Answer
Question Nov 3 In Question 5, is there an error on the y axis on page 60 of the 2000 notes? I can't make the equation and the graph agree.
Answer In the 2001 notes, on page 60, I corrected the Y axis of the plot to be |epsilon/(Delta_omega/omega_n)|. If you have the 2000 notes you need to make this correction.
Question Nov 3 For question 4, what was that I heard about calculating two values for omega_n?
Answer There are two conditions you have to meet for this question. One is for maximum settling time, the other for maximum phase error. Both depend on omega_n in some way. Bigger omega_n will mean faster settling time and lower phase error for a given frequency step. Thus, it should be possible to pick an omega_n big enough so that both settling and phase error conditions are just met. One way to do this is to solve for the minimum omega_n to meet each condition by itself, then pick the omega_n which will meet both.
Question Nov 3 I am getting numbers for my filter components that are somewhat larger than the values you quoted in class. I don't know where I went wrong. Maybe if you could comment on my values of tau3 and tau2 that might give me a hint as to where the problem might be? I had tau3 between 100 ms and 1 s depending on which phase detector I used.
Answer I can't remember exact values, but it seems to me that tau_3 was in the tens of milliseconds and tau_2 was less than a millisecond, somewhere in the hundreds of microseconds. This is of course assuming you are using the assignment value for Kvco, not your own. If you are using your own, your numbers may be different, see the next question.
So your value of tau 3 sounds a bit high. In the calulation of it, you divide K by the square of omega_n. Make sure you have rememberd 2 pi in all the right places. Note that (omega_n)^2 = (2pi*f_n)^2 so you end up with a pi^2.
Question Nov 3 I had a measured value of Kvco about twice as high as the one listed in the assignment. Have I made a mistake?
Answer There is quite a range of possible answers. My number was based on estimating that the frequency varied from DC to 1 MHz in about 2.5 V for a slope of 400 kHz/V or Kvco = 2pi*400krad/s/V. Some people had 2 MHz as the top frequency, others had the complete swing over less than 2V (maybe 1.5V), some had it over 3V, (maybe 3.5V), so there is quite a range possible, and all of these can be valid. For all these numbers listed, frequency 1 to 2 MHz, voltage range from 1.5 to 3.5V the Kvco would range from 2pi*286krad/s/V to 2pi*1333krad/s/V.
Question Nov 2 For question 5, I can figure out the damping constant and natural frequency, but I am having trouble with the last part. How can I find the frequency for which phase error is less than 0.25 rad?
Answer In the course notes, there is a an equation for error versus frequency on one page and a curve of the same on the next page. You can look at the curve, and noting the peak has an actual error of 0.53 radians look for which frequency the phase error is about half of this value (peak is 0.53, you are looking for 0.25). This will tell you roughly what the frequency is. Then you can use the equation, for example iteratively you can try this frequency, calculate the phase error, if it is more than 0.25, decrease the frequency a bit. A few trials should get you fairly close.
Question Nov 2 For the opamp filter, in 97.359 I read that if there was an input voltage applied to the positive input, then it was a non-inverting amplifier. We have a voltage source on the positive input, shouldn't the function be positive?
Answer You are correct, about the positive input, however there is only a DC voltage at the positive input. The small signal, or AC input is applied to the negative input so the transfer function from ve to vc has a negative sign on it.
Further to that point, we could express the output as a sum of the two inputs, however, since the input to the positive terminal is DC only, it will only affect the DC component. We could write the equation:
vc = VDC * (1+ZF/ZI) - ve * Zf/Zi.
At Dc this becomes :
vc = Vdc * (1+R3/R1) - ve * R3/R1.
or Vdc is at 2.5 Volts and if ve is at its nominal value of 2.5V,
vc = 2.5 * 3 - 2.5 * 2 = 2.5V
Thus, all voltages are nominally at 2.5V, so we can simply call this the bias point, then ignore it.
Question Nov 2 So, does the filter have a negative transfer function, or should we remove it and why?
Answer The filter has a negative transfer function, but in order to keep the loop stable, we then have to use a positive input on the phase detector. Thus the phase detector has a positive input and the loop transfer function is then:
theta_out/theta_in = F.G./(1 - L.G.) = K F(s)/(s - K F(s))
Then, if F(s) is negative, it cancels the negative sign in the denominator, and all is positive once more. This is equivalent to using the regular loop assuming the filter is positive and using the negative input to the phase detector.