Sample Problems for the 9th Quiz

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1 Sample Problems for the 9th Quiz. Draw the line coded signal waveform of the below line code for (a Unipolar nonreturn-to-zero (NRZ signaling (b Polar nonreturn-to-zero (NRZ signaling (c Unipolar return-to-zero (RZ signaling (d Bipolar return-to-zero (BRZ signaling (e Split-phase (Manchester code See Slide Here, I wish you to learn how to associate the waveforms with the naming. So it is not a problem to test the ability of memorization but the ability of association.. Do the follow subproblems. (a Find the Fourier transform S(f ofs(t a n g(t nt, where G(f g(te ıπft dt. (b Find the Fourier transform S NT (f of the truncated signal of s(t, i.e., s NT (t N n N a n g(t nt (c Given {a n } is i.i.d., show that the PSD of s(t isequalto σ T G(f + µ T G(f where E[a n ]µ and E[a n ]σ + µ. Hint: PSD(f lim N (d Continue from (c. Since p(t µ T G(f µ T ( δ f n T } {{ } p(t NT S(fS NT(f. ( n G T ( δ f n T δ we can conclude that p(t 0 if, and only if, ( f n T, (i µ 0, or (ii G(n/T 0 for every integer n.

2 Give an example of the signaling design (including statistics of {a n } and g(t that gives µ 0. Give an example of the signaling design that gives G(n/T 0 for every integer n. (a See Slide (b See Slide (c See Slide (d See the slides and do it yourself. 3. Suppose for a switch, its nominal output bit rate is f out 5M bits/second and its nominal input bit rate per input stream is f in.m bits/second. There are 4 input streams. Here, M 0 6. (a During a period of 5 milli-second, how many bits are nominally received? How many bits should be nominally sent out? (b Suppose in the output frame design, we can have one bit-stuffing indicator per L 88 bits as shown below: In other words, the switch must receive at least L 87 bits for one frame time period (which is indicated by (C C C3 (000 and at most L 88 bits for one frame time period (which is indicated by (C C C3 (. Determine the nominal bit stuffing rate S. (a 4 streams. 0 6 seconds bits/seconds 4, 440 bits for input, and seconds bits/seconds 5, 000 bits for output. (b 76 5 S ( S.. S

3 4. (a Use the below components to draw the functional diagram of delta modulation (DM. Please separately picture the functional diagrams of transmitter and receiver. Please also indicate where the below messages are located, including the sampled message m[n] at the transmitter, the delta modulate message m q [n] at both the transmitter and the receiver, the error message e[n] m[n] m q [n ] at the transmitter, and the quantized error message e q [n] at both the transmitter and the receiver. Note: Please add Encoder, Decoder, Low-pass Filter functional blocks when necessary. (b What is slope overload condition for DM? (c Again, use only one comparator, one one-bit quantizer and one accumulator to draw the functional diagram of delta-sigma modulation. Note: Please add Encoder, Decoder, Low-pass Filter functional blocks when necessary. (d Replace z in the Accumulator by a prediction filter and replace one-bit quantizer by a quantizer. Draw the functional diagram of DPCM. Note: Please add Encoder, Decoder, Low-pass Filter functional blocks when necessary. (a See Slide 3-0. (b See Slide 3-. (c See Slide 3-5. (d See Slide

4 5. For the linear predictor below. prove that the optimal solution w [w w w p ] that minimizes E[(x[n] ˆx[n] ] satisfy the Wiener-Hopf equations, i.e., R X [0] R X [] R X [p ] w R X [] R X [] R X [0] R X [p ] w R X []. R X [p ] R X [p ] R X [0] R X [p] where R X [ ] is the autocorrelation function of WSS process x[n]. See Slides (a Find the impulse response h(t such that the output signal-tonoise ratio (at the output of the sampler is maximized. w p Hint: Apply Cauchy-Swartz inequality to G(fH(fexp{ıπf T}df, where G(f andh(f are respectively the Fourier transforms of g(t andh(t. (b Can the solution in (a (which is referred to as matched filter be used to eliminate intersymbol interference (ISI? 4

5 (c Assume that antipodally modulated signal s(t I g(t, where Pr[I ] Pr[I +], is fed into the system below. Show that the error rate under λ 0 is equal to Q( γ, where γ E g /N 0,E g g(t dt, and the PSD of the white noise w(t isn 0 /. Hint: Determine the distributions of y Y (T /k given I and I. (a From the figure, we learn that y(t g(t h(t tt + w(t h(t tt G(fH(fexp(ıπfTdf + n(t, where n(t w(t h(t. Note that n(t (which is the output due to WSS input w(t andltifilterh(t is WSS and S n (f S w (f H(f.Thus, E[n (T ] R n (0 S n (fdf N 0 H(f df. The output SNR can then be represented as the η-formula in Slide 4-7. By applying Cauthy-Schwarz s inequality (cf. Slide 4-8, we obtain H(f k G (fexp{ ıπft}. The optimal impulse response is then given by the derivation on Slide 4-9. (b The matched filter only considers the transmission of one symbol (here, g(t and ignore entirely the ISI effect between consecutive symbols; so, it cannot be used to resolve the ISI. 5

6 (c First show that y I E g +n as did in Slide 4-4. Then argue that the pdfs of y given I andi +areequaltoφ (t and φ + (t in Slide 4-5, respectively. Finally, copying the derivations in Slides , we obtain BER Φ 0 µ µ/σ φ (ydy + 0 { exp y πσ σ π exp ( µ, σ { z } dz φ + (ydy } dy (Let z y/σ. where Φ( is the cdf of the standard normal distribution. The proof is completed by noting that µ E g, σ E g N 0 /and Φ( x Q(x. 7. (a Prove that p(nt b exp( ıπnt b f T b P (f ntb, where P (f is the Fourier transform of p(t. (b For the system below, we suppose b k {0, }, a k b k, s(t a kg(t nt b andp(t g(t h(t c(t. Prove that if {, if n 0 p(nt b 0, otherwise then the sample y(it b is nothing to do with..., a i, a i, a i+, a i+,..., and hence, no ISI occurs. 6

7 (c Show that is equivalent to p(nt b T b {, if n 0 0, otherwise P (f ntb. (d A matched receiver filter design that matches both the transmitter and channel filters is Can this design avoid ISI? (a See Slides (b From the system, we learn that C(f G(f H (fexp{ ıπft b }. y(it b s(t h(t c(t titb + w(t c(t ( titb a k g(t nt b h(t c(t + w(t c(t titb tit b a k (g(t nt b h(t c(t + w(t c(t titb tit b a k p(t nt b + w(t c(t titb tit b where g(t nt b h(t c(t p(t nt b was proved in Slide 4-7. Continue the derivation: y(it b a k p(t nt b + w(t c(t titb tit b a k p(it b nt b +w(t c(t titb a k p((i nt b +w(t c(t titb a i + w(t c(t titb. 7

8 Thus, y(it b is only a function of a i but is nothing to do with all the remaining {a j } j,j i. (c A consequence of the result in (a. (d P (f G(fH(fC(f G(fH(f[G(f H (fexp{ ıπft}] G(f H(f exp{ ıπft b } Then P (f ntb (f G n H T b exp{ ıπft b } (f ntb { } exp ıπ (f ntb T b (f G n H T b which is a real number only when f is a multiple of (f ntb, P (f ntb T b for every f R T b, and hence, cannot be valid, which implies that ISI cannot be avoided. 8

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