EECE 442 Homework, Spring 2008

LECE444 Hsmusrrk 3 SxloTan iro Problem 3.13 fa) 0, otherwise Equivalently, we may write f att) = cos} (mt) |Arect| ri \T,/ \Ty where rect(t) is a rectangular function of unit amplitude and unit duration. The Fourier transiorm of g(7) is given by GY) = Sar, f- 4) +5 (745) * sinc(fT,) where A denotes the pulse amplitude and * denotes convolution in the frequency domain. These Py” aye Kobievede Za Hag bin Sladen. Pleniatl Using the replication property of the delta function 5(/), we get Gf) = SF sine rf 7) +me(7{s+7))] Using Eq. (1.52) of the textbook, the power spectral density of the binary data stream is Ion T, S(f) 4 apf sine { Tf - 7)) + sinc (nls + 7) | -#2sine( Ty f- 2) sine (r{r+ 7) o , . ‘Note that the two spectral components sine( 7, f- 2 ) and sine( 7, ft 7)) overlap in u F the frequency interval -(1/T,) < f < (I/T;), hence the presence of cross-product terms in Eq. (1). Figure | plots the normalized power spectral density S(PKAP7T,/4) versus the normalized frequency fT. The interesting point to note in this figure is the significant reduction in the power spectrum of the pulse-shaped data stream x(r) in the interval -1/T, <f< 1/7; (c) The power spectral density of the standard form of polar NRZ signaling is S(f) = A’Tysine (fT) (2) Comparing this expression with that of Eq. (1), we observe the following differences: Polar NRZ signals using Polar NRZ signals using cosine pulses rectangular pulses f=0 A’T, f= UT, A’T,/4 9 3/10 1.4 —<— ——~ oO CARAT Co) : i eeec al eseedae ve : + : : : : t L J 1 n L 1 io x ~ Q- © < x 4 e So f=) fo] fem 3.2 The quantizer has the following input-output curve: Outpur volts At the sampling instants we have: t 3/8. -1/8 41/8 +3/8 a the coded waveform is (assuming on-off signaling): Problem 3.22 m(t) <3V2 ~342 372 372 aie Se eee ne | ttre cade 0011 0011 1100 1100 2 8 4 8 0 ' 8 The transmitted code words are: t/T, AU fw a= code 001 0710 014 100 101 110 3 g Time (Seconds) The sampled analog signal is —— . a 3% 6% O9T 27% 8B ET Problem 3.25 mit) = Atann{ Br) To avoid siope overload, we require a> max|72)! qa) T, i ae: dm(t 2, ; mit) ABsech (Br) (2) dt Hence. using Eq. (2) in (1): A2max(ABsech {82)) xT, (3) i Since sech( Br} = ——.— Br sosh{ Br) 2 ot Bt +e or it follows that the maximum value of sech(Br) is 1. which occurs at time r= 0. Hence, from Eq. (3) we find that A > ABT,. flo Problem 3.26 The modulating wave is mt) = An eos(2nf t) The slope of m{t) is dm(t) dt = Went AL sin¢ anf t) The maximum slope of m(t) is equal to ant Ane Tne maximum average slope of the approximating signal m,(t) produced by the delta modulator is 8/T. where 6 is the step size and TS is the sampling period. The limiting value of a is therefore given by 6 ent a T or Assuming a load of 1 ohm, the transmitted power is a2, Therefore, the maximum . ‘ 2 pg Cplme power that may be transmitted without slope-overload distortion is equal to 6 18x fas? Problem 3.27 fs = L0fNyquist Fxyquist = 6.8 kHz f,=10x6.8x 10? =6.8 x 10* Hz A jdm(t) —2> F 2 max) For the sinuscidal signal m(t) = A,,sin(27f,,7), we have a m(t) a? 20 f pAmCOS(2Tf pt) Hence, jdm(t - : | A Mo 7 [PFS nA max or, equivalently, A 77 PRs A m | max Therefore. i) ue A \Anlmax = FX Inx f, _ Af; ~ Ae oni in _ OLX 6.8 x 10" 2n x 10° 1.08V 7M Tits Steg Fes gS RRR SMG cong oe do atte Deke fiche garding yore oe pot TR “fo (a) From the solution to Problem 3.27, we have _ Af, _ 2nf A 1 A oF. or A= F, ; ()) 1 2 So(f) = A’/3 f, —-W<fsw ‘ 0, otherwise where W is the bandwidth of the reconstruction filter at the demodulator output. Hence, the average quantization noise power is wv “ ._ 2aW ~ N= J Soifar = SO 2) : -w Substituting Eq. (2) into (1), we get SNR A> ZXRX 10> x1 50 x 10° = 0.126V (b) (SNR)... = out 16x f > 16x" = 475 In decibels, (SNR)oy, = 1Olog 9475 = 26.8 dB Problem 3.30 (a) For linear delta modulation, the maximum amplitude of a sinusoidal test signal that can be Cmte 3. (30x 10°)" Xe 10° x5x 10 used without slope-overload distortion is 0.1 x 60x 10° 2mx ix 10° 0.95V (b) (1) Under the pre-filtered condition, it is reasonable to assume that the granular quantization noise is uniformiy distributed between -A and +A. Hence, the variance of the quantization noise is f, = 2x3xi0x/9 ho wa A 3 The signai-to-noise ratio under the pre-filtered condition is therefore (SNR) (ii)The signal-to-noise ratio under the post-filtered condition is f S) \N) postfiltered prefiltered = Hl i lt = 135 = 21.3 dB 2 3 f; fnW lon 3. (60) 3 2 lén” (1)°x3 1367 31.3 dB fol fo The filtering gain in signal-to-noise ratio due to the use of a reconstruction filter at the demodulator output is therefore 31.3 - 21.3 = 10 dB.