N10-z Transform - Lecture notes 6 PDF

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Notes on transformation functions,...

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Part 10 – Discrete Signals and z-Transforms • Sampling of continuous time signals – Sequences

• z-transform – Definition – Main properties

• z- transforms of typical signals – Step – Ramp – Exponential

• Inverse z-transform 1

1

z-Transforms for discrete sequences The z-transform of a sequence {fk} is defined as below, where z is the transform variable

𝑓𝑘 = 𝑓0, 𝑓1, 𝑓2, 𝑓3, … , 𝑓𝑛, … ¥

F(z) = å f k z -k = f 0 + f 1 z-1 + f 2 z -2 + ! k=0

2

2

1

Z definition Use this example to illustrate

x[ n] = {1, 2, 1, 3, ...}

…eq-1

x[n] 3 2 1

1

x[-2] x[-1] x[0] x[1]

X[ z] = 1 + 2 × z -1 +1 × z -2 + 3 × z -3 + ...

…

…eq-2

x[3]

x[2]

x1[ n] = x[ n - 1] = {0, 1, 2, 1, 3, ...} …eq-3

x[ n - 1]

3

2 1 x[-1] x[0] x[1]

…

1 x[2]

x[3]

-1

X 1[ z ] = 0 + 1z + 2z

-2

+ 1z

-3

4 + 3z - + ...

= z - 1 (1 + 2z - 1 + 1z - 2 + 3z - 3 + ...) x[4]

= z -1 × X [ z ]

…eq-4

3

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z-Transforms – Signals with finite non-zero samples, e.g. 𝑓𝑘 = 0, 3, 2, 0, 0, … …

…eq-1

𝐹 𝑧 = 0 ∗ 𝑧 ! + 3 ∗ 𝑧 "# + 2 ∗ 𝑧 "$ + 0 …

…eq-2

𝐹 𝑧 = 𝑧 "# (3 + 2 ∗ 𝑧 "# )

…eq-3

𝐹 𝑧 =

3𝑧 + 2 𝑧2

Zero at z = -2/3; Double pole at z = 0

…eq-4

4

4

2

z-Transforms – unit pulse

fk =

k =0 ì1 í0 all other k . î

…eq-1

¥

F(z) = å f k z -k = 1 k =0

…eq-2

5

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z-Transforms – unit step 𝑓𝑘 = 1,

𝑘 = 0,1,2, …

…eq-1

F(z) = 1 + z - 1 + z - 2 +! =

F(z) =

Zero at z=0; Pole at z=1

1 1 - z -1

z z -1

…eq-2

…eq-3

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6

3

z-Transforms – unit ramp 𝑓𝑘 = 𝑘𝑇,

𝑘 = 0,1,2, …

…eq-1

F ( z) = T z -1+ 2T z -2 + 3T z -3 + != Tz -1 (1 + 2 z -1+ 3 z -2 + !) …eq-2

F ( z) =

Tz T z-1 = 2 -1 2 (1 - z ) ( z - 1 )

…eq-3

Zero at z=0; A double pole at z=1

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z-Transforms – exponential 𝑓(𝑡) = 𝑒 %( 𝑓𝑘 = 𝑒 %&' ,

F(z)=

𝑘 = 0,1,2, …

1 1 - eaT z

Zero at z=0; Pole at z = e aT

= -1

z z - eaT

𝑓(𝑡) = 𝑒 "%( 𝑓𝑘 = 𝑒 "%&' , 𝐹 𝑧 =

𝑘 = 0,1,2, … 1

1−

𝑒 #$%𝑧 #&

=

𝑧 𝑧 − 𝑒 #

%$Zero at z=0; Pole at z = e -aT Note: pole is actually a number, as a & T are constants

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4

Linearity of z-Transforms

Z{ ( f k

If

()

Z { a (f

then

Z { ( gk ) } =

) }= F z k

G(z)

()

()

)+ b (g k ) } = a F z + b G z ¥

Z{a ( f

) + b ( g k) } = å( a f k + b gk ) z-k k k =0 ¥

= a å f k z -k + b

¥

åg

k

z -k

k=0

k=0

= a F (z) + b G(z )

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Example:- sine function fk =

fk

F(z) =

ìcos(wkT ) k = 0,1,2,.. í 0 k 0

zp >1

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Stability of digital systems:- General conclusions • Transfer function of discrete-time system • Zeros and Poles • If all poles of the z-transform of a signal have magnitude less than 1, the signal decays to 0 • If at least one pole has magnitude greater than 1 the signal is unbounded • In the s-domain (simple) poles on the j –axis correspond to oscillatory responses. In the z-domain this is the unit circle.

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Stability:- s-plane vs z-plane

Im Stable region

Unstable region

Unstable region -1

Stable region

Re

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Example - Find the zeros and poles of the following discrete transfer function and determine the system stability

𝐹1 𝑧 =

𝑧+2 (𝑧 − 0.3)

…eq-1

𝑧𝑜 = −2

…eq-2

𝑧𝑝 = +0.3

…eq-3

𝑧𝑝 = 0.3 < 1

…eq-4

Therefore, the system STABLE

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9

Example - Find the zeros and poles of the following discrete transfer function and determine the system stability

𝑧2 𝐹2 𝑧 = 2 (𝑧 + 8𝑧 + 25)

…eq-1

𝑧𝑜 = 0, 0

…eq-2

−8 ± 8$ − 4 > 1 > 25 𝑧𝑝 = 2>1

…eq-3

𝑧𝑝 = −4 ± 𝑗3

…eq-4

𝑧𝑝 = 5 > 1

…eq-5 19

Therefore, the system UNSTABLE 19

Example - Find the zeros and poles of the following discrete transfer function and determine the system stability

1 (1 + 0.2𝑧 "# ) 1 𝑧 𝐹3 𝑧 = × (1 + 0.2𝑧 "# ) 𝑧 𝑧 𝐹3 𝑧 = (𝑧 + 0.2)

𝐹3 𝑧 =

𝑧𝑜 = 0

…eq-1

…eq-2 …eq-3 …eq-4

𝑧𝑝 = −0.2

…eq-5

𝑧𝑝 = 0.2 < 1

…eq-6

Therefore, the system STABLE

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Part 6 – Discrete Signals and z-Transforms Inverse z-transforms (i.e. to obtain a time signal from its z-transform)

• Split into partial fractions and obtain standard inverses from the z-transforms of well known signals (often given in tables) • Use polynomial division to obtain numerical values of first few samples

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Example:-

F (z )=

F (z ) =

1 (1 - z -1)(1 - 2 z -1)

-1 2 + -1 1- z 1 - 2z 1

1 Û1 1 - z -1

…eq-1

…eq-2

1 Û eakT aT -1 1- e z

f k = - 1+ 2 × 2k = - 1+ 2k +1

eaT = 2

…eq-3 22

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Practice P10-B: For the discrete transfer functions below, find the zeros and poles and determine if the corresponding system is stable or not

𝐹1 𝑧 =

𝑧−1 (𝑧 + 1.2)

𝐹2 𝑧 =

𝑧+2 (𝑧 + 0.6)

𝐹3 𝑧 =

𝑧2 (𝑧2 + 8𝑧 + 25)

𝐹4 𝑧 =

1 (1 +

0.8𝑧 "# )(1

− 0.5𝑧 "# )

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