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

{ 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:

{ 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,

{ sinh n } = 1 2 { e n } 1 2 { 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

{ 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,

{ cosh α n } = 1 2 { e α n } + 1 2 { 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

{ 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

{ e i ω n } = z z e + i ω { 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

{ cos ω n } = 1 2 { e i ω n } + 1 2 { 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 { cosh α n } and { cos ω n } .)

Task!

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

We have

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

{ 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

{ cos ω n } = z 2 z cos ω z 2 2 z cos ω + 1 { 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

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

Use the results for { cos ω n } and { 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

    { 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

    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 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!)