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    Sampling signals, periodicity, Z transforms

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    two part question and solution

    PART1
    Finding the periodicity in N of the discrete time signal x(n) = Cos(3pi*Ts.n + pi/4) and then plotting the discrete signal. From this the condition on the sampling period is determined

    PART2
    It is shown also how to determine the z transform of the composite discrete time signal given by

    y(n) = y(n-1) -0.5*y(n-2) +x(n) + x(n-1)

    From this the output function due to a step input x(n) = u(n) = 1 is determined as

    Y(z) = z^2*(z + 1)/{(z - 1)*(z^2 - z + 0.5)}

    Further it is shown how to do the partial expansion of Y(z) into its partial fractions.

    From this the inverse z transform to find the output in the time domain due to the step input is determined

    © BrainMass Inc. brainmass.com October 10, 2019, 3:28 am ad1c9bdddf
    https://brainmass.com/engineering/electrical-engineering/sampling-signals-periodicity-z-transforms-421206

    Solution Preview

    A two part question and solution

    PART1
    Finding the periodicity in N of the discrete time signal x(n) = Cos(3pi*Ts.n + pi/4) and then plotting the discrete signal. From this ...

    Solution Summary

    A two part question and solution

    PART1
    Finding the periodicity in N of the discrete time signal x(n) = Cos(3pi*Ts.n + pi/4) and then plotting the discrete signal. From this the condition on the sampling period is determined

    PART2
    It is shown also how to determine the z transform of the composite discrete time signal given by

    y(n) = y(n-1) -0.5*y(n-2) +x(n) + x(n-1)

    From this the output function due to a step input x(n) = u(n) = 1 is determined as

    Y(z) = z^2*(z + 1)/{(z - 1)*(z^2 - z + 0.5)}

    Further it is shown how to do the partial expansion of Y(z) into its partial fractions.

    From this the inverse z transform to find the output in the time domain due to the step input is determined

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