# Sampling signals, periodicity, Z transforms

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

#### 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