TSKS04 Digital Communication, Continuation Course

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1 SKS04 Digital Communication, Continuation Course Problems for utorial 1 1. Let the input to a digital modulator consist of independent, equally probable bits. Determine the power-spectral densities in the following cases. a On-Off Keying with s 0 t = 0 and E s 1 t =, 0 t <. b Bipolar signalling with and s 1 t = s 0 t. c Orthogonal signalling with s 0 t = E, 0 t <, and s 0 t = s 1 t = E, 0 t <, E, E 0 t < /,, / t <.. Compare the results from asks 1a and 1b. What is the reason that there is an impulse in the PSD in one of the cases, but not in the other? Can you formulate that reason as a general rule? 3. A delay of a signal corresponds to a multiplication by a complex exponential in the transform domain. hat means that the delay does not affect the absolute value of the transform of the signal. hus, it can be convenient to first shift the signal so that the signal interval is from / to / before calculating the transform. hat especially holds if the signal has some symmetry around the center of the signal as in subtasks a and b below. Determine the energy spectra of the following basis functions: 1 3 a φ 1 t = rectt/ b φ t = trianglet/

2 4. here are several ways to define the bandwidth of a signal. hose definitions try to capture to what extent our signal distorts other signals, to what extent other signals causes distortion to us, or to what extent we are sensitive to BP-filtering. Whatever bandwidth definition we use, we are usually interested in keeping the bandwidth small, given that all other parameters are fixed. Which of the two basis functions in ask 3 should we prefer based on the following bandwidth definitions? a he bandwidth is the smallest positive frequency where Φf is zero. b he bandwidth is the smallest positive frequency B such that Φf is at least 0 db less than the maximum value of Φf for all frequencies above B. 5. he eight signals in the following signal constellation are equally probable, and subsequent symbols are independent. 3A A A A A 3A Determine the power-spectral density if the two basis functions are φ 1 t = cosπf ct, 0 t <, and φ t = sinπf ct, 0 t <, where f c is a positive integer.

3 Hints 1. a Interprete the input as a time-discrete process A[n], consisting of independent, equally probable 0 and 1. hen PAM that using s 1 t as the pulse shape. he standard spectral relation for PAM can then be used: R S f = 1 S1 f R A [f], where S 1 f is the Fourier transform of s 1 t, and where St is the output process. b he same as in a, but the input alphabet is ±1, and the pulse shape is s 0 t. c Notice that the signals are orthogonal. We have a two-dimensional situation, and the PSD is given by R S f = 1 S 1f,Sf R A1,A 1 [f] R A,A 1 [f] S1 f, R A1,A [f] R A,A [f] S f where A 1 [n] and A [n] are the two component processes for the two dimensions. Identify the two component processes and how they are related, and determine the auto-correlation and cross-correlation functions.. Compare the formulas. rack the factor in front of the impulse through your calculations. 3. he energy spectrum of a signal xt is Xf, where Xf is the Fourier transform of xt. he function triangleat is a convolution of two rectat. 4. Basing the calculations on a sinc can be a bit complicated. herefore simplify the situation and try to bound the bandwidths from above and from below. herefore, consider 1/πx instead of sincx. 5. A two-dimensional situation. he PSD is given by R S f = 1 Φ 1f,Φ R S1,S 1 [f] R S,S 1 [f] f R S1,S [f] R S,S [f] Φ1 f Φ f where S 1 [n] and S [n] are the two component processes for the two dimensions, and whereφ 1 fandφ farethefouriertransformsofthetwo basisfunctionsφ 1 tand φ t. Identify the two component processes and how they are related, and determine the auto-correlation and cross-correlation functions.,

4 Answers 1+ 1 δf sinc f 1. a R S f = E b R S f = E sinc f c R S f = E 4 sinc f+ 1 δf+sinc f sin π f + modd 1 πm δ f m. It is the square of the mean of the resulting signal, which is a general rule. 3. a Φ 1 f = sinc f b Φ f = 3 4 sinc4 4. a We prefer φ 1 t. b We prefer φ t. 5. R S f = 9 4 A sinc f +f c +sinc f f c. f

5 Solutions 1. a Alphabet: 0, 1, equally probable. We interprete the input as a time-discrete process A[n], consisting of independent, equally probable 0 and 1. hen the modulation is PAM using s 1 t as the pulse shape. he standard relation for PAM can then be used: R S f = 1 S1 f R A [f], where S 1 f is the Fourier transform of s 1 t, and where St is the output process. he non-zero signal can be written as Its Fourier transform is s 1 t = E rect t /. S 1 f = F {s 1 t} = E sincfe jπf We need the ACF of A[n]. First the case k = 0: r A [0] = E { A [n] } = = 1 For k 0 we use the independence and get r A [k] = E { A[n]A[n+k] } = E { A[n] } E { A[n+k] } = = 1 4 otally, r A [k] = δ[k]. he PSD is the Fourier transform of the ACF. hus: R A [θ] = 1 1+ δθ m 4 m Finally, we plug everything into the standard formula for PAM: R S f = 1 S1 f R A [f] = E 1+ 1 δ f m sinc f = E sinc f+ E δf Notice that only one of the impulses in the sum of R A [f] survives. All the other impulses are cancelled since they occur exactly where S 1 f is zero. m

6 b his can be solved exactly the same way as part a. he only difference is that the input alphabet is ±1. Now, we pulse-amplitude modulate the input process A[n] with s 0 t as pulse shape. hat signal can be written as E t / s 0 t = rect and its Fourier transform is S 0 f = F { s 0 t } = E sincf e jπf he ACF of A[n]. For k = 0, we have r A [0] = E { A [n] } = = 1 For k 0, again using the independence, we get otally: Its PSD: r A [k] = E { A[n] } E { A[n+k] } = And combine everything: R S f = 1 r A [k] = δ[k] R A [θ] = = 0 S1 f R A [f] = E sinc f c he input bits, say 0 and 1, are mapped on the vectors 1,0 and 0,1, respectively. So, we have two component processes, A 0 [n] and A 1 [n], which are the boolean inverses of each other. When one of them is 0, the other is 1, and vice versa. hen A 0 [n] modulates s 0 t, while A 1 [n] modulates s 1 t. his is clearly a two-dimensional case. We use the realation R S f = 1 S 0 f,s 1 f R A0 [f] R A1,A 0 [f] R A0,A 1 [f] R A1 [f] S0 f S 1 f where S 0 f and S 1 f are the Fourier transforms of the two orthogonal signals s 0 t and s 1 t. Our signals: E t / s 0 t = rect E t /4 t 3/4 s 1 t = rect rect / /

7 heir spectra: S 0 f = F { s 0 t } = E sincfe jπf S 1 f = F { s 1 t } = j f E sinc sin π f e jπf he two component processes behave exactly as the input process in part a. hus, we have r A0 [k] = r A1 [k] = δ[k]. his gives us the PSDs R A0 [θ] = R A1 [θ] = 1 δθ m+1. 4 We also need the cross-spectra. herefore, we are interested in the crosscorrelation function of the component processes, which is defined as For k = 0, we have r A0,A 1 [k] = E { A 0 [n]a 1 [n+k] }. r A0,A 1 [0] = E { A 0 [n]a 1 [n] } = = 0. Samples from the two component processes are independent for k 0, since the input consists of independent bits. So, for k 0, we have r A0,A 1 [k] = E { A 0 [n] } E { A 1 [n+k] } = m = 1 4 otally, we have and similarily Cross spectra: r A0,A 1 [k] = 1 1 δ[k] 4 r A1,A 0 [k] = 1 1 δ[k]. 4 R A0,A 1 [θ] = R A1,A 0 [θ] = 1 δθ m 1 4 m he PSD of the output is given by R S f = 1 S 0 f,s 1 f R A0 [f] R A1,A 0 [f] R A0,A 1 [f] R A1 [f] S0 f S 1 f

8 We notice that S 0fR A1,A 0 [f]s 1 f = S 1fR A0,A 1 [f]s 0 f holds, which means that the off-diagonal terms cancel out, and all we have left is R S f = 1 S 0 fr A0 [f]s 0 f+s1fr A1 [f]s 1 f = 1 S0 f R A0 [f]+ S1 f R A1 [f] Plugging in everything into that equation, we get R S f = E f sinc f+sinc sin π f 1+ δf m 4 m = E f sinc f+δf+sinc sin π f πm δf m modd = E sinc f+ 1 f sin 4 δf+sinc π f + 1 f πm δ m modd Notice that only one of the impulses in the sum of R A0 [f] survives, since the other impulses occur exactly where S 0 f is zero. For R A1 [f], the situation is different. Here the impulses for all even m in the sum are cancelled out by zeros in S 1 f, which leaves us with the sum over odd m in the expression above.. he factor in front of the impulse is r A [k] for k 0, which is m A. Generally, we can write the ACF as r A [k] = λ A [k]+m A, where λ A [k] is the auto-covariance function. he term m A will result in the term δθ m in the PSD of A[n]. Any impulses in that sum that coinside with m A m zeros in S 1 f will be cancelled in the subsequent expression R S f = 1 S1 f R A [f]. In ask 1a, m A is non-zero, resulting in the impulse δf. In ask 1b, m A is zero, resulting in no impulse.

9 3. We use ables and Formulas for Signal heory from the Signal heory course. a Page 19 gives us { } 1 t Φ 1 f = F rect = from which we get the energy spectrum Φ1 f = sinc f. b Page 19 gives us { } 3 t Φ f = F triangle = 3 f = sinc, from which we get the energy spectrum Φ f = 3 4 sinc4 f 1 sincf = sincf, 3 f sinc. 4. From ask 3, we have Φ1 f = sinc f, Φ f = 3 4 sinc4 f. a he first zero in Φ1 f occurs at f = 1, which gives us f = 1/. he first zero in Φ f occurs at f/ = 1, which gives us f = /. hus, according to the bandwidth definition used in this sub-task, we should prefer the basis function φ 1 t. b he drop of 0 db corresponds to a power drop by the factor 10 0/10 = 100. hus, in the two cases we are interested in Case 1: sinc f < 10, Case : sinc 4 f/ < 10. Since these expressions are a bit complicated to deal with, we aim at bounding the bandwidth both from above and from below instead of calculating it exactly. Our hope is that the intervals found for the two cases will not overlap, which is enough to make a certain decision. Recall that we have sincx = sinπx πx for x 0.

10 We simplify the expressions above based on sincx 1/πx. hen we are interested in Case 1: πf < 10, Case : πf/ 4 < 10, which will give us an upper bound on the bandwidth. We rewrite that as and further to Case 1: πf > 10, Case : πf/ > 10, Case 1: Case : f > 10/π 3./, f > 10/π.0/. hus, we know that for all frequencies above those calculated values, the drop is at least 0 db. So for the bandwidth B, we have Case 1: Case : B 10/π 3./, f 10/π.0/. o find a lower bound on B, we observe that the situation is especially simple when the argument to the sinc function is k/, where k is an odd integer. hen we have sinck/ = /kπ. hus, in case 1 the bandwidth B is larger than.5/, which is the largest such frequency that is smaller than 3./. In case, the corresponding reasoning gives us the lower bound 1/. o conclude this, we have bounded the bandwidths as follows: Case 1: Case :.5/ < B < 3./, 1/ < B <.0/. Obviously, we should prefer φ t in this case, and that is enough to solve the given problem. Note: o get at a more exact value for the bandwidth in those two cases, we should use numerical methods to study the energy spectra between our bounds. If we do that, we find that we have Case 1: Case : B.68/, B 1.48/,

11 As an illustration, here follows two graphs of sinc f case 1 and sinc 4 f/ case. he first one uses linear scales on both axes, while the second plots them in db i.e. logarithmic scale against a linear frequency scale. he frequency scale is normalized with respect to 1/, i.e. the horizontal axis corresponds to f Case 1 Case Case 1 Case

12 For the interested reader, we also have a graph of the two cases in a log-log graph. Here we have also included the simplified expressions that we used in the reasoning, namely πf for case 1 and πf/ 4 for case. hese are the two dotted straight lines that are drawn as tangents to the main lobe and all the side lobes Case 1 Case his is a two-dimensional case. he input process is an infinite sequence of independent stochastic variables, each having eight equally probable symbols. his input processismappedontotwocomponentprocesses, S 1 [n]ands [n], oneforeachdimensionof the signal constellation. hus, the sample space of S 1 [n] is{ A,A,0,A,A}, with probabilities /8,1/8,/8,1/8,/8, respectively. he sample space of S [n] is { 3A,0, 3A}, with probabilities 3/8,/8,3/8, respectively. he PSD of the output process St is given by R S f = 1 Φ 1 f,φ f R S1 [f] R S1,S [f] R S,S 1 [f] R S [f] Φ1 f Φ f where Φ i f is the Fourier transform of the basis function φ i t. hus, we first need to determine the auto-correlation and cross-correlation functions in order to Fourier transform them into PSDs and cross-spectra. We wish to determine the auto-correlation and cross-correlation functions r S1 [k] = E { S 1 [n+k]s 1 [n] }, r S [k] = E { S [n+k]s [n] }, r S1,S [k] = E { S 1 [n+k]s [n] } = r S,S 1 [ k].,

13 Since subsequent symbols are independent, we can identify two cases. One is k = 0 and the other is k 0. For k = 0 we have r S1 [0] = E { S 1 [n]} = 1 8 A + A +A + A + 0 = 9 4 A, r S [0] = E { S [n]} = A +3 3A + 0 = 9 4 A, r S1,S [0] = E { S 1 [n]s [n] } For k 0 we have = 1 8 A 3A A 3A+ 0 ± 3A+ ±A 0 = 0. r S1 [k] = E { S 1 [n+k] } E { S 1 [n] }, r S [k] = E { S [n+k] } E { S [n] }, r S1,S [k] = E { S 1 [n+k] } E { S [n] }, where we have used the independence. Obviously, we need E { S 1 [n] } and E { S [n] }. For those, we have E { S 1 [n] } = 1 8 A+ A+A+ A+ 0 = 0, E { S [n] } = A+3 3A+ 0 = 0. otally, we have the correlation functions r S1 [k] = r S [k] = 9 4 A δ[k], r S1,S [k] = r S,S 1 [ k] = 0, and the corresponding spectra R S1 [θ] = R S [θ] = 9 4 A, R S1,S [θ] = R S,S 1 [θ] = 0. hese observations give us the following simplified expression of the resulting PSD: R S f = 1 Φ1 R S 1 [f] f + Φ f As we can see, we need the Fourier transforms of the basis functions. Using standard properties of the Fourier transform, we find that we have Φ 1 f = sinc f +f c e jπf+fc +sinc f f c e jπf fc, Φ f = j sinc f +f c e jπf+fc sinc f f c e jπf fc,

14 which we rewrite as Φ 1 f = Φ f = j sinc f +f c e jπfc +sinc f f c e jπfc e jπf, sinc f +f c e jπfc sinc f f c e jπfc e jπf. But, we are interested in the corresponding energy spectra. hose we find as Φ1 f = Φ 1 fφ 1f, which after plugging in the spectra becomes Φ1 f = sinc f +f c +sinc f f c Φ f = Φ fφ f, +sinc f +f c sinc f f c e jπfc +e jπfc, Φ f = sinc f +f c +sinc f f c which we rewrite as Φ 1 f = Φ f = sinc f +f c sinc f f c e jπfc +e jπfc, sinc f +f c +sinc f f c +sinc f +f c sinc f f c cosπf c, sinc f +f c +sinc f f c sinc f +f c sinc f f c cosπf c, using Euler s formula. According to the problem formulation, f c is a positive integer. hus we have cosπf c = ±1, and we get Φ1 f = Φ f = sinc f +f c +sinc f f c ±sinc f +f c sinc f f c, sinc f +f c +sinc f f c sinc f +f c sinc f f c, Combining everything, we get R S f = 9 4 A sinc f +f c +sinc f f c.