1. Since we are free to choose the VCO characteristics, we can use the fact that, for a simple lag compensator: For the simple RC lag compensator k A = 1. k  is given as 4/  volts/rad, and  n ~ (2  ) 200 x 10 3 r/s. Thus So we can solve for  c and k 0 which will give the desired response with a simple RC lag compensator. Lag Compensation

2. For the simple RC lag compensator, H(s) =  c /(s+  c ). The closed loop transfer functions to the phase detector output (T 1 ) and the VCO input (T 2 ) are: To find the step responses, multiply the transfer functions by 1/s and take the inverse Laplace Transforms:

3. Lag Compensated Phase Detector Response Lag Compensated VCO Response

4. Lead-Lag Compensation Since we are free to choose the VCO characteristics we must choose VCO gain to satisfy Surprise! For the simple RC lead-lag compensator k A = 1. k  is given as 4/  volts/rad, and  n ~ (2  ) 200 x 10 3 r/s. Thus We can now step through the lead-lag compensator design per the method presented in class. It is interesting to note that if we set k 0 =  n /2  k , as we did with the simple lag compensator, the zero (  2 ) moves to infinity, and  1 =  c from the lag approach.

5. For the simple RC lead/lag compensator, H(s) = (  1 /  2 )(s+  2 )/(s+  1 ). The closed loop transfer functions to the phase detector output (T 1 ) and the VCO input (T 2 ) are: To find the step responses, multiply the transfer functions by 1/s and take the inverse Laplace Transforms. Note that the form of T 1 and T 2 is exactly the same as T 1 for the simple lag network phase detector output with  1 or  2 substituted for  c.

6. Lead-Lag Compensated Phase Detector Response Lead-Lag Compensated VCO Response

7. Answers to In-class Exam 1aEnough information exists in that bandwidth to make voice intelligible. 1b6.6 k-samples/sec 1cAvoid aliasing 1d52,800 bits/sec 1eNonlinear gain characteristic where low levels are boosted before A/D conversion and restored after D/A conversion at the receiver. Improves S/N for low volume levels. 2a25 channels 2b100 channels 2cBaud rate is symbol rate, symbols may contain multiple bits/symbol. 2a-biphase: 1200Kbaud, 1200K bit/sec 2b-16 QAM: 1200Kbaud, 4800K bit/sec 2dTakes a much larger noise phasor to cause an error in biphase than 16 QAM 3BW = 37.5 Khz for all. 2-phase: 75K bit/sec, 3-phase: 118.9 K bit/sec, 4-phase: 150K bit/sec 4FEC adds redundant information to a digital message so that transmission errors can be corrected at the receiver. The theory tells us we can construct FEC codes to correct an arbitrary number of errors, given enough redundancy, reducing the probability of error to near zero. 5-- 01 10 11 11 10 01 0 0 10 00 01 11 0 1 01 00 10 01 –- 1 0 0 1 1 1 1 0 1 0 0 1 0 1 1 1 6a“COMM IS KOOL” 6bEven Parity 6c4 th bit in “K”.. Should be “C” 6d80.77%