Tutorial 6 z-transforms

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Electrical Engineering

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Oct 30, 2023

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Tutorial 6: discrete time & the z-transform Discrete time : • Fixed intervals of time (nTs) • Magnitude continuous (can adopt any value) • Reduced data set • Less processing • Less storage • First step in creating digital signal • Data between samples missing • Interpolation used to ‘fill in’ missing data Discretisation process: • Choose number of samples to be taken (N) • Construct an integer index (n) • n=0:1:N-1 • Identify Nyquist frequency ( 𝐹𝐹 𝑛𝑛 ) • Calculate Nyquist rate ( 𝐹𝐹 𝑛𝑛𝑛𝑛 ) • This is the MINIMUM rate of sampling • 𝐹𝐹 𝑛𝑛𝑛𝑛 = 2 ∗ 𝐹𝐹 𝑛𝑛 • Decide sampling frequency ( 𝐹𝐹 𝑠𝑠 ) • Normally at least five time Nyquist rate • Calculate sampling period ( 𝑇𝑇 𝑠𝑠 ) • Multiply integer index (n) and sampling period 𝑇𝑇 𝑠𝑠 to give ( 𝑛𝑛𝑇𝑇 𝑠𝑠 ) • Replace ( t ) with ( 𝑛𝑛𝑇𝑇 𝑠𝑠 ) in system equation and calculate Example: convert the following to discrete sequence Take the first eight samples. 𝑥𝑥 𝑡𝑡 = sin 2 ∗ 𝑝𝑝𝑝𝑝 ∗ 8 ∗ 𝑡𝑡 + sin(2 ∗ 𝑝𝑝𝑝𝑝 ∗ 5 ∗ 𝑡𝑡 ) Number of samples: 𝑁𝑁 = 8 Integer index: 𝑛𝑛 = [0 1 2 3 4 5 6 7] Nyquist frequency: 𝐹𝐹 𝑛𝑛 = 8𝐻𝐻𝐻𝐻 Nyquist rate: 𝐹𝐹 𝑛𝑛𝑛𝑛 = 2 ∗ 8𝐻𝐻𝐻𝐻 = 16𝐻𝐻𝐻𝐻 Let us sample at five times the Nyquist rate… Sampling frequency: 𝐹𝐹 𝑠𝑠 = 5 ∗ 16𝐻𝐻𝐻𝐻 = 80𝐻𝐻𝐻𝐻 Sampling period: 𝑇𝑇 𝑠𝑠 = 1 𝐹𝐹 𝑠𝑠 = 1 80𝐻𝐻𝐻𝐻 = 0. 0125𝑠𝑠 Construct discrete time base 𝑛𝑛𝑇𝑇 𝑠𝑠 … 𝑛𝑛𝑇𝑇 𝑠𝑠 = [0 0.0125 0.025 0.0375 0.05 … 0.0625 0.075 0.0875 ] 𝑥𝑥 [ 𝑛𝑛 ] = sin 2 ∗ 𝑝𝑝𝑝𝑝 ∗ 8 ∗ 𝑛𝑛𝑇𝑇 𝑠𝑠 + sin(2 ∗ 𝑝𝑝𝑝𝑝 ∗ 5 ∗ 𝑛𝑛𝑇𝑇 𝑠𝑠 ) Replace ( t ) with ( 𝑛𝑛𝑇𝑇 𝑠𝑠 ) … Calculate!

If we want to sample more of the signal, just increase the number of samples N!

Z-transforms • Discrete version of Laplace transform • Uses difference equations • Derivative law still applies • ‘z’ rather than ‘s’ • ‘ 𝐻𝐻 2 ’ rather than ‘ 𝑠𝑠 2 ’ • First order initial conditions • f ( n+1 )→ 𝐻𝐻𝐹𝐹 𝐻𝐻 − 𝐻𝐻𝑧𝑧 (0) • Second order initial conditions • f ( n+2 )→ 𝐻𝐻 2 𝐹𝐹 𝐻𝐻 − 𝐻𝐻 2 𝑧𝑧 0 − 𝐻𝐻𝑧𝑧 (1) • Solve using same process as Laplace… General process • transform difference equation to the z-domain • apply the derivative law • insert any initial conditions • solve algebraic equation to produce X(z) • perform inverse z-transform to solve • Use tables of z-transforms if possible • Calculate if not • difference equation expressed in discrete time domain Example: transform the following discrete sequence x [ n ] using a z- transform 𝑥𝑥 𝑛𝑛 = 𝛿𝛿 𝑛𝑛 + 2𝛿𝛿 𝑛𝑛 − 1 + 5𝛿𝛿 𝑛𝑛 − 2 + 3𝛿𝛿 [ 𝑛𝑛 − 3] Remember that 𝑛𝑛 = 0 ≡ 𝐻𝐻 0 , 𝑛𝑛 = 1 ≡ 𝐻𝐻 −1 , 𝑛𝑛 = 2 ≡ 𝐻𝐻 −2 etc ∴ 𝑋𝑋 𝐻𝐻 = 1 𝐻𝐻 0 + 2 𝐻𝐻 −1 + 5 𝐻𝐻 −2 + 3 𝐻𝐻 −3 Example: determine the discrete sequence x[n] for the following z-transform X(z) by the application of an inverse z-transform 𝑋𝑋 𝐻𝐻 = 𝐻𝐻 −1 + 𝐻𝐻 −2 + 2 𝐻𝐻 −3 + 2 𝐻𝐻 −4 + 3 𝐻𝐻 −5 + 𝐻𝐻 −6 − 4 𝐻𝐻 −7 Remember that 𝑛𝑛 = 0 ≡ 𝐻𝐻 0 , 𝑛𝑛 = 1 ≡ 𝐻𝐻 −1 , 𝑛𝑛 = 2 ≡ 𝐻𝐻 −2 etc 𝑥𝑥 𝑛𝑛 = 𝛿𝛿 𝑛𝑛 − 1 + 𝛿𝛿 𝑛𝑛 − 2 + 2𝛿𝛿 𝑛𝑛 − 3 + 2𝛿𝛿 𝑛𝑛 − 4 … + 3𝛿𝛿 𝑛𝑛 − 5 + 𝛿𝛿 𝑛𝑛 − 6 − 4𝛿𝛿 [ 𝑛𝑛 − 7] When expressed in terms of power series, forward and inverse z-transforms are very straightforward! Maintain position!

Tutorial 6 z-transforms

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