3 Linearity property and applications

3.1 Linearity property

This simple property states that if { v n } and { w n } have z-transforms V ( z ) and W ( z ) respectively then

ℤ { a v n + b w n } = a V ( z ) + b W ( z )

for any constants a and b .

(In particular if a = b = 1 this property tells us that adding sequences corresponds to adding their z-transforms).

The proof of the linearity property is straightforward using obvious properties of the summation operation. By the z-transform definition:

ℤ { a v n + b w n } = ∑ n = 0 ∞ ( a v n + b w n ) z − n = ∑ n = 0 ∞ ( a v n z − n + b w n z − n ) = a ∑ n = 0 ∞ v n z − n + b ∑ n = 0 ∞ w n z − n = a V ( z ) + b V ( z )

We can now use the linearity property and the exponential sequence { e − α n } to obtain the z-transforms of hyperbolic and of trigonometric sequences relatively easily. For example,

sinh n = e n − e − n 2

Hence, by the linearity property,

ℤ { sinh n } = 1 2 ℤ { e n } − 1 2 ℤ { e − n } = 1 2 z z − e − z z − e − 1 = z 2 z − e − 1 − ( z − e ) z 2 − ( e + e − 1 ) z + 1 = z 2 e − e − 1 z 2 − ( 2 cosh 1 ) z + 1 = z sinh 1 z 2 − 2 z cosh 1 + 1

Using α n instead of n in this calculation, where α is a constant, we obtain

ℤ { sinh α n } = z sinh α z 2 − 2 z cosh α + 1

Task!

Using cosh α n ≡ e α n + e − α n 2 obtain the z-transform of the sequence { cosh α n } = { 1 , cosh α , cosh 2 α , … }

We have, by linearity,

ℤ { cosh α n } = 1 2 ℤ { e α n } + 1 2 ℤ { e − α n } = z 2 1 z − e α + 1 z − e − α = z 2 2 z − ( e α + e − α ) z 2 − 2 z cosh α + 1 = z 2 − z cosh α z 2 − 2 z cosh α + 1

3.2 Trigonometric sequences

If we use the result

ℤ { a n } = z z − a | z | > | a |

with, respectively, a = e i ω and a = e − i ω where ω is a constant and i denotes − 1 we obtain

ℤ { e i ω n } = z z − e + i ω ℤ { e − i ω n } = z z − e − i ω

Hence, recalling from complex number theory that

cos x = e i x + e − i x 2

we can state, using the linearity property, that

ℤ { cos ω n } = 1 2 ℤ { e i ω n } + 1 2 ℤ { e − i ω n } = z 2 1 z − e i ω + 1 z − e − i ω = z 2 2 z − ( e i ω + e − i ω ) z 2 − ( e i ω + e − i ω ) z + 1 = z 2 − z cos ω z 2 − 2 z cos ω + 1

(Note the similarity of the algebra here to that arising in the corresponding hyperbolic case. Note also the similarity of the results for ℤ { cosh α n } and ℤ { cos ω n } .)

Task!

By a similar procedure to that used above for ℤ { cos ω n } obtain ℤ { sin ω n } .

We have

ℤ { sin ω n } = 1 2 i ℤ { e i ω n } − 1 2 i ℤ { e − i ω n } (Don’t miss the i factor here!)

∴ ℤ { sin ω n } = z 2 i 1 z − e i ω − 1 z − e − i ω = z 2 i − e − i ω + e i ω z 2 − 2 z cos ω + 1 = z sin ω z 2 − 2 z cos ω + 1
Key Point 6

ℤ { cos ω n } = z 2 − z cos ω z 2 − 2 z cos ω + 1 ℤ { sin ω n } = z sin ω z 2 − 2 z cos ω + 1

Notice the same denominator in the two results in Key Point 6.

Key Point 7

ℤ { cosh α n } = z 2 − z cosh α z 2 − 2 z cosh α + 1 ℤ { sinh α n } = z sinh α z 2 − 2 z cosh α + 1

Again notice the denominators in Key Point 7. Compare these results with those for the two trigonometric sequences in Key Point 6.

Task!

Use Key Points 6 and 7 to write down the z-transforms of

  1. sin n 2
  2. cos 3 n
  3. { sinh 2 n }
  4. { cosh n }
  1. ℤ sin n 2 = z sin 1 2 z 2 − 2 z cos 1 2 + 1
  2. ℤ { cos 3 n } = z 2 − z cos 3 z 2 − 2 z cos 3 + 1
  3. ℤ { sinh 2 n } = z sinh 2 z 2 − 2 z cosh 2 + 1
  4. ℤ { cosh n } = z 2 − z cosh 1 z 2 − 2 z cosh 1 + 1
Task!

Use the results for ℤ { cos ω n } and ℤ { sin ω n } in Key Point 6 to obtain the z-transforms of

  1. { cos ( n π ) }
  2. sin n π 2
  3. cos n π 2

Write out the first few terms of each sequence.

  1. With ω = π

    ℤ { cos n π } = z 2 − z cos π z 2 − 2 z cos π + 1 = z 2 + z z 2 + 2 z + 1 = z z + 1

    { cos n π } = { 1 , − 1 , 1 , − 1 , … } = { ( − 1 ) n }

    We have re-derived the z-transform of the unit alternating sequence. (See Task on page 17).

  2. With ω = π 2

    ℤ sin n π 2 = z sin π 2 z 2 − 2 z cos π 2 + 1 = z z 2 + 1

    where sin n π 2 = { 0 , 1 , 0 , − 1 , 0 , … }

  3. With ω = π 2 ℤ cos n π 2 = z 2 − cos π 2 z 2 + 1 = z 2 z 2 + 1

    where cos n π 2 = { 1 , 0 , − 1 , 0 , 1 , … }

    (These three results can also be readily obtained from the definition of the z-transform. Try!)