h

Part A: Chapter 3(C):

ja
ira
Digital Signal Processing

a
lia ik
3.1 Introduction to Digital Signal Processing

b
3.2 Analogue to Digital Conversion Process

tra m
3.3 Quantisation and Encoding
3.4 Sampling of Analogue Signals

us . A
3.5 Aliasing
3.6 Digital to Analogue Conversion
3.7 Introduction to digital filters
,A E3.7.1 Non-Recursive Digital filters
3.7.2 Recursive Digital filters
or
3.8 Digital filter Realisation
3.8.1 Parallel Realisation
SW ss

3.8.2 Cascade Realisation

3.9 Magnitude and Phase Responses
U ofe

3.10 Minimum/Maximum/Mixed Phase Systems

3.11 All-pass Filters
3.12 Second Order Resonant Filter
Pr

3.13 Stability of a Second Order Filter

N

3.14 Digital Oscillators

3.15 Notch Filters
 h
Example

ja
(a) State an application where a constant group delay filter is used.

ira
• Filtering music signals or video signals

a
lia ik
(b) Draw a parallel and a cascade realisation of the filter

b
2
H ( z) =

tra m
1 −1 1
(1 − z )(1 − z −1 )
2
x(n)
y(n)

us . A
+ +
2 4
cascade z-1 z-1
,A E +
4
0.5 0.25
or
z-1
x(n)
+ y(n)
SW ss

2 4 2
= − 0.5

⎛ 1 −1 ⎞⎛ 1 −1 ⎞ ⎛ 1 −1 ⎞ ⎛ 1 −1 ⎞
⎜1 − z ⎟⎜1 − z ⎟ ⎜1 − z ⎟ ⎜1 − z ⎟
U ofe

⎝ 2 ⎠⎝ 4 ⎠ ⎝ 2 ⎠ ⎝ 4 ⎠
-2
parallel
z-1
Pr

0.25
N
 h
3.10 Minimum-phase, Maximum-phase

ja
and Mixed phase systems [11]

a ira
Let us consider two FIR filters:

lia ik
b
1 −1

tra m
H1 ( z) = 1 + z

us . A
2
1
H 2 ( z ) = + z −1
,A E
2
or
SW ss

ρ = -0.5
U ofe

|z|=1
1
Pr

= −2
ρ |z|=1
N
 h
ja
„ H2(z) is the reverse of the system H1(z). This is due

ira
to the reciprocal relationship between the zeros of

a
H1(z) & H2(z).

lia ik
b
1 − jθ 1
H 1 (θ ) = 1 + e & H 2 (θ ) = + e − jθ

tra m
2 2

us . A
5
| H 1 (θ ) |=| H 2 (θ ) |= + cosθ
,A E
4
or
SW ss

„ The magnitude characteristics for the two filters are

U ofe

identical because the roots of H1(z) & H2(z) are

reciprocal.
Pr
N
 h
sin θ

ja
−1
„ Phase: φ2 (θ ) = tan

ira
1
+ cos θ
2

a
lia ik
sin θ
φ1 (θ ) = tan −1

b
2 + cos θ

tra m
us . A
φ2 (θ) φ1(θ)
,A E π
π
or
SW ss

-π π θ -π π θ
U ofe

-π -π
Pr
N
 h
Note: If we reflect a zero z = ρ that is inside

ja
„

ira
1
the unit circle into a zero z = outside the unit

a
ρ

lia ik
circle the magnitude characteristic of the system is

b
tra m
us . A
unaltered, but the phase response changes.
,A E
„ We observe that the phase characters φ1(θ) begins
or
at zero phase at frequency θ = 0 and terminates at
SW ss

zero phase at the frequency ω = π. Hence the net

U ofe

phase change.
Pr

Minimum phase filter

φ1 (π ) − φ1 (0) = 0
N
 h
ja
„ On the other hand, the phase characteristic for the

ira
filter with the zero outside the unit circle undergoes a
net phase change

a
lia ik
φ2 (π ) − φ1 (0 ) = −π radians

b
tra m
„ As a consequence of these different phase

us . A
characteristics, we call the first filter a minimum-
,A E
phase system and the second system is called a
or
maximum-phase system.
SW ss

If a filter with M zeros has some of its zeros inside

U ofe

the unit circle and the remaining outside the unit

Pr

circle, it is called a mixed-phase system.

N
 h
ja
„ A minimum-phase property of FIR filter

ira
carries over to IIR filter.

a
B( z )

lia ik
Let us consider H ( z ) =

b
„
A( z )

tra m
us . A
„ is called minimum phase if all its poles and
zeros are inside the unit circle.
,A E
or
|z|=1
SW ss
U ofe

Minimum phase
Re(z)
Pr
N
 h
ja
If all the zeros lie outside the unit circle, the

ira
„

system is called maximum phase.

a
lia ik
b
tra m
|z|=1

us . A
,A E
Maximum phase
or
Re(z)
SW ss
U ofe
Pr
N
 h
ja
If zeros lie both inside and outside the unit

ira
„

circle, the system is called mixed-phase.

a
lia ik
b
tra m
us . A
,A E
Mixed phase
Re(z)
or
SW ss

|z|=1
U ofe
Pr
N
 h
ja
Note: For a given magnitude response,

ira
the minimum-phase system is the

a
lia ik
causal system that has the smallest

b
tra m
magnitude phase at every frequency

us . A
(θ). That is, in the set of causal and
,A E
stable filters having the same
or

magnitude response, the minimum-

SW ss
U ofe

phase response exhibits the smallest

deviation from zero phase.
Pr
N
Example:

h
ja
ira
„ Consider a fourth-order all-zero filter containing a
double complex conjugate set of zeros located at

a
lia ik
π
±j

b
z = 0 .7 e.4

tra m
The minimum-phase, mixed phase and maximum

us . A
„

phase system pole-zero patterns having identical

,A E
magnitude response are shown below.
or
|z|=1 |z|=1
|z|=1
SW ss

2
2
ρ
U ofe

4
4 4
2
Pr

ρ=0.7 1/ρ 2
N

Minimum-phase mixed-phase maximum-phase

 h
„ The magnitude response and the phase response of

ja
the three systems are shown below: The minimum-

ira
phase system seems to have the phase with the

a
smallest deviation from zero at each frequency.

lia ik
|H(θ)|

b
tra m
us . A
π θ
,A E
φ(θ) minimum phase
or
SW ss

θ
-π
U ofe

-2π mixed-phase (In the case linear phase)

Pr

-3π
N

-4π
maximum phase
 h
Example:

ja
ira
„ A third order FIR filter has a transfer function
G(z) given by

a
G ( z ) = (6 − z −1 − 12 z −2 )(5 + 2 z −1 )

lia ik
b
From G(z), determine the transfer function of an

tra m
„

FIR filter whose magnitude response is identical

us . A
to that of G(z) and has a minimum phase
,A E
response.
or
G ( z ) = (2 − 3 z )(3 + 4 z )(2 + 5 z −1 )
−1 −1
SW ss
U ofe

3 −1 4 −1 5 −1
G(z) = 12(1 − z )(1 + z )(1 + z )
2 3 2
Pr
N

>1
 h
ja
ira
lm(z)

a
lia ik
2 3

b
3 2

tra m
- 5 4 3 2 Re(z)
− − − −

us . A
-
2 3 4 5
,A E |z|=1
or
SW ss
U ofe

2 −1 3 −1 2 −1
The Minimum phase filter P ( z ) = k (1 − z )(1 + z )(1 + z )
3 4 5
Pr
N
 h
3.11 All-Pass Filters [11]

ja
ira
„ An all-pass filter is one whose magnitude response is

a
constant for all frequencies, but whose phase

lia ik
response is not identically zero.

b
tra m
[The simplest example of an all-pass filter is a pure

us . A
„

delay system with system function H(z) = z-k]

,A E
or
„ A more interesting all-pass filter is one that is
SW ss

described by −1 − L +1
a L + a L −1 z + L + a1 z + a0 z − L
U ofe

H ( z) =
1 + a1 z −1 + L + a L z − L
Pr

where a0 = 1 and all coefficients are real.

N

„
 h
ja
ira
„ If we define the polynomial A(z) as

a
lia ik
L
A( z ) = ∑ ak z

b
−k
a0 = 1

tra m
k =0

us . A
A( z −1 )
∴ H ( z ) = z −L
,A E ⇒| H (θ ) |2 = H ( z ) ⋅ H ( z −1 ) | z =e jθ = 1
A( z )
or
SW ss

i.e. all pass filter,

U ofe
Pr
N
 1

h
Furthermore, if z0 is a pole of H(z), then is a

ja
„
z0
zero of H(z) {ie. the poles and zeros are reciprocals

ira
of one another}. The figure shown below illustrates

a
lia ik
typical pole-zero patterns for a single-pole, single-

b
zero filter and a two-pole, two-zero filter.

tra m
|z|=1

us . A
|z|=1
,A E (1/r, θ0)
r
θ0
or

0 a 0
SW ss

1 (r, -θ0)
a
U ofe

(1/r, -θ0)
All-pass filter All pass filter
Pr

1 −1
1− z
N

H ( z) = a |a| < 1 for stability

1 − az −1
 h
We can easily show that the magnitude response is

ja
„

constant.

ira
H (θ ) | 2 = H (θ ) ⋅ H * (θ ) = H ( z ) ⋅ H ( z −1 ) | z =e− jθ

a
lia ik
b
1 − jθ 1 jθ 2 1
1− e 1− e 1 − cos θ + 2

tra m
a a a a = a2
= ⋅ =

us . A
1 − ae − jθ 1 − ae jθ 1 − 2a cos θ + a 2
„
,A E
Phase response:
or
1 − jθ
1− e
SW ss

jθ
a 1 − ae 2 − ( a + a −1
) cos θ − j ( a − a −1
) sin θ
H (θ ) = ⋅ =
U ofe

− jθ jθ
1 − ae 1 − ae 1 − 2a cos θ + a 2
⎡ − ( a − a −1
) sin θ ⎤
Pr

∴φ (θ ) = tan ⎢
−1
⎥
⎣ 2 − (a + a ) cos θ ⎦
−1
N
 h
φ(θ)

ja
ira
π
a = 0.5

a
π

lia ik
θ

b
tra m
a = -0.5 a= -0.8
-π

us . A
When 0 < a < 1, the zero lies on the positive real
„
,A E
axis. The phase over 0 ≤ θ ≤ π is positive, at θ = 0 it
or

is equal to π and decreases until ω = π, where it is

SW ss

zero.
U ofe

When -1< a < 0, the zero lies on the negative real

Pr

axis. The phase over 0 ≤ θ ≤ π is negative, starting

N

at 0 for θ = 0 and decreases to -π at ω = π.

 h
3.12 A second Order Resonant Filter

ja
ira
x[n] y[n]

a
+ r p1 θ0

lia ik
b
-b1 z-1

tra m
p2

us . A
,A E z-1
-b2
p1 = re jθ 0 = r cos θ 0 + jr sin θ 0
or

p2 = re − jθ 0 = r cos θ 0 − jr sin θ 0
SW ss
U ofe

1 z2
H ( z) = −1 −2
= 2 (A)
1 + b1 z + b2 z z + b1 z + b2
Pr
N

All pole system has poles only (without counting the

zeros at the origin)
 h
z2 z2 z2

ja
H ( z) = 2 = =
z + b1 z + b2 ( z − p1 )( z − p2 ) ( z − re jθ 0 )( z − re − jθ 0 )
−1

a ira
z2 z2
H ( z) = =

lia ik
jθ 0 − jθ 0
(B)
z − r (e
2
+ e )z + r 2
z 2 − 2r cos θ 0 z + r 2

b
tra m
Comparing (A) and (B), we obtain

us . A
b1 = −2r cos θ 0 ,
,A E b2 = r 2

∴
or
− b1 2πf 0
Cosθ 0 = θ0 =
SW ss

2 b2 fs
U ofe

θ0 = resonant frequency
Pr
N
 h
3.13 Stability of a second-order filter

ja
a ira
„ Consider a two-pole resonant filter given by

lia ik
b
z2 1

tra m
H ( z) = 2 =
z + b1 z + b2 1 + b1 z −1 + b2 z − 2

us . A
,A E
b1 & b2 are coefficients
or
SW ss

„ This system has two zeros at the origin and poles at

U ofe

b1 b1 − 4b2
Pr

p1 , p 2 = − ±
N

2 2
 h
ja
„ The filter is stable if the poles lies inside the unit

ira
circle i.e. |p1| < 1 & |p2| < 1

a
lia ik
For stability b2 < 1. If b2 = 1 then the system is an

b
„

tra m
oscillator (Marginally stable)

us . A
Assume that the poles are complex
„
,A E
or

i.e. b12 – 4b2 < 0 ⇒ b12 < 4b2 and b1 < ±2 b2 , b2 > 0
SW ss
U ofe

„ If b12 – 4b2 ≥ 0 then we get real roots.

Pr
N
 h
„ The stability conditions define a region in the

ja
coefficient plane (b1, b2) which is in the form of a

ira
triangle (see below)

a
lia ik
The system is only stable if and only if the point

b
„

tra m
(b1, b2) lie inside the stability triangle.

us . A
2
b
parabola b2 = 1
,A E 4
b2
b2 = 1
or
SW ss

1
U ofe

Complex Conjugate Poles

Pr

-2 -1 0 1 2
Real Poles b1
N

b2 = −b1 − 1 -1 b2 = b1 − 1
 h
ja
Stability Triangle

ira
If the two poles are real then they must have

a
„

lia ik
a value between -1 and 1 for the system to

b
be stable.

tra m
us . A
− b1 ± b12 − 4b2
,A E −1 < <1
2
or
− 2 + b1 < ± b12 − 4b2 < 2 + b1
SW ss

∴ −2 + b1 < − b12 − 4b2 and b12 − 4b2 < 2 + b1

U ofe

(−2 + b1 ) 2 > b12 − 4b2 and b12 − 4b2 < (2 + b1 ) 2

b1 − b2 − 1 < 0 and b1 + b2 + 1 > 0
Pr
N
 h
2
b

ja
parabola b2 = 1
4

ira
b2
b2 = 1

a
1

lia ik
Complex Conjugate Poles

b
tra m
-2 -1 0 1 2
Real Poles b1

us . A
b 2 = − b1 − 1 -1 b 2 = b1 − 1
,A E
„ The region below the parabola (b12 > 4b2)
or
corresponds to real and distinct poles.
SW ss

The points on the parabola (b12 = 4b2) result in real

U ofe

and equal (double) poles.

Pr
N

„ The points above the parabola correspond to

complex-conjugate poles.
 h
3.14 Digital Oscillators

ja
ira
„ A digital oscillator can be made using a second order

a
discrete-time system, by using appropriate

lia ik
coefficients. A difference equation for an oscillating

b
tra m
system is given by
p[n] = A cos(nθ )

us . A
,A E
„ From the table of z-transforms we know that the
or
z-transform of p[n] above is
SW ss
U ofe

−1
1 − cos θz
P( z ) =
1 − 2 cos θz −1 + z −2
Pr
N
 h
ja
Y ( z) 1 − cos θz −1
Let P( z ) = =

ira
X ( z ) 1 − 2 cos θz −1 + z − 2

a
lia ik
Taking inverse z-transform on both sides, we obtain

b
tra m
y[n] − 2 cosθy[n − 1] + y[n − 2] = x[n] − cosθx[n − 1]

us . A
,A E
or
No Input term for an oscillator
SW ss

x[n] = 0, x[n-1] = 0
U ofe

So the equation of the digital oscillator becomes

Pr

y[n] = 2 cosθy[n − 1] − y[n − 2]

N
 h
ja
So the equation of the digital oscillator becomes
y[n] = 2 cosθy[n − 1] − y[n − 2]

a ira
and its structure is shown below.

lia ik
b
tra m
y[n] = A cos(nθ)

us . A
y[n-2]
,A E z-1 y[n-1] z-1
+
or

b1 = 2cosθ
SW ss
U ofe
Pr
N

b2 = -1
 h
ja
„ To obtain y[n] =Acos(nθ), use the following initial

ira
conditions:

a
lia ik
y[0] = A cos(0.θ) = A

b
y[-1] = A cos(-1.θ) = A cosθ

tra m
us . A
The frequency can be tuned by changing the coefficient
,A E
b1 (b2 is a constant). The resonant frequencyθ of the
or

oscillator is,
SW ss

− b1
U ofe

b1
cos θ = = − (For an oscillator b2 = 1)
2 b2 2
Pr
N
 h
Example:

ja
ira
„ A digital sinusoidal oscillator is shown below.

a
lia ik
x[n] y[n] = A sin(n+1)θ

b
+

tra m
z-1

us . A
,A E -b1 z-1
or
SW ss
U ofe

(a) Assuming θ0 is the resonant frequency of the

digital oscillator, find the values of b1 and b2 for
Pr

sustaining the oscillation.

N
 h
ja
ira
K K
H ( z) = −1 −2
= jθ 0 − jθ 0
1 + b1 z + b2 z ( z − re )( z − re )

a
lia ik
K

b
=

tra m
z 2 − r ( e jθ 0
+ e − jθ 0 ) z + r 2

us . A
K
= 2
,A E
z − 2r cos θ 0 z + r 2
or
SW ss

∴b1 = -2 r cosθ
U ofe

0 ; b 2 = r2

∴b
Pr

For oscillation b2 = 1 ⇒ r = 1 1 = -2 cosθ0

N
 h
ja
ira
(b). Write the difference equation for the above figure.

a
Assuming

lia ik
x[n] = (Asinθ0)δ[n], and y(-1) = y(-2) = 0.

b
tra m
us . A
Show, by analysing the difference equation, that the
,A E
application of an impulse at n = 0 serves the purpose of
or
beginning the sinusoidal oscillation, and prove that the
SW ss

oscillation is self-sustaining thereafter.

U ofe
Pr
N
 h
y[n] = −b1 y[n − 1] − y[n − 2] + x[n]

ja
ira
y[n] = 2 cosθ 0 y[n − 1] − y[n − 2] + A sin θ 0δ [n]

a
lia ik
n=0

b
y[0] = 2cosθ0 y[-1] – y[-2] + A sinθ0 δ[0]

tra m
us . A
0 0 1

y[0] = A sinθ0
,A E
or
n=1 0 0
SW ss

y[1] = 2cosθ0 y[0] –y[-1] + A sinθ0 δ[1]

U ofe

y[1] = 2cosθ0 , Asinθ0 = A sin2θ0

Pr
N
 h
ja
n=2

ira
y[2] = 2 cosθ0 y[1] – y[0] + A sinθ0δ[2]

a
= 2 cosθ0 Asin2θ0 – A sinθ0

lia ik
= 2A cosθ0 [2 sinθ0cosθ0] - sinθ0

b
tra m
= A sinθ0 [4 cos2θ0 –1 ] = A[3sinθ0 – 4 sin3θ0]

us . A
where sin3θ0 = 3sinθ0 – 4 sin3θ0
,A E
or
y[2] = A sin3θ0 and so forth.
SW ss
U ofe
Pr
N
 h
ja
„ By setting the input to zero and under certain initial

ira
conditions, sinusoidal oscillation can be obtained
using the structure shown above. Find these initial

a
lia ik
conditions.

b
tra m
y[n] = 2 cosθy[n − 1] − y[n − 2] + x[n]

us . A
,A E
(x[n] = 0 for an oscillator)
or
n=0 y[0] = 2cosθ0 y[-1] – y[-2]
SW ss

for oscillation, y[-1] = 0 ⇒ no cosine terms

U ofe

y[0] = -y[-2] ⇒ y[-2] = -Asinθ0 (sine term is equired)

y[0] = 0-(-A sinθ0) = A sinθ0
Pr

Initial conditions: y[-1] =0; y[-2] = -Asinθ0

N
 h
Sine and cosine oscillators [1]

ja
ira
„ Sinusoidal oscillators can be used to deliver the
carrier in modulators. In modulation schemes, both

a
lia ik
sines and cosines oscillators and needed. A structure

b
that delivers sines and cosines simultaneously is

tra m
shown below:

us . A
z-1
,A E cosθ
or
+ y[n]= cos(nθ)
sinθ
SW ss
U ofe

-sinθ
cosθ
Pr

+
x[n]= sin(nθ)
N

z-1
Proof:

h
ja
Trigonmetric equation for cos(n+1)θ is:

ira
cos(n+1)θ = cos(nθ)cosθ - sin(nθ) sinθ
Let y[n] = cos(nθ) and x[n] = sin(nθ)

a
∴ y[n+1] = cosθ y[n] – sinθ x[n]

lia ik
b
tra m
Replace n by n-1
y[n] = cosθ y[n-1] – sinθ x[n-1]

us . A
(A)
,A E
Similarly
sin(n+1)θ = sinθ cos(nθ) + sin(nθ) cosθ
or

∴ x[n+1] = sinθ y[n] + x[n] cosθ

SW ss

Replace n → n-1
U ofe

x[n] = sinθ y[n-1] + x[n-1] cosθ (B)

Pr
N

Using equations A & B above, the structure shown above can

be obtained.
Exercise:

h
ja
ira
„ An oscillator is given by the following coupled
difference equations expressed in matrix

a
lia ik
form.

b
⎡ yc [n]⎤ ⎡cos θ 0 − sin θ 0 ⎤ ⎡ yc [n − 1]⎤

tra m
⎢ y [n]⎥ = ⎢ sin θ ⎥ ⎢
cosθ 0 ⎦ ⎣ ys [n − 1]⎦ ⎥

us . A
,A E⎣ s ⎦ ⎣ 0

„ Draw the structure for the realisation of this

or

oscillator, where θ0 is the oscillation

SW ss

frequency. If the initial conditions

U ofe

yc[-1] = Acosθ0 and ys[-1] = -Asinθ0, obtain

Pr

the outputs yc[n] & ys[n] using the above

N

difference equations.
 h
yc [n] = cosθ 0 yc [n − 1] − sin θ 0 y s [n − 1]

ja
ira
y s [n] = sin θ 0 yc [n − 1] + cosθ 0 y s [n − 1]

a
lia ik
b
z-1

tra m
cosθ0

us . A
,A E + yc[n]
sinθ0
or
SW ss

-sinθ0
U ofe

cosθ0
+
Pr

ys[n]
N

z-1
 h
ys[0] = sinθ0 (A cosθ0) + cosθ0 (-Asinθ0) = 0

ja
n=0

ira
n=0 yc[0] = cosθ0 (Acosθ0) - sinθ0(-Asinθ0) = A

a
lia ik
b
n=1 yc[1] = cosθ0.A - sinθ0 .0 = Acosθ0

tra m
us . A
n=1 ys[1] = A sinθ0 + 0 = A sinθ0
,A E
n=2 yc[2] = cosθ0 yc[1] - sinθ0 ys[1]
or

= cosθ0 A cosθ0 - sinθ0 A sinθ0 = A cos2θ0

SW ss
U ofe

n=n yc[n] = A cos (nθ0)

Pr

similarly ys[n] = A sin(nθ0)

N
 h
Question 5

ja
The transfer function of a second order digital oscillator with an oscillation frequency of θ0 is

ira
given by z
H ( z) = 2 ; b1 = 2 cosθ 0
z − b1 z + 1

a
(a) Show that the impulse response corresponding to the system function H(z) is given by

lia ik
sin( nθ 0 ) u (n)
h( n) =

b
[3 marks]
sin(θ 0 )

tra m
z sin(θ 0 ) sin(nθ 0 )u (n) z
From z - transform table : Z{sin(nθ 0 )u (n)} = ⇒ → 2 = H ( z)

us . A
z − 2 z cos(θ 0 ) + 1
2
sin(θ 0 ) z − 2 z cos(θ 0 ) + 1
sin(nθ 0 )u (n)
⇒ h( n) =
,A E sin(θ 0 )

(c) The frequency of oscillation of the system H(z) can be determined using b1. Show that
or
− f s Δb1
Δf 0 =
SW ss

f [4 marks]
4π sin(2π 0 )
fs
where fs=sampling frequency and f0 =desired frequency of oscillation and Δ f0 is the
U ofe

frequency resolution b = 2 cos(θ ) 1 0

2πΔf 0
Δb1 = −2 sin(θ 0 )Δθ 0 = −2 sin(θ 0 )
Pr

fs
N

− f s Δb1 − f s Δb1
∴ Δf 0 = =
4π sin(θ 0 ) 4π sin( 2π f 0 )
fs
 h
3.15 Notch filters [4]

ja
ira
„ When a zero is placed at a given point on the

a
z-plane, the frequency response will be zero

lia ik
at the corresponding point. A pole on the

b
other hand produces a peak at the

tra m
corresponding frequency point.

us . A
,A E
„ Poles that are close to the unit circle give rise
or
large peaks, where as zeros close to or on
SW ss

the unit circle produces troughs or minima.

Thus, by strategically placing poles and zeros
U ofe

on the z-plane, we can obtain sample low

Pr

pass or other frequency selective filters

N

(notch filters).
 h
Example:

ja
ira
„ Obtain, by the pole-zero placement method,
the transfer function of a sample digital notch

a
lia ik
filter (see figure below) that meets the

b
following specifications: [4]

tra m
us . A
„ Notch Frequency: 50Hz
„
,A E3db width of the Notch: ±5Hz
„ Sampling frequency: 500 Hz
⎛ Δf ⎞
or
The radius , r of the poles is determined by : r = 1 − ⎜⎜ ⎟⎟π
SW ss

⎝ fs ⎠
|H(f)|
U ofe
Pr
N

0 50 250 f (Hz)
 h
„ To reject the component at 50Hz , place a pair of

ja
complex zeros at points on the unit circle corresponds to

ira
50
50Hz. i.e. at angles of 3600 × = ±360 = ±0.2π

a
500

lia ik
To achieve a sharp notch filter and improved amplitude

b
„

tra m
response on either side of the notch frequency , a pair of

us . A
complex conjugate zeros are placed at a radius r < 1.
,A E |z| =1
⎛ Δf ⎞
⎟π = 1 − ⎛⎜
10 ⎞
r = 1 − ⎜⎜ ⎟π = 0.937
or
⎟
⎝ fs ⎠ ⎝ 500 ⎠
SW ss
U ofe

360
Pr

360
N

0.937
 h
(z − e )(z − e )

ja
− j 0.2π j 0.2π

ira
H ( z) =
(z − 0.937e − j 0.2π
)(z − 0.937e j 0.2π
)

a
lia ik
j 0.2π − j 0.2π
z + 1 − (e +e

b
2
)
= 2

tra m
j 0.2π − j 0.2π
z + 0.878 − 0.937(e +e )z

us . A z + 1 − 2 cos(0.2π )
,A E 2
= 2
or

z + 0.878 − 2 × 0.937 cos(0.2π )

SW ss

−1 −2
1 − 1.6180 z + z
U ofe

= −1 −2
1 − 1.5161z + 0.878 z
Pr
N
 h
Summary of part A Chapter 3

ja
ira
„ At the end of this chapter, it is expected that you should know:

a
lia ik
„ A block diagram of the conversion from analog to digital and

b
back to analog form, including descriptions of the blocks

tra m
us . A
„ Analog to digital conversion, in particular amplitude
,A E
quantization and quantization error. Be able to calculate the
signal-to-quantization error ratio.
or
SW ss

„ Sampling of analog signals, in particular deriving the

U ofe

mathematical relationship between the analog and digital spectra.

Pr

„ The sampling theorem and the Nyquist frequency.

N
 h
ja
„ How to demonstrate the effect of aliasing using

ira
sketched magnitude spectrum plots.

a
lia ik
„ Digital to analog conversion and the role of the

b
reconstruction filter. Show your understanding using

tra m
both mathematical and hand-sketched explanations.

us . A
„
,A E
Calculations of aliased frequencies: f0 = fk –kfs, where
or
fk is the frequency outside the Nyquist frequency.
SW ss

Types of digital filters (FIR/IIR) and their properties.

U ofe

„
Pr

„ Conversion from FIR/IIR difference equations to

N

transfer functions and back again.

 h
ja
„ FIR (Non Recursive, all-zero) Filters

ira
„ Understanding of phase delay and group delay

a
„ Definition of linear phase filters

lia ik
b
tra m
„ IIR (Recursive, all-pole or pole-zero) Filters

us . A
„ Cascaded, parallel, and canonic structures
,A E
or
„ Calculation of 3 dB cut-off frequency and 3 dB
SW ss

bandwidth for a simple first-order FIR and IIR filter

U ofe

„ Be able to plot the magnitude response of a simple first-

Pr

order FIR and IIR filter.

N
 h
„ Be able to distinguish between lowpass and highpass

ja
filters based on the difference equations or transfer

ira
functions for both FIR and IIR

a
lia ik
b
„ Given an FIR filter (difference equation or transfer

tra m
function), be able to draw the magnitude and phase

us . A
responses, and be able to explain the relationship
,A E
between magnitude and phases responses.
or
SW ss

„ The differences between minimum, maximum and

U ofe

mixed phase filters.

Pr

„ The similarity in magnitude responses when filter

N

zeroes are reflected about the unit circle.

 h
ja
„ All-pass filters: be able to show that their magnitude

ira
response is constant but their phase response is non-zero.

a
lia ik
„ Be able to derive the transfer function for a second order

b
tra m
resonator filter, and be able to analyse its stability

us . A
properties using the stability triangle and pole positions.
,A E
Be able to understand the range the filter coefficients can
or
take in order to preserve stability.
SW ss
U ofe

„ Principles of stable, marginally stable and unstable filters

and equations for digital oscillators.
Pr
N