DIGITAL DATA TRANSMISSION - II

LABORATORY PROCEDURE FSK involves frequency shifting a carrier between known fixed frequencies to convey digital data.. The resultant modulated signal may be regarded as the sum of two a

LAB REPORT

EXPERIMENT 6

DIGITAL DATA TRANSMISSION - II

SUBMITTED BY

PURPOSE

The Objectives of this laboratory are:

1 To examine the frequency spectrum of an FSK signal

2 To examine the correlation between two FSK signals

3 To implement and investigate a coherent FSK detection system EQUIPMENT LIST

1 PC wth Matlab and simulink

LABORATORY PROCEDURE

FSK involves frequency shifting a carrier between known fixed frequencies to convey digital data

Normally, FSK modulation is achieved using just two frequencies to represent binary data For example, a frequency of 2000 Hz may represent a binary 1

(mark) and 1000 Hz a binary 0 (space) The energy of the signal alternates

between the mark and space frequencies to convey the digital message

FSK describes the modulation of a carrier (or two carriers) by using a different frequency for a 1 or 0 The resultant modulated signal may be regarded as the sum

of two amplitude modulated signals of different carrier frequencies:

Figure 1a: Frequency shift keying

Figure 1b: Frequency shift keying frequency domain

A The circuit was set up according to figure 6A shown below It is a basic assembly to generate a FSK signal

Figure 6a Spectrum of a FSK signal – FSK generation

Figure 6bInput signal for generating FSK signal

The output of the FSK generator is shown below

The FSK Modulator module consists of f0 and f1 connectors, which are frequency

adjustable signals of the form;

sx(t) = A cos2Πfxt where x = 0 or 1 and the module is in FREE – RUNNING However, when the switch is

in the SYNC position , a time restriction is placed on the waveforms:

sx(t) = A cos2Πfxt for 0 < t < T where T = bit rate = 1/fs (selected by the Master Clock)

In this mode, a cosine begins at the start of every data period and proceeds at the frequency fx until the end of the data period Then the process repeats

Figure 6c The FSk Modulated signal

B Cross – Correlation of the Two FSK Signals

A FREQUENCY SHIFT KEYING - ERROR PROBABILITY:

The Probability of error of an FSK system depends on the separation of “distance” d, between two (or more) signals as given by the equation below:

P.E = ½ erfc(√(Eb / N)d)

Thus the probability of error is minimum when the distance of separation is maximum i.e larger the d, smaller is the P.E

The distance d is given by:

d = (1 - ρ) / 2 where

ρ = (1 / E) 0∫T s0(t) s1(t) dt

The integral is the cross-correlation function over the period 0 to T ρ is restricted to values between +1 and –1

Figure B (a) The FSk Modulated signal

Figure B (b) The FSk signal

Figure B (c) Block parameters of the Digital Signal source

Figure B (d) The FSk signal – Block parameters of sample and hold

Figure B (e) The block parameters of Pulse generator

Figure B (f) The FSk multiplied modulated signal

The DC meter is implemented using a Display block

MINIMUM SHIFT KEYING:

Minimum shift keying occurs when ρ = 0 or , or (ω1 - ω0 ) T = Π or the frequency between a ‘1’ and a ‘0’ = 1 / 2T

i.e (f 1 - f 0 ) = 1 / 2T

Most negative ρ occurs when (ω1 - ω0 ) T = 3Π / 2 or (f 1 - f 0 ) = 3 / 4 T

B Figure 6B was implemented for this part of the lab

The various values for f1,f0 and fs were as follows:

fs = 1khz = 1 / T

f0 = 5.0 khz

f 1 (in khz) Vd.c (in Volts) ρ f 1 - f 0 (f 1 - f 0 ) T

Graph of ρ vs (f1 - f0 ) T is as shown below:

The orthogonal frequencies (frequency at which ρ = 0) from the graph are found to be 0.57 kHz, 1.1kHz, 1.57khz

The distance d is greatest at (f 1 - f 0 ) = 0.74 kHz

x 10-3 -0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Plot of rho v/s (f1 - f0)T

distance d

B Theoretically,optimum spacing i.e maximum d would occur when ρ is at it’s

maximum negative i.e at ρ = -1.In the experiment the maximum negative value of ρ was seen to be – 0.2427 and the corresponding spacing was found to be 0.62135

It is observed that for incoherent demodulation (with the FSK modulator in the free – running mode) the distance d required is larger for a given value of P.E In other words the error probability for incoherent demodulation systems is higher that that of coherent systems for the same value of d

III Coherent FSK Detection :

A The circuit is set up as shown in figure 6C The settings are as follows:

f s = 1 khz;

f 0 = 5 khz

B The output of the digital summer as well as that of the Digital Signal Source was observed on the oscilloscope The graphs are as shown

It is seen that there is no change in the output signal when f1 is varied, this is because the output of the circuit is the modulating signal which when detected after

demodulation would be the same for all values of f1

C The signal voltage at the output of the 50Ω feedthru terminator is found to be

Vs = 0.4658 V and T = 1ms

Therefore energy per bit Eb = (Vs)2 T = (0.4658) 2 * 10-3

N0 = (Vrms)2 / BW = (142.6 mV)2 / 191 Hz

d for f1 = 6.0 khz is seen to be = 0.54

Therefore P.E = ½ erfc(√(Eb / N)d) = 0.069

D For f1 = 5.75 ;

The signal voltage at the output of the 50Ω feedthru terminator is found to be

Vs = 0.4566V and T = 1ms

Therefore energy per bit Eb = (Vs)2 T = (0.4566) 2 * 10-3

N0 = (Vrms)2 / BW = (142.6 mV)2 / 191 Hz

d for f1 = 5.75 khz is seen to be = 0.62135

Therefore P.E = ½ erfc(√(Eb / N)d) = 0.0594

For f1 = 5.5 ;

The signal voltage at the output of the 50Ω feedthru terminator is found to be

Vs = 0.45 V and T = 1ms

Therefore energy per bit Eb = (Vs)2 T = (0.45) 2 * 10-3

N0 = (Vrms)2 / BW = (142.6 mV)2 / 191 Hz

d for f1 = 5.5 khz is seen to be = 0.44685

Therefore P.E = ½ erfc(√(Eb / N)d) = 0.0961

III Real Life FSK :

A Actual FSK systems use “continuos phase” signals wherein one signal starts where the other signal stops A VCO is used to generate such signals

Figure 6D (a)FSK using Phase lock loop

C The VCO generates the continuos phase FSK signal centered at 5khz

The fsk modulated signal looks like this

Figure 6D (b)FSK modulated signal

Figure 6D (c) Output of LPF

Figure 6D-(d)The output

D the circuit was set up as shown in figure 6D and the output was detected as shown in the graphs

Figure 6D B -(a)FSK with noise and improved output

The input signal parameters

Figure 6D B -(b) block parameters of the sine wave

The filter characteristics are also shown below

Figure 6D B -(c) block parameters of the butter worth filter

Figure 6D B -(d)Output of the LPF filter

Figure 6D B -(e)Output

Figure 6D B -(f)Output of the second LPF filter

IV Result :

The experiment gives an insight to the FSK modulation and demodulation schemes and emphasizes the effect of the distance of separation of the two signal components

on the error probability

The results of the prelab and those from the experiment were seen to be matching

APPENDIX - PRELAB

Consider the FSK signal set:

1 s1(t) may be thought of as a continuous cosine wave multiplied by a pulse of duration T and repeated every T seconds Or:

where * denotes convolution and indicates the “pulse” notation, unit magnitude, centered at xi, and witdth τ

Find the positive frequency spectrum of s1(t) (magnitude only)

Since

Therefore, the fourier transform of s 1 (t), S 1 (f) = Sx(f)Sd(f) where

By integrating the two integrands, we get

T t t

πf T

E t

T t t

πf T

E t

b

b

≤

≤

=

≤

≤

=

0 for ) 2 cos(

) / 2 ( )

(

s

0 for ) 2 cos(

) / 2 ( )

(

s

0 0

1 1

∑∞

−∞

=

−

∗

⎟

⎠

⎞

⎜

⎝

⎛ − Π

×

=

n

t t

πf T

E

(

⎟

⎠

⎞

⎜

⎝

⎛ = Π τ

2 /

i

x x

∑∞

−∞

=

−

∗

⎟

⎠

⎞

⎜

⎝

⎛ − Π

×

=

n

t t

πf T

E

(

∫ ∑

∫

∞

∞

−

−

∞

−∞

=

−

−

=

=

dt e nT t f

dt e t πf T

E f

ft j n

T

ft j b

π

π

2 d

0

2 1 x

) δ(

)

(

S

) 2 cos(

) / 2 ( )

(

S

) ( S of version sampled

the

is

which

) / δ(

1 ) (

2

1 )

( 2

) 1 (

2

/ 2 )

(

S

) / δ(

1

)

(

S

) (

2

1 )

( 2

) 1 (

2

/ 2 )

(

S

x

1

) ( 2 1

) ( 2 1

d

1

) ( 2 1

) ( 2 x

1 1

1 1

f

T n f T

f f

e f

f j e T b

E f

T n f T

f

f f

e f

f j e T b

E f

n

T f f j T

f f j n

T f f j T

f f j

∑

∑

∞

−∞

=

+

−

−

−

∞

−∞

=

+

−

−

−

−

⋅

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

− +

−

−

−

=

∴

−

=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

− +

−

−

−

=

π π

π π

π π

π π

Sx(f) has very high peak at f=f1,-f1 and decay rapidly away from the peak

The following is the sketch for S1(f) when f1=5000, T=0.001

2 The PE for the FSK system using an optimal filter is:

where

= correlation coefficient

For the FSK signal set given, find ρ Assume f o + f 1 >> (1/T)

x 104 0

1 2 3 4 5

6

x 10-3

frequenc y (Hz )

S 1

M agnitude s pec trum of s

1 (t)

2

) 1 ( ) / ( erfc 2

1

∫

= T s t s t dt

E1 0 0( ) 1( )

ρ

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

−

− +

+

+

=

− +

+

=

=

∫

∫

T T

T

T

f f

t f f f

f

t f f

dt t f f t

f f

dt t f t

f T

E E

0 0 1

0 1 0

1 0

1 0

0

0 1 1

0

0

0 1

) (

2

) ) (

2 sin(

) (

2

) ) (

2 sin(

T

1

) ) (

2 cos(

) ) (

2 cos(

T

1

) 2 cos(

) 2 cos(

2 1

π

π π

π

π π

π π

ρ

3 Plot ρ as a function of frequency separation (f1-f0)T

4 If f 0 =5 kHz and 1/T = 1 kHz, what is the optimal value of f 1 for the smallest PE?

Solve the equation dρ/dt =0 to get the (f 1 -f 0 )T that corresponds to the most negative ρ because the most negative ρ gives the smallest PE

T f f

T f f

T f f T

f

f

f f

T f f T

f f

T f f T

) (

2

) ) (

2 sin(

zero ely approximat is

above first term the

so 1 ) (

then

1 if

) (

2

) ) (

2 sin(

1 )

( 2

) ) (

2 sin(

1

0 1

0 1

1 0 1

0

0 1

0 1 1

0

1 0

−

−

=

∴

>>

+

>>

+

−

− +

+

+

=

π

π ρ

π

π π

π

-0.4

-0.2

0 0.2 0.4 0.6 0.8 1

(f

1 -f

0 )T

rho vers us frequenc y s eparation

Solve the equation above and get (f1-f0)T = 0.7151 This value is confirmed from the graph above

If (f1-f0)T = 0.7151 then

f1-f0 = 0.7151*1000 = 715.1 Hz

f1 = 715.1 + 5000 = 5.7151 kHz

Therefore, the optimal value of f1 for the smallest PE is 5.7151 kHz

0 )

( 2

) ) (

2 cos(

2 )

) ((

2

) ) (

2 sin(

)

0 1 2

0 1

0 1 0

1

=

−

− +

−

−

−

=

T f f T

f f

T f f T

f f

d

d

π

π π

π π ρ