fsk modulation demodulation
1.1 ABSTRACT:
2 .1 BLOCK DIAGRAM:
Fig 2.1 Block diagram for FSK modulation and de-modulation
2.2.1 PIN DIAGRAM- IC 565 :
Fig 2.2 Pin diagram for IC 565 2.2.2 PIN DIAGRAM -IC 555: Fig 2.3 Pin diagram for IC 555
3.1 DESIGN CALCULATION:
3.1.1 FSK MODULATOR:
ON time TH=0.693RBC
OFF time TL=0.693(RA+RB)C
Total time T=TH+TL=0.693(RA+2RB)C
1 f1= -------------------------- → (1) 0.69(RA+2RB) C1
1 f2= ------------------------------------ →(2)
0.69(RA+2RB) C 1C2 ----------- C1+C2
Duty cycle D=ON time/Total time=(RB)/(RA+2RB)=0.3
RB= 0.3RA+0.6RB
RB=0.75RA
Let f1= 1050 Hz, f2= 1250 Hz, C1=0.01μf
From (1)
1
1050 = ---------------------------------
0.69(RA+2*0.75RA) 0.01μf
1 =---------------------------------
0.69*2.5RA*0.01μf
RA=55.2KΩ
RB=0.75RA
RB=41.4KΩ
From (2)
1
1250 = ------------------------------------------------
0.69(55.2KΩ+2(41.4KΩ)) 0.01μf*C2 ---------------- 0.01μf+C2
0.01μf+C2 C2 = ----------------------------------- 0.69(138KΩ)1250*0.01μf
0.01μf+C2 C2 = --------------- 1.19
1.19C2 - C2 = 0.01μf
0.19C2 = 0.01μf
C2 = 52.63 nf
3.1.2 FSK DEMODULATOR:
Upper cut off frequency of RC ladder circuit fH=1/(2πRC)
Assume R2=R3=R4=R C2=C3=C4=C fH=(key in frequency+2 maximum frequency)/2
=(150+2(1250))/2=1325Hz
Let C=0.02μf then R=1/(2πCfH)=1/(2π*1325Hz*0.02μf )=7K Ω f0=0.3/(R1C1) f0=(f1+f2)/2=(1050+1250)/2=1150Hz
Let C1=0.01μf
R1=0.3/(1150*0.01μf)=26K Ω
flock=8f0/10=(8*1150)/10=920Hz
fcapture should be less than flock.
Choose fcapture =400Hz
fcapture =(1/2π)(2π*flock)/(R0C0)
2 .1 BLOCK DIAGRAM:
Fig 2.1 Block diagram for FSK modulation and de-modulation
2.2.1 PIN DIAGRAM- IC 565 :
Fig 2.2 Pin diagram for IC 565 2.2.2 PIN DIAGRAM -IC 555: Fig 2.3 Pin diagram for IC 555
3.1 DESIGN CALCULATION:
3.1.1 FSK MODULATOR:
ON time TH=0.693RBC
OFF time TL=0.693(RA+RB)C
Total time T=TH+TL=0.693(RA+2RB)C
1 f1= -------------------------- → (1) 0.69(RA+2RB) C1
1 f2= ------------------------------------ →(2)
0.69(RA+2RB) C 1C2 ----------- C1+C2
Duty cycle D=ON time/Total time=(RB)/(RA+2RB)=0.3
RB= 0.3RA+0.6RB
RB=0.75RA
Let f1= 1050 Hz, f2= 1250 Hz, C1=0.01μf
From (1)
1
1050 = ---------------------------------
0.69(RA+2*0.75RA) 0.01μf
1 =---------------------------------
0.69*2.5RA*0.01μf
RA=55.2KΩ
RB=0.75RA
RB=41.4KΩ
From (2)
1
1250 = ------------------------------------------------
0.69(55.2KΩ+2(41.4KΩ)) 0.01μf*C2 ---------------- 0.01μf+C2
0.01μf+C2 C2 = ----------------------------------- 0.69(138KΩ)1250*0.01μf
0.01μf+C2 C2 = --------------- 1.19
1.19C2 - C2 = 0.01μf
0.19C2 = 0.01μf
C2 = 52.63 nf
3.1.2 FSK DEMODULATOR:
Upper cut off frequency of RC ladder circuit fH=1/(2πRC)
Assume R2=R3=R4=R C2=C3=C4=C fH=(key in frequency+2 maximum frequency)/2
=(150+2(1250))/2=1325Hz
Let C=0.02μf then R=1/(2πCfH)=1/(2π*1325Hz*0.02μf )=7K Ω f0=0.3/(R1C1) f0=(f1+f2)/2=(1050+1250)/2=1150Hz
Let C1=0.01μf
R1=0.3/(1150*0.01μf)=26K Ω
flock=8f0/10=(8*1150)/10=920Hz
fcapture should be less than flock.
Choose fcapture =400Hz
fcapture =(1/2π)(2π*flock)/(R0C0)