3 Initial and final value theorems of z-transforms

These results are important in, for example, Digital Control Theory where we are sometimes particularly interested in the initial and ultimate behaviour of systems.

3.1 Initial value theorem

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If f n is a sequence with z-transform F ( z ) then the ‘initial value’ f 0 is given by

f 0 = lim z → ∞ F ( z ) (provided, of course, that this limit exists).

This result follows, at least informally, from the definition of the z-transform:

F ( z ) = f 0 + f 1 z − 1 + f 2 z − 2 + …

from which, taking limits as z → ∞ the required result is obtained.

Task!

Obtain the z-transform of

f ( n ) = 1 − a n , 0 < a < 1

Verify the initial value theorem for the z-transform pair you obtain.

Using standard z-transforms we obtain

Z { f n } = F ( z ) = z z − 1 − z z − a = 1 1 − z − 1 − 1 1 − a z − 1

hence, as z → ∞ : F ( z ) → 1 − 1 = 0

Similarly, as n → 0

f n → 1 − 1 = 0

so the initial value theorem is verified for this case.

3.2 Final value theorem

Suppose again that { f n } is a sequence with z-transform F ( z ) . We further assume that all the poles of F ( z ) lie inside the unit circle in the z − plane (i.e. have magnitude less than 1) apart possibly from a first order pole at z = 1 .

The ‘final value’ of f n i.e. lim n → ∞ f n is then given by lim n → ∞ f n = lim z → 1 ( 1 − z − 1 ) F ( z )

Proof: Recalling the left shift property

Z { f n + 1 } = z F ( z ) − z f 0

we have

Z { f n + 1 − f n } = lim k → ∞ ∑ n = 0 k ( f n + 1 − f n ) z − n = z F ( z ) − z f 0 − F ( z )

or, alternatively, dividing through by z on both sides:

( 1 − z − 1 ) F ( z ) − f 0 = lim k → ∞ ∑ n = 0 k ( f n + 1 − f n ) z − ( n + 1 )

Hence ( 1 − z − 1 ) F ( z ) = f 0 + ( f 1 − f 0 ) z − 1 + ( f 2 − f 1 ) z − 2 + …

or as z → 1

lim z → 1 ( 1 − z − 1 ) F ( z ) = f 0 + ( f 1 − f 0 ) + ( f 2 − f 1 ) + … = lim k → ∞ f k

Example

Again consider the sequence f n = 1 − a n 0 < a < 1 and its z-transform

F ( z ) = z z − 1 − z z − a = 1 1 − z − 1 − 1 1 − a z − 1

Clearly as n → ∞ then f n → 1 .

Considering the right-hand side

( 1 − z − 1 ) F ( z ) = 1 − ( 1 − z − 1 ) 1 − a z − 1 → 1 − 0 = 1 as z → 1.

Note carefully that

F ( z ) = z z − 1 − z z − a

has a pole at a ( 0 < a < 1 ) and a simple pole at z = 1 .

The final value theorem does not hold for z-transform poles outside the unit circle

e.g. f n = 2 n F ( z ) = z z − 2

Clearly f n → ∞ as n → ∞

whereas

( 1 − z − 1 ) F ( z ) = z − 1 z z ( z − 2 ) → 0 as z → 1

Exercises
  1. A low pass digital filter is characterised by

    y n = 0.1 x n + 0.9 y n − 1

    Two such filters are connected in series. Deduce the transfer function and governing difference equation for the overall system. Obtain the response of the series system to (i) a unit step and (ii) a unit alternating input. Discuss your results.

  2. The two systems

    y n = x n − 0.7 x n − 1 + 0.4 y n − 1

    y n = 0.9 x n − 1 − 0.7 y n − 1

    are connected in series. Find the difference equation governing the overall system.

  3. A system S 1 is governed by the difference equation

    y n = 6 x n − 1 + 5 y n − 1

    It is desired to stabilise S 1 by using a feedback configuration. The system S 2 in the feedback loop is characterised by

    y n = α x n − 1 + β y n − 1

    Show that the feedback system S 3 has an overall transfer function

    H 3 ( z ) = H 1 ( z ) 1 + H 1 ( z ) H 2 ( z )

    and determine values for the parameters α and β if H 3 ( z ) is to have a second order pole at z = 0.5 . Show briefly why the feedback systems S 3 stabilizes the original system.

  4. Use z-transforms to find the sum of squares of all integers from 1 to n :

    y n = ∑ k = 1 n k 2

    [Hint: y n − y n − 1 = n 2 ]

  5. Evaluate each of the following convolution summations (i) directly (ii) using z-transforms:
    1. a n ∗ b n a ≠ b
    2. a n ∗ a n
    3. δ n − 3 ∗ δ n − 5
    4. x n ∗ x n where x n = 1 n = 0 , 1 , 2 , 3 0 n = 4 , 5 , 6 , 7 …
  1. Step response: y n = 1 − ( 0.99 ) ( 0.9 ) n − 0.09 n ( 0.9 ) n

    Alternating response: y n = 1 361 ( − 1 ) n + 2.61 361 ( 0.9 ) n + 1.71 361 n ( 0.9 ) n

  2. y n + 0.3 y n − 1 − 0.28 y n − 2 = 0.9 x n − 1 − 0.63 x n − 2
  3. α = 3.375 β = − 4
  4. ∑ k = 1 n k 2 = ( 2 n + 1 ) ( n + 1 ) n 6
    1. 1 ( a − b ) ( a n + 1 − b n + 1 )
    2. ( n + 1 ) a n
    3. δ n − 8
    4. { 1 , 2 , 3 , 4 , 3 , 2 , 1 }