Electronic Communication Systems Design - Assignment Example

Electronic communication systems design Your name Name of Assignment 25th June , 2012 1. For a binary stream of 10kbps, use Simulink to generate the following three pulses a) Rectangular pulse of width 10%, 20%, 50%, 90%, respectively; b) sinc pulse; c) raised cosine pulse with rolloff factor 0.2, 0.4, and 0.6, respectively. 2. Based on Task 1, build a Simulink model to transmit the three pulse signals through a channel with bandwidth 5kHz. Use scope and spectrum scope to observe the time-domain and frequency-domain representation of the output signals for these 3 pulses. ASSIGNMENT 2 Using a rectangular pulses of width 50% to modulate the data of a given random binary data stream of rate 1kbps’ the following signaling schemes: Unipolar NRZ >> b=[1 0 1 0 1 1]; >> x=wave_gen(b,'unipolar_nrz',1 ); >> waveplot(x,b,'unipolar_nrz', 1) There are no transitions when there is 0 and 1. This may lead to power wastage. The following is the normalised frequency of the pulses Polar NRZ In this case there will be no transitions and DC in long strings and DC making it difficult for recovery of time. The following is signaling scheme for Polar NRZ using rectangular pulses of width 50% to modulate the data >>x=wave_gen(b,'___polar_nrz',1 ); >>waveplot(x,b,'___polar_nrz', 1) >> subplot(212), channel ( b, ’___polar_nrz’, 1, 0.01, 1.0 ); >> subplot(211), waveplot(x,b,'___polar_nrz', 1) >> b=[1 0 0 0 1 1 0 1 1 1]; >> x=wave_gen(b,'___polar_nrz',1 ); >> subplot(321),waveplot(x,b,'___polar_nrz',1 ); >> subplot(322),channel(b,’___polar_nrz’,1,0,1.0); Unipolar RZ In this case it has problem with transition long string 0 but with some less problems with 1. This makes it difficult to recover time and power may be wasted. Polar RZ a) Manchester Pulses render the contents of electronic transmissions unreadable by anyone who might intercept them. Modern encryption technologies provide a high degree of protection against the vast majority of potential attackers. Legitimate recipients can decrypt transmission contents by using a piece of data called a “key”. The recipient typically possesses the key for decryption as a result of a previous interaction. Like passwords, keys must be kept secret and protected from social engineering, physical theft, insecure transmission, and a variety of other techniques hostile forces use to obtain them. Encryption does little good if the key that decrypts is available to attackers. Encryption does not conceal everything about a network transmission. Hackers still can gain useful information from the pattern of transmission, the lengths of messages, or their origin or destination addresses. Encryption does not prevent attackers from intercepting and changing the data in a transmission. The attackers may not know what they are changing, but subtle changes can still wreak havoc, especially if the intended recipient is a computer that expects data to arrive in particular format. Matched filter: This is a receiver filter which filters out signal-to-noise power ratio; it is referred as the optimum receiver. It is realized by the time reverse h(−t) of the transmitter filter. a simple threshold scheme where a target has been declared to exist and during sampling time, the threshold (T) is exceeded. Under the conventional assumptions of a pulsed system then, and with an additional independent Gaussian noise, there can be two scenarios. Either, the presence of a target causes the threshold detector to recognize a signal-to-noise Gaussian distribution. The distribution has a mean value of voltage equal to V, and a power variance equal to N Otherwise, when a target is absent, the threshold detector recognizes a noise Gaussian distribution with a mean value of zero and a noise (power) variance, equal to N. In the first situation, interest is on the probability of detection, P (d). This is the probability of detecting a present target. In the second scenario, the computation is for the false alarm probability, P (fa). This is the probability that when a target is absent, one can incorrectly conclude that a target is present. This happens when, during sampling, the noise alone is so large so that it exceeds the set threshold. Matched filtering is used in communications BPSK modulation and an optimum receiver with matched filtering modulation scheme (ASK,PSK,QAM, arbitrary) 2-PSK (∆0=2) 4-PSK (∆0=) Exact result For ASK: . Q, ∆0,ASK For PSK and QAM: Not available Q Q. Tight PSK bound 2Q. ∆0,PSK=2sin(π/) 2Q 2Q Tight QAM bound 4Q. , ∆0,QAM 4Q. Not ingenious 4Q. Not ingenious General weak bound exp EXP EXP It is important to note that each singular term used to denote the different complex notations such as the terms point, sample, value and signal constitutes both a real part and an imaginary part. Therefore, the basic principle of operation in the binary is the decomposition of a single point time domain signal into two t time domain signals that each contains a single point. What follows this initial decomposition stage is the calculation of the frequency spectra that corresponds to the given t time domain signals. The final stage involves the synthesizing of the calculated t spectra into a single frequency spectrum. A given a typical 16 point signal, it can be decomposed through four stages that are each distinct. The first stage of decomposition results in two signals that each has 8 points while the second stage of decomposition will result in four signals that each has 4 points. The idea is to continue this decomposition by half until a given number of signals, in this case 16, that each contains a single point remain. It should be noted that each decomposition stage is interlaced in nature and this means that the two samples which result from a decomposition are even and odd numbered respectively. An important feature to note concerning the decomposition process is that it is what makes it possible for the samples in the original signal to be reordered. However, the re-ordering of these samples is supposed to follow a specific pattern which is determined by the binary equivalents of each sample. This algorithm that involves the rearrangements of the order of the t time domain samples through the counting in binaries that have been flipped from left to right. After the bit reversal sorting stage of the algorithm, the next step is the finding of the frequency spectra which belongs to the 1 point time domain signals at the end of the last decomposition phase. This is a very easy process since the frequency spectrum of a 1 point signals is equal to itself and therefore there is virtually nothing to be done at this stage. Also, it should be noted that the final 1 point signals are no longer time domain signals but rather, a frequency spectrum. Lastly, the frequency spectra are then supposed to be recombined in the reverse order that is exactly similar to the order followed during the decomposition of the time domain. Since bit reversal formula is not applicable for this recombination, the reverse process is performed one stage at a time. The reverse process is known as synthesis and this is done stage wise from the single point spectra and finally forming the 16 point frequency spectrum. The different flow diagrams used to represent this synthesis is referred to as a butterfly and it is also what forms the basic computational element of the binary as it transforms two complex points into two different but still complex points. The process of synthesizing the frequency domain from the separate single signals undergoes three main essential loops which are also concentric in nature. The outermost loop is the one that runs through the Log2N stages while the middle loop is the one which moves through each of the individual frequency spectra that are in the stage currently being worked on. The final loop which is also the innermost is what now uses the butterfly diagram mentioned earlier in the calculation of the points that are in each frequency spectra. The three loops are the three main stages that constitute the transformation of a given data from the time domain data into the frequency domain data and vice versa. The exact symbol-error rates for the cases of 2-PSK AND 4-PSK. For 2-PSK the deviation is fairly small: according to table 2.2 the tight PSK bound has an error of a factor of 2. However, for 4-PSK the tight bound is very exact, as can be seen in table 2.2 as well as in figure 2.10 where the tight bound cannot even be distinguished from the exact curve. Further computer simulations of the symbol-error rate show only small deviations for QAM and extremely small deviations for PSK from the tight bounds. So, for most practical applications and for all modulation schemes except 2-PSK, the tight bounds can be considered as approximately exact results. The reason for this is fairly simple. Since the in phase and quadrature phase of 4-PSK are statistically independent, one 4-PSK signal is equivalent to two consecutive 2-PSK signals. The bit-error rates Pb of 2-PSK and 4-PSK are actually identical in reference to Ee/N0 as will now become clear. 2 use simulink blocks to implement BPSK and 16QAM. Transmit the modulated signals through an AWGN channel with E b /No varying in the range 0:3:12Db. Measure the error rate of these modulations use BER tool to plot bit error rate vs. E b/No of the AWGN channel. The results derived previously are summarized as a compendium of formulas in table 2.2, and shown as performance curves for 2-PSK, 4-PSK, 8-PSK, 16-PSK, 16-PSK as well as 16-QAM and 64-QAM, each for the energy per symbol in figure 2.10 and for the energy per bit in figure 2.11. Since 2-PSK and 2-PSK as well as 4-PSK and 4-QAM are completely identical modulation schemes, these two cases are addressed as PSK schemes. In the figure above the tight bounds as shown with solid lines for PSK and the exact symbol-error rate is shown with dotted lines for 2-PSK and 4-PSK and for 16-QAM and 64-QAM. The exact symbol-error rate for 8-PSK and 16-PSK is within the line width of the concerned graphs for the tight bounds. Appendix Unipolar U=a; n= length(a); %POLAR P=a; for k=1:n; if a(k)==0 P(k)=-1; end end %Bipolar B=a; f = -1; for k=1:n; if B(k)==1; if f==-1; B(k)=1; f=1; else B(k)=-1; f=-1; end end end %Mark M(1)=1; for k=1:n; M(k+1)=xor(M(k), a(k)); end %Space S(1)=1; for k=1:n S(k+1)=not(xor(S(k), a(k))); end %Plotting Waves subplot(5, 1, 1); bpuls(U) axis([1 n+2 -2 2]) title('Unipolar NRZ') grid on subplot(5, 1, 2); bpuls(P) axis([1 n+2 -2 2]) title('Polar NRZ') grid on subplot(5, 1, 3); bpuls(B) axis([1 n+2 -2 2]) title('Bipolar NRZ') grid on subplot(5, 1, 4); bpuls(M) axis([1 n+2 -2 2]) title('NRZ-Mark') grid on subplot(5, 1, 5); bpuls(S) axis([1 n+2 -2 2]) title('NRZ-Space') grid on %Spetra Calculation, Taking 512 points FFT f=1000*(0:256)/512; %unipolar yu=fft(U, 512); pu=yu.*conj(yu)/512; %polar yp=fft(P, 512); pp=yp.*conj(yp)/512; %bipolar yb=fft(B, 512); pb=yb.*conj(yb)/512; %Mark ym=fft(M, 512); pm=ym.*conj(ym)/512; %Space ys=fft(S, 512); ps=ys.*conj(ys)/512; %Power spectra PLOT figure; plot(f,pu(1:257),'-k','linewidth',2) hold on title('Power Spectra Plot') plot(f,pp(1:257),'-b','linewidth',2) plot(f,pb(1:257),'-g','linewidth',2) plot(f,pm(1:257),'-r','linewidth',2) plot(f,ps(1:257),'-m','linewidth',2) grid on legend('Unipolar','Polar','Bipoalr','Mark','Space') hold off size(f) function bpuls(a) n=length(a); b=a; b(n+1)=b(n); %retaining last value for entire last duration stairs(b,'linewidth',2) axis([1 length(b) min(b)-0.5 max(b)+0.5]) Read More