![Lecture 16 Slide 1
PYKC 10-Mar-11 E2.5 Signals & Linear Systems
Lecture 16
More z-Transform
(Lathi 5.2,5.4-5.5)
Peter Cheung
Department of Electrical & Electronic Engineering
Imperial College London
URL: www.ee.imperial.ac.uk/pcheung/teaching/ee2_signals
E-mail: p.cheung@imperial.ac.uk
Lecture 16 Slide 2
PYKC 10-Mar-11 E2.5 Signals & Linear Systems
Shift Property of z-Transform
If
then
which is delay causal signal by 1 sample period.
If we delay x[n] first:
If we ADVANCE x[n] by 1 sample period:
L5.2 p508
Lecture 16 Slide 3
PYKC 10-Mar-11 E2.5 Signals Linear Systems
Convolution property of z-transform
If h[n] is the impulse response of a discrete-time LTI system, then
then
If
Then
That is: convolution in the time-domain is the same as multiplication in
the z-domain.
Therefore, we can derive the input-output relationship fo any LTI
systems in z-domain:
L5.2 p511
Lecture 16 Slide 4
PYKC 10-Mar-11 E2.5 Signals Linear Systems
More Properties of z-Transform
For all these cases, we assume:
Scaling Property:
Multiply by n property:
Time reversal property:
Initial value property:
L5.2 p512](https://image.slidesharecdn.com/lecture16-morez-transform-231005072924-fe055ece/85/Lecture-16-More-z-transform-pdf-1-320.jpg)
![Lecture 16 Slide 5
PYKC 10-Mar-11 E2.5 Signals Linear Systems
Summary of z-transform properties (1)
L5.2 p514
Lecture 16 Slide 6
PYKC 10-Mar-11 E2.5 Signals Linear Systems
Summary of z-transform properties (2)
L5.2 p514
Lecture 16 Slide 7
PYKC 10-Mar-11 E2.5 Signals Linear Systems
Discrete LTI System and Difference Equation
Consider a discrete time system where the input-output relation is
described by:
This is known as a difference equation, where current output is dependent
on current input x[n], and two previous inputs and outputs x[n-1], x[n-2],
y[n-1] and y[n-2].
Take z-transform on both sides and assume zero-state condition:
The transfer function of a general Nth order causal discrete LTI system is:
[ ] 5 [ 1] 6 [ 2] [ ] 3 [ 1] 5 [ 2]
y n y n y n x n x n x n
− − + − = + − + −
L5.4 p525
1 2 1 2
( ) 5 ( ) 6 ( ) ( ) 3 ( ) 5 ( )
Y z z Y z z Y z X z z X z z X z
− − − −
− + = + +
1 2 1 2
(1 5 6 ) ( ) (1 3 5 ) ( )
z z Y z z z X z
− − − −
⇒ − + = + +
1 2
1 2
( ) 1 3 5
( )
( ) 1 5 6
Y z z z
H z
X z z z
− −
− −
+ +
⇒ = =
− +
1 1
0 1 1
1 1
1 1
......
[ ]
1 ......
N N
N N
N N
N N
b b z b z b z
H z
a z a z a z
− − + −
−
− − + −
−
+ + + +
=
+ + + +
Lecture 16 Slide 8
PYKC 10-Mar-11 E2.5 Signals Linear Systems
Realization of LTI System – Direct Form I
The general transfer function H(z) can be realised using Direct Form I as
follows:
L5.4 p525
1 1
0 1 1
1 1
1 1
......
[ ] [ ] [ ] [ ]
1 ......
N N
N N
N N
N N
b b z b z b z
Y z H z X z X z
a z a z a z
− − + −
−
− − + −
−
+ + + +
= =
+ + + +
( )
1 1
0 1 1
1 1
1 1
1
...... [ ]
1 ......
N N
N N
N N
N N
b b z b z b z X z
a z a z a z
− − + −
−
− − + −
−
⎛ ⎞
= + + + +
⎜ ⎟
+ + + +
⎝ ⎠
1 1
1 1
1
( )
1 ...... N N
N N
W z
a z a z a z
− − + −
−
⎛ ⎞
= ⎜ ⎟
+ + + +
⎝ ⎠](https://image.slidesharecdn.com/lecture16-morez-transform-231005072924-fe055ece/85/Lecture-16-More-z-transform-pdf-2-320.jpg)
![Lecture 16 Slide 9
PYKC 10-Mar-11 E2.5 Signals Linear Systems
Realization of LTI System – Direct Form II
Or use Canonical Director Form II:
L5.4 p525
1 1
0 1 1
1 1
1 1
......
[ ] [ ] [ ] [ ]
1 ......
N N
N N
N N
N N
b b z b z b z
Y z H z X z X z
a z a z a z
− − + −
−
− − + −
−
+ + + +
= =
+ + + +
( )
1 1
0 1 1 1 1
1 1
1
...... [ ]
1 ......
N N
N N N N
N N
b b z b z b z X z
a z a z a z
− − + −
− − − + −
−
⎛ ⎞
= + + + + ⎜ ⎟
+ + + +
⎝ ⎠
( )
1 1
0 1 1
...... ( )
N N
N N
b b z b z b z W z
− − + −
−
= + + + +
Lecture 16 Slide 10
PYKC 10-Mar-11 E2.5 Signals Linear Systems
Realization of LTI System – Transposed Direct Form II
Or the transposed version:
L5.4 p525
1 1
0 1 1
1 1
1 1
......
[ ] [ ] [ ] [ ]
1 ......
N N
N N
N N
N N
b b z b z b z
Y z H z X z X z
a z a z a z
− − + −
−
− − + −
−
+ + + +
= =
+ + + +
( ) ( )
1 1 1 1
1 1 0 1 1
[ ] ...... [ ] ...... [ ]
N N N N
N N N N
Y z a z a z a z Y z b b z b z b z X z
− − + − − − + −
− −
+ + + + = + + + +
1
( [ ] [ ])
N N
b X z a Y z z−
−
( ) ( )
1 1 1 1
0 1 1 1 1
[ ] ...... [ ] ...... [ ]
N N N N
N N N N
Y z b b z b z b z X z a z a z a z Y z
− − + − − − + −
− −
= + + + + − + + +
1
1 1
( [ ] [ ]) ( [ ] [ ])
N N N N
b X z a Y z z b X z a Y z
−
− −
− + −
Lecture 16 Slide 11
PYKC 10-Mar-11 E2.5 Signals Linear Systems
Examples
Direct Form II
L5.4 p527
Transposed
1
1
2 2
[ ]
5 1 5
z
H z
z z
−
−
= =
+ +
1
1
4 28 4 28
[ ]
1 1
z z
H z
z z
−
−
+ +
= =
+ +
1
2 1 2
4 28 4 28
[ ]
6 5 1 6 5
z z
H z
z z z z
−
− −
+ +
= =
+ + + +
Lecture 16 Slide 12
PYKC 10-Mar-11 E2.5 Signals Linear Systems
Remember that for a continuous-time system, the system response to an
input ejωt is H(jω) ejωt.
The response to an input cos ωt is |H(jω)| cos (ωt + ∠ H(jω)).
Now, consider a discrete-time system with z-domain transfer function H[z].
Let z = ejΩ, the system response to an input ejΩn is H[ejΩ] ejΩn .
The response to an input cos Ωn is |H[ejΩ]| cos (cos Ωn + ∠ H[ejΩ]).
Frequency Response of Discrete-time Systems
L5.5 p531
Continuous-time
System H(jω)
j t
e ω
cos t
ω
( ) j t
H j e ω
ω
Discrete-time
System H [ejΩ]
j n
e Ω
cos n
Ω
[ ]
j j n
H e e
Ω Ω
[ ] cos( [ ])
j j
H e n H e
Ω Ω
Ω +∠
(.) continuous-time
[.] discrete-time
Ω = ωTS where TS is the sampling period.](https://image.slidesharecdn.com/lecture16-morez-transform-231005072924-fe055ece/85/Lecture-16-More-z-transform-pdf-3-320.jpg)
![Lecture 16 Slide 13
PYKC 10-Mar-11 E2.5 Signals Linear Systems
Frequency Response Example (1)
For a system specified by the following difference equation, find the
frequency response of the system.
Take z-transform on both sides to find the transfer function:
Therefore the frequency response is:
L5.5 p533
[ 1] 0.8 [ ] [ 1]
y n y n x n
+ − = +
[ ] 0.8 [ ] [ ]
zY z Y z zX z
− =
1
[ ] 1
[ ]
[ ] 0.8 1 0.8
Y z z
H z
X z z z−
⇒ = = =
− −
Lecture 16 Slide 14
PYKC 10-Mar-11 E2.5 Signals Linear Systems
Frequency Response Example (2)
Amplitude response
Phase response
L5.5 p533
Lecture 16 Slide 15
PYKC 10-Mar-11 E2.5 Signals Linear Systems
Since where T = 2π/ωs
we can map the s-plane to the z-plane as below:
Mapping from s-plane to z-plane
( )
sT j T T j T
z e e e e
σ ω σ ω
+
= = =
jω
Re( )
s σ
=
jω
−
s-plane z-plane
/ 2
s
jω
/ 2
s
jω
−
Re( )
z
Im( )
z
1
+
1
−
1
−
1
+
Ω = ωT
jω
Lecture 16 Slide 16
PYKC 10-Mar-11 E2.5 Signals Linear Systems
Since where T = 2π/ωs
we can map the s-plane to the z-plane as below:
Mapping from s-plane to z-plane
( )
sT j T T j T
z e e e e
σ ω σ ω
+
= = =
jω
Re( )
s σ
=
jω
−
s-plane z-plane
/ 2
s
jω
/ 2
s
jω
−
Re( )
z
Im( )
z
1
+
1
−
1
−
1
+
Ω = ωT](https://image.slidesharecdn.com/lecture16-morez-transform-231005072924-fe055ece/85/Lecture-16-More-z-transform-pdf-4-320.jpg)
Lecture 16 - More z-transform.pdf
- 1. Lecture 16 Slide 1 PYKC 10-Mar-11 E2.5 Signals & Linear Systems Lecture 16 More z-Transform (Lathi 5.2,5.4-5.5) Peter Cheung Department of Electrical & Electronic Engineering Imperial College London URL: www.ee.imperial.ac.uk/pcheung/teaching/ee2_signals E-mail: p.cheung@imperial.ac.uk Lecture 16 Slide 2 PYKC 10-Mar-11 E2.5 Signals & Linear Systems Shift Property of z-Transform If then which is delay causal signal by 1 sample period. If we delay x[n] first: If we ADVANCE x[n] by 1 sample period: L5.2 p508 Lecture 16 Slide 3 PYKC 10-Mar-11 E2.5 Signals Linear Systems Convolution property of z-transform If h[n] is the impulse response of a discrete-time LTI system, then then If Then That is: convolution in the time-domain is the same as multiplication in the z-domain. Therefore, we can derive the input-output relationship fo any LTI systems in z-domain: L5.2 p511 Lecture 16 Slide 4 PYKC 10-Mar-11 E2.5 Signals Linear Systems More Properties of z-Transform For all these cases, we assume: Scaling Property: Multiply by n property: Time reversal property: Initial value property: L5.2 p512
- 2. Lecture 16 Slide 5 PYKC 10-Mar-11 E2.5 Signals Linear Systems Summary of z-transform properties (1) L5.2 p514 Lecture 16 Slide 6 PYKC 10-Mar-11 E2.5 Signals Linear Systems Summary of z-transform properties (2) L5.2 p514 Lecture 16 Slide 7 PYKC 10-Mar-11 E2.5 Signals Linear Systems Discrete LTI System and Difference Equation Consider a discrete time system where the input-output relation is described by: This is known as a difference equation, where current output is dependent on current input x[n], and two previous inputs and outputs x[n-1], x[n-2], y[n-1] and y[n-2]. Take z-transform on both sides and assume zero-state condition: The transfer function of a general Nth order causal discrete LTI system is: [ ] 5 [ 1] 6 [ 2] [ ] 3 [ 1] 5 [ 2] y n y n y n x n x n x n − − + − = + − + − L5.4 p525 1 2 1 2 ( ) 5 ( ) 6 ( ) ( ) 3 ( ) 5 ( ) Y z z Y z z Y z X z z X z z X z − − − − − + = + + 1 2 1 2 (1 5 6 ) ( ) (1 3 5 ) ( ) z z Y z z z X z − − − − ⇒ − + = + + 1 2 1 2 ( ) 1 3 5 ( ) ( ) 1 5 6 Y z z z H z X z z z − − − − + + ⇒ = = − + 1 1 0 1 1 1 1 1 1 ...... [ ] 1 ...... N N N N N N N N b b z b z b z H z a z a z a z − − + − − − − + − − + + + + = + + + + Lecture 16 Slide 8 PYKC 10-Mar-11 E2.5 Signals Linear Systems Realization of LTI System – Direct Form I The general transfer function H(z) can be realised using Direct Form I as follows: L5.4 p525 1 1 0 1 1 1 1 1 1 ...... [ ] [ ] [ ] [ ] 1 ...... N N N N N N N N b b z b z b z Y z H z X z X z a z a z a z − − + − − − − + − − + + + + = = + + + + ( ) 1 1 0 1 1 1 1 1 1 1 ...... [ ] 1 ...... N N N N N N N N b b z b z b z X z a z a z a z − − + − − − − + − − ⎛ ⎞ = + + + + ⎜ ⎟ + + + + ⎝ ⎠ 1 1 1 1 1 ( ) 1 ...... N N N N W z a z a z a z − − + − − ⎛ ⎞ = ⎜ ⎟ + + + + ⎝ ⎠
- 3. Lecture 16 Slide 9 PYKC 10-Mar-11 E2.5 Signals Linear Systems Realization of LTI System – Direct Form II Or use Canonical Director Form II: L5.4 p525 1 1 0 1 1 1 1 1 1 ...... [ ] [ ] [ ] [ ] 1 ...... N N N N N N N N b b z b z b z Y z H z X z X z a z a z a z − − + − − − − + − − + + + + = = + + + + ( ) 1 1 0 1 1 1 1 1 1 1 ...... [ ] 1 ...... N N N N N N N N b b z b z b z X z a z a z a z − − + − − − − + − − ⎛ ⎞ = + + + + ⎜ ⎟ + + + + ⎝ ⎠ ( ) 1 1 0 1 1 ...... ( ) N N N N b b z b z b z W z − − + − − = + + + + Lecture 16 Slide 10 PYKC 10-Mar-11 E2.5 Signals Linear Systems Realization of LTI System – Transposed Direct Form II Or the transposed version: L5.4 p525 1 1 0 1 1 1 1 1 1 ...... [ ] [ ] [ ] [ ] 1 ...... N N N N N N N N b b z b z b z Y z H z X z X z a z a z a z − − + − − − − + − − + + + + = = + + + + ( ) ( ) 1 1 1 1 1 1 0 1 1 [ ] ...... [ ] ...... [ ] N N N N N N N N Y z a z a z a z Y z b b z b z b z X z − − + − − − + − − − + + + + = + + + + 1 ( [ ] [ ]) N N b X z a Y z z− − ( ) ( ) 1 1 1 1 0 1 1 1 1 [ ] ...... [ ] ...... [ ] N N N N N N N N Y z b b z b z b z X z a z a z a z Y z − − + − − − + − − − = + + + + − + + + 1 1 1 ( [ ] [ ]) ( [ ] [ ]) N N N N b X z a Y z z b X z a Y z − − − − + − Lecture 16 Slide 11 PYKC 10-Mar-11 E2.5 Signals Linear Systems Examples Direct Form II L5.4 p527 Transposed 1 1 2 2 [ ] 5 1 5 z H z z z − − = = + + 1 1 4 28 4 28 [ ] 1 1 z z H z z z − − + + = = + + 1 2 1 2 4 28 4 28 [ ] 6 5 1 6 5 z z H z z z z z − − − + + = = + + + + Lecture 16 Slide 12 PYKC 10-Mar-11 E2.5 Signals Linear Systems Remember that for a continuous-time system, the system response to an input ejωt is H(jω) ejωt. The response to an input cos ωt is |H(jω)| cos (ωt + ∠ H(jω)). Now, consider a discrete-time system with z-domain transfer function H[z]. Let z = ejΩ, the system response to an input ejΩn is H[ejΩ] ejΩn . The response to an input cos Ωn is |H[ejΩ]| cos (cos Ωn + ∠ H[ejΩ]). Frequency Response of Discrete-time Systems L5.5 p531 Continuous-time System H(jω) j t e ω cos t ω ( ) j t H j e ω ω Discrete-time System H [ejΩ] j n e Ω cos n Ω [ ] j j n H e e Ω Ω [ ] cos( [ ]) j j H e n H e Ω Ω Ω +∠ (.) continuous-time [.] discrete-time Ω = ωTS where TS is the sampling period.
- 4. Lecture 16 Slide 13 PYKC 10-Mar-11 E2.5 Signals Linear Systems Frequency Response Example (1) For a system specified by the following difference equation, find the frequency response of the system. Take z-transform on both sides to find the transfer function: Therefore the frequency response is: L5.5 p533 [ 1] 0.8 [ ] [ 1] y n y n x n + − = + [ ] 0.8 [ ] [ ] zY z Y z zX z − = 1 [ ] 1 [ ] [ ] 0.8 1 0.8 Y z z H z X z z z− ⇒ = = = − − Lecture 16 Slide 14 PYKC 10-Mar-11 E2.5 Signals Linear Systems Frequency Response Example (2) Amplitude response Phase response L5.5 p533 Lecture 16 Slide 15 PYKC 10-Mar-11 E2.5 Signals Linear Systems Since where T = 2π/ωs we can map the s-plane to the z-plane as below: Mapping from s-plane to z-plane ( ) sT j T T j T z e e e e σ ω σ ω + = = = jω Re( ) s σ = jω − s-plane z-plane / 2 s jω / 2 s jω − Re( ) z Im( ) z 1 + 1 − 1 − 1 + Ω = ωT jω Lecture 16 Slide 16 PYKC 10-Mar-11 E2.5 Signals Linear Systems Since where T = 2π/ωs we can map the s-plane to the z-plane as below: Mapping from s-plane to z-plane ( ) sT j T T j T z e e e e σ ω σ ω + = = = jω Re( ) s σ = jω − s-plane z-plane / 2 s jω / 2 s jω − Re( ) z Im( ) z 1 + 1 − 1 − 1 + Ω = ωT
- 5. Lecture 16 Slide 17 PYKC 10-Mar-11 E2.5 Signals Linear Systems Mapping from s-plane to z-plane Since where T = 2π/ωs we can map the s-plane to the z-plane as below: ( ) sT j T T j T z e e e e σ ω σ ω + = = = jω Re( ) s σ = jω − s-plane z-plane / 2 s jω / 2 s jω − Re( ) z Im( ) z 1 + 1 − 1 − 1 + Ω = ωT