DIGITAL DATA TRANSMISSION - II

LAB REPORT EXPERIMENT DIGITAL DATA TRANSMISSION - II SUBMITTED BY PURPOSE The Objectives of this laboratory are: To examine the frequency spectrum of an FSK signal To examine the correlation between two FSK signals To implement and investigate a coherent FSK detection system EQUIPMENT LIST PC wth Matlab and simulink LABORATORY PROCEDURE I Spectrum of an FSK Signal : 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 (mark) and 1000 Hz a binary (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 or 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 = or 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 < 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) / where ρ = (1 / E) 0∫T s0(t) s1(t) dt The integral is the cross-correlation function over the period 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 = or , or (ω1 - ω0 ) T = Π or the frequency between a ‘1’ and a ‘0’ = / 2T i.e (f1 - f0 ) = / 2T Most negative ρ occurs when (ω1 - ω0 ) T = 3Π / or (f1 - f0 ) = / 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 = / T f0 = 5.0 khz f1 (in khz) 5.0 5.25 5.5 5.75 6.0 6.25 6.5 6.75 Vd.c (in Volts) + 3.3075 + 2.613 + 0.3518 - 0.803 - 0.29 + 0.402 + 0.149 - 0.353 ρ + 1.0 + 0.79 + 0.1063 - 0.2427 - 0.0876 + 0.1215 + 0.045 - 0.1067 f1 - f0 0.25 0.5 0.75 1.0 1.25 1.5 1.75 (f1 - f0 ) T 0.25 e -03 0.5 e -03 0.75 e -03 1.0 e -03 1.25 e -03 1.5 e -03 1.75 e -03 Graph of ρ vs (f1 - f0 ) T is as shown below: Plot of rho v/s (f1 - f0)T Correlation coefficient rho 0.8 0.6 0.4 0.2 -0.2 -0.4 0.2 0.4 0.6 0.8 distance d 1.2 1.4 1.6 1.8 -3 x 10 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 (f1 - f0 ) = 0.74 kHz 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: fs = khz; f0 = 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) * 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) * 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) * 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: s1 (t ) = (2 Eb / T ) cos(2πf1t ) for ≤ t ≤ T s (t ) = (2 Eb / T ) cos(2πf t ) for ≤ t ≤ T s1(t) may be thought of as a continuous cosine wave multiplied by a pulse of duration T and repeated every T seconds Or: ∞ s1 (t ) = (2 Eb / T ) cos( 2πf 1t ) × Π ⎛⎜ t − T/ ⎞⎟ ∗ ∑ δ(t − nT ) ⎝ T ⎠ n = −∞ where * denotes convolution and centered at xi, and witdth τ ⎛ x = xi / ⎞ Π⎜ ⎟ τ ⎝ ⎠ indicates the “pulse” notation, unit magnitude, Find the positive frequency spectrum of s1(t) (magnitude only) Since ∞ s1 (t ) = (2 Eb / T ) cos( 2πf 1t ) × Π ⎛⎜ t − T/ ⎞⎟ ∗ ∑ δ(t − nT ) ⎝ T ⎠ n = −∞ Therefore, the fourier transform of s1(t), S1(f) = Sx(f)Sd(f) where T S x ( f ) = ∫ (2 E b / T ) cos( 2πf1t )e − j 2πft dt ∞ Sd ( f ) = ∞ ∫ ∑ δ(t − nT )e − j 2πft dt − ∞n = −∞ By integrating the two integrands, we get 2E / T b Sx ( f ) = Sd ( f ) = ⎛ − (e − j 2π ( f − f1 )T − 1) e − j 2π ( f + f1 )T − ⎞ ⎜⎜ ⎟ + 2π ( f + f ) ⎟⎠ ⎝ j 2π ( f − f1 ) ∞ ∑ δ( f − n / T ) T n = −∞ 2E / T b ⎛ − (e − j 2π ( f − f1 )T − 1) e − j 2π ( f + f1 )T − ⎞ ∞ ⎜⎜ ⎟ ⋅ ∑ δ( f − n / T ) + 2π ( f + f1 ) ⎟⎠ T n = −∞ ⎝ j 2π ( f − f ) which is the sampled version of S x ( f ) ∴ S1 ( f ) = 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 x 10 M agnitude s pectrum of s (t) -3 S (f) 0 0.5 1.5 2.5 frequenc y (Hz ) 3.5 (1 − ρ ) erfc ( Eb / N o ) 2 where ρ= T s (t ) s1 (t )dt = correlation coefficient E ∫0 For the FSK signal set given, find ρ Assume fo + f1 >> (1/T) T 2E ρ= ∫ cos(2πf 1t ) cos(2πf t )dt E0 T T = cos(2π ( f + f1 )t ) + cos(2π ( f − f )t )dt T ∫0 sin( 2π ( f − f )t ) ⎛ sin( 2π ( f + f )t ) = ⎜ + T ⎜ 2π ( f + f1 ) 2π ( f − f ) ⎝ T 4.5 x 10 The PE for the FSK system using an optimal filter is: PE = T ⎞ ⎟ ⎟ ⎠ = sin( 2π ( f + f )T ) sin(2π ( f1 − f )T ) + T T 2π ( f − f ) 2π ( f + f1 ) then ( f + f )T >> so the first term above is approximately zero T sin( 2π ( f1 − f )T ) ∴ρ = 2π ( f − f )T if f + f >> Plot ρ as a function of frequency separation (f1-f0)T rho vers us frequenc y s eparation 0.8 0.6 rho 0.4 0.2 -0.2 -0.4 0.2 0.4 0.6 0.8 (f -f )T 1.2 1.4 1.6 1.8 If f0=5 kHz and 1/T = kHz, what is the optimal value of f1 for the smallest PE? Solve the equation dρ/dt =0 to get the (f1-f0)T that corresponds to the most negative ρ because the most negative ρ gives the smallest PE − sin(2π ( f1 − f )T ) 2π cos(2π ( f1 − f )T ) dρ = + =0 d ( f1 − f )T 2π ( f1 − f )T 2π (( f1 − f )T ) 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 ... 0.3518 - 0.803 - 0.29 + 0.402 + 0.149 - 0.353 ρ + 1.0 + 0.79 + 0.1063 - 0.2427 - 0.0876 + 0.1215 + 0.045 - 0.1067 f1 - f0 0.25 0.5 0.75 1.0 1.25 1.5 1.75 (f1 - f0 ) T 0.25 e -0 3 0.5 e -0 3 0.75 e -0 3... 0.75 e -0 3 1.0 e -0 3 1.25 e -0 3 1.5 e -0 3 1.75 e -0 3 Graph of ρ vs (f1 - f0 ) T is as shown below: Plot of rho v/s (f1 - f0)T Correlation coefficient rho 0.8 0.6 0.4 0.2 -0 .2 -0 .4 0.2 0.4 0.6... occurs when ρ = or , or (ω1 - ω0 ) T = Π or the frequency between a ‘1’ and a ‘0’ = / 2T i.e (f1 - f0 ) = / 2T Most negative ρ occurs when (ω1 - ω0 ) T = 3Π / or (f1 - f0 ) = / T B Figure 6B was