units of cycles/second while omega has units of radians/second - neither cycles nor radians are official units, so both are often labeled as having units of 1/seconds. The different between the two is simply a unitless constant of 2 pi.
Question Dec 5 In assignment 2, Q6, when N=40, fout =10.01MHz (TA's answer), rather than fout=250k*40+1k*40=10M+40K=10.04MHz (my answer), could you please explain it?
Answer The frequency drift is at the crystal, not at the phase detector. Thus, the crystal oscillator is at 1 MHz + 1KHz, then after you divide by 4, at the phase detector the frequency is at 250kHz + 250Hz, then you multiply by 40 to get 10MHz + 10kHz.
Question Dec 5 For year 2005, Q5, part a) the phase deviation range is 0 to 180 degrees for XNOR gate, how about XOR gate? Is it from - pi/2 to pi/2?
Answer It is also from 0 to 180 degrees.
Question Dec 5 In Exam 2005, Q5 part c), what.s the loop parameters, tau_1 and tau_2 and damping factor, anything else? If the maximum phase error is larger than specified pi/2, should we change the damping factor, or some other parameters also need to be changed?
Answer Larger damping factor can help, but choosing larger natural frequency will probably be the best way to reduce the phase error (since peakerror=0.46xDelta omega/omega_n, a larger omega_n will directly help.
Question Dec 5 In Exam 2005 Q5 part d), why is our system stable if the phase margin is greater than 45 degrees? What's the relationship between phase margin and stability? 45 degree phase margin stable is only for first order PLL or both first and second order?
Answer Instability is when the phase is 180 degrees when the loop gain goes to unity. Phase margin measures how far away the phase is from 180 degrees, so further away (bigger phase margin) means it is more stable. A similar argument can be used for any order, but we are only talking about second order systems here.
Question Dec 5 In assignment 2, question 4, the phase detector is an exclusive-or gate and the loop filter is an integrator with phase lead correction. I do not know why you put this sentence here, is it significant, or what point is being shown here? I don't see how it could be used to solve the question.
Answer Knowing that it is an exclusive-or gate indicates what the maximum phase error can be. In comparison, the tri-state phase detector can have up to plus or minus 2 pi of phase error. Also, standard equations and curves use the integrator with phase lead correction so you know you can use the standard answers and curves. Other filters (e.g., LPF with phase lead as in lab 3) would give approximately the same answers for natural frequency and damping constant, but other parameters like steady state phase error, or lock range could be different.
Question Dec 4 I thought beta >= X1/X2, but you write beta >= C2/C1. Was that a mistake?
Answer No, x1 = 1/(omega C1) and x2 = 1/(omega C2) so X1/X2 = C2/C1.
Question Dec 3 Here are some questions about Class C amplifiers from last year's exam (and before):
Answer
Question Dec 3 In the 2002 tuned amplifier solution it states that the single-tuned bandwidth is 700KHz. The question requires the -3db bandwidth for the entire system to be 450 450KHz. How do we go from 450 kHz double tuned to 700 kHz single tuned?
Answer For a double-tuned amplifier, -3dB bandwidth will occur where the attenuation of a single-tuned amplifier is -1.5 dB and as stated in the question, this occus at 0.642 times the -3 dB bandwidth. Thus, the single tuned bandwidth must be 450 kH/0.642 or 700 kHz.
Question Dec 3 For year 2005, question 6, you told us the current gain is 1 at 10GHz and 10 at 1GHz, but how do you know the slope is 1? since there is a "aim for a factor of 2" then we pick up beta =2, could you please explain "aim for a factor of 2" a little bit more?
Answer This is a first order system with a current source gm x vpi and capacitors c_pi and c_mu, as you have seen in ELEC 3509. Thus, rolloff is like a first order filter in which current gain is inversely proportional to frequency. We solve for the condition for gain which tells us that beta should be equal to or greater than C2/C1 at 1 GHz. For a margin of 2 we choose beta to be 2C2/C1 instead of just C2/C1. Since beta at 1 GHz is equal to 10, we choose C2/C1 to be equal to 5.
Question Dec 3 Yes, but how do you know the transistor is first order or second order, or what?
Answer In a transistor, there is only one capacitor (since C_pi is parallel to C_mu, since for ft measurements, the collector is grounded - see 3509). With only one capacitor it can be no more than a first order filter.
Question Dec 3 For year 2003, question 6, how do we know that current gain equals 1 at fT=100MHz? and how do we know that there is a flat region (beta =100)between 0 to 1MHz?
Answer As you have seen in ELEC 3509, beta is constant at its low frequency value up to f_beta. After that, beta drops of and beta is inversely proportional to frequency, so that for example at 4 times f_beta, beta is down by a factor of 4 times. At ft, beta has dropped to 1. So starting from f_beta we can work out how much beta drops with frequency. Or starting at ft, we can work backwards and see how big beta will be at some lower frequency. For example, at ft/5, beta is 5 and at ft/10, beta is 10, etc. Thus if ft were 100 MHz - then at 10 MHz (our operating frequency), beta would be equal to 10. If ft were only 50 MHz, then beta at 10 MHz would only be 5. Thus, if ft is more than 50 MHz then beta will be bigger than 5 at 10 MHz, (beta is equal to or bigger than 5 was the condition for oscillation). To complete this example, if low frequency beta is equal to 100, then f_beta is at ft/100, and if ft happened to be at 100 MHz, then this would be at 1 MHz. Before that, beta would be constant at its low frequency value of 100.
Question Dec 3 For year 2005, question 4, we need to draw the time domain waveforms, I do not know how to get the output peried?
Answer For a double sideband suppressed carrier, the envelope is the period of the modulating signal. This envelope goes from plus V to minus V. When one sideband is filtered, the envelope no longer crosses over from positive to negative so it is now more like a double frequency (or half the period).
Question Dec 3 In the phase lock loop section, the Greek symbol zeta is used on page 65 and in Assignment 2, Question 5, and in some exams. However, on page 63 of the notes, the figure shows a family of curves with the Greek symbol Xi. Do these different Greek symbols in fact have different meanings?
Answer No, my carelessness, both are intended to be damping constant.
Question Dec 3 For year 2000 exam, Q6, part d), I do not really know what.s the break point? I think the magnitude plots is easy, but cannot really draw the phase plot, could you please give me some hints?
As well, I do not really understand the stability part lecture notes on page 67, so do not know how to explain the stability part in part d). So could you please give us more hints and explain it a little bit more?
Answer Note: I went over this exam problem in one of the later lectures, second-last one, I believe.
The break point is the point where the slope changes from -2 to -1 or from -1 to -2. In other words, the line has been broken, or at least bent at these points. And, if you have the equation for each straight line, you have both the magnitude and the phase. That is, for a loop gain with equation K/s (1+s tau_2)/(1+s tau_1) the first part of the curve is K/s or replacing s with j omega, the equation is K/(j omega) or -jk/omega and both the magnitude and phase are obvious (magnitude decreases with a slope of -1, and phase is -90 degrees). In the second region, the equation is K/(s2 tau_1) = -K/(omega2 tau_1). Again magnitude and phase are obvious (magnitude decreases with a slope of -2 and phase is at -180). As a conclusion, if the magnitude has a slope of 0, phase is 0. Slope of -1 means phase of -90. Slope of -2 means phase of -180. At the break point between two slopes, the phase is half way between the two phases.
Continuing from above, at the break point between the slope of -2 and -1, the phase is at -135. This is 45 degrees away from 180, so if this happened to be the unity gain magnitude, the phase margin would be 45 degrees. If the unity gain magnitude occurs at higher frequency, then the phase margin will be even higher. So, we need to determine where the unity gain magnitude occurs to tell us if the system is stable. We can solve for unity gain on any of the straight-line approximations, but the one with a slope of -2 is possibly the easiest to remember how to do. As I said in part 3, the gain on this line is -K/(omega2 tau_1), but K/tau_1 is equal to omega_n2, so the magnitude of the gain becomes omega_n2/omega2. By equating this to unity, we get omega = omega_n. Thus this straight line segment crosses unity gain where omega is equal to omega_n. We have already solved for omega_n, so if this is larger than the break point frequency (1/tau_2) then the unity gain point is to the right of the break point and our phase margin will be bigger than 45 degrees and our system will be stable.
Question Dec 2 Is the formula sheet going to be exactly like the 2005 version you have in the course pack ? if not, can you post it on your page?
Answer The new formula sheet is very nearly, but not quite the same as the previous one - the PLL transient curves have been made darker and the TV part has been removed. Where the TV part used to be are now transistor equations (these were also on last year's exam, except they were on the first page). Since it is so similar, I probably won't post it, but I'll see.
Question Dec 2 also, are we responsible for material that isn't covered by the formula sheet ? like for the PLL section, you provided one of the filter functions F(s)= (sT2+1)/sT2 so are we reponsible for knowing the other transfer functions and their respective magnitude and phase graph as well ? thanks for your response
Answer Regarding the formula sheet, you are responsible for everything in the course notes (except for the TV section, or parts marked bonus section) and all the material in the labs, whether or not they are on the formula sheet. The formula sheet is intended to provide you with most of the formulas you might need, and in some cases to be starting points to derive things that are not directly on the formula sheet. For instance, using your example, the formula sheet does not give you any plots of magnitude and phase, you are supposed to be able to work these out. And, the particular filter you mention F(s)=(1+stau_2)/stau_1) is the most important one - most other filters can be approximated by this one, as you discovered in assignment 2. However, there are situations where this is not enough, and you should be able to deal with that - for example, a frequency synthesizer has an extra factor or 1/N in the transfer function - that is not on the formula sheet, but it is an easy extension. Similarly, other filters would mean you have to substitute for F(s), but that should also be straight forward if you understand the concept.
Question Dec 1 Question 5 on the 2003 final... I cant seem to figure out how you get the value of K. In the solutions, it says K=3000*2pi...but I dont seem to get that. Any hints?
Answer The loop remains locked from 4k to 10, which is a range of 6k, or we could think of it as the centre frequency of 7k plus and minus 3k. Lock range for an analog phase detector is plus and minus K, hence K is 2pi x 3k.