ES 442 Homework #8 Solutions (Spring 2018 Due April 16, 2018 ) Print out homework and do work on the printed pages.. Problem 1 ASCII Code (20 points)

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Toby Moore
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1 Hoework 8 NAME olutions E 44 Hoework #8 olutions (pring 018 Due April 16, 018 ) Print out hoework and do work on the printed pages.. Proble 1 ACII Code (0 points) he Aerican tandard Code for Inforation Interexchange (ACII) has 18 binarycoded characters. ACII codes represent text in coputers, telecounications equipent, and other devices. If a coputer generates 1,000 characters per second, deterine the following: (a) he nuber of (binary digits) required per character. For 18 levels we have L, where n = nuber of in PCM sybol. We ust represent 18 characters which corresponds to n 7 = = 18 sybols (1 sybol = 1 character). n = 7. (b) he nuber of per second required to transit the coputer output, and the iniu bandwidth required to transit this signal. We are told that we ust transit 1,000 characters per second. For 7 per sybol, the bit rate will be character Bit rate = 7 1,000 character second Bit rate = 87, bps (/ second) But the bandwidth is one-half the bit rate, therefore, = 437,00 Hz (note the units are hertz) B (c) For single error detection capability, an additional bit (called a parity bit) is added to the code for each character. Modify your answers for part (a) and (b) to accoodate the addition of the parity bit to each character. Now we add another bit ( ie.., the parity bit) so we require n 8 per character. his increases the bit rate to 1,000,000 bps (= 1 Mb/second) Bandwidth calculation: herefore, B = 00,000 Hz 1

2 Hoework 8 Proble CD Audio (30 points) A copact disc (CD) records audio signals digitally using PCM. he audio baseband signal s bandwidth is 1 khz. (a) If the Nyquist saples are uniforly quantized into L = 6,36 levels and then binary-coded; deterine the nuber of binary digits () n required to encode a saple. For 1 khz as the upper frequency of signal t ( ), the Nyquist sapling rate = 1 khz = 30 KHz. his eans we ust take 30,000 saples/second. We have 6,36 levels which corresponds n to 16 of resolution (We get this fro 6,36 = ; so solving for n gives n = 16. hus, the transission rate is 16 /saple 30,000 saples/ second = 480,000 /second (b) If the audio signal has an average power of 0.1 watt and a peak voltage of 1 volt. Find the resulting ratio of signal-to-quantization (QNR) of the unifor quantizer output in part (a). Given the signal power to be P = 0.1 watt and 1 volt. Using the equation for signal-to-quantization noise ratio, P P 0.1 QNR 3L 3L 3(6.36) Nq p p (1) QNR d 9 B log db p (c) Deterine the nuber of binary digits per second (bps) required to encode and transit the audio signal. Given the sapling rate to be 30,000 saples/second and with 16, the bit rate becoes Bit rate = f N 30,000 Hz ,000 bps Nyquist

3 Hoework 8 (d) In practice CD usic is actually sapled at 44,100 saples/second, which is well above the Nyquist rate. till retaining the nuber of levels L to be 6,36 levels, deterine the bit rate needed to support this higher sapling rate. What is the iniu bandwidth needed to support this higher data rate. Bit rate = f N 44,100 Hz 16 70,600 bps Nyquist hus, the iniu bandwidth B is one-half the sapling rate 1 70,600 B f Nyquist N Hz 3,800 Hz Proble 3 Bit Rate and Baud Rate (10 points) A ulti-level digital counication syste sends one of 3 possible levels over a channel every 0.6 illiseconds. (a) What is nuber of corresponding to each level? 3 levels corresponds to levels, so the nuber of = (b) What is the Baud rate? he Baud rate corresponds to the nuber of sybols that can be sent per unit tie. If the sybol period is 0.6 illiseconds, the sybol 1 frequency is 3 1,666.7 Hz (c) What is the bit rate? he bit rate corresponds to the nuber of that can be sent per unit tie. If 1,666.7 sybols can be sent each second, and given that there are /sybol, then 1,666.7 bps = 8,333 bps Proble 4 Delta Modulation (10 points) Consider a sinusoidal signal (t) = Acos(t), where f =, that is applied to a delta odulator with step size. how that slope overload distortion will occur when 3

4 Hoework 8 f A f where f = 1/ is the sapling frequency and is the period between saples. d() t tarting with ( t) A cos( t); then -A sin( t) dt d() t or A because the axiu value of sin( t) is unity. dt ax When A equals or exceeds the axiu slope of ( t), slope overload does not occur. lope overload occurs when below this liit. herefore, if A >, slope overload will occur. Proble Delta Modulation (10 points) A delta odulation syste is designed to operate at 3 ties the Nyquist rate for a signal with a 3 khz bandwidth. he quantization step size is 0 v. Find the axiu aplitude of a 1 khz input sinusoidal for which the delta odulator does not show slope overload. he threshold for slope overload is A =. For this proble the Nyquist sapling rate f is 3 ties B, where B is the bandwidth 3 khz; hence, f 3 3 khz 18 khz. he signal frequency is 1 khz = 6,83 radian/sec. 0. Maxiu aplitude A = f 18, volt 683 4

5 Hoework 8 Proble 6 Delta Modulation (0 points) For this proble you have a delta odulation (DM) syste for transitting voice signals. Assue the sapling frequency is 6 khz and the voice bandwidth 3.4 khz. he syste is designed to have a axiu signal aplitude of 10 volts. Find the following paraeters: (a) What is the iniu quantization step size allowed for this DM syste? f 3400 rad/sec = 1,363 rad/sec; A 10 volts 6 6 khz iplies second 6 A 1, volt in (b) Deterine the average power of the granular noise. We know that the granular noise power is given by Nq 3 (0.834) herefore, Nq 0.3 watt 3 3 Reeber this is noralized to a resistance of one oh. (c) Deterine the iniu channel bandwidth *in per second) required to transit a DM signal using this syste. Finally we deterine the channel bandwidth required. bit saples ransission bandwidth BW = 1 6,000 saple second ransission bandwidth BW = 6,000 second But bandwidth B is one-half of this value.

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) 3.75 sin 2 10 t 25 sin(6 10 t ) Hoework NAME Solutions EE 442 Hoework #6 Solutions (Spring 2018 Due April 2, 2018 ) Print out hoework and do work on the printed pages. Proble 1 Tone-Modulated FM Signal (12 points) A 100 MHz carrier wave