Lag Compensation
Since we are free to choose the VCO characteristics, we can use the fact that,
for a simple lag compensator:
c
1

2 k k A ko
n  k k Akoc
For the simple RC lag compensator kA = 1. k is given as
4/p volts/rad, and n ~ (2p) 200 x 103 r/s. Thus
n2
 koc
k
4 k 
2
c
ko
So we can solve for c and k0 which will give the desired response with a
simple RC lag compensator.
 c 
  2
2


koc    c    4 k
 k 
 ko 
c  2n
2
n


2
 ko 



koc    k02   n  12
 k   4 k 
 c 
ko 
n
2k 
For the simple RC lag compensator, H(s) = c/(s+c). The closed loop transfer
functions to the phase detector output (T1) and the VCO input (T2) are:
T1  s  
k
 c   1 
1  k 
 k0  
s


s
c 

 
T2  s   T1  s   c
 s  c

sk  s  c 
s 2  sc  k k0c

skc
 2
 s  sc  k k0c
To find the step responses, multiply the transfer functions by 1/s and take the
inverse Laplace Transforms:
 T1  s  

n t  c  n
s1  t   L 
sin r t   cos r t  
  k e

 r

 s 

n t  c
1  T2  s  
s2  t   L 
  k e
 sin r t  
 r

 s 
1
n  k koc

c
2n
r  n 1   2
Lag Compensated Phase Detector Response
2.5
2
1.5
PD-200K
1
0.5
0
-0.5 0
PD-20K
PD Step-200K
0.000002
0.000004
0.000006
0.000008
0.00001
0.000012
0.000014
-1
PD Step-20K
-1.5
-2
Lag Compensated VCO Response
1
0.8
0.6
0.4
VCO-200K
0.2
VCO-20K
0
-0.2 0
-0.4
-0.6
-0.8
-1
0.000002
0.000004
0.000006
0.000008
0.00001
0.000012
0.000014
VCO Step-200K
VCO Step-20K
Lead-Lag Compensation
Since we are free to choose the VCO characteristics we must choose VCO gain
to satisfy
n
k k A ko 
2
For the simple RC lead-lag compensator kA = 1. k is
given as 4/p volts/rad, and n ~ (2p) 200 x 103 r/s. Thus
n
ko 
2k
Surprise!
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 k0 = n/2k, as we did with the simple lag
compensator, the zero (2) moves to infinity, and 1 = c from the lag approach.
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 (T1) and the VCO input (T2) are:
T1  s  
k
 s  2  1  1 
1  k 
 k0  
s


s

1  2

sk  s  1 
 k k 
s 2  s1 1   0   k k01
2 

sk  s  2 
 s  2  1  1 
T2  s   T1  s  


 
 s  1  2  2  s 2  s 1  k k0   k k 

1
2   0 1

To find the step responses, multiply the transfer functions by 1/s and take the
inverse Laplace Transforms. Note that the form of T1 and T2 is exactly the
same as T1 for the simple lag network phase detector output with 1 or 2
substituted for c .

n t  1  n
1  T1  s  
s1  t   L 
sin r t   cos r t  
  k e

s


r



  1 
n t  2  n
1  T2  s  
s2  t   L 

k
e
sin

t

cos

t




 

r
r 

s



r
 2 


n  k ko1
1  n 1 


2  2 n 
 
r  n 1   2
Lead-Lag Compensated Phase Detector Response
2.5
2
1.5
1
PD-200K
0.5
PD-20K
0
-0.5 0
0.000002
0.000004
0.000006
0.000008
0.00001
0.000012
PD Step-200K
0.000014
PD Step-20K
-1
-1.5
-2
-2.5
Lead-Lag Compensated VCO Response
1
0.04
0.8
0.03
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
0.02
0.01
VCO-200K
0
VCO Step-200K
-0.01
-0.02
-0.8
-0.03
-1
-0.04
VCO-20K
VCO Step-20K
Answers to In-class Exam
1a
1b
1c
1d
1e
Enough information exists in that bandwidth to make voice intelligible.
6.6 k-samples/sec
Avoid aliasing
52,800 bits/sec
Nonlinear 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 .
2a
2b
2c
2d
25 channels
100 channels
Baud rate is symbol rate, symbols may contain multiple bits/symbol.
2a-biphase: 1200Kbaud, 1200K bit/sec 2b-16 QAM: 1200Kbaud, 4800K bit/sec
Takes a much larger noise phasor to cause an error in biphase than 16 QAM
3
BW = 37.5 Khz for all. 2-phase: 75K bit/sec, 3-phase: 118.9 K bit/sec, 4-phase: 150K bit/sec
4
FEC 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 00 10 00 01 11 01 01 00 10 01 –1 0 0 1 1 1 1 0 1 0 0 1 0 1 1 1
6a
6b
6c
6d
“COMM IS KOOL”
Even Parity
4th bit in “K” . . Should be “C”
80.77%