DigitalModSolvedProblems

# DigitalModSolvedProblems - 1 Digital Modulation Solved...

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Unformatted text preview: 1 Digital Modulation Solved Problems Data Rate Limits 1. What is the bandwidth required to transmit at a rate of 10 Mbps in the presence of a 28‐dB S/N ratio? Solution: Convert S/N as a power value: S S 28 = 10 log( ); = 630.957 N N Substituting: S C = B log 2 (1 + ); 10 x10 6 = B log 2 (1 + 630.957 ) N B = 1.075 MHz 2. Find the minimum allowable bandwidth for a binary signal with a bit rate of 200 kbps. Solution: For a binary communications channel: R = 2B; B = (R/2) = (200K/2) = 100 KHz 3. A broadcast television channel has a bandwidth of 6 MHz. Calculate the maximum data rate that could be carried in a TV channel using a 16‐level code. Ignore noise. Solution: C = 2 B log 2 M = 2 (6 x10 6 ) log 2 16 = 48 Mbps 4. Suppose that the spectrum of a channel is between 3 MHz and 4 MHz and the SNR is 24 dB. Assuming theoretical limit is achievable, how many signaling levels are required? Solution: B = 4 MHz – 3 MHz = 1 MHz Convert S/N as a power value: S S 24 = 10 log( ); = 251.19 N N Glenn Ople Avendaño Digital Modulations Solved Problems 2 S ); C = 1x10 6 log 2 (1 + 251.19) = 7.98 Mbps N From: C = 2 B log 2 M ; 7.98 x10 6 = 2(1x10 6 ) log 2 M ; M = 15.88; hence M = 16 C = B log 2 (1 + Digital/Analog Modulation – Keying Techniques 1. The bit rate of a digital signal is 3000 bps. If each signal element carries 6 bits, what is the baud? Solution: bits 3000 signal element sec Baud = = 500 = 500 baud bits sec ond 6 signal element 2. Determine the minimum bandwidth for an FSK signal with a mark frequency of 99 KHz, a space frequency of 101 KHz and a bit rate of 10 Kbps. Solution: f − fs 99 K − 101K Δf = m = = 1K 2 2 B = 2( f b + Δf ) = 2 (10 K + 1K ) = 22 KHz 3. A constellation diagram consists of 8 equally spaced points on a circle. If the bit rate is 4800 bps, find the baud. Solution: For 8 equally spaced points: implies 8‐PSK For 8‐PSK: 23 = 8; 3 bits/symbol bits 4800 sec = 1600 symbols = 1600 baud Baud = bits sec ond 3 symbol 4. A modem uses 16 different phases and 4 different amplitudes. How many bits does it transmit per symbol? Solution: No. of available symbols, M = (16) (4) = 64 Hence: 2n = M; 2n = 64; n = 6 bits/symbol Glenn Ople Avendaño Digital Modulations Solved Problems 3 5. Compute the baud for a 36 Kbps, 64‐QAM signal. Solution: For 64‐QAM: 26 = 64; 6 bits/symbol bits 36000 sec = 6000 symbols = 6000 baud Baud = bits sec ond 6 symbol 6. What is the bit rate of a 1000‐baud, 16‐QAM signal? Solution: For 16‐QAM: 24 = 16; 4 bits/symbol symbols bits bits Bit Rate = (1000 )( 4 ) = 4000 sec sec ond symbol 7. For a 16‐QAM modulator with an input bit rate equal to 20 Mbps, determine the minimum bandwidth required for transmission. Solution: For all D/A signals (except FSK): Min BW = baud For 16‐QAM: 24 = 16; 4 bits/symbol Mbits 20 sec = 5 Msymbols = 5 Mbaud Baud = bits sec ond 4 symbol Hence: BW = 5 MHz 8. Determine the minimum bandwidth required to achieve a Eb / No of 14 dB for an 8‐PSK system operating at 20 Mbps with a C/N of 11 dB. Solution: Convert C/N as a power value: C C 11 = 10 log( ); = 12.589 N N Convert Eb / No as a power value: E E 14 = 10 log( b ); b = 25.12 No No Substituting: Eb ⎛ C ⎞ ⎛ B ⎞ B ⎞ = ⎜ ⎟ ⎜ ⎟ ; 25.12 = (12.589 ) ⎛ ⎜ ⎟ ; B = 39.9 MHz ⎜f ⎟ No ⎝ N ⎠ ⎝ b ⎠ ⎝ 20 x10 6 ⎠ Glenn Ople Avendaño Digital Modulations Solved Problems ...
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## This note was uploaded on 02/08/2012 for the course ITT 650 taught by Professor Dewey during the Spring '11 term at UNC Asheville.

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