Exam 8: Designing Signals for Multiple-Access Systems

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An FDMA data signal duration TFDMA =0.25 ms\mathrm { T } _ { \text {FDMA } } = 0.25 \mathrm {~ms} The fundamental frequency equals f f1=____kHz.\mathrm { f } _ { 1 } =\_\_\_\_\mathrm { kHz. }

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f1=1/TFOMA =1/0.25 ms=4kHz\mathrm { f } _ { 1 } = 1 / \mathrm { T } _ { \text {FOMA } } = 1 / 0.25 \mathrm {~ms} = 4 \mathrm { kHz }

An FDMA data signal duration is nx=100\mathrm { n } _ { \mathrm { x } } = 100 samples, the maximum number of users equals nu=____\mathrm { n } _ { \mathrm { u } } =\_\_\_\_

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f1=1/nx.f _ { 1 } = 1 / n _ { x } . The maximum frequency by Nyquist criterion equals fmax⁡<nx/2.\mathrm { f } _ { \max } < \mathrm { nx } / 2. Hence, the maximum  number of harmonics =nx/2−1=49.\text { number of harmonics } = n _ { x } / 2 - 1 = 49.

If an FDMA data signal interval generates nx=100n _ { x } = 100 samples with sampling period T5=0.1\mathrm { T } _ { 5 } = 0.1 ms, the duration of the interval TFDMA =___ms.\mathrm { T } _ { \text {FDMA } } =\_\_\_\mathrm { ms }.

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TFDMA=nxTs=10\mathrm { T } _ { \mathrm { FDMA } } = \mathrm { n } _ { \mathrm { x } } \mathrm { T } _ { \mathrm { s } } = 10

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