z-transforms - engineering mathematics filez-transforms 4.1 introduction â transform plays an...
TRANSCRIPT
Page | 1
Chapter 4
Z-TRANSFORMS 4.1 Introduction
ðâ Transform plays an important role in discrete analysis and may be seen as discrete
analogue of Laplace transform. Role of ðâ Transforms in discrete analysis is the same as that
of Laplace and Fourier transforms in continuous systems.
Definition: The ðâTransform of a sequence ð¢ð defined for discrete values ð = 0,1,2,3, ⊠and (ð¢ð = 0 for ð < 0 ) is defined as ð ð¢ð = ð¢ðð§âðâ
ð=0 . ðâ Transform of the sequence
ð¢ð i.e. ð ð¢ð is a function of ð§ and may be denoted by ð(ð§)
Remark:
ðâ Transform exists only when the infinite series ð¢ðð§âðâð=0 is convergent.
ð ð¢ð = ð¢ðð§âðâð=0 is termed as one sided transform and for two sided ð â
transform ð ð¢ð = ð¢ðð§âðâð=ââ
Results on ðâ Transforms of standard sequences
1.
ð ðð = ððð§âðâð=0
= 1 +ð
ð§+
ð2
ð§2+
ð3
ð§3+ ⯠+
ðð
ð§ð+ â¯
=1
1âð
ð§
, ð
ð§ < 1
⎠ð ðð =ð§
ð§âð ,
ð
ð§ < 1
2.
ð 1 =ð§
ð§â1 Putting ð = 1 in Result 1
3.
ð (â1)ð =ð§
ð§+1 Putting ð = â1 in Result 1
4.
ð ð = ðð§âðâð=0 = ð ð§âðâ
ð=0
= ð 1 +1
ð§+
1
ð§2+
1
ð§3+ ⯠+
1
ð§ð+ â¯
⎠ð ð =ðð§
ð§â1
5. Recurrence formula for ðð:
ð ðð =ð
ð â ð
ð ð =ð
ð â ð
ð (âð)ð =ð
ð + ð
ð ð =ðð
ð â ð
ð ðð = âðð
ð ðð ððâð
Page | 2
ð ðð = ððð§âðâð=0 , ð is a positive integer âŠâ
ð ððâ1 = ððâ1ð§âðâð=0 ⊠â¡
Differentiating â¡ w.r.t. ð§, we get
ð
ðð§ð ððâ1 = ððâ1(âð)ð§âðâ1â
ð=0
= âð§â1 ððð§âðâð=0
âð
ðð§ð ððâ1 = âð§â1ð ðð usingâ
â ð ðð = âð§ð
ðð§ð ððâ1
6. Multiplication by ð:
ð ðð¢ð = ðð¢ðð§âðâð=0
= âð§ ð¢ð(âð)ð§âðâ1âð=0
= âð§ ð¢ðð
ðð§ð§âðâ
ð=0
= âð§ ð
ðð§(ð¢ðð§âðâ
ð=0 )
= âð§ð
ðð§ ð¢ðð§âðâ
ð=0
= âð§ð
ðð§ð ð¢ð
7.
ð ð = âð§ð
ðð§ð ð0 using Recurrence Result 5 or 6
= âð§ð
ðð§ð 1
= âð§ð
ðð§
ð§
ð§â1 using result 2
â ð ð =ð
ðâð ð
8.
ð ð2 = âð§ð
ðð§ð ð using Recurrence Result 5 or 6
= âð§ð
ðð§
ð
ðâð ð using Result 7
â ð ð2 =ð§2+ð§
ð§â1 3
9. , ð ð = ð, ð < 0ð, ð ⥠ð
is Unit step sequence
ð ð¢(ð) = ð¢ ð ð§âðâð=0 = 1ð§âðâ
ð=0
= 1 +1
ð§+
1
ð§2+
1
ð§3+ ⯠+
1
ð§ð+ â¯
â ð ð¢(ð) =ð
ðâð
ð ð =ð
ð â ð ð
ð ðð =ðð + ð
ð â ð ð
ð ð ð = ð, ð < 0ð, ð ⥠ð
=ð
ð â ð
ð ððð = âðð
ð ðð ðð
Page | 3
10. ð¹ ð = ð, ð = ðð, ð â ð
is Unit impulse sequence
ð ð¿ ð = ð¿ ð ð§âðâð=0
= 1 + 0 + 0 + â¯
â ð ð¿ ð = 1
4.2 Properties of Z-Transforms
1. Linearity: ð ððð + ððð = ðð ðð + ðð ðð
Proof: ð ðð¢ð + ðð£ð = ðð¢ð + ðð£ð ð§âðâð=0
= ð ð¢ðð§âð + ð ð£ðð§âðâð=0
âð=0
= ðð ð¢ð + ðð ð£ð
2. Change of scale (or Damping rule):
If ð ðð â¡ ðŒ(ð), then ð ðâððð â¡ ðŒ(ðð) and ð ðððð â¡ ðŒ ð
ð
Proof: ð ðâðð¢ð = ðâðð¢ðð§âðâð=0
= ð¢ð(ðð§)âð â¡ ð(ðð§) âð=0
Similarly ð ððð¢ð â¡ ð ð§
ð
Results from application of Damping rule
i.
Proof: ð ð =ð§
ð§â1 2â¡ ð(ð§) say
⎠ð ððð â¡ ð ð§
ð =
ð§
ð
ð§
ðâ1
2 =ðð§
ð§âð 2
ii.
Proof: ð ð2 =ð§2+ð§
ð§â1 3â¡ ð(ð§) say
⎠ð ððð2 â¡ ð ð§
ð =
ð§
ð
2+
ð§
ð
ð§
ð â1
3 =ð(ð§2+ðð§)
ð§âð 3
iii.
Proof: We have ð ðâððð = ð ððð âð
= ð ððð âð
. 1
Now ð 1 =ð§
ð§â1
⎠ð ððð âð
. 1 =ð§ð ðð
ð§ð ðð â1 âµ ð ðâðð¢ð â¡ ð(ðð§)
=ð§
ð§âðâðð
=ð§(ð§âð ðð )
ð§âðâðð ð§âð ðð
ð ð¹ ð = ð, ð = ðð, ð â ð
= ð
ð ððð =ðð
ð â ð ð
ð ðððð =ððð + ððð
ð â ð ð
ð ððšð¬ððœ =ð ðâððšð¬ ðœ
ððâððððšð¬ ðœ+ð , ð ð¬ð¢ð§ ððœ =
ð ð¬ð¢ð§ ðœ
ððâððððšð¬ ðœ+ð
Page | 4
=ð§ (ð§âððð ð âðð ððð )
ð§2âð§ ð ðð +ðâðð +1 âµ ððð = ððð ð + ðð ððð
=ð§ (ð§âððð ð âðð ððð
ð§2â2ð§ððð ð +1 âµ ððð ð =
ð ðð +ðâðð
2
⎠ð ðâððð =ð§(ð§âððð ð )
ð§2â2ð§ cos ð+1â ð
ð§ sin ð
ð§2â2ð§ cos ð+1
â ð cos ðð â ð sin ðð =ð§(ð§âððð ð )
ð§2â2ð§ cos ð+1â ð
ð§ sin ð
ð§2â2ð§ cos ð+1
⎠ð cos ðð =ð§ ð§âcos ð
ð§2â2ð§ cos ð+1 âŠâ¢
and ð sin ðð =ð§ sin ð
ð§2â2ð§ cos ð+1 âŠâ£
iv.
By Damping rule, replacing ð§ by ð§
ð in â¢and â£, we get
ð ðð cos ðð =ð§ ð§âðcos Ξ
ð§2â2ðð§ cos ð+ð2 and ð ðð sin ðð =
ðð§ sin ð
ð§2â2ð§ cos ð+ð2
3. Right Shifting Property
For ð ⥠ð, ð ððâð = ðâðð ðð , ð is positive integer
Proof: ð ð¢ðâð = ð¢ðâðð§âðâð=0
= ð¢âðð§0 + ð¢1âðð§â1 + ⯠+ð¢â1ð§âð+1 + ð¢0ð§âð + ð¢1ð§â(ð+1) + ð¢2ð§â(ð+2) + â¯
= 0 + ð¢0ð§âð + ð¢1ð§â(ð+1) + ð¢2ð§â(ð+2) + ⯠ⵠð¢ð = 0 for ð < 0
= ð¢ðâðð§âðâðâð=0
= ð¢ðð§âðâðâð=0
= ð§âð ð¢ðð§âðâð=0
= ð§âð ð¢ðð§âðâð=0
= ð§âðð ð¢ð
4. Left Shifting Property
If ð is a positive integer ð ðð+ð = ðð ð ðð â ðð âðð
ðâ
ðð
ððâ ⯠â
ððâð
ððâð
Proof: ð ð¢ð+ð = ð¢ð+ðð§âðâð=0
= ð§ð ð¢ð+ðð§â(ð+ð)âð=0
= ð§ð ð¢ðð§âð + ð¢1+ðð§â(1+ð) + ð¢2+ðð§â(2+ð) + â¯
= ð§ð ð¢0 + ð¢1ð§â1 + ð¢2ð§â2 + ⯠+ ð¢ðâ1ð§â(ðâ1) + ð¢ðð§âð + â¯
âð§ð ð¢0 + ð¢1ð§â1 + ð¢2ð§â2 + ⯠+ ð¢ðâ1ð§â(ðâ1)
= ð§ð ð¢ðð§âðâð=0 â ð¢ðð§âððâ1
ð=0
= ð§ð ð¢ðð§âðâð=0 â ð¢ðð§âððâ1
ð=0
= ð§ð ð ð¢ð â ð¢0 âð¢1
ð§â
ð¢2
ð§2â ⯠â
ð¢ðâ1
ð§ðâ1
In particular for ð = 1,2,3
ð ð¢ð+1 = ð§ ð ð¢ð â ð¢0
ð ð¢ð+2 = ð§2 ð ð¢ð â ð¢0 âð¢1
ð§
ð ððððšð¬ððœ =ð ðâðððšð¬ ð
ððâðððððšð¬ ðœ+ðð , ð ðð ð¬ð¢ð§ ððœ =
ðð ð¬ð¢ð§ðœ
ððâððððšð¬ ðœ+ðð
Page | 5
ð ð¢ð+3 = ð§3 ð ð¢ð â ð¢0 âð¢1
ð§â
ð¢2
ð§2
5. Initial Value theorem:
If ð ðð = ðŒ(ð), then ðð = ð¥ð¢ðŠ ðââ
ðŒ(ð)
ðð = ð¥ð¢ðŠ ðââ
ð ðŒ ð âðð
ðð = ð¥ð¢ðŠ ðââ
ðð ðŒ ð âðð âðð
ð
â®
Proof: By definition ð ð§ = ð ð¢ð = ð¢ðð§âðâð=0
â ð ð§ = ð¢0 +ð¢1
ð§+
ð¢2
ð§2+
ð¢3
ð§3+ ⯠âŠâ€
⎠ð¢0 = lim ð§ââ
ð ð§ = limð§ââ
ð¢0 +ð¢1
ð§+
ð¢2
ð§2+
ð¢3
ð§3+ â¯
= ð¢0 + 0 + 0 + 0 + ⯠= ð¢0
Again fromâ€, we get
ð ð§ âð¢0 =ð¢1
ð§+
ð¢2
ð§2+
ð¢3
ð§3+ â¯
â ð§ ð ð§ âð¢0 = ð¢1 +ð¢2
ð§+
ð¢3
ð§2+ â¯
â limð§ââ
ð§ ð ð§ âð¢0 = limð§ââ
ð¢1 +ð¢2
ð§+
ð¢3
ð§2+ ⯠= ð¢1
Similarly ð¢2 = lim ð§ââ
ð§2 ð ð§ âð¢0 âð¢1
ð§
Note: Initial value theorem may be used to determine the sequence ð¢ð from the given
function ð ð§
6. Final Value theorem:
If ð ðð = ðŒ(ð), then ð¥ð¢ðŠ ðââ
ðð = ð¥ð¢ðŠ ðâð
(ð â ð)ðŒ(ð)
Proof: ð ð¢ð+1 â ð¢ð = ð¢ð+1 â ð¢ð ð§âðâð=0
â ð ð¢ð+1 â ð ð¢ð = ð¢ð+1 â ð¢ð ð§âðâð=0
â ð§ ð ð¢ð â ð¢0 â ð ð¢ð = ð¢ð+1 â ð¢ð ð§âðâð=0
By using left shifting property for ð = 1
â ð§ â 1 ð ð¢ð â ð¢0 = ð¢ð+1 â ð¢ð ð§âðâð=0
or ð§ â 1 ð(ð§) â ð¢0 = ð¢ð+1 â ð¢ð ð§âðâð=0 âµ ð ð¢ð = ð(ð§)
Taking limits ð§ â 1 on both sides
lim ð§â1
ð§ â 1 ð ð§ â ð¢0 = ð¢ð+1 â ð¢ð âð=0
or lim ð§â1
ð§ â 1 ð ð§ â ð¢0 = lim ðââ
ð¢1 â ð¢0 + ð¢2 â ð¢1 + ⯠+ ð¢ð+1 â ð¢ð
= lim ðââ
ð¢ð+1 â ð¢0
â lim ð§â1
ð§ â 1 ð ð§ = ð¢â
or lim ð§â1
ð§ â 1 ð ð§ = lim ðââ
ð¢ð
Note: Initial value and final value theorems determine the value of ð¢ð for ð = 0 and
for ð ⶠâ from the given function ð ð§ .
7. Convolution theorem
Convolution of two sequences ð¢ð and ð£ð is defined as ð¢ð â ð£ð = ð¢ðð£ðâððð=0
Convolution theorem for ð-transforms states that
If ðŒ ð = ð ðð and ðœ ð = ð ðð , then ð ðð â ðð = ðŒ ð . ðœ ð
Proof: ð ð§ . ð ð§ = ð ð¢ð . ð ð£ð = ð¢ðð§âðâ
ð=0 . ð£ðð§âðâð=0
Page | 6
= ð¢0 +ð¢1
ð§+
ð¢2
ð§2+ ⯠+
ð¢ð
ð§ð+ ⯠. ð£0 +
ð£1
ð§+
ð£2
ð§2+ ⯠+
ð£ð
ð§ð+ â¯
= ð¢0ð£0 + ð¢0ð£1 + ð¢1ð£0 ð§â1 + ð¢0ð£2 + ð¢1ð£1 + ð¢2ð£0 ð§â2 + â¯
= ð¢0ð£ð + ð¢1ð£ðâ1 + ð¢2ð£ðâ2 + ⯠+ ð¢ðð£0 ð§âðâð=0
= ð¢ðð£ðâððð=0 ð§âðâ
ð=0
â ð ð§ . ð ð§ = ð ð¢ðð£ðâððð=0 âµ ð¢ðð§âð = ð ð¢ð â
ð=0
â ð ð§ . ð ð§ = ð ð¢ð â ð£ð âµ ð¢ð â ð£ð = ð¢ðð£ðâððð=0
Example1 Find the ðâtransform of 2ð + 3 sinðð
4â 5ð4
Solution: By linearity property
ð 2ð + 3 sinðð
4â 5ð4 = 2ð ð + 3ð sin
ðð
4 â 5ð ð4
= 2ð ð + 3ð sinðð
4 â 5ð4ð 1
=2ð§
ð§â1 2+
3ð§ ðððð
4
ð§2â2ð§ððð ð
4+1
â5ð4ð§
ð§â1
âµ ð ð =ð§
ð§â1 2 , ð sin ðð =
ð§ sin ð
ð§2â2ð§ cos ð+1 , ð 1 =
ð§
ð§â1
⎠ð 2ð + 3 sinðð
4â 5ð4 =
2ð§
ð§â1 2+
3ð§
2
ð§2â 2 ð§+1â
5ð4ð§
ð§â1
Example2 Find the ðâtransform of the sequence 4, 8, 16, 32, âŠ
Solution: ð¢ð = 2ð+2 , ð = 0,1,2,3 âŠ.
ð 2ð+2 = ð 222ð
= 4 ð 2ð =4ð§
ð§â2 ,
2
ð§ < 1 âµ ð ðð =
ð§
ð§âð ,
ð
ð§ < 1
Example3 Find the ðâtransform of (ð + 1)2
Solution: ð (ð + 1)2 = ð ð2 + 2ð + 1 = ð ð2 + 2ð ð + ð 1
=ð§2+ ð§
(ð§â1)3+
2ð§
(ð§â1)2+
ð§
ð§â1
âµ ð ð2 =ð§2+ ð§
(ð§â1)3, ð ð =
ð§
ð§â1 2, ð 1 =
ð§
ð§â1
Example4 Find the ðâtransform of
i. ð cos ðð ii. sin ðð
2+
ð
4 iii. ðâ2ð iv. ð ð
Solution: i. ð cos ðð =ð§ (ð§âððð ð )
ð§2â2ð§ððð ð +1
⎠ð ðcos ðð = âð§ð
ðð§
ð§ ð§âððð ð
ð§2â2ð§ððð ð +1
âµ ð ðð¢ð = âð§ð
ðð§ð ð¢ð
= âð§ âð§2ððð ð +2ð§âððð ð
ð§2â2ð§ððð ð +1 2
=ð§3ððð ð â2ð§2+ ð§ððð ð
ð§2â2ð§ððð ð +1 2
ii. ð sin ðð
2+
ð
4 = ð sin
ðð
2cos
ð
4+ cos
ðð
2sin
ð
4
= cosð
4ð sin
ðð
2 + sin
ð
4ð cos
ðð
2
Page | 7
=1
2
ð§ sinð
2
ð§2â2ð§ cosð
2+1
+ð§ ð§âððð
ð
2
ð§2â2ð§ ððð ð
2+1
âµ ð sin ðð =ð§ sin ð
ð§2â2ð§ cos ð+1 , ð cos ðð =
ð§ ð§âcos ð
ð§2â2ð§ cos ð+1
=1
2
ð§
ð§2+1+
ð§2
ð§2+1 =
1
2 ð§+ð§2
ð§2+1
iii. ð ðâ2ð = ð ðâ2 ð
=ð§
ð§âðâ2 âµ ð ðð = ð§
ð§âð
=ð§ð2
ð§ð2â1
iv. ð ð = ðâð , ð < 0ðð , ð ⥠0
Taking two sided ðâ transform: ð ð ð = ð ð ð§âðâð=ââ
⎠ð ð ð = ðâðð§âð + ððð§âðâ0
â1ââ
= ⊠+ð3ð§3 + ð2ð§2 + ðð§ + 1 +ð
ð§+
ð2
ð§2+
ð3
ð§3+ â¯
=ðð§
1âðð§+
1
1âðð§
, ðð§ < 1 and ð
ð§ < 1
=ðð§
1âðð§+ ð§
ð§âð , ð§ <
1
ð and ð < ð§
=ð§ 1âðð§
1âðð§ ð§âð , ð < ð§ <
1
ð
â ð ð ð =ð§
ð§âð , ð < ð§ <
1
ð
Example5 Find the ðâtransform of ð¢ð = 2ð , ð < 03ð , ð ⥠0
Taking two sided ðâ transform: ð ð¢ð = ð¢ðð§âðâð=ââ
⎠ð ð¢ð = 2ðð§âð + 3ðð§âðâ0
â1ââ
= ⊠+ð§3
23+
ð§2
22+
ð§
2 + 1 +
3
ð§+
32
ð§2+
33
ð§3+ â¯
=ð§2
1âð§2
+1
1â3ð§
, ð§
2 < 1 and
3
ð§ < 1
=ð§
2âð§+
ð§
ð§â3 , ð§ < 2 and 3 < ð§
=ð§ ð§â3 +ð§ 2âð§
2âð§ ð§â3 , 3 < ð§ < 2
â ð ð¢ð =ð§
ð§2â5ð§+6 , 3 < ð§ < 2
⎠ðâtransform does not exist for ð¢ð = 2ð , ð < 03ð , ð ⥠0
as the set 3 < ð§ < 2 is infeasible.
Example6 Find the ðâtransform of
i. ðð¶ð 0 †ð †ð ii. ð+ðð¶ð iii. 1
ð+ð ! iv.
1
ðâð !
Solution: i. ð ðð¶ð = ðð¶ð ð§âððð=0
= 1 + ðð¶1 ð§â1 + ðð¶2 ð§â2 + ⯠+ ðð¶ð ð§âð
= 1 + ð§â1 ð
ii. ð ð+ðð¶ð = ð+ðð¶ð ð§âðâð=0
= 1 + ð+1ð¶1 ð§â1 + ð+2ð¶2 ð§â2 + ð+3ð¶3 ð§â3 + â¯
= 1 + (ð + 1) ð§â1 +(ð+2)(ð+1)
2! ð§â2 +
(ð+3)(ð+2)(ð+1)
3! ð§â3 + â¯
Page | 8
= 1 + âð â 1 (â ð§â1) + âðâ1 âðâ2
2! âð§â1 â2
+(âðâ1)(âðâ2)(âðâ3)
3! âð§â1 â3 + â¯
= 1 â ð§â1 âðâ1
Example 7 Find the ðâtransform of
i. 1
ð ! ii.
1
ð+ð ! iii.
1
ðâð !
i. ð 1
ð ! =
1
ð ! ð§âðâ
ð=0 = 1 +1
1! ð§â1 +
1
2! ð§â2 +
1
3! ð§â3 + â¯
= 1 +1
1!
1
ð§+
1
2!
1
ð§2+
1
3!
1
ð§3+ ⯠= ð
1
ð§
â ð 1
ð ! = ð
1
ð§
ii. ð 1
ð+ð ! =
1
ð+ð ! ð§âðâ
ð=0
Now ð 1
ð ! = ð
1
ð§
Also from left shifting property ð ð¢ð+ð = ð§ð ð ð¢ð â ð¢0 âð¢1
ð§â
ð¢2
ð§2â ⯠â
ð¢ðâ1
ð§ðâ1
⎠ð 1
ð+ð ! = ð§ð ð
1
ð§ â 1 â1
ð§â
1
2! ð§2â ⯠â
1
ðâ1 !ð§ðâ1
In particular 1
ð+1 ! = ð§1 ð
1
ð§ â 1
1
ð+2 ! = ð§2 ð
1
ð§ â 1 â1
ð§
â®
iii. ð 1
ðâð ! =
1
ðâð ! ð§âðâ
ð=0
Now ð 1
ð ! = ð
1
ð§
Also from right shifting property, ð ð¢ðâð = ð§âðð ð¢ð , ð is positive integer
⎠ð 1
ðâð ! = ð§âðð
1
ð§
In particular 1
ðâ1 ! = ð§â1ð
1
ð§
1
ð+2 ! = ð§â2ð
1
ð§
â®
Example8 Find ð ð¢ð+2 if ð ð¢ð =ð§
ð§â1+
ð§
ð§2+1
Solution: Given ð ð¢ð = ð ð§ =ð§
ð§â1+
ð§
ð§2+1
From left shifting property ð ð¢ð+2 = ð§2 ð ð¢ð â ð¢0 âð¢1
ð§ âŠâ
Now from initial value theorem ð¢0 = lim ð§ââ
ð ð§
= lim ð§ââ
ð§
ð§â1+
ð§
ð§2+1
Page | 9
= lim ð§ââ
1
1â1
ð§
+ 1
ð§
1+1
ð§2
= 1 + 0
⎠ð¢0 = 1 âŠâ¡
Also from initial value theorem ð¢1 = lim ð§ââ
ð§ ð ð§ âð¢0
= lim ð§ââ
ð§ ð§
ð§â1+
ð§
ð§2+1â 1
= lim ð§ââ
ð§ 2ð§2âð§+1
(ð§2+1)(ð§â1) = 2
⎠ð¢1 = 2 ⊠â¢
Using â¡and ⢠in â , we get ð ð¢ð+2 = ð§2 ð§
ð§â1+
ð§
ð§2+1â 1 â
2
ð§
â ð ð¢ð+2 =ð§ ð§2âð§+2
ð§â1 ð§2+1
Example9 Verify convolution theorem for ð¢ð = ð and ð£ð = 1
Solution: Convolution theorem states that ð ð¢ð â ð£ð = ð ð§ . ð ð§
We know that ð¢ð â ð£ð = ð¢ðð£ðâððð=0
= ððð=0 . 1
= 0 + 1 + 2 + 3 + ⯠+ ð =ð ð+1
2
â ð ð¢ð â ð£ð = ð ð+1
2
âð=0 ð§âð =
1
2 ð2â
ð=0 ð§âð + ðâð=0 ð§âð
=1
2 ð ð2 + ð ð
=1
2 ð§ ð§+1
ð§â1 3+
ð§
ð§â1 2 =
1
2 ð§ ð§+1 +ð§(ð§â1)
ð§â1 3 =
1
2
2ð§2
ð§â1 3
⎠ð ð¢ð â ð£ð =ð§2
ð§â1 3 âŠâ
Also ð ð§ = ð ð =ð
ðâð ð and ð ð§ = ð 1 =
ð§
ð§â1
â ð ð§ . ð ð§ =ð§2
ð§â1 3 ⊠â¡
From â and â¡ ð ð¢ð â ð£ð = ð ð§ . ð ð§
Example10 If ð¢ð = ð¿ ð â ð¿ ð â 1 , ð£ð = 2ð¿ ð + ð¿ ð â 1 , Find the ð -transform of
their convolution.
Solution: ð ð§ = ð ð¿ ð â ð¿ ð â 1 , ð ð§ = ð 2ð¿ ð + ð¿ ð â 1
Now ð ð¿ ð = 1 and ð ð¿ ð â 1 = ð§â1 âµ ð ð¢ðâð = ð§âðð ð¢ð
⎠ð ð§ = 1 â ð§â1 and ð ð§ = 2 + ð§â1
Also ð ð¢ð â ð£ð = ð ð§ . ð ð§
â ð ð¢ð â ð£ð = 1 â ð§â1 2 + ð§â1 = 2 â ð§â1 â ð§â2
4.3 Inverse Z-Transform
Given a function ð ð§ , we can find the sequence ð¢ð by one of the following methods
Inspection (Direct inversion)
Direct division
Partial fractions
Page | 10
Residues (Inverse integral)
Power-Series
Convolution theorem
4.3.1 Inspection (Direct inversion) method
Sometimes by observing the coefficients in the given series ð ð§ , it is possible to find
the sequence ð¢ð as illustrated in the given examples.
Example 11 Find ð¢ð if ð ð§ = 1 +1
2ð§â1 +
1
4ð§â2 +
1
8ð§â3 + â¯
Solution: Given that ð ð§ = 1
2
ð
ð§âðâð=0 ⊠â
Also by the definition of Z-transform ð(ð§) = ð¢ðð§âð âð=0 âŠâ¡
.Comparing â and â¡, we get ð¢ð = 1
2
ð
Example12 Find ð¢ð if ð ð§ =ð§3
ð§â1 3
Solution: ð ð§ =ð§3
ð§â1 3=
ð§â1
ð§
â3
= 1 â1
ð§
â3
⎠ð ð§ = 1 + 3
ð§+
6
ð§2+
10
ð§3+ â¯
âµ 1 â ð¥ âð = 1 + ðð¥ +ð ð+1
2!ð¥2 +
ð(ð+1)(ð+2)
3!ð¥3 + â¯
â ð ð§ = ð+1 (ð+2)
2ð§âðâ
ð=0
Comparing with ð(ð§) = ð¢ðð§âðâð=0 , we get ð¢ð =
ð+1 (ð+2)
2
Example13 Find inverse ð-transform of 3 +2ð§
ð§â1â
ð§
2ð§â1
Solution: Given that ð ð§ = 3 +2ð§
ð§â1â
ð§
2ð§â1
⎠ð¢ð = 3ðâ1 1 + 2ðâ1 ð§
ð§â1 â
1
2ðâ1
ð§
ð§â1
2
â ð¢ð = 3ð¿ ð + 2ð¢ ð â1
2
1
2
ð
= 3ð¿ ð + 2ð¢ ð â 1
2
ð+1
âµ ðâ1 1 = ð¿ ð , ðâ1 ð§
ð§â1 = ð¢(ð) , ðâ1
ð§
ð§âð = ðð where ð¿ ð and
ð¢(ð) are unit impulse and unit step sequences respectively.
Example14 Find inverse ð-transform of 2ð§â2 âð§â3
ð§â1+
2ð§â5
2ð§â1
Solution: Given that ð ð§ = 2ð§â2 âð§â3
ð§â1+
2ð§â5
2ð§â1
⎠ð¢ð = 2ðâ1 ð§â2. 1 â ðâ1 ð§â4.ð§
ð§â1 + ðâ1 ð§â6 ð§
ð§â1
2
â ð¢ð = 2ð¿ ð â 2 â ð¢ ð â 4 + 1
2
ðâ6
ð¢ ð â 6
âµ From Right shifting property ðâ1 ð§âðð(ð) = ð¢ðâð
4.3.2 Direct division method
Direct division is one of the simplest methods for finding inverse ð -transform and can
be used for almost every type of expression given in fractional form.
Example15 Find the inverse ð-transform of ð§
ð§2â3ð§+2
Page | 11
Solution: Given that ð ð§ =ð§
ð§2â3ð§+2
By actual division, we get
ð§â1 + 3ð§â2 + 7ð§â3
ð§2 â 3ð§ + 2 ð§
ð§ â 3 + 2ð§â1
3 â 2ð§â1 3 â 9ð§â1 + 6ð§â2
7ð§â1 â 6ð§â2
7ð§â1 â 21ð§â2 + 14ð§â3
15ð§â2 â 14ð§â3
â® â ð ð§ = ð§â1 + 3ð§â2 + 7ð§â3 + â¯
= 2ðâ 1 ð§âðâð=0
⎠ð¢ð = 2ð â 1
Example16 Find the inverse ð -transform of 4ð§2+2ð§
2ð§2â3ð§+1
Solution: Given that ð ð§ =4ð§2+2ð§
2ð§2â3ð§+1 , by actual division, we get
2 + 4ð§â1 + 5ð§â2
2ð§2 â 3ð§ + 1 4ð§2 + 2ð§
4ð§2 â 6ð§ + 2
8ð§ â 2 8ð§ â 12 + 4ð§â1
10 â 4ð§â1
10 â 15ð§â1 + 5ð§â2
11ð§â1 â 5ð§â2
â® â ð ð§ = 2 + 4ð§â1 + 5ð§â2 + â¯
= 6 â 22âð ð§âðâð=0
⎠ð¢ð = 6 â 22âð
⎠ð¢ð = 6â4 1
2
ð
4.3.3 Partial fractions method
Partial fractions method can be used only if order of expression in the numerator is less
than or equal to that in the denominator. If order of expression in the numerator is
greater, then the fraction may be brought to desired form by direct division. Partial
fractions are formed of the expression ð(ð§)
ð§ as demonstrated in the examples below.
Example17 Find the inverse ð -transform of ð§
6ð§2â5ð§+1
Solution: Given that ð ð§ =ð§
6ð§2â5ð§+1
⎠ð(ð§)
ð§=
1
6ð§2â5ð§+1=
1
2ð§â1 (3ð§â1)
âð(ð§)
ð§=
2
2ð§â1â
3
3ð§â1
â ð ð§ =ð§
ð§â 1
2
âð§
ð§â 1
3
⎠ð¢ð = 1
2
ð
â 1
3
ð
âµ ð ðð =ð§
ð§âð or ðâ1
ð§
ð§âð = ðð
Page | 12
i.e. ð¢ð= 2âðâ 3âð
Example18 Find the inverse ð -transform of 4ð§2+2ð§
2ð§2â3ð§+1
Solution: Given that ð ð§ =4ð§2â2ð§
2ð§2â3ð§+1=
2ð§(2ð§â1)
2ð§â1 (ð§â1)
⎠ð(ð§)
ð§=
2(2ð§+1)
2ð§â1 (ð§â1)
By partial fractions, we get
ð(ð§)
ð§=
â8
2ð§â1+
6
ð§â1
â ð ð§ =â8ð§
2ð§â1+
6ð§
ð§â1
⎠ð¢ð = â4ðâ1 ð§
ð§â1
2
+ 6ðâ1 ð§
ð§â1
â ð¢ð = â4 1
2
ð
+ 6 1 ð âµ ðâ1 ð§
ð§âð = ðð
i.e. ð¢ð= â4 1
2
ð
+ 6
Example19 Find the inverse ð -transform of 1
1âð§â1 (2âð§â1)
Solution: Given that ð ð§ = 1
1âð§â1 (2âð§â1)
Multiplying and dividing by ð§2, we get
ð ð§ =ð§2
ð§ 1âð§â1 ð§(2âð§â1)=
ð§2
ð§â1 (2ð§â1)
⎠ð(ð§)
ð§=
ð§
ð§â1 (2ð§â1)
By partial fractions, we get
ð(ð§)
ð§=
1
ð§â1 â
1
(2ð§â1)
â ð ð§ =ð§
ð§â1 â
ð§
(2ð§â1)
⎠ð¢ð = ðâ1 ð§
ð§â1 â
1
2ðâ1
ð§
ð§â1
2
â ð¢ð = 1 ð â1
2
1
2
ð
âµ ðâ1 ð§
ð§âð = ðð
i.e. ð¢ð= 1 â 1
2
ð+1
Example20 Find the inverse ð -transform of 4ð§2â2ð§
ð§3â5ð§2+8ð§â4
Solution: Given that ð ð§ =4ð§2â2ð§
ð§3â5ð§2+8ð§â4=
2ð§(2ð§â1)
ð§â1 (ð§â2)2
⎠ð(ð§)
ð§=
2(2ð§â1)
ð§â1 (ð§â2)2
By partial fractions, we get
ð(ð§)
ð§=
2
ð§â1â
2
ð§â2+
6
ð§â2 2
â ð ð§ =2ð§
ð§â1â
2ð§
ð§â2+
6ð§
ð§â2 2
Page | 13
⎠ð¢ð = 2ðâ1 ð§
ð§â1 â 2ðâ1
ð§
ð§â2 + 3ðâ1
2ð§
ð§â2 2
â ð¢ð = 2 1 ðâ 2 2 ð + 3ð 2 ð âµ ðâ1 ð§
ð§âð = ðð and ðâ1
ðð§
ð§âð 2 = ððð
i.e. ð¢ð= 2â 2ð+1 + 3ð. 2ð
4.3.4 Method of residues (Inverse integral)
By using the theory of complex variables, it can be shown that the inverse ð-transform
is given by ð¢ð =1
2ðð ð ð§ ð§ðâ1ðð§ =ð
sum of residues of ð ð§
where c is the closed contour which contains all the insolated singularities of ð ð§ in
the region of convergence.
Method of residues is one of the most efficient methods and can be used to find the
inverse ð- transform where partial fractions are tedious to find.
Example21 Find the inverse z-transform of ð§
ð§2+7ð§+10
Solution: ð ð§ =ð§
ð§2+7ð§+10
Now ð¢ð =1
2ðð ð ð§ ð§ðâ1ðð§ð
â ð¢ð =1
2ðð
ð§
ð§2+7ð§+10ð§ðâ1ðð§
ð
=1
2ðð
ð§ð
ð§2+7ð§+10ðð§
ð
=1
2ðð
ð§ð
ð§+2 (ð§+5)ðð§
ð
There are two simple poles at ð§ = â2 and ð§ = â5
Residue at ð§ = â2 is given by limð§ââ2
ð§ + 2 ð§ð
ð§+2 (ð§+5)=
â2 ð
3
Residue at ð§ = â5 is given by limð§ââ5
ð§ + 5 ð§ð
ð§+2 (ð§+5)=
â5 ð
â3
⎠ð¢ð= sum of residues = â2
ð
3+
â5 ð
â3=
1
3 â 2
ðâ â 5
ð
Example22 Find the inverse z-transform of ð§2+ð§
ð§â1 (ð§2+1)
Solution: ð ð§ =ð§2+ð§
ð§â1 (ð§2+1)
Now ð¢ð =1
2ðð ð ð§ ð§ðâ1ðð§ð
â ð¢ð =1
2ðð
ð§2+ð§
ð§â1 (ð§2+1)ð§ðâ1ðð§
ð
=1
2ðð
ð§ð (ð§+1)
ð§â1 (ð§+ð)(ð§âð)ðð§
ð
There are three simple poles at ð§ = 1, ð§ = âð and ð§ = ð
Residue at ð§ = 1 is given by limð§â1
ð§ â 1 ð§ð (ð§+1)
ð§â1 (ð§+ð)(ð§âð)= 1
Residue at ð§ = âð is given by limð§ââð
ð§ + ð ð§ð (ð§+1)
ð§â1 (ð§+ð)(ð§âð)= â
1
2 âð ð
Residue at ð§ = ð is given by limð§âð
ð§ â ð ð§ð (ð§+1)
ð§â1 (ð§+ð)(ð§âð)= â
1
2ðð
⎠ð¢ð= sum of residues = 1 â1
2 âð ð â
1
2ðð = 1 â
1
2 âð ð + ðð
Page | 14
Example23 Find the inverse z-transform of ð§(ð§+1)
(ð§â1)3
Solution: ð ð§ =ð§(ð§+1)
(ð§â1)3
Now ð¢ð =1
2ðð ð ð§ ð§ðâ1ðð§ð
â ð¢ð =1
2ðð
ð§ð (ð§+1)
(ð§â1)3ðð§
ð
Here ð§ = 1 is a pole of order 3
Residue at ð§ = 1 is given by 1
2!limð§â1
ð2
ðð§ 2
(ð§â1)3ð§ð (ð§+1)
(ð§â1)3
=1
2!limð§â1
ð2
ðð§ 2 ð§ð (ð§ + 1)
=1
2!limð§â1
ð
ðð§ (ð + 1)ð§ð + ðð§ðâ1
=1
2!limð§â1
(ð + 1)ðð§ðâ1 + ð(ð â 1)ð§ðâ2
= ð2 + ð + ð2 â ð = ð2
⎠ð¢ð = Sum of residues = ð2
4.3.5 Power series method
In this method, we find the inverse ð - transform by expanding ð ð§ in power series.
Example 24 Find ð¢ð if ð ð§ = logð§
ð§+1
Solution: Given ð ð§ = logð§
ð§+1= log
ð§+1
ð§
â1
= â logð§+1
ð§= â log 1 +
1
ð§
⎠ð ð§ = â log 1 + ðŠ Putting 1
ð§= ðŠ
= âðŠ +ðŠ2
2â
ðŠ3
3+
ðŠ4
4â â¯
âµ log 1 + ð¥ = ð¥ âð¥2
2+
ð¥3
3â
ð¥4
4+ â¯
â ð ð§ = â1
ð§+
1
2ð§2â
1
3ð§3+
1
4ð§4â ⯠ⵠðŠ =
1
ð§
â ð ð§ = 0 + (â1)ð
ðð§âðâ
ð=1
Comparing with ð(ð§) = ð¢ðð§âðâð=0 , we get
ð¢ð = 0 for ð = 0
(â1)ð
ð, ðð¡âððð€ðð ð
4.3.6 Convolution theorem method
Convolution theorem for ð-transforms states that:
If ð ð§ = ð ð¢ð and ð ð§ = ð ð£ð , then ð ð¢ð â ð£ð = ð ð§ . ð ð§
â ðâ1 ð ð§ . ð ð§ = ð¢ð â ð£ð
Example25 Find the inverse z-transform of ð§2
ð§â1 (2ð§â1) using convolution theorem.
Solution: Let ð ð§ = ð ð¢ð =ð§
ð§â1 and ð ð§ = ð ð£ð =
ð§
2ð§â1 =
1
2
ð§
ð§â1
2
Clearly ð¢ð = 1 ð and ð¢ð =1
2
1
2
ð
âµ ðâ1 ð§
ð§âð = ðð
Now by convolution theorem ðâ1 ð ð§ . ð ð§ = ð¢ð â ð£ð
Page | 15
â ðâ1 ð§2
ð§â1 (2ð§â1) = 1 ð â
1
2
ð+1
We know that ð¢ð â ð£ð = ð¢ðð£ðâððð=0
= 1 ð 1
2
ð+1âððð=0
= 1
2
ð+1
+ 1
2
ð
+ 1
2
ðâ1
+ ⯠+1
2
=1
2 1 +
1
2+
1
2
2
+ ⯠+ 1
2
ð
=1
2
1
1â1
2
1 â 1
2
ð+1
âµ ðð =ð
1âð 1 â ðð
=1
2 2 1 â
1
2
ð+1
= 1 â 1
2
ð+1
Exercise 4A
1. Find the ðâtransform of ð¢ð = 2ð , ð < 03ð , ð ⥠0
2. Find the ðâtransform of ð¢ð =1
ðâð !
3. Find the inverse ðâtransform of ð¢ð =2ð§
ð§â1 (ð§2+1)
4. Solve the difference equation ðŠð¥+2 + 4ðŠð¥+1 + 3ðŠð¥ = 3ð¥ , ðŠ0 = 0, ðŠ1 = 1
using ðâtransforms
Answers
1. 2ð§
ð§2â8ð§+15 , 3 < ð§ < 5
2. ð§âðð1
ð§
3. 1 â ððð ðð
2â ðð ðð
ðð
2
4. ðŠð¥ =1
243ð¥ â
5
12(â3)ð¥ +
3
8(â1)ð¥