1999 Telecom Questions

Answer

Question Dec 22

Answer

Question Dec 21

Answer

• Given answer on page 169 is 2 pi x 4350
• The correct answer should be 2 pi x 2174

Steps are:

Look at transient curve given on formula sheet, maximum normalized phase error is 0.46 for damping of 0.707. Actual phase error epsilon is normalized by dividing by (Delta omega / omega_n) so equation at the peak is: :

epsilon/(Delta omega/omega_n) = 0.46

Given maximum allowed phase error (epsilon = 0.5) and omega_n = 2 pi x 2000 can solve for Delta omega.

Delta omega = epsilon x omega_n / 0.46 = 2 pi x 2174

Factor of 2 error on page 169.

Question Dec 20

Answer

I did this in the Friday review session.

I calculated the gain at 1 kHz where the gain is:

   gain = (Kvco/omega x Kphase x A0) = 100 or 40 dB

by using the known gain of the phase detector and VCO and the known total gain (given in the plot). The only unknown is A0.

Question Dec 20

Answer

The drift can bring you closer to the sidebands, meaning you must reduce the lock range to compensate to avoid getting too close to the sidebands.

Details:

It is desired to have the lock range cover the range from 16 kHz to 22 kHz to avoid locking on to the sidebands.

When calculating lock range, the standard formula is

    lock range = +/- Kphase X Kvco X A0 / N. 

This is for lock range at the phase detector. Without drift, this could be used directly to note that

    K/n = 2 pi X 3000

So, without drift, the maximum VCO input voltage is

    vc = Kphase X A0 

at which point the VCO output frequency is

    omega_VCO = omega_nominal + Kphase X A0 X Kvco. 

If the VCO drifts upwards, for the same input voltage the VCO output frequency will be 1 kHz higher or at the phase detector after the divide-by-2 the output would be not at 22 kHz but at 22.5 kHz. Thus to allow for the possibility that there may be drift, one must design the lock range to be less.

In other words, without drift omega_noimina is 38 kHz, with drift, it could be 37 kHz or 39 kHz, (18.5 or 19.5 kHz after divide-by-2) then the deviation away from the nominal must be less to prevent the possiblility of getting too close to the sideband.

Question Dec 20

When we were doing Assignment 1, most of us proceded to do the noise figure calculations as follows:

   NF = 10Log(Nout/GNin)

• For G, we used V(3)/V(7);
• For Nout we used the file "total output noise voltage";
• For Nin we used the file "equivalent input noise at V(7).

We were, of course, advised that this was wrong, and proceded to use 10Log(Total OP noise voltage/Noise at Rin)

One last time, why was our first assumption wrong?

Answer

    NF = 10Log(Nout/GNin)

However to make it clearer let us re-write that as:

    NF = 10Log(Nout_total / G Nin_source)

• where Nout_total is the total output noise
• Nin_source is the output noise due to the source only

What you have used is not correct becsaue:

What is labelled "equivalent input noise at V(7)" could be labelled Nin but is actually more correctly Nin_total that is the total output noise divided by the gain while what you need is Nin_source, which is not given anywhere in the SPICE output deck.

Also what you refer to as "Noise at Rin" is in fact not noise at Rin but "noise at the output due to Rin" which is exactly what I referred to above as "Nout_source" which is equal to "G Nin_source"

Note your calculation if done very accurately would result in Nout_total /G Nin_total = 1. Thus you would get NF = 0 dB.

Question Dec 19

Answer

I am not quite sure what you are getting at, and I have reworded your question to what I think you might have been asking. I note that you seem to be looking for formulas to use in answering questions. Rather than memorizing formulas, I would rather you have understanding about what is going on.

When a transformer is used for impedance matching as in the following rather poor ascii diagram, there will be an impedance seen looking into the secondary side of

          ZY = RY = RS X (N2/N1)^2.

Impedance matching is simply making this values of RY equal to the secondary side impedance. The secondary impedance if all the reactive elements (L and C) have cancelled out is R2 which is the parallel combination of Rp and Rt:

              R2 = Rp || Rt.  

where:

• Rp is the transformer parallel resistance and
• Rt is the input parallel resistance of the transistor amplifier.

Note in a real life circuit there might also be bias resistors which could also be in parallel with the above.

So when looking into the transformer from the secondary side, you should see a transformed resistance equal to R2. When doing a calculation on the secondary side only, such as for calculating the bandwidth, this transformed resistance is included in the model of the secondary side. Thus the resistance seen by the tank circuit on the secondary side is not simply R2 but it will also see the transformed resistance RY.

So do you use resistors once or twice? You check by drawing the model of the circuit you are trying to analyse as below. It should be obvious that impedance matching requires RS be matched to R2 while bandwidth calculations require R2 parallel with RY.

                       N1  N2
    ------/\/\/---------    -------------------------------------
            RS          D  C                             |       |
                        D  C  L                          /       |
                        D  C                             \       |
ZX = RX          2      D  C <=== ZY               2     /       |
   = R2 X (N1/N2) ===>  D  C    = RY = RS X (N2/N1)      \ R2    = C 
                        D  C                             /       |
                        D  C                             |       |
   ---------------------    -------------------------------------

Question Dec 15

Answer

1/(\sqrt(1.2*1.1))=0.87
1/(\sqrt(0.8*0.9))=1.18

These numbers are reasonably representative of integrated inductors and capacitors. Capacitors are less accurate because of the dependence on oxide thickness. Note absolute values are not accurate, but matching between two similar capacitors can be very good, down to 0.1%. Inductors are formed by a metal spiral, so depends mainly on the precision of the lithography so can be much more accurate.

Question Dec 10

Answer

      _                                                              _
     /  The double ended spectrum is convenient mathematically,       \
    |   and for this reason, mathematicians tend to use it. Because    |
    |   they don't usually worry about physical explanations, and they |
    |   don't tyically use spectrum analyzers or oscilloscopes, they   |
    |   are quite happy with the concept of negative frequency.        |
     \_                                                              _/

So, in looking at the equation:

> v=cos(Wc*t)-beta*sin(Wc*t)*sin(Wm*t)

The first term is v=cos(Wc*t) is simply a sine wave with an gain of one, therefore the result is one volt. This agrees with what you would see on a spectrum analyzer (the single-ended spectrum).

The second term is beta*sin(Wc*t)*sin(Wm*t) when viewed on the scope would look like a double-sindeband suppressed carrier signal with a peak amplitude of beta. This can be broken into two sinewaves of different frequencies each with half the amplitude.

Note by using trigonometry as in class, the DSBSC forms two components at omega_c plus and minus omega_m each with half the amplitude. This is the single-ended spectrum. Only when you express sines and cosines in exponential form you get plus and minus frequencies each with a further reduction to half the amplitude. Both ways of looking at it are valid, but single-ended is what we usually use if we really are trying to find output voltage.