DSP-1-Introduction 1
Mathematische Methoden des Visual Computing (Signaltheorie)
VU (1+1h) 186.849
http://ti.tuwien.ac.at/rts/teaching/courses/mmvc
Herbert Grünbacher Institut für Technische Informatik (E182)
DSP-1-Introduction 2
Exercises & Exam
• (Home) exercises based on Matlab – Have exercises ready for (oral) exam
• Registration for exam:
Matlab Campus (Win/Linux/Mac)
• INTU (€ 18,00): MATLAB /Simulink including 14 selected tool boxes (DVD)
• Download (€13,90): www.zid.tuwien.ac.at/studentensoftware
DSP-1-Introduction 3
Alternative solution public domain http://www.scilab.org
Books • James H. McClellan, Ronald Schafer, Mark Yoder:
Signal Processing First Prentice Hall
• John G. Proakis, Dimitris G. Manolakis Digital Signal Processing Prentice Hall
• Vinay K. Ingle und John G. Proakis Digital Signal Processing Using MATLAB Thomson Learning
DSP-1-Introduction 4
DSP-1-Introduction 5
(Digital) Signal (Processing)
• Signal A physical quantity varying with time, space, or any other independent variable. E.g.: Temperature, pressure, audio, …
DSP-1-Introduction 6
Representation of Signals • Signals appear in different forms and
representations, e.g. speech / acoustic signal Microphone / electric signal Magnetization / magnetic signal Sequence of numbers / digital signal (CD)
cont
inuo
us
disc
rete
DSP-1-Introduction 7
Composition of Signals
• Impulse Function Representation in the Time Domain
• Fourier Series, Fourier Transformation Representation in the Frequency Domain a.k.a. spectral representation
DSP-1-Introduction 8
Processing / System
• A system is a facility which influences, changes, records, plays back, transmits… signals, e.g. CD player
• Systems operate on signals and generate new signals or signal representations
DSP-1-Introduction 9
Digital • Representation by numbers
– Sampling: Quantization in the time domain – Digital: Quantization in the amplitude domain
Time Amplitude
continuous discrete
Analog (T&A cont.) digital signal (T&A discrete)
DSP-1-Introduction 10
Discretization
• In Time Sampling (Shannon-) Theorem
• In Amplitude no similar theorem but quality requirements
DSP-1-Introduction 11
Representation
• We represent Signals and Systems in mathematical form to – describe and understand Signals und
Systems and – design and implement systems with
requested properties
Content of lecture Tuesday 16:00 – 17:30
DSP-1-Introduction 12
16.10. Introduction, Complex numbers (13+30) 23.10. Systems, Time functions (16+20) 30.10. Signals & Spectrum (75) 06.11. Sampling (50) 13.11. FIR filters (90) 20.11. z-Transform (30) 27.11. IIR filters (60)
Complex Numbers
Quadratic Equations
Complex Numbers 2
( ) ( )
( ) ( ) ( )
( )
21,2 1 2
21,2 1 2
21
2
,2 1,2
2 3 0 1 1 3 3; 1
2 1 0 1 1 1 1; 1
2 5 0 1 1 5 1 2
3 1 0
1 1 1 0
1 2
x x x x x
x x x x x
x x
x x
x x x x j
x
x j x
− + =
− − = − =
− − −
− − = = ± + = = −
− + = = ± − = =
− + = = ± − =
±
two real roots
double root
two conjugate complex roots
( )1 2 0j+ =
Complex Numbers 3
Real and Imaginary Part
( , ) Re Im z x y x jy z j z= + = +
Complex Numbers 4
Phasor (Zeiger) ≠ Vector (Vektor)
• Vector: Physical Quantity with Direction e.g. Force, Acceleration, Impulse
• Phasor: Representation of a Complex Number
• Algebra only partially the same (e.g. Addition), but not Multiplication (inner and outer product)
Complex Numbers 5
Magnitude and Angle (Phase) z r ϕ= ∠
compass(z)
ϕ
r
Complex Numbers 6
Angle: Calculation Imagination
• We calculate using radian measure • We visualize using degrees
360 2
Representation: [rad] [°]=45 0.785180
ππα
° =
⋅ =
Complex Numbers 7
Cartesian Polar Representation
( )2 2
cos sin cos sinImaginary Part tan arctan
Real Part
x r y r z r j ry yr x yx x
ϕ ϕ ϕ ϕ
ϕ ϕ
= = = +
= + = = =
Complex Numbers 8
Cartesian Polar
• Addition • Subtraction • Conjugate
• Multiplication • Division • Power • Root • Conjugate
Complex Numbers 9
Attention Phase (1)
[ ][ ]
Imaginary PartReal Part
11
11
sin( /180)
arctan Division by Zero
defined only for 90 90
De
a
gree/Radian:
Real Part 0:
arctarctan 0.7854 45
arctan( ) 0.7854 45
n
rad
rad
α π
ϕ
ϕ
ϕ
ϕ
−−
= =
° ⋅
= ⇒
− ° ≤
= °
°
= =
≤
= ° = −135°
Complex Numbers 10
Attention Phase (2) Phase defined only from 0 2 but physical quantities often have: 2n
ϕ πϕ π ϕ
≤ ≤+ ⋅ =
unwrap(phase)
Complex Numbers 11
Attention Phase (3)
.Runding errors may lead tophase jum
and represent the same phase
Magnit
ps .
If (or very small)ude is Zero Phase has no me !g
,anin
π π
π π→
−
−
Complex Numbers 12
Magnitude is positive!
Complex Numbers 13
Matlab (1) MATLAB knows (only) complex numbers:
3 + 4i oder 3 + 4j
Avoid using i or j as variable: 4+3*i 4+3i i=3; i = 4+3*i 13 but 4+3i 4.00 + 3.00i Restore i as imaginary unit: i = sqrt(-1)
Use notation 4 + 3*1i to avoid troubles.
Complex Numbers 14
Matlab (2)
real(z) Real Part of z real(3-4i) 3 imag(z) Imaginary Part of z imag(3-4i) -4 abs(z) Magnitude of z abs(3-4i) 5 angle(z) Phase of z angle(3-4i) -0.9273 conj(z) Conjugate of z conj(3-4i) 3+4i
angle(z) defined from -180° to 180°
Attention transpose z'
Complex Numbers 15
z' is conjugate transpose z.' is non-conjugate transpose
[3+4i, −1 − 1i, 10 − 10i]' [3+4i, − 1 − 1i, 10 − 10i].'
Complex Numbers 16
Euler (1)
z r ϕ= ∠ cos sinje jϕ ϕ ϕ= +
Complex Numbers 17
Euler (2)
c
2 3 4 5
2 3 4 2 4 3
2
os s n
4 3
i
12! 3! 4! 5!
5 51 1 ( )2! 3! 4! 5! 2! 4! 3! 5!
5cos 1 sin 2! 4! 3! 5!
j
x
j
x x x xe x
e j j j j
ϕ ϕ
ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕϕ ϕ
ϕ ϕ ϕ ϕϕ ϕ ϕ
+
= + + + + + +
= + − − + + + = − + − + − + −
= − + − = − + −
Proof algebraic!
Algebra
Geometry
Complex Numbers 18
Euler (3) cos sincos sin2cos
j
j
j j
e je j
e e
ϕ
ϕ
ϕ ϕ
ϕ ϕ
ϕ ϕ
ϕ
−
−
= +
= −
+ =
cos2
sin2
j j
j j
e e
e ej
ϕ ϕ
ϕ ϕ
ϕ
ϕ
−
−
+=
−=
Complex Numbers 19
1je π = −
Complex Numbers 20
2*exp(1i*45*pi/180) 1.4142 + 1.4142i 3.6056*exp(2.1588i)*10.8167*exp(-0.9828i) 15.00 +36.00i compass(exp((i*30*pi/180)*(0:11)))
0.2
0.4
0.6
0.8
1
30
210
60
240
90
270
120
300
150
330
180 0
MATLAB results always in Cartesian form, i is imaginary unit; input in Euler form possible, i or j as imaginary unit allowed.
abs(3*exp(1i*45*pi/180)) 3 (180/pi)*angle (3*exp(1i*45*pi/180)) 45.00
Complex Numbers 21
Complex Algebra (1)
Addition Subtraction
1 2 1 1 2 2
1 2 1 2
( ) ( ) ( ) ( )z z x jy x jy
x x j y y+ = + + +
= + + +1 2 1 1 2 2
1 2 1 2
( ) ( ) ( ) ( )z z x jy x jy
x x j y y− = + − +
= − + −
Complex Numbers 22
Complex Algebra (2)
Multiplication Division
1 2
1 2
1 2 1 2( )
1 2
j j
j
z z r e r e
r r e
ϕ ϕ
ϕ ϕ+
× = ×
=
1
2
1 2
11 2
2
( )1
2
j
j
j
r ez zr er er
ϕ
ϕ
ϕ ϕ−
÷ =
=
Complex Numbers 23
Complex Algebra (3)
1
2
3
4
5
30
210
60
240
90
270
120
300
150
330
180 0
Conjugate Complex, transpose
1 1 1 1
1 1
( ) ( )z z x jy
x jy
′∗ ∗= = += −
( )1
1
1 1
1
j
j
z r e
re
ϕ
ϕ
∗∗
−
=
=
Complex Numbers 24
Complex Algebra (4)
( )NN j N jNz re r eϕ ϕ= =
Power
Complex Numbers 25
Complex Algebra (4)
2
1 0,1, 2,..., 1
j nNN e
n N
π
== −
Root
2( )
0,1, 2,..., 1
nN NjN j Nz re re
n N
ϕ πϕ += == −
1 0Nz − =
Complex Numbers 26
je ϕ
je ϕ−
cosϕ cosϕ
sinj ϕ
cos sin2cos
j
j j
e je e
ϕ
ϕ ϕ
ϕ ϕ
ϕ −
= +
= +
Complex Numbers 27
0j te ω
0cos tω
0sinj tω
0 0( ) 2t t f tϕ ω π= = ⋅
ϕ
t
90° 180° 270°
Complex Function
Complex Numbers 28
Rotating Phasor
-1 0 1-1
-0.5
0
0.5
1
Realteil
Imag
inär
teil
0 0.5 1-1
-0.5
0
0.5
1
t
Imag
inär
teil
-1 0 10
0.2
0.4
0.6
0.8
1
Realteil
t
0
0
0 0cos sin
cos Re
j t
j t
e t j t
t e
ω
ω
ω ω
ω
= +
=
Complex Numbers 29
2cos j t j tt e eω ωω −= +
DSP-3-Systems 1
Systems
DSP-3-Systems 2
Block diagram
1 1
2 2
m n
y xy x
y x
= ⋅
A
Excitation Response
DSP-3-Systems 3
Tasks : Excitation & System
? : Response System Analysis : Excitation & Response
? : System System Synthesis
: System & Response ? : Excitation Measurement
Cause Effect
DSP-3-Systems 4
Linearity (1) Cause EffectCause Effectk k
→⋅ → ⋅
homogeneous
Cause1 Effect1Cause2 Effect2
Cause1+Cause2 Effect1+Effect2
→→→
Superposition principle
1 2 1 2Cause1 + Cause2 Effect1 + Effect2k k k k⋅ ⋅ → ⋅ ⋅
No interaction between causes.
additive
DSP-3-Systems 5
Mental arithmetic
x 4 1024 ?
1000 x 4 4000 20 x 4 80 4 x 4 16
4096
DSP-3-Systems 6
Linearity (2) 1. »Complicated« signals composed of simple
(basic) signals
2. Compute system responses for all basic signals
3. Sum of all system responses of basic signals is system response of complicated signal
Important basic signals: Sinusoidal Function und Impulse Function
Nonlinear Systems (1)
DSP-3-Systems 7
0 0.5 1 1.5 2 2.5 3 3.5 40
2
4
6
8
10
12
14
16
x y
y = x2
2
2 2
27 20 720 400
27 7297 49
400 49 449
= +
= =
≠ =+ =
Superposition principle violated Nonlinear System
DSP-3-Systems 8
Nonlinear systems (2)
0 0.5 1 1.5 2 2.5 3 3.5 40
2
4
6
8
10
12
14
16
x y
y = x2
[ ]
( )( ) ( ) ( )
( )12cos cos( ) cos co
2 2 2
2
2
2
s
2
cos ( ) 2 cos( ) cos( ) cos ( )1 11 cos 2 2 cos( )cos( ) 1 cos 2
cos( ) cos( )
2 2t t t t
A t A t B t B t
A t B t
A t AB t t B tω ω ω
ω ω
ω
ω ωΩ = Ω− + Ω+
= + Ω + Ω =
= + + ⋅ Ω + + Ω
+ Ω =
DSP-3-Systems 9
0 5 100
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Spektrum Eingangssignal0 5 10
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Spektrum Ausgangssignal
0
ω
Ω
2Ω
2ω
Ω-ω Ω+ω
2 2 2[ cos( )] [ cos( )] [ cos( ) cos( )]A t B t A t B tω ω+ Ω + Ω≠
Spectral representation
DSP-3-Systems 10
( ) ( )
( ) ( ) ( )
( )
( )
222 12
2 2212
2 2 212
2cos2
12
cos
2
2
1 cos2
j t j t
A B
j t j t j t j t
j t j t
t
t e e
e e e e
e e
t
ω ω
ω ω ω ω
ω ω
ω
ω
ω
−
+
− −
−
= +
= + +
= + +
= +
Complex Algebra
DSP-3-Systems 11
( ) ( )( ) ( )2 2
2
2 3 31 12 2 2 2
2 23 3 3 31 1 1 12 2 2 2 2 2 2 2
2
(cos 3cos )
2
j t j t j t j t
A B
j t j t j t j t j t j t j t j t
ABA B
t t e e e e
e e e e e e e e
ω ω
ω ω ω ω
ω − Ω − Ω
− − Ω − Ω Ω − Ω
+ Ω = + + + =
+ + + + + +
( ) ( ) ( ) ( )
[ ]( ) ( )
3 31 12 2 2 2
2 2 22 21 1 1 12 2 2 2
cos2
( ) ( ) ( ) ( )3 3 3 31 1 1 12 2 2 2 2 2 2 2
cos(
212
)
2
cos( )
2
2
3 32 2
2
2 cos 2
2
3 cos( ) cos( )
2
2j t j t
t
j t j t j t j t
t t
e e t
e e e e
t t
ω ω
ω
ω ω ω ω
ω ω
ω
ω ω
−
Ω+ − Ω+ Ω− − Ω−
Ω+ Ω−
+ + = +
+ + + =
= Ω+ + Ω−
+
( ) ( )( ) ( )
2 22 23 232
22 312
2
2
4.5 cos 2
2 2 5
2j t j te e tΩ − Ω+ = + Ω
=+
DSP-3-Systems 12
Linear Systems (3)
• Electric circuits consisting of R, C and L • Amplifiers and Filters • Mass-Spring-Damping-Systems • Series and parallel resonance • Differentiation und Integration • Propagation of electromagnetic and acoustic
waves in isotropic media
Linear systems are described by linear differential or difference equations
DSP-3-Systems 13
Linearity (4)
Linearity only within certain physical limits
Beyond those limits systems become nonlinear (e.g. clipping) or get destroyed
DSP-3-Systems 14
Nonlinear Systems (3)
• Electrical power P = RI2 • Electronic circuits like comparators,
discriminators, … • Clipping, nonlinear amplification • (Magnetic) hysteresis • Multiplication of signals • Digital logic gates • Propagation of electromagnetic and acoustic
waves anisotropic media
Nonlinear systems are described by nonlinear differential or difference equations
DSP-3-Systems 15
LinearTimeInvariant-Systems
Delay n0 System
System Delay n0
x[n] x[n-n0] w[n]
y[n] y[n-n0]
TI if: w[n] = y[n-n0]
(L)TI: input signal delayed by n0 causes output signal delayed by n0
DSP-3-Systems 16
• Linear : Nonlinear
• Time independent : Time dependent
• Dynamic : Non-dynamic
• Causal : Non-causal
• Lumped : Distributed
• Time continuous : Time discrete
• Analog : Digital
• Stable : Non-stable
DSP-4-Time functions 1
Time functions
DSP-4-Time functions 2
• It is not possible to represent signals (voice, music, physical values, ...) as (simple) time functions.
• Therefore we use simple »test signals«, which can be represented mathematically and investigate the response of the system to the test signal.
• We compose real signals of those test signals (remember the superposition principle in linear systems) and add (superpose) the responses of the test signals to gain the system response of the real signal.
DSP-4-Time functions 3
Elementary Functions
1
0 0( )
1 0t
tt
δ−
<= ≥
0
0
( ) 0 0
( ) 1
t t
t dt
δ
δ∞
−∞
= ≠
=∫
u(t)
o
DSP-4-Time functions 4
Impulse “Function“ • The Impuls Function is no true function and
doesn‘t describe a distinct function either. • The Impulse Function is zero, but at t = 0, but
it is not defined at t = 0 either. • The impact of the Impulse Function is the
action on other functions: We are not interested in how the Impulse Function looks like or what it is, we are only interested how the Impulse Function acts on other functions – the signal functions s(t).
DSP-4-Time functions 5
1 1"form( ) ( ) ( ) ( )al"t
i i i idt d t tdt
δ δ τ τ δ δ+ +−∞
= =∫
-2 -1 1 2
0.2
0.4
0.6
0.8
1Signal Plot
-2 -1 1 2
0.2
0.4
0.6
0.8
1Signal Plot
-2 -1 1 2
0.5
1
1.5
2Signal Plot
-2 -1 1 2
0.5
1
1.5
2Signal Plot
δ0 δ-1
δ-3 δ-2
DSP-4-Time functions 6
Impulse Step
1 0
0 0( ) ( )
1 0
t tt dt
tδ δ τ−
−∞
<= = ≥∫
DSP-4-Time functions 7
Convolution (1) 1 2 1 2 1 2( ) ( ) ( ) ( ) ( ) ( )g t g t g g t d g t g dτ τ τ τ τ τ∞ ∞
−∞ −∞
∗ = − = −∫ ∫
DSP-4-Time functions 8
Convolution (Folding) (2)
• g1(t), g2(t) g1(τ), g2(τ )
• t is independent Variable hence τ
• g1(-τ) »Folding«
• g1(-τ) delayed - by t g1(t - τ) - multiplied - integrated new value of g1∗ g2
1 2 1 2 1 2( ) ( ) ( ) ( ) ( ) ( )g t g t g g t d g t g dτ τ τ τ τ τ∞ ∞
−∞ −∞
∗ = − = −∫ ∫
DSP-4-Time functions 9
Convolution (3) 1 2
1 2 1
1 2
2
( ), ( ) [ ], [ ] kDiscrete:
[ ] [ ] [ ] [ ]k
g t g t
g n g n g
g
k
n
n g
n
k
gτ
∞
=−∞
∗
⇒⇒
= −∑
DSP-4-Time functions 10
Convolution with elementary function
0
10
Step:
Impulse: 0 ( ) ( )
1 ( ) ( ) ( )
( )t
i g t
i g t t g t
t g d
δ
δ τ τ+
−−= − ∗ =
= ∗ =
∫
The Impulse Function δ0(t) has the property of a sampling function: Convolving with δ0(t) delivers a sample of g(t) at t.
DSP-4-Time functions 11
Mathematical representation of the sampling process
0 00
( ) ( ) ( ) ( ) ( )t
g t g t t g t dδ τ δ τ τ+
−
= ∗ = −∫
DSP-4-Time functions 12
Sinusoidal Function
0
00
( ) sin( )12 2
s t A t
fT
ω ϕ
ω π π
= +
= =
0
2 3( ) sin4
s t A tTπ π
=
−
DSP-4-Time functions 13
Complex Exponential
0
0
( )0 0
( )0
( ) cos sin
( R) ce os
j t
j t
x t Ae A t jA t
x t Ae A t
DSP-4-Time functions 14
Euler ( )( ) cos( ) sin( )oj t
o os t Ae A t jA tω ϕ ω ϕ ω ϕ+= = + + +
-1 0 1-1
-0.5
0
0.5
1
Realteil
Imag
inär
teil
0 0.5 1-1
-0.5
0
0.5
1
tIm
agin
ärte
il
-1 0 10
0.2
0.4
0.6
0.8
1
Realteil
t
( )( ) Re cos( )oj tos t Ae A tω ϕ ω ϕ+= = +
DSP-4-Time functions 15 15
12cos( ) j t j tt e e
Signale
Left and right turning phasors generate (real) cosine function. Positive and negative frequencies!
Inverse Euler
Why Cosine?
• Cosine allows representation of DC component (mean value)
DSP-4-Time functions 16
cos(2 ) cos( )A f t A t Aπ = =0
DSP-4-Time functions 17
Complex Exponential ( )
( ) ;
( )
o
st j
j t
s t Ke K Ae
s t Ae
s j
ϕ
ϕ
ω
σ ω
+=
= = = +
( )oe sR cst tKe K e tσ ω ϕ= +
DSP-4-Time functions 18
Time delay Phase delay
1
0 3
1
0 3
1
0 3
2 5
-1 2
( )x t
( 2)x t
( 1)x t
t
t
t
12
0( ) 3 1 3
0 else
t tx t t t
Delay towards »right« (positive time delay)
1( )x t t
negative sign
DSP-4-Time functions 19
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Phase = -72°
0( ) cosx t A t
Frequency (Period)
0 00
10.025 40T fT
Maximum of Cosine if argument zero
00 11 0 ; 2 40 0.005 1.2566 72
t t
DSP_5-Signals&Spectrum 1
Signals & Spectrum
DSP_5-Signals&Spectrum 2
• A signal is a physical quantity varying with time, carrying information. E.g. voice, audio, video, temperature, …
• Signals can be changed, stored, transmitted Energy consumption
• Signals appear in different forms and representations, e.g. voice/acoustical signal microphone/electrical signal magnetization/magnetic signal series of numbers/digital signal (CD)
DSP_5-Signals&Spectrum 3
• Signals can be processed (analog and/or digital) by signal processing systems, e.g. cassette recorder, CD-player,…
• Systems have signals as input(s) and generate new signals or signal representations as output(s)
DSP_5-Signals&Spectrum 4
• Signals and systems are represented in mathematical form to describe and analyze them to design and implement systems with specified behavior.
• There is an infinite number of signals. To simplify analysis and design, signals are decomposed in simple elementary (basic) signals.
DSP_5-Signals&Spectrum 5
Composition of »sounds«
DSP_5-Signals&Spectrum 6
ASP
ASP
DSP
DSP_5-Signals&Spectrum 7
Synthesis of periodic signals
20
1 1
( ) cos(2 ) k
k
N Nj f t
k k k o
jk
kk k
k
x t A A f
X A
t X e X e
e
Periodic signals can be composed of sinusoids consisting of the fundamental component f0 = 1/T0 and harmonic components. Harmonic components are integer multiples of the fundamental component.
DSP_5-Signals&Spectrum 8
»Complex« Magnitude Phase in real representation
2 2
2
cos(2 )
kk k k
k
k k
j f t j f tk
j f
k
jk
kt
A f t
A e A e
X
e
e
Complex magnitude in complex exponential
DSP_5-Signals&Spectrum 9
Two-sided spectrum Based on inverse Euler we get:
*2 2
1
( )2 2
k k
Nj f t j f tk k
ok
X Xx t X e e
x(t) generated by right and left turning phasors Two-sided spectrum
1, 21, 2, 21
21
,1 12 2
( ,0), , , , ,1 12 2
...o X f X ffX X X f
Spectral representation
• Signals can be composed of sinusoidal components
• The waveform of sinusoids is known, therefore it is sufficient to represent only magnitude(s) and phase(s) as function of frequency spectral lines
DSP_5-Signals&Spectrum 10
DSP_5-Signals&Spectrum 11
Time domain and frequency domain (spectral) representation
t
s(t)
S(f)
DSP_5-Signals&Spectrum 12
Analysis of periodic signals
DSP_5-Signals&Spectrum 13
1. Odd periodic signals can be represented by sine-functions
01 0 2 0 0
00
1
( ) sin sin 2 ... sin2
sin( )2
n
N
kk
as t b t b t b n t
a b k t
DSP_5-Signals&Spectrum 14
2. We multiply signal s(t) by sinω0t and average over period
DSP_5-Signals&Spectrum 15
We remember from mathematics
2
20
0 für sin sin
für
1sin 1 cos2
2 2
T
o o
T
T T
m nn t m t dt
m n
t dt t dtT
0
sin sin 3 0T
t tdt
DSP_5-Signals&Spectrum 16
2 1sin 1 cos2
21 1
2o o
T T
tT
t dt t d
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-1
-0.5
0
0.5
1
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-1
-0.5
0
0.5
1
T
DSP_5-Signals&Spectrum 17
3. All integrals but sin2ω0 t are zero multiplication by sinω0 t and integration „sucks“ spectral component ω0 from s(t)
4. For further spectral components we multiply by sin(kω0 t ), k = 2, 3,…
0
2( )sin
T
k ob s t k t dt
T
Real representation of Fourier series
DSP_5-Signals&Spectrum 18
• We repeat for even signals and get in real-valued representation
0
2( )cos
T
k oa s t k t dt
T
• Signals which are neither even nor odd must be composed of sine and cosine components and we need to compute ak and bk .
19
Complex-valued Fourier series
0 0
00
2
2
00
( )
1 ( )
jk tjk t T
k kk k
j k tT Tk
x t X e X e
X x t e dtT
DSP_5-Signals&Spectrum
20
Σ X0 e j k ω0t
k = −∞
∞
DSP_5-Signals&Spectrum
21
Σ Xk e j k ω0t
k = −∞
∞
DSP_5-Signals&Spectrum
22
Σ Xk e j k ω0t
k = −∞
∞
022 fTπω π= =
DSP_5-Signals&Spectrum
23
0
00
00
( )
1 ( )
j kk
k
t
kj tkT
x t X e
X x t e dtT
DSP_5-Signals&Spectrum
Phasor interpretation
DSP_5-Signals&Spectrum 24
X1
X2
X3
Xn
ω0
2ω0
3ω0
nω0
DSP_5-Signals&Spectrum 25
01 0 2 0 0
1 0 2 0
00
0
0
0
( ) cos cos2 ... cos2
+ sin sin 2 ... sin
2 ( )cos = mean valu
2 (
e
)
of
s
( )2
in
n
nT
k
k
as t a t a t a n t
b t b
a s
t b n t
a f t k tdtT
b f t k tdT
t
0
0
0
0 1 0 1 2 0 2 0
2 2
12
( ) cos cos 2 ... cos
, tan
( )
1 ( )
T
n n
kk k k k
k
kjk t
kk
jk tk
t
s t c c t c t c n tbc a b
a
s t X e
aX f t e dt
T
012
0 12
for 0( ) for 0
( ) for 0
T
k k
k k
ka jb k
a jb k
real: sine + cosine
real: cosine + phase
complex
DSP_5-Signals&Spectrum 26
Fourier coefficients Fourier coefficients are often based on period 2π
0
13
2 22
4( ) sin sin
1
...1 3
Ta
y t t t
DSP_5-Signals&Spectrum 27
Scaling of time axes
1
0
13
22
2
4( ) sin sin ...3
f
t t
ay t t t
DSP_5-Signals&Spectrum 28
Spectral analysis
Both signals have the same sound.
Hearing doesn‘t recognize phase!
http://www.jhu.edu/~signals/listen-new/listen-newindex.htm
Hearing • Audio band: 16 Hz − 20 kHz • 16 Hz: vibration
3.500 Hz: fundamental frequency Piccolo Flute • Threshold of hearing 20 µPascal (frequency dependent)
Threshold of pain 130 dB • Hearing properties: resolution1/60 of a whole step
= 3 Hz (medium frequency range)
• Logarithmic magnitude characteristic, ear differentiates approx. 325 sound levels
29 DSP_5-Signals&Spectrum
DSP_5-Signals&Spectrum 30
Spectrum numerical
01 ( ) , 0, 1, 2,...jk t
kT
X s t e dt kT
ω−= = ± ±∫
Ts
T
/ stripssN T T
DSP_5-Signals&Spectrum 31
0
0
1
00
1 ( ) , 0, 1, 2,...
1lim ( ) s
s
jkk
TN
jks sT
n
t
nT
X s t e dt kT
s nT e TT
st nT
DSP_5-Signals&Spectrum 32
0
1
00
1 (m )lis
Nn
nT
kk
jsX s nT e
N
Discrete Fourier Transform (DFT)
Neglecting error of numerical integration leads to
DSP_5-Signals&Spectrum 33
% Fourier coefficients |sinus| with FFT N = 256; t = linspace (0,1,(N+1)); t = t([1:N]); s = abs(sin(pi*t)); subplot (311), plot(t,s) d = fft(s)/N; % complex spectrum dMag = abs(d); dPhase=angle(d); % magnitude, phase M = 10; % Fourier coefficients of M spectral lines d0 = dMag(1); dM=2*dMag(2:M); % single-sided spectrum cMag = [d0,dM]; cPhase=dPhase(1:M); subplot (312), stem((0:(M-1)),cMag), xlabel 'Betragsspektrum' subplot (313), stem((0:(M-1)),cPhase), xlabel 'Phasenspektrum'
DSP_5-Signals&Spectrum 34
2 / 2)T T
DSP_5-Signals&Spectrum 35
2 4 cos cos cos( ) ...1.3 3.5 5.7
0.6366 0.4244cos2 0.0849cos4 0.0364cos6 ..
2 4 6
.
t t ts t
t t t
2 / 2T T
sin is function Cosine tereven ms onlyt
DSP_5-Signals&Spectrum 36
Magnitude Phase Magnitude Phase (FFT) (FFT) (Series) (Series)
0.6366 0 0.6366 0
0.4244 3.1416 0.4244 π
0.0849 3.1416 0.0849 π
0.0364 3.1416 0.0364 π
0.0202 3.1416 0.0202 π
0.0129 3.1416 0.0129 π
0.0089 3.1416 0.0089 π
0.0065 3.1416 0.0065 π
0.0050 -3.1416 0.0050 π ≡ - π
0.0040 3.1416 0.0039 π
DSP_5-Signals&Spectrum 37
Dirichlet conditions • For the series to exist, the coefficients must be finite, which is
the case if f(t) is absolutely integrable over one period (weak Diriclet condition):
( )T
s t dt
• s(t) have only a finite number of maxima and minima in one period, and only a finite number of discontinuities in one period (strong Diriclet condition)
• Any periodic waveform generated in a lab satisfies strong Diriclet conditions.
Discrete periodic signals
DSP_5-Signals&Spectrum 38
Signal periodic if [ ] [ ] for all (Shifting property)
A discrete complex exponential can be periodic with period ,but the frequency must be an integer multiple of 2 /
x n N x n n
NNπ
− =
( )
0 0 0 0 0
0
0
1
20
2
1 2
only N different frequencies 0,1,2, , 1
j n N j n j n j N j n
j N j k
kj nj n N
e e e e e
e e N k
e ek N
ω ω ω ω ω
ω π
πω
ω π
− −
=
− −
= → =
= = → =
=→ = −
DSP_5-Signals&Spectrum 39
Non-periodic signals
DSP_5-Signals&Spectrum 40
Periodic extension
Non-periodic signal converted to periodic signal (to use Fourier series) and let tPeriod ∞.
DSP_5-Signals&Spectrum 41
Fourier Transform
( ) ( )
( ) ( )
1( ) ( )2
j t
j t
f t F
F f t e dt
f t F e d
DSP_5-Signals&Spectrum 42
• The Fourier Transform (non-periodic signals) delivers continuous spectra (spectral density).
• The Fourier Series (periodic signals) delivers
discrete spectra.
DSP_5-Signals&Spectrum 43
DSP_5-Signals&Spectrum 44
/ 2
/ 2 / 2
/ 2
2sin1 2
sin2
(
)
s2
in
2
c
j t j je dt eF ej
DSP_5-Signals&Spectrum 45
Spectrum Impulse function
0 )( 1() ( ) j tF tF t e dt
The impulse contains all frequencies with equal magnitude and is therefore an excellent (theoretical) test signal for systems.
DSP_5-Signals&Spectrum 46
Understanding Fourier
• Dk or rather F(ω) magnitude
• Sum if discrete spectral lines kω0
• Integral if spectral density ω
00 0
0
00
1
0
1
1 ( )
1 ( )
( )
(
( )
( ) ( ) 2
)
T
k
N
k k
j n
jk t jk t
sn
j t
kk
D D
D
dtT
s nT eN
f t f t
f dt
e e
e f tF F
j tde t
FS:
FT:
DSP_5-Signals&Spectrum 47
Discrete Fourier transform (DFT) periodic
discrete [n], discrete (ω or k)
1(2 / )
0
1 1
0 0
1(2 / )
0
0,1,
1[ ] [ ]
22 ;
1[ ] [
]
2,...,[
0,1,2,...,
] [ ]
X
[ ] [
1
1
]k n k n
k
n
k s n
N Nj t j t
k
Nj n
N
n
j
k
N
n
k Nx X
k kf t nTN NT
x n X k e k x
X x n
k eN
n
k e k N
N
n
n
eN
π
ω
π
ω
πω π
− −
−−
−
= =
=
−
=
= = =
= ⇔
=
= −
=
=
=
−
∑ ∑
∑
∑
DSP_5-Signals&Spectrum 48
(2 0 / )
(2 / )
(2 2 / )
(2 ( 1) / )
0,1,2,..., 1
1[ ] [0]
1 + [1]
1 [2]
1 + [ 1]
------------------------
---------------------------[ ]
---
-- -0
-[
n
n
n
j N
j N
j N
nj N N
x X eN
X eN
X eN
X
n n N
N
xkX
N e
π
π
π
π −
= = −
=
+
−
(2 0 / )
(2 1/ )
(2 2 / )
(2 ( 1) / )
] [1]
[2]
[ 1
0,1,2,..
]
1
.,
j N
j
k
k
k
k
N
j N
j N N
ex ex e
x
k
N e
Nπ
π
π
π
−
−
−
− −
= −
+
+
+ −
1
0
N
k
−
=∑
1
0
N
n
−
=∑
1(2 / )
0
1[ ] [ ] nN
j k N
knx X k e
Nπ
−
=
= ∑
DSP_5-Signals&Spectrum 49
2
[ ] , 0,1,2, , 1kj n
Nx n e k Nπ
= = −
Re
0 5-1
0
1
k = 0
0 5-1
0
1
k = 10 5
-1
0
1
k = 20 5
-1
0
1
k = 30 5
-1
0
1
k = 4
0 5-1
0
1
k = 70 5
-1
0
1
k = 60 5
-1
0
1
k = 5
0 5-1
0
1
k = 80 5
-1
0
1
k = 90 5
-1
0
1
k = 100 5
-1
0
1
k = 11
N = 8 Sample of next period
1 ↔ 7 2 ↔ 6 3 ↔ 5
8 ↔ 0 9 ↔ 1 10 ↔ 2 11 ↔ 3
DSP_5-Signals&Spectrum 50
( )
( ) ( )
12 /
01
0
cos si[ ] [ ]
[ ] [ ] cos 2 / sin
n
2 /
jakN
j n N
nN
n
k
k
eX x n e
X k
j
x n n N j n Nk
π α α
π π
−−
=
−
=
=
= −
= +∑
∑
( ) ( )3
0
[ ] [ ] cos 2 / 4 sin 2 / 4
[ ] [0]cos(2 0 / 4) [0]sin(2 0 / 4)
0 0 00 [1]cos(2 1 / 4) [1]sin(2 1 / 4)
[2]cos(2 2 / 4) [2]sin(2 2 / 4) [3]cos(2 3
000 /
0
nX x n n j n
X x jxx jxx jxx
k k kπ π
π ππ ππ ππ
=
= −
= ⋅ ⋅ − ⋅ ⋅ ++ ⋅ ⋅ − ⋅ ⋅ ++ ⋅ ⋅ − ⋅ ⋅ ++ ⋅ ⋅
∑
4) [3]sin(2 3 / 4)[ ] [0]cos(2 0 / 4) [0]sin(2 0 / 4)
[1]cos(2 1 / 4) [1]sin(1 1 1
1 11
2 1 / 4) [2]cos(2 2 / 4) [2]sin(2 2 / 4) [3]cos(2 3 / 4) [3]si
1n(1 3
0
2
jxX x jx
x jxx jxx jx
ππ ππ ππ ππ π
− ⋅ ⋅= ⋅ ⋅ − ⋅ ⋅ ++ ⋅ ⋅ − ⋅ ⋅ ++ ⋅ ⋅ − ⋅ ⋅ ++ ⋅ ⋅ − ⋅ ⋅ / 4)
[ ] [0]cos(2 0 / 4) [0]sin(2 0 / 4) [1]cos(2 1 / 4) [1]sin(2 1 / 4) [2]cos(2 2 / 4) [2]sin(2 2 / 4) [3]cos(2 3 / 4) [3]sin
2
(2 3 / 4)[ ] [0]
2 22
1
c
22 22
s2
3 o
X x jxx jxx jxx jx
X x
π ππ ππ ππ π
= ⋅ ⋅ − ⋅ ⋅ ++ ⋅ ⋅ − ⋅ ⋅ ++ ⋅ ⋅ − ⋅ ⋅ ++ ⋅ ⋅ − ⋅ ⋅= (2 0 / 4) [0]sin(2 0 / 4)
[1]cos(2 1 / 4) [1]sin(2 1 / 4)
3 33 33[2]cos(2 2 / 4) [2]sin(2 2 / 4)
[3]cos(2 3 / 4) [3]sin(2 3 / )3 43
3
jxx jxx jxx jx
π ππ ππ ππ π
⋅ ⋅ − ⋅ ⋅ ++ ⋅ ⋅ − ⋅ ⋅ ++ ⋅ ⋅ − ⋅ ⋅ ++ ⋅ ⋅ − ⋅ ⋅
1
0
N
n
−
=∑
1
0
N
n
−
=∑
1
0
N
n
−
=∑
1
0
N
n
−
=∑
DSP_5-Signals&Spectrum 51
Each X[k] DFT-term is the sum of the point-to-point product of the input sequence x[n] and the complex exponential sequence in the form cos(ϕ) – j sin(ϕ).
The “frequency” [k] depends on
• fs the sampling frequency of the original signal and • N the number of samples.
For a sampling frequency of 480 Hz and a 16 point-DFT the frequency spacing is fs/N = 480/16 = 30 Hz and the frequency component are:
X[0]= 0 Hz, X[1]= 30 Hz, X[2]= 60 Hz, …, X[15]= 450 Hz
DSP_5-Signals&Spectrum 52
Matrix representation of DFT
1(2 / )
1(2 / )
0
1
0
2 /0
usi
0,1,2,..., 1
0,1,2,
[ ] [ ] 0,1,2,..., 1
0,1,2,...
[ ] [ ]
1[ ] [ ]
1[ ] [ ]
ng leads to:
..., 1
j
Nj k N
k
Nk
Nk
Nj n N
nN
N
n
Nn
n
k
k
n n N
n n N
x
X
X k eN
x X k WN
x
W e
n e
X n
k k N
k kWx
π
π
π
−
=
−
−−
=
−
−
=
=
=
=
= −
=
=
= −
= −
=
∑
∑
∑1
0
, 1N
nN
−
=
−∑
DSP_5-Signals&Spectrum 53
( )
1 2 1
2 12 4
1 1
[0] [0] 1 [1] 1 [2] 1 [ 1] 1[1] [0] 1 [1] [2] [ 1]
[2] [0] 1 [1] [2] [ 1]
[ 1] [0] 1 [1] [2] [ 1]__________________________
NN N N
NN N N
N NN N
X x x x x NX x x W x W x N W
X x x W x W x N W
X N x x W x W x N W
−
−
− −
= ⋅ + ⋅ + ⋅ + − ⋅
= ⋅ + ⋅ + ⋅ + − ⋅
= ⋅ + ⋅ + ⋅ + − ⋅
− = ⋅ + ⋅ + ⋅ + − ⋅
0
_____________________1NW =
1
0
[ ] [ ] 0,1,2,..., 1N
Nkn
nk kn NX x W
−
=
= −= ∑
DSP_5-Signals&Spectrum 54
2 /j NNW e π−=
1 2 1
2 4 2( 1)
1 2( 1) 1) ( 1)
1 1 1 1[0] [0]1[1] [1]1[2] [2]
1[ 1] [ 1]
NN N N
NN N N
N N N x NN N N
X xW W WX xW W WX x
W W WX N x N
−
−
− − − −
= ⋅ − −
NW
= NX W x
1 1= N
− −= 1 *N NNx W WW X
0.2
0.4
0.6
0.8
1
30
210
60
240
90
270
120
300
150
330
180 0
N=8
DSP_5-Signals&Spectrum 55
Matlab support for WN function DFTmatrixDarst sig=[1 1 1 1 0 0 0 0]; N=length(sig);
SIG=sig*dftmtx(N); sigrueck=(1/N)*(SIG*conj(dftmtx(N))); figure subplot(1, 2 ,1) stem(sig) title 'Signal' subplot(1, 2, 2) stem(abs(sigrueck)) title 'Signal transf & rücktransf'
0 2 4 60
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Signal0 2 4 6
0
0.5
1
1.5
2
2.5
3
3.5
4
Betragsspektrum0 2 4 6
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Signal => transf => rücktransf
DSP_5-Signals&Spectrum 56
34( ) sin(2 . . ) 0.5cos(2 . . )
2 2( ) [ ] cos
1
sin
2x t t t
X k x n kn j knN N
π π π
π π
= + +
= − ∑
Example:
See Excel-sheet
DSP_5-Signals&Spectrum 57
0 1 2 3 4 5 6 7 8-1.5
-1
-0.5
0
0.5
1
1.5
[ ] [0.3536 0.3536 0.6464 1.0607 0.3536 -1.0607 -1.3536 -0.3536 0.3536]x n =
DSP_5-Signals&Spectrum 58
x[n] "2pi/8" n k=1 cos sin x[n].cos x[n].sin
0,3535 0,785398 0 0,000 1,000 0,000 0,354 0,000
0,3535 0,785398 1 0,785 0,707 0,707 0,250 0,250
0,6464 0,785398 2 1,571 0,000 1,000 0,000 0,646
1,0607 0,785398 3 2,356 -0,707 0,707 -0,750 0,750
0,3535 0,785398 4 3,142 -1,000 0,000 -0,354 0,000
-1,0607 0,785398 5 3,927 -0,707 -0,707 0,750 0,750
-1,3535 0,785398 6 4,712 0,000 -1,000 0,000 1,354
-0,3535 0,785398 7 5,498 0,707 -0,707 -0,250 0,250
0,000 -4,000 4 ∠ -90 Grad
DSP_5-Signals&Spectrum 59
x = 0.3535 0.3535 0.6464 1.0607 0.3535 -1.0607 -1.3535 -0.3535
Matlab: X = fft(x)
Magnitude spectrum
34( ) sin(2 . . ) 0.5cos(21 2. . )x t t tπ π π= + +
1 2
[1] 4 [2] 2 1 0.5X X
A A= =
= =
real2 complex
rA
cA
NX A
X AN
=
=
Attention!
Real- & Imaginary part of DFT
DSP_5-Signals&Spectrum 60
0 1 2 3 4 5 6 7-4
-2
0
2
4
Realteil
0 1 2 3 4 5 6 7-4
-2
0
2
4
Imaginärteil
Matlab
DSP_5-Signals&Spectrum 61
FFT Discrete Fourier transform. FFT(X) is the discrete Fourier transform (DFT) of vector X. For matrices, the FFT operation is applied to each column. For N-D arrays, the FFT operation operates on the first non-singleton dimension.
IFFT Inverse discrete Fourier transform. IFFT(X) is the inverse discrete Fourier transform of X.
• fft(x) represents spectrum from 0 to 2π
• Use Matlab function fftshift(X) to shift zero-frequency component to center of spectrum
DSP_5-Signals&Spectrum 62
DSP_5-Signals&Spectrum 63
• All types of FT are based on the complex exponential function (sequence).
• The complex exponential reaches from -∞ to ∞.
• The synthesis of a (continuous) aperiodic signals requires ∞ frequency components (which cancel out by superposition if signal is zero)
• It is not possible to store ∞ components in a digital computer
• The DFT represents discrete, periodic signals.
DSP_5-Signals&Spectrum 64
(Complex) DFT
0 N-1 0 N/2 N-1
0 N-1 0 N/2
N-1
Time domain Frequency domain
Realteil Realteil
Imaginärteil Imaginärteil
Symmetry of DFT
DSP_5-Signals&Spectrum 65
1 bis 1 2 2N Nk k k= = − >
Frequency components
[ ] *[ ]X k X N k= −
Real part even, Imaginary part odd
Frequency spacing of DFT
DSP_5-Signals&Spectrum 66
After sampling the frequency axes is „lost“
ˆ[ ] ( ) cos( ) cos( )s sx n x nT A nT A nω ϕ ω ϕ= = + = +ˆ snTω =
scontinous
k ffN
= Spacing of spectral lines sfN
No frequency components outside spacing raster!
(Circular) Time-shifting
DSP_5-Signals&Spectrum 67
[ ] [ ]x n X k⇔2
shifted[ ] [ ] [ ]kj m
Nx n m X k X k eπ
−− ⇔ =
Each spectral component is shifted by
Phase shift is proportional to frequency (linear phase shift)
2 kj mNeπ
−
68
[ ] [1.0607 0.3536 -1.0607 -1.3536 -0.3536 0.3536 0.3536 0.3536 0.6464 ]x n =
1 90 , 2 45k k= → − ° = → + °
-3 -2 -1 0 1 2 3 4 5 6 7-1.5
-1
-0.5
0
0.5
1
1.5
0 1 2 3 4 5 6 7-4
-2
0
2
4
Betrag
0 1 2 3 4 5 6 7
-50
0
50
Phase
2 138
1352 23
8
270
90 45
+45 315 45
j
j
e
e
π
π
°
°
⇒ − ° + = + °
⇒ ° + = ° = − °
[ ] [0.3536 0.3536 0.6464 1.0607 0.3536 -1.0607 -1.3536 -0.3536 0.3536]x n
DSP_5-Signals&Spectrum
DFT
DSP_5-Signals&Spectrum 69
21
0
[ ] [ ]nN j k
N
nX k x n e
π−
=
= ∑−
21
0
1[ ] [ ]kN j n
N
kx n X e
Nk
π−
=
= ∑
iDFT
Computation of iDFT with DFT
DSP_5-Signals&Spectrum 70
2 2
2
*1 1
* *
0 0*
1*
0
1 1[ ] [ ] [ ]
1[ ] [ ]
N N
N
N Nj kn j kn
k k
Nj kn
k
x n X k e X k eN N
x n X k eN
-1FFT
-1
1/N
1/N
Re [ ]X k
Im [ ]X k
Re [ ]x n
Im [ ]x n
( )*
*
*
1.
2.
3. Scaling by 1 /(4. not required for real sequences)
FFT
N
→X X
X
x
DSP_5-Signals&Spectrum 71
FFT
1/N
1/N
Re [ ]X k
Im [ ]X k
Re [ ]x n
Im [ ]x n
Non-periodic signals (Zero Padding)
DSP_5-Signals&Spectrum 72
DFT based on periodic sequences signal representation periodic in time and frequency domain
0 5 10 150
0.2
0.4
0.6
0.8
1Zeitbereich L= 16 N= 16
0 5 10 150
2
4
6
8
Frequenzbereich
20.5 [1 cos 0 1[ ]
0 sonst
n n Lx n L
π − ≤ ≤ − =
DSP_5-Signals&Spectrum 73
0 5 10 15 20 25 30 350
0.2
0.4
0.6
0.8
1Zeitbereich L= 16 N= 32
0 5 10 15 20 25 30 350
2
4
6
8
Frequenzbereich
L = 16 N = 32
DSP_5-Signals&Spectrum 74
0 20 40 60 80 100 120 1400
0.2
0.4
0.6
0.8
1Zeitbereich L= 16 N= 128
0 20 40 60 80 100 120 1400
2
4
6
8
Frequenzbereich
L = 16 N = 128
DSP_5-Signals&Spectrum 75
iscrete ime
(
[ ]) j n
n
N
x n eX ωω∞
−
=−∞
→ ∞
= ∑
(D T FT)
DSP_5-Signals&Spectrum 76
1DTF
1[ ] 0.5 [ ] ( )
1T
0.5
n
jx n n X
e
-10 -8 -6 -4 -2 0 2 4 6 8 10
0.8
1
1.2
1.4
1.6
1.8
2
Betrag
-10 -8 -6 -4 -2 0 2 4 6 8 10-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
Phase
0 0.5 1 1.5 20
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
DSP_6-FIR 1
Digital Analog
1. Digital signals less sensitive to noise than analog signals (signal state only 0 or1)
2. Coding (compression or security) and processing easier as for analog signals
DSP_6-FIR 2
3. Duplication of digital data is easy, no loss of quality, copying analog data (audio tape, film) leads to reduced quality due to noise
4. Storing digital signals easier and more cost effective than analog signals
5. Digital hardware (chip technology) easier and more cost effective to build than analog HW
DSP_6-FIR 3
7. Analog signal processing still required (aliasing and interpolation)
8. Most signals are analog 9. Sampling theorem and A/D- and conversion
limit digital applications (chip technology) 10. Today: audio signal processing digital, video
signal processing a good portion DSP
DSP_6-FIR 4
A/D conversion
S&H circuit A/D converter
DSP_6-FIR 5
Finite Impulse Response Filter
Discrete-Time System
Input Output
[ ] [ ]y n x n[ ]x n
DSP_6-FIR 6
3 point running average
13
13
13
[0] ( [0] [1] [2])[1] ( [1] [2] [3])
...[ ] ( [ ] [ 1] [ 2])
y x x xy x x x
y n x n x n x n
Filter equation
Output before input: non causal!
n n < -2 -2 -1 0 1 2 3 4 5 6 7 n > 7
x[n] 0 0 0 2 4 6 4 2 0 0 0 0
y[n] 0 2/3 2 4 14/3 4 2 2/3 0 0 0 0
DSP_6-FIR 7
Causality
0 5 10 15 20 250
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Now
Past | Future
13[ ] ( [ ] [ ] [ ])
backward vs foreward avera
1 2
ge
y n x n x n x n
DSP_6-FIR 8
General FIR Filter
0
1 1 13 3 3
[ ] [ ]
Number of filter coefficients: Length of filter 1
e.g.: [ ] [ ] [ 1] [ 2
Order of f
]
ilter M,
M
kk
y n b x n k
L M
y n x n x n x n
DSP_6-FIR 9
Example
DSP_6-FIR 10
• We observe a M-point run-into phase (and run-off phase (not shown)).
• Size of higher frequency component has been reduced Low pass behavior
• Output signal phase-shifted to the right
DSP_6-FIR 11
• FIR-Application (1): Suppress high frequency components, ==> low pass
• FIR-Application (2): Suppress low frequency components, ==> high pass
DSP_6-FIR 12
13[ ] ( [ ] [ 1] [ 2])y n x n x n x n
1 1 13 3 3( [ 1] [ 2] [ 3[ 2] ] ][ )x n x n xx ny nn
Av subtracted from aerage ctual va vlue alue
bk = -1/3 2/3 -1/3
bk = 1/3 1/3 1/3
DSP_6-FIR 13
Unit (Im)Pulse Impulse function, Dirac function
1 0[ ]
0 0n
nn
n … -2 -1 0 1 2 3 δ[n] 0 0 0 1 0 0 0 δ[n-2] 0 0 0 0 0 1 0
-4 -3 -2 -1 0 1 2 3 40
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Einheitsimpuls
DSP_6-FIR 14
Representing signals by impulse functions
[ ] [ ] [ 1] [ 2]2 4 6 [ 3 2 44 ] [ ]x n n n n n n
[ ][ ] [ ]k
x kx n n k
DSP_6-FIR 15
Unit Impulse Response Input: Unit Impulse Output: Impulse Response
0
0
[ ] [ ]
[ ]0,1,2,...,
[ ] [ ]0 el e
[ ]
[ ]s
M
kk
Mn
kk
y n b x n k
x nb n M
y n n kbh n
n
bk ][[
]x nn ]
[[
]y nh n
DSP_6-FIR 16
Example 1
0 1 2 3
1 2
Running Avarage Filter [ ] [ ] [ 1] [ 2]
o
b b by n b n b n b n
Impulse response is sequence of filter coefficients
[ ] 0 for 0 and for h n n n M
Length of impulse response ist finite finite impulse response
0 0.5 1 1.5 20
0.05
0.1
0.15
0.2
0.25
0.3
0.35
n (samples)
Am
plitu
de
Impulse Response
B = [1 1 1]/3 A = 1 impz(B,A)
DSP_6-FIR 17
Delay Filter
0
1 2
Filter coefficients 0,0,1
[ ] [ ] [ 1] [ 2]
[ ] [ ]
k
o
by n b x n b x n b x n
y n x n n
0 0 1 [ 2]x n
[ ] [ 2]y n x n Delay by two time steps
0 0.5 1 1.5 20
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
n (samples)
Am
plitu
de
Impulse Response
impz([0 0 1],1)
DSP_6-FIR 18
Convolution (Faltung)
0
[ ] [ ]k
M
k
y n x n kb
Impulse response Filter coefficients
0
replaced by [ ]
[ ][ ] [ ]M
k
k
y n xh k
b h
k
k
n
DSP_6-FIR 19
x = [1 2 3 -2 2 1] h = [3 -1 2 1]
0 1 2 3-1
0
1
2
3
h[k]0 2 4 6
-2
-1
0
1
2
3
x[k]
-6 -4 -2 0-2
-1
0
1
2
3
x[-k]
folded
DSP_6-FIR 20
-5 0 5 10-202
n = 00 5 10
05
1015
x[n]*h[n] n = 0
-5 0 5 10-202
n = 10 5 10
05
1015
x[n]*h[n] n = 1
-5 0 5 10-202
n = 30 5 10
05
1015
x[n]*h[n] n = 3
-5 0 5 10-202
n = 80 5 10
05
1015
x[n]*h[n] n = 0
h = [3 -1 2 1] x = [1 2 3 -2 2 1]
h = [3 ] * x = [1] 3
h = [3 -1 ] * x = [2 1] 5
h = [3 -1 2 1] * x = [ -2 3 2 1 ] - 4
3 5 9 -4 16 0 1 4 1
DSP_6-FIR 21
Superposition of input pulses Weight Weight x (delayed Impulse Response)
x[0] = 1 y[0] = [1]h[n-0] 3 -1 2 1 0 0 0 0 0 0
x[1] = 2 y[1] = [2]h[n-1] 0 6 - 2 4 2 0 0 0 0 0
x[3] = 3 y[3] = [3]h[n-3] 0 0 9 - 3 6 3 0 0 0 0
x[4] = -2 y[4] = [-2]h[n-4] 0 0 0 - 6 2 - 4 - 2 0 0 0
x[5] = 2 y[5] = [2]h[n-5] 0 0 0 0 6 - 2 4 2 0 0
x[6] = 1 x[6] = [1]h[n-6] 0 0 0 0 0 3 - 1 2 1 0
y[n] = ∑x[n]h[n-k] 3 5 9 - 4 16 0 1 4 1 0
Each input sample generates is own impulse response. The system response is the superposition of weighted und time-delayed impulse responses.
DSP_6-FIR 22
Composition of output signal
h[0] = 3 y[0] = [3]x[n-0] 3 6 9 - 6 6 3 0 0 0 0
h[1] =-1 y[1] = [-1]x[n-1] 0 - 1 - 2 - 3 2 - 2 - 1 0 0 0
h[3] = 2 y[3] = [2]x[n-3] 0 0 2 4 6 - 4 4 2 0 0
h[4] = 1 y[4] = [3]x[n-4] 0 0 0 1 2 3 - 2 2 1 0
y[n] = ∑h[k]x[n-k] 3 5 9 - 4 16 0 1 4 1 0
Output signal is sum of weighted and time-delayed input signals. Input signal weighted by filter coefficients.
DSP_6-FIR 23
Convolution and Matlab x = [1 2 3 -2 2 1] h = [3 -1 2 1] conv(h,x) 3 5 9 -4 16 0 1 4 1
1 2 3 4 5 6 7 8 9-4
-2
0
2
4
6
8
10
12
14
16
stem(conv(h,x))
DSP_6-FIR 24
Willkommen!
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5x 104
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.5 1 1.5 2 2.5x 104
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
DSP_6-FIR 25
DSP_6-FIR 26
Remember: Multiplication of polynomials
2 21 2 1 2
21 0 2 0
2 31 1 1 2 1
2 3 42 1 2 2 2
( )( )
o o
o o
o
o
a a x a x b b x b x
a b a b x a b x
a b x a b x a b x
a b x a b x a b x
A = [a0,a1,a2] B = [bo,b1,b2] conv(A,B)
DSP_6-FIR 27
Implementation of FIR-Filters
0
[ ] [ ]M
kk
y n b x n k
Addition Multiplication Delay (Storage)
x[n] ×
y[n]
b
+
x1[n]
y[n] x2[n]
x[n] y[n] Unit Delay
DSP_6-FIR 28
Block diagram (direct form) 3
0
0 1 2 3
[ ] [ ]
[ ] [ ] [ 1] [ 2] [ 3]
kk
y n b x n k
y n b x n b x n b x n b x n
×
Unit Delay
Unit Delay
× × ×
+ + +
0b 1b 2b3b
[ ]x n [ 1]x n [ 2]x n [ 3]x n
[ ]y n
Unit Delay
DSP_6-FIR 29
Block diagram (transposed form)
×
Unit Delay
Unit Delay
Unit Delay
× × ×
+ + +
3b 2b 1b0b
[ ]x n
[ ]y n
1[ ]v n2[ ]v n3[ ]v n
0 1
1 1 2
2 2 3
3 3
[ ] [ ] [ 1][ ] [ ] [ 1][ ] [ ] [ 1][ ] [ ]
y n b x n v nv n b x n v nv n b x n v nv n b x n
2 2 3
1 1 2 3
0 1 2 3
[ ] [ ] [ 1][ ] [ ] [ 1] [ 2][ ] [ ] [ 1] [ 2] [ 3]
v n b x n b x nv n b x n b x n b x ny n b x n b x n b x n b x n
DSP_6-FIR 30
Convolution and LTI systems [ ][ ]
x nn
[ ][ ]
y nh nLTI System
[ ] [ ] [ ] general case l
x n x l n l l
Input signal is a sequence of weighted and time-delayed impulses. The system response is:
[0] [0] [0] [ ][1] [ 1] [1] [ 1][2] [ 2] [2] [ 2][ ] [ ] [ ] [ ]
x x h nx n x h nx n x h nx l n l x l h n l
DSP_6-FIR 31
Convolution and LTI Systems (2)
Application of superposition principle leads to:
[ ] [ ] [ ] [ ] [ ] [ ]l l
x n x l n l y n x l h n l Most general case: Neither x[n] nor h[n] are of finite duration:
[ ] [ ] [ ]l
y n x l h n l
A LTI-System can be represented by its convolution sum.
DSP_6-FIR 32
Properties of LTI systems
0
1 2
[ ] [ ] [ ] [ ] [ ]
Convolution operator
Convolution with an impulse [ ] [ ] [
Commutativ:
]
[ ] [ ] [ ] [ ]
[ ] [ ] [ ] [ ] [ ]
[ ] [ ]Associative:
l
o
l
y n x n h n x l h n l
x n n n x n n
x n h n h n x n
y n h n x n h l x n l
x n x n x
3 1 2 3[ ] [ ] [ ] [ ]n x n x n x n
DSP_6-FIR 33
Cascaded LTI Systems LTI 1 LTI 2
LTI
LTI 2 LTI 1
1 2 1 2
2 1 2 1
[ ] ( [ ] [ ]) [ ] [ ] [ ] [ ]
[ ] [ ] [ ] ( [ ] [ ]) [ ]
y n x n h n h n x n h n h n
x n h n h n x n h n h n
1[ ] [ ]x n h n[ ]x n 1 2( [ ] [ ]) [ ]x n h n h n
2[ ] [ ]x n h n[ ]x n 2 1( [ ] [ ]) [ ]x n h n h n
1 2[ ] [ ]h n h n
1[ ]h n 2[ ]h n
1[ ]h n2[ ]h n
Frequency Response of FIR Filters (Frequenzgang)
DSP_6-FIR 35
Sinusoidal input signals
ˆ is a dimension-less
ˆ[ ] ( ) cos( ) cos( )ˆ ˆ is the normalized frequenc
uniy
t.
s s
s
x n x nT A nT A nT
The index in x[n] is also dimensionless. A sampled (continous) signal x(t)carries no time information : A time - discrete signal is only a sequence ofnumbers and doesn't contain information
To reconstruct the signal the s
about the sampl
ampling period m
i
u
ng
st
int
be
erva
kn
ll.
own!
DSP_6-FIR 36
ˆ
ˆ (
0
)
0
0
( ) sampling continous discrete
ˆˆ /2
[ ]
[ ]
[ ] []
]
[
M M
j j t
s s
j j n
jM
jk k
k
kn
kk k
j
b k
x t Ae e
f fT
x n Ae e
x n x n
Ae e
f f
y n
b e
h k k
b
ˆˆ
0
( )ˆ
j jM
k
k
nAe e
H
DSP_6-FIR 37
The complex exponential is the only signal passing the linear system without changing the wave form. Only amplitude and phase (complex amplitude) change, but the sinusoidal property remains. The expression is only valid for the input signal and is meaningless for any other signals.
ˆ[ ] [ ]( )y n x n= Hˆj j nAe e
DSP_6-FIR 38
Frequency response ˆ
ˆ ˆ
0 0
ˆ( )
ˆ( )
ˆ ˆ( ) describes response of the LTI-
[ ]
Sys
[
tems for ev
]
ery .
j j n
M Mj k j k
k kk k
y n Ae e n
b e h k e
= H
H
Hˆ ˆj n j ne e A B
same frequency
but different (complex) amplitudes
DSP_6-FIR 39
Example
ˆ ˆ 2
ˆ
ˆ
ˆ ˆ
1 2 1
ˆ( ) 1 2 ( 2 ) ˆ2 2cos
ˆ
(ˆ
) ˆ) (
j
k
j j
j j j
b
e ee
ee e
H
H
ˆ
0
ˆ( )M
j kk
kb e
H
DSP_6-FIR 40
Frequency response (2)
ˆ( )
ˆ( ) ... complex funktionˆ ˆ ˆ( ) ( ) ( )
ˆ ( ) j
e j m
e
H
H
H H
H
H
Magnitude, Betrag Gain, Verstärkung
Phase
DSP_6-FIR 41
Bode diagram
0.5 ½
fsampling π
DSP_6-FIR 42
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9-180
-160
-140
-120
-100
-80
-60
-40
-20
0
Normalized Frequency (×π rad/sample)
Phas
e (d
egre
es)
Phase Response
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.90
0.5
1
1.5
2
2.5
3
3.5
4
Normalized Frequency (×π rad/sample)
Mag
nitu
de
Magnitude Response
Low pass bk = [1 2 1]
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-200
-150
-100
-50
0
Normalized Frequency (×π rad/sample)
Pha
se (d
egre
es)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-100
-80
-60
-40
-20
0
20
Normalized Frequency (×π rad/sample)
Mag
nitu
de (d
B)
Magnitude in logarithmic representation
freqz([1 2 1],1)
fvtool([1 2 1],1)
1
Nomalized Frequency (∙ π rad/sample) 1
1
1
DSP_6-FIR 43
Logarithmic representation
log [Bel]out
in
P aP
2
2log 2log [Bel] 20log [dB]out out out
in in in
U U UU U U
To describe properties of filters one is interested in the ratio of input power to output power. Because this power ratio spans several powers of ten logarithmic representation is appropriate. The ratio is measured in Bel.
Comparing (voltages) amplitudes leads to:
10 20 dB
1/1000 -60 dB 1/100 -40 dB 1/10 -20 dB
-3 dB 1 0 dB
3 dB 2
1/ 2
DSP_6-FIR 44
Example FIR Filter
ˆ
ˆ ˆ
ˆ ˆ ˆ2 3 4
ˆ ˆ2 2
ˆ2
ˆ2
ˆ2
2 ˆˆ ˆ2 ˆ
1, 2,4, 2,1
ˆ( ) 1 2 4 2after separating we get
ˆ(
ˆ2c
2 2
ˆ2 2 4cosˆ4cos
) 4
ˆ( )ˆ( ) 4
os2 ˆ2cos
4
42
j j
j j
j j
j j
k
j j j j
j
j
j
ee e
b
H e e e ee
H e
H e
e
e ee
H
e
ˆ ˆ ˆ ˆcos 2cos2 ( ) 2
DSP_6-FIR 45
bk = [1 -2 4 -2 1]
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.451
2.8
4.6
6.4
8.2
10
Frequency (Hz)
Mag
nitu
deMagnitude and Phase Responses
-7
-5.6
-4.2
-2.8
-1.4
0
Phas
e (r
adia
ns)
DSP_6-FIR 46
In example [1 2 1] and [1 -2 4 -2 1] the phase ˆ ˆ ˆ ˆ( ) ( ) 2
is a linear function of
FIR-Filters have linear phase if the filter coefficients are:
[1 -2 4 -2 1]
für 0,1,...,k M kb b k M
0
*0, 0 *
0, 0
If is a zero of H( ) is, than1 1 , ,
are also zeros of ( ).
z z
z zz z
H z
DSP_6-FIR 47
Meaning of linear phase
0 0.5 1 1.5 2 2.5 3 3.5-1
-0.5
0
0.5
1
0 0.5 1 1.5 2 2.5 3 3.5-1
-0.5
0
0.5
1
0 0.5 1 1.5 2 2.5 3 3.5-1
-0.5
0
0.5
1
0 0.5 1 1.5 2 2.5 3 3.5-1.5
-1
-0.5
0
0.5
1
1.5
0 0.5 1 1.5 2 2.5 3 3.5-1
-0.5
0
0.5
1
0 0.5 1 1.5 2 2.5 3 3.5-1
-0.5
0
0.5
1
Phase shift constant ϕ1 = ϕ 2 = ϕ3 = 60°
Phase linear (proportional to frequency) ϕ1 = 30 ° ϕ2 = 60° ϕ3 = 90°
Linear phase maintains wave form of signal
DSP_6-FIR 48
Sounds are phase independent: All 4 wave forms sound the same
0 20 40 60 80 100 120 140 160 180 200-3
-2
-1
0
1
2
3
0 20 40 60 80 100 120 140 160 180 200-3
-2
-1
0
1
2
3
0 20 40 60 80 100 120 140 160 180 200-3
-2
-1
0
1
2
3
0 20 40 60 80 100 120 140 160 180 200-3
-2
-1
0
1
2
3
Test signal for linear phase
DSP_6-FIR 49
DSP_6-FIR 50
0 200 400 600 8000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Tscheb 5. Ordnung, 1 dB
200 400 600 800
0.5
1
1.5
0 200 400 600 8000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Eingangssignal
200 400 600 800
0.5
1
1.5
0 200 400 600 8000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
LinPhase 40.Ordnung, max. flach
200 400 600 800
0.5
1
1.5
Ringing
DSP_6-FIR 51
Superposition (Überlagerung)
*
01
01
*
0
0
ˆ ˆ
ˆ ˆ
1
[ ]2 2
cos
... (for real sequences [ ]ˆ ˆ( ) ( )
ˆ ˆ(0
)
[ ]2 2
) ( ) ( )
( 0)
ˆ
k k
k k
Nk k
k
N
k kk
Nk k
k
j n j n
k
j n j nk k
X Xx n X
X X X
x n
X Xy n X
e
n
X
e
e e
*H H
H H H
H 1
ˆˆ ˆ( ) (cos )N
k kk kk
k X n X
H H
DSP_6-FIR 52
Example
4
910
1,2,1
[ ]
(0)
4
3.414
94 3cos 2cos4 2 10
9
4
0.097
4 3.414 0.03cos 2cos
49
4
10
[ ]2
94
7
k
j
j
b
x n
e
e
y n
n n
n
H
H
H
00 119n
DSP_6-FIR 53
Using Matlab
[H,w] = freqz(B,A,w) [H,w] = freqz( [1 2 1],1, [0,(pi/4),(pi*9/10)]) [4,(2.4142-2.4142i),(-0.0931-0.0302i] abs(H) und angle(H) [4ej0,3.41420e-j0.7854,0.0979e-j2.8274] [4,3.41420e-jπ/4,0.0979e-j9π /10]
DSP_6-FIR 54
FREQZ Digital filter frequency response. H = FREQZ(B,A,W) returns the frequency response at frequencies designated in vector W, in radians/sample (normally between 0 and pi) of the filter:
ˆ ˆ0 1
ˆ ˆ0 1
....ˆ( )....
Numerator and denominator coefficients given in vectors B and A.
j j mm
j j nn
b b e b eHa a e a e
A = [1] for FIR-Filters
DSP_6-FIR 55
Frequency Domain
ˆ( )HSpectrum
of signals is modified by the
frequency response of the system
Single sampli NOTng values considered.
DSP_6-FIR 56
Properties of frequency response
ˆ
0 0
ˆ 2
0 1
ˆ
ˆ ˆ( ) (
ˆ[ ] [ ] [ ] ( ) [ ]
ˆ ( 2 )
ˆ ˆ( ) ( )
) wenn reel
l
M Mj k
k kM
j k j kk
k
k
h n h k n k h k e
b e e
b
*
H H
H
H
H H
H(ω) Spectrum periodic
ˆ( )
ˆe ( ) ... ˆ ˆ ˆ( ) ( ) ( )
ˆe ( )ˆm ( )
ˆFrequency respose only needed for 0 .
ˆm ( ) . .
..
H
H
H
H
H
H
Magnitude ... evenPhase ... odd
Re evenIm odd
. Symmetry reasons
DSP_6-FIR 57
Cascaded LTI Systeme
1 2
ˆ ˆ ˆ ˆ ˆ( ) ( ) ( ) ( ) ( )ˆ ˆ[ ] [ ] ( ) ( )h n h n
1 2 2 1
1 2
H H H H H
H H
Convolution in time domain Multiplication in frequency domain
DSP_6-FIR 58
ˆ ˆ ˆ ˆ2 3 4
ˆ ˆ ˆ2 3
Cascaded systemsˆ( ) 2 4 6 4 2ˆ( ) 1 2 2
ˆ ˆ ˆ( ) ( ) ( )
j j j j
j j j
e e e ee e e
1
2
1 2
H
H
H H H
w=-pi:(pi/500):pi; b1=[2 4 6 4 2]; b2=[1 -2 2 -1]; b=conv(b1, b2); H1=freqz(b1,1,w); H2=freqz(b2,1,w); H=freqz(b,1,w);
-4 -3 -2 -1 0 1 2 3 40
5
10
15
20
-4 -3 -2 -1 0 1 2 3 40
2
4
6
-4 -3 -2 -1 0 1 2 3 40
5
10
15
H1
H2 H3
DSP_6-FIR 59
First Difference System bk =[1 -1]
ˆ
[ ] [ ] [ 1]ˆ ˆ ˆ[ ] 1 1 cos sinˆ ˆ[ ] 2 sin( / 2)
ˆsinˆ[ ] arctanˆ1 cos
j
y n x n x ne j
H
H
H
DSP_6-FIR 60
High pass bk =[1 -1]
-15 -10 -5 0 5 10 150
0.5
1
1.5
2
ω
Bet
rag
-15 -10 -5 0 5 10 15-2
-1
0
1
2
ω
Pha
se
2π 4ππ
DSP_6-FIR 61
Running-Average Filter
1
01
ˆ
0
1
0
ˆ ˆ ˆ/2 /2 /2ˆ1ˆ
ˆ ˆ ˆ ˆ/2 /2 /20
1[ ] [ ]
( )
L-point running averager
1( )
1Using 1
1 1 1 11
L
kL
j k
kLL
k
k
j L j L j Lj LLj k
j j j jk
y n x n kL
eL
e e eeeL L e L e e e
H
H
Phase
Magnitude
ˆ 1 /2
ˆsi
n / 2
ˆs n
/ 2=
i
LjLL
e
DSP_6-FIR 62
ˆ ( 1)/2ˆ( )ˆsin / 2
ˆsin /
ˆ( )
ˆ(
ˆ(
2)
)
j
L
L
LLD
Le
L
H
D
D
diric
is called Diriclet - or periodic sinc -Function,the corresponding Matlab function is .
DSP_6-FIR 63
x=-pi:(pi/200):pi; plot (x,diric(x,11));
Periodic sinc-Function from -3π to 3π -4 -3 -2 -1 0 1 2 3 4
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-10 -8 -6 -4 -2 0 2 4 6 8 10-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
π -π
DSP_6-FIR 64
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.90
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Normalized Frequency (×π rad/sample)
Mag
nitu
de
Magnitude Response
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9-200
-150
-100
-50
0
50
100
Normalized Frequency (×π rad/sample)
Phas
e (d
egre
es)
Phase Response
X: 0.422 Y: -20.2
X: 0.418 Y: -16.6
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9-900
-800
-700
-600
-500
-400
-300
-200
-100
0
Normalized Frequency (×π rad/sample)
Con
tinuo
us P
hase
(deg
rees
)
Continuous Phase Response
Phase ±180° Phase continuous
Run
ning
Avg
.-Filt
er b
k =
ones
(1,1
1)/1
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9-90
-80
-70
-60
-50
-40
-30
-20
-10
0
Normalized Frequency (×π rad/sample)
Mag
nitu
de (d
B)
Magnitude Response (dB)
Amplitude linear Amplitude in dB
DSP_6-FIR 65
Transient behavior (1) • Frequency domain analysis only for
sinusoidal signals, i.e. input (output) signal reaches from -∞ to ∞.
• But real signals are switched on/off. Steady state of FIR filter only after all delay elements are »filled« with values from input signal.
• Until this state is reached analysis of the system behavior must be done in the time domain (transient analysis).
DSP_6-FIR 66
Transient behavior (2) bk = [1,2,4,2,1]
DSP_6-FIR 67
Low pass – high pass
DSP_6-FIR 68
Band pass – band stop
DSP_6-FIR 69
Multiband
DSP_6-FIR 70
From ideal to real low pass filter
ˆ1ˆ( )
ˆ0c
iLPc
Hω ω
ωω ω π
≤= < ≤
π ωc
ˆ1 1 sinˆ ˆ( ) , 2 2
c cj n j nj n c
iLP iTPe e nh H e d n
jn jn n
π ω ωω
π
ωω ωπ π π
−
−
= = − = −∞ < < ∞
∫
Impulse response not causal!
Fourier transform
DSP_6-FIR 71
-10 -5 0 5 10-0.2
0
0.2
0.4
0.6
0.8
1
1.2
Time (sec)
Am
plitu
deSinc Function
-M M
DSP_6-FIR 72
( )( )
sin 0 1, 2 1
0 else
c n Mn M
LP
n N N Mh
ωπ
− −
≤ ≤ − = +=
Truncating at – M and M and shifting to the right by M produces a (different but) causal low pass filter
Computing frequency response of filter hLP (again by using the Fourier transformation)
DSP_6-FIR 73
Above expression corresponds to multiplication with rectangular window
LP idealLP Rh h w= ×
( )( )
sin 0 1, 2 1
0 else
c n Mn M
LP
n N N Mh
ωπ
− −
≤ ≤ − = +=
DSP_6-FIR 74
LP idealLP Rh h w= ×
Frequency domain
idealLPh ⇒ Ideal filter
ˆ(2 1)2ˆˆ2
1 0[ ]
0 else
sinˆ( )
sin
R
MMj n
Rn M
n Mw n
W eω
ωω
ω+
−
=−
≤ ≤=
= =∑
DSP_6-FIR 75
ˆ( )1ˆ ˆ ˆ( ) ( ) ( ) ( ) ( )2
j jLP idealLP R idealLP RH H W H e W e d
πφ ω φ
π
ω ω ω φπ
−
−
= ∗ = ∫
LP idealLP Rh h w= ×Multiplication in the time domain ↓ convolution in the frequency domain
DSP_6-FIR 76
DSP_6-FIR 77
Gibbs‘ phenomenon
DSP_6-FIR 78
Windowing (1)
Time
Frequency
DSP_6-FIR 79
Windowing (2)
Hamming2[ ] 0.54 0.46cos
2 1nw n M n M
Mπ = + − ≤ ≤ +
Time
Frequency
DSP_6-FIR 80
( )( )
20
Kaiser0
1 /,
I n Mw M n M
I
β
β
− = − ≤ ≤
0 50 1000
0.2
0.4
0.6
0.8
1
β = 00 50 100
0
0.2
0.4
0.6
0.8
1
β = 2
0 50 1000
0.2
0.4
0.6
0.8
1
β = 40 50 100
0
0.2
0.4
0.6
0.8
1
β = 6
DSP_6-FIR 81
Properties of windows
Window B A Rectangular 4π/(2M+1) -13 dB Hann 8π/(2M+1) -32 dB Hamming 8π/(2M+1) -43 dB Blackmann 12π/(2M+1) -58 dB
B
A
Sharper edge from pass to stop band Lower attenuation A in stop band
DSP_6-FIR 82
10 20 30 40 50 600
0.2
0.4
0.6
0.8
1
Samples
Ampl
itude
Time domain
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9-140
-120
-100
-80
-60
-40
-20
0
20
40
Normalized Frequency (×π rad/sample)
Mag
nitu
de (d
B)
Frequency domain
Rechteck
von Hann
Dreieck
Hamming
Comparing windows in Matlab with wintool
DSP_6-FIR 83
Matlab function FIR1
B = FIR1(N,Wn) designs an N'th order lowpass FIR digital filter and returns the filter coefficients in length N+1 vector B. The cut-off frequency Wn must be between 0 < Wn < 1.0, with 1.0 corresponding to half the sample rate. The filter B is real and has linear phase. The normalized gain of the filter at Wn is -6 dB.
DSP_6-FIR 84
B = FIR1(N,Wn,'high') designs an N'th order highpass filter. You can also use B = FIR1(N,Wn,'low') to design a lowpass filter. If Wn is a two-element vector, Wn = [W1 W2], FIR1 returns an order N bandpass filter with passband W1 < W < W2. You can also specify B = FIR1(N,Wn,'bandpass'). If Wn = [W1 W2], B = FIR1(N,Wn,'stop') will design a bandstop filter.
DSP_6-FIR 85
B = FIR1(N,Wn,WIN) designs an N-th order FIR filter using the N+1 length vector WIN to window the impulse response. If empty or omitted, FIR1 uses a Hamming window of length N+1. For a complete list of available windows, see the help for the WINDOW function.
DSP_6-FIR 86
% Lowpass filter
% Ordnung 20
%
b=fir1(20,0.2);
disp (b);
[h,w]=freqz(b,1,1024);
mag=20*log10(abs(h));
plot(w/pi,mag);grid 0 0.2 0.4 0.6 0.8 1-120
-100
-80
-60
-40
-20
0
fir (20, 0.2)
DSP_6-FIR 87
Practical filter design
Design software allows specification of pass and stop bands (tolerance diagrams). Filter characteristic are very sensitive to location of poles and zeros.
DSP_6-FIR 88
Matlab fdatool
DSP_6-FIR 89
Summary(1) • FIR filters use only values from input sequence x[n]. • Filter coefficients b[k] determine behavior of filter (low pass, high pass, band pass, band stop). • Impulse response and frequency response to describe properties of (FIR-) filters. • Impulse response (and transient response) has finite duration.
DSP_6-FIR 90
Summary (2)
• FIR filters may have (and usually have) linear phase, a property important for maintaining the signal wave form. • FIR filters are always stable and easy to realize in hardware and software. • FIR filters of the same order as IIR filters are less selective. In other words: To realize a given selectivity using FIR filters a higher filter order (higher cost) is required.
z-Transform
z-Transform 2
Why z-Transform? The z-Transform introduces polynomials and rational functions in the analysis of linear time-discrete systems and has a similar importance as the Laplace transform for continuous systems Convolution becomes a multiplication of polynomials Algebraic operations like division, multiplication and factoring correspond to composing and decomposing of LTI systems Carrying out the z-transform in general leads to functions consisting of nominator and denominator polynomials The location of the roots of these polynomial determins important properties of digital filters
z-Transform 3
ˆ, , Domainn z
ˆ Frequency dom
Impulse response, difference equations, "real" signal domain
Frequency response, spectral representation, analysis of sou
a
nd
Opera Doma
Tim
in
e do
ton
ma
ir
in
z z
n
s, poles and zeros,
mathematical analysis and synthesis
z-Transform 4
Input signals
Impulse Complex exponential function
• x[n] = zn
ˆj nAe
z-Transform 5
z-Transform and FIR- filter
0
0
[ ] [ ] (using filter coefficients)
or notation using convolut
for a
ion sum
[ ] [ ] [ ] [ ] [ ]
We now use the "signal" [ ] ll n ,
M
kk
M
kk
n
y n b x n k
y n x n h n h n b n k
x n z z
0
0
0 0 0
0
arbitrary (complex) numberas input signal
[ ] [ ]
...[ ] system fu( ) nction
M M Mn k n k
Mk
kk
M
nk k k
k
Mk
k
kk
k
k
k
y n b x n k b z b z z b z
b z h k z
z
H z
ˆ
Remember:jz e
z-Transform 6
0 0
[ ] [ ] (
T
)
For the input signal we get[ ] [ ] ( )( ) is the Transform
he system function ( ) is the Transformof the impulse response:
of [ ].
M Mk
k kk k
n
n n
h n b n k H z b z
zy n h n z H z zH z z h n
H z z
z-Transform 7
0 1 2 1 20 1 2
2
2 2
[ ] ( )1,2,1 ( ) 1 2 1
2 1 2 1 =1+z
k
h n H zb H z b z b z b z z z
z zz z
The system function is a rational function numerator polynomial
denominator polynomial
2 2 1z z
2z
z-Transform 8
0 1 2
21 2 0 1 2
0 1 2 2
Impulse response: [ ] [ ] [ 1] [ 2]
System function: ( )
h n b n b n b n
b z b z bH z b b z b zz
z - Transform
z-Transform 9
Representation of signals
0
A signal of finite lenght can be represented as
[ ] [ ] [ ]
the Transform of this signal is
is an arbitrary complex number, e.g., is the independent(complex) variable of
0( ) [ ]
N
k
N
x n x k n k
z
k
kz z
X z x k z
1
01
the -Transform.Alternatively we may write
( ) [ ]( )
which means, that X( ) is polynomial of order N of the variable .
Nk
k
z
X z x k z
z z
0 0
... System function( ) [ ] M M
k kk
k kH h kz b z z
z-Transform 10
Transform The transition from n z is called z-Transform of x[n].
X [z] is a polynomial in z-1 , the coefficients are the values of the sequence x[n]. e.g.:
n n < -1 -1 0 1 2 3 4 5 n > 5x[n] 0 0 2 4 6 4 2 0 0
1 2 3 4( ) 2 4 6 4 2X z z z z z
z-Transform 11
0 0
[ ] [ ] [ ] ( ) [ ]
[ ] ( )
N Nk
k k
n Domain z Domain
x x k kn n X
zXn
x kz
x
z
0
Example:
[ ] [ ]
[ ]o
n
x n n n
X z z
0
[ ]
of [ ]
j
Nj k j k
kk
z e
z re x k r e
x k r
ω
ω ω− −
=
−
= →
= →∑
DFT
DFT
z-Transform 12
The Transform of a FIR-Filters is apo
The system function ( ) is a function
lynom of degree and ha
of the comp
s zeros(fundame
lex
nta
variable
l theore
m of algebra).
. z
M M
H zz
1 11 2 1 1 3 2
2
1 13 2
Example:[ ] 6 [ ] 5 [ 1] [ 2]
( ) 6 5 (3 )(2 ) 6
Zeros at and
y n x n x n x nz z
H z z z z zz
z-Transform 13
Properties of the z-Transform
1 2 1 2
1
[ ] [ ] ( ) ( )
in the domain correponds to a time de
Linea
layof 1 in the domain (time do
rity
Time
mai
-delay
n)
ax n bx n aX z bX z
z zn
n n < 0 0 1 2 3 4 5 n > 5 x[n] 0 3 1 4 1 5 9 0
1 2 3 4 5
1
0 1 2 3 4 5 6
( ) 3 4 5 9( ) ( )
= (0 ) 3 4 5 9
X z z z z z zY z z X z
z z z z z z z
z-Transform 14
z-Transform as Operator
1
1
1
1 1
The ex
Unit-Delay
pression
Operator [ ] [ ] [ 1][ ] for all [ ] [ ] [ ]
Nervertheless it is common to use as nota[ ] is only valid for [ ]
ti !
o
n
n n
n
n
z
y n x n x nx n z ny n x n z z z z z
x
x n
zn x n z
DD
D D
1
n for theunit-delay operator sym
bol .
[ ] [ ] [ 1]y n z x n x nD
z-Transform 15
Unit Delay
x[n] x[n-1] z -1
X(z) z-1X(z)
×
z-1
× × ×
+ + +
0b 1b 2b3b
[ ]x n [ 1]x n [ 2]x n [ 3]x n
[ ]y n
z-1 z-1
z-Transform 16
Convolution and z-Transform
Domain Domain[ ] [ ] [ ] ( ) ( ) ( )
n zy n h n x n Y z H z X z
Cascading of systems
LTI 1 LTI 2 1[ ] [ ]x n h n[ ]x n 1 2( [ ] [ ]) [ ]x n h n h n
1[ ]h n 2[ ]h n
1 2 1 2[ ] [ ] [ ] ( ) ( ) ( )h n h n h n H z H z H z
z-Transform 17
1 2 3
12 1
1 2 31 2
2 11
1
1
1 2
Example:( ) 1 2 2
One root of this polynomoial is at , and we can write( ) ( )(1 ) ( ) ( )
( ) 1 2 2 1( ) 1
( ) 1 1
11
H z z z z
H z H z z H zH z z z zH z zH z z
H z
z
z z
z
z
11( ) (1 )H z z 1 2
2 ( ) 1H z z z
Partitioning
z-Transform 18
ˆ
ˆ
0 0
ˆ( )
ˆ
(
..
)
.
M Mk j k
k kk k
jz
H z b
e
z H b e
ω
ωω− −
= =
=
= =
⇔
⇔∑ ∑
z - dom
Unit
ain
circle in
the
co
ω-
mplex
domain
plane
z-Transform 19
Poles and Zeros
3 3
3 3
1 2 3
1 1 1
3 2
3 3
e.g.: ( ) 1 2 2
(1 )(1 )(1 )
2 2 1 ( 1)( )( )
j j
j j
H z z z z
z e z e z
z z z z z e z ez z
ο
ο
ο
x3
Triple pole at zero
z-Transform 20
What does a zero mean?
3
1 2 33 3 3 3
23
3
[ ] [ ] 2 [ 1] 2 [ 2] [ 3]
[ ]
[ ] 2 2
(1 2 2 )
1 1 3 1 3 1 0
n
n n n n
n
n
j
j j j j
j j j j
j
y n x n x n x n x n
x n e
y n e e e e
e e e e
e j j
3 31 2 3
The complex input signals
( ) 1, (z ) , (z )get supressed.
n nj jn n nz e e
z-Transform 21
Nulling Filter
0 0
0ˆ ˆ1 1
0 2 2
Zeors in the plan "remove" signalsof the form [ ] .
ˆTo remove a cosine signal cos( )both complex signals must be removed.The zeros are conjugate complex.
n
j n j n
zx n z
n e e
z-Transform 22
PN-Video
z-Transform 23
21 11
10
10 109
1 90
( ) (1 )
10-point running average 101 1( )1 ( 1)
j kL Lk L
kk
k
k
H z z e z
Lz zH z zz z z
z-Transform 24
3 21 2 3
3
2 2 1( ) 1 2 2 z z zH z z z zz
− − − − + −= − + − =
-1 -0.5 0 0.5 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
3
Real Part
Imag
inar
y Pa
rt
3 2 /3 /3
/30
2 2 1 ( 1)( ),
01
( )
j j
j
z z z z z e z ez e
π π
π
−− + −
= ±
= − − − =
B = [1 –2 2 -1] A = [1] zplane(B,A)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
0
1
2
3
4
5
6
Normalized Frequency ( ×π rad/sample)
Mag
nitu
de
Magnitude Response
z-Transform 25
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
0
1
2
3
4
5
6
Normalized Frequency ( ×π rad/sample)
Mag
nitu
de
Magnitude Response
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
-20
-15
-10
-5
0
5
10
15
Normalized Frequency ( ×π rad/sample)
Mag
nitu
de (
dB)
Magnitude Response (dB)
lin log
z-Transform 26
Inverse Filtering (1)
Channel
Output signal known
Channel known
Input signal wanted
[ ] [ ] [ ][ ] ?
y n h n x nx n
= ∗= Inverse convolution
Deconvolution
z-Transform 27
Inverse Filtering (2) Solution in the time domain difficult or impossible
Solution in the z-domain
2Correction
1 2 2
1Ch
1
annel
( )( ) ( ) ( ) ( )
1( ) ( ) 1 )
)
(( )
(Y z X z H z X z
H z H z H
z
zH z
zH H= =
= ⇒ =
z-Transform 28
Inverse Filtering (3) 1 2
1
2 1 21
( ) (1 0.5 )1 1H (z) = =
H ( ) (1 0.5 )
H z z z
z z z
− −
− −
= − +
− +
Zeros in denominator (outside of zero) Polstellen
-1 -0.5 0 0.5 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
2
Real Part
Imag
inar
y Pa
rt
z-Transform 29
-1
-0.5
0
0.5
1-1
-0.50
0.51
0
2
4
6
8
10
12
14
16
ImRe
H(z
)
11
1 0.9( )
zH z −−
=
-1 -0.5 0 0.5 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Real Part
Imag
inar
y Pa
rtPole/Zero Plot
0 10 20 30 40 50 60 70 80 900
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Samples
Impulse Response
Am
plitu
de
PN-diagram
Impulse response
z-Transform 30
11
1 0.9( )
zH z −−
=
-1 -0.5 0 0.5 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Real Part
Imag
inar
y Pa
rt
0
π
DSP-8-IIR 1
Infinite Impulse Response-Filter
DSP-8-IIR 2
First order IIR-Filter 1 1[ ] [ 1] [ ] [ 1]oy n a y n b x n b x n
FIR block
z-1
x
x + +
x
z-1
b0
b1 a1
v[n] x[n] y[n]
y[n-1] Feed-forward block (FIR-Filter)
Feed-back block
DSP-8-IIR 3
Example 1[ ] 0.8 [ 1] 5 [ ] ( 0)
[ ] 2 [ ] 3 [ 1] 2 [ 3]y n y n x n bx n n n n
Input assumed to be zero prior to starting time n0, x[n] = 0 for n < n0 Output assumed to be zero before starting time of the signal y[n] = 0 for n < n0 System initially at rest
DSP-8-IIR 4
0.8 [ 1] 5(2) 0.8 (0) 5 ( ) 0.8 [0] 5 [1] 0.8 (10) 5 ( ) 0.8 [1] 5 [2] 0.8 ( 7) 5 ( ) 0.8 [2] 5
[0] 10[1] 7[2
[3] ] 5.6
[3]
23
0
yyy
yy x
xxy
yy
5.52[4 4.4416
0.8 ( 5.6) 5 ( ) ] 0.8 [3] 5 [4] 0.8 (5.52) 5 ( )
0.8 [4] 5 [5] 0.8 (4.4416) 5 ( ) 0.8 [5] 5 [6] 0.8
[5] 3.5328
2
(3
0
50
[6] .
yy
y xy xyy x
328) 5 ( ) 2. 6 20 82
0 5 10 15-8
-6
-4
-2
0
2
4
6
8
10
Eingang0 5 10 15
-8
-6
-4
-2
0
2
4
6
8
10
Ausgang
3
3 (Input zero)[ ] 0.8[ ]
[ 1][3](0.8)ny n y
ny n y n
B = [5]; A = [1 -0.8] filter(B,A,x)
[ ] 0.8 [ 1] 5 [ ]y n y n x n
DSP-8-IIR 5
01
[[ ] [ ]]N M
klkl
b xa y n ly kn n
FIR: Output is f(Input)
Feed-forward
Output is f(Output)
Feed-back
M is order of filter for FIR filters, N is order of filter for IIR filters.
DSP-8-IIR 6
Linearity, Time-invariance
IIR-Filter 01
[[ ] [ ]]N M
klkl
b xa y n ly kn n
are linear und time-invariant
DSP-8-IIR 7
Impulse response of a 1st order system
The response to a unit pulse characterizes a LTI system completely.
Input signal represented as superposition of weighted and delayed impulses, output signal constructed from weighted and delayed impulse responses:
[ ] [ ] [ ]k
y n x k h n k
DSP-8-IIR 8
1 0 1 0
1
1
1
1
0
Proof by evaluating:[0] [ 1] [0]
[ ] [ 1] [ ]of impulse response
The
( )(0) 1for 1[
[ ] [ 1] [ ]
( ) für 0[ ]
0 für 0
soluti
] [
on is:o
no
o
h n a h n b n
b
y n a y n
a nh
b
nn
x n
h a h b a b bn
h n a h n
Difference equation
0
11 1 1 0 1
1] [ ]
( )n n no o
b n
b a a b a b a
0
1.
2.
DSP-8-IIR 9
Notation using step response
1 1
1
1 for 0[ ]
0 for 0hen[ ] ( ) [ ]
cen
oh a
nn
n
n b n
DSP-8-IIR 10
Step response 1[ ] [ 1] [ ]oy n a y n b x n
Iterate difference equation to calculate output sequence one sample at a time:
0
0 0 12
0 0 1 0 1
2 30 1 1 1
[ ] [ ] 1 1 ( )
01
2
3
1 ( ) ( )
1 1
x n y nbb b ab b a b a
b a
n
a a
DSP-8-IIR 11
21 1 1 0
1
11
0 1
k
1
10
L
=0
[ ] (1 ... )
We recall
1 11
1 1
for 0, if 1[ ]1
1
no
nk
k
n
L
ky n b a a a b
r rr
L
ay
r
nb
a
ana
r
DSP-8-IIR 12
We identify three cases:
11
11
0
1
1
1If 1 , than decays
If 1 , than dominate
to zero as
stab
s and [ ] getslarger without bound
l
1 [ ] ( 1) output [
e system lim [
==> unstable syst
]1
emn
n
n
o
a a y n
ay n n
a a nby n
a
b y
] grows as 1
[ ] if even [ ] 0 if odd
o
n na
y n b ny n n
11
01
1[ ]1
nay n ba
DSP-8-IIR 13
0 5 100
0.5
1
1.5
2
a = 0.5 ... stabil
Am
plitu
deStep Response
0 10 200
50
100
a = 1.1 ... instabil
Am
plitu
de
Step Response
0 5 100
0.5
1
a = -1 ... Grenzfall
Am
plitu
de
Step Response
0 5 100
5
10
a = 1 ... instabil
Am
plitu
de
Step Response
DSP-8-IIR 14
System function of an IIR filter
Domain Domain[ ] [ ] [ ] ( ) ( ) ( )
n zy n h n x n Y z H z X z
The system function of an FIR system is always a polynomial in z
-1. If the difference equation has feedback as in IIR systems, the system function is the ratio of two polynomials (rational function).
DSP-8-IIR 15
1 11 1
1
11
1
1
11
11 1
[ ] [ 1] [ ] [ 1]
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( )( )( 1) ( )
o
o
o
o
y n a y n b x n b x n
Y z a z Y z b X z b z X z
Y z a z Y z b X z b z X z
Y z B zH zX
b b zz z Aa z
Numerator polynomial: Feed-forward coefficients Denominator polynomial: 1 + negative feed-back coefficients
DSP-8-IIR 16
Block diagram 1st Direct form
1 11 1
111
11( ) ( )1
1 1 ( )o
oa z a zb b zH z B z
Ab b z
z
DSP-8-IIR 17
Block diagram 2nd Direct form
1 1( ) ( )( ) ( )
B zz
BA A
zz
DSP-8-IIR 18
Delay elements combined
DSP-8-IIR 19
Poles and Zeros 1
1 11
1 1
1
1
( )1
Zero
Pole
o o
o
b b z b z bH za z z a
bzb
z a
DSP-8-IIR 20
Poles and Stabilität
1
0
0
11 1 1
11
11 1
11 1 11 1
0 00
1
The system function
( )( )
1leads to the impulse response
[ ] ( ) ( )
[ ] [ 1]
no
bo bo
n no
nb n
b b a a n
b zb b zH za z z a
h n b a bn na
DSP-8-IIR 21
1
1
1
The impulse response is proportional to for 1.
The response if and 1 .
The impulsThe location of pole(s) indicat
e responsees decaying
or growing impulse respo
if 1
se.S
.
n
na n
n a
a
decayes
grows
ystems with decaying impuls response are stable systems.
The location of poles of stable system functions is strictly inside the unit circle of the z-plane.
DSP-8-IIR 22
Frequency response of an IIR-filter
11( )
1 0.8H z
z
-1
-0.5
0
0.5
1-1
-0.50
0.51
0
2
4
6
8
10
12
14
16
ImRe
H(z
)
ˆ
ˆ
ˆ
ˆ[ ] ( )ˆ( ) ( ) ( ) j
j n
jz e
y n eH e H z
H
H
If sinus sequence meets pole at the unit circle: bounded input unbounded output (resonance)
DSP-8-IIR 23
z-domain
Frequency response (lin)
Time-domain
DSP-8-IIR 24
ˆ[ ] ( ) ( )jh n H z H e
Poles, Zeros H(z)
Frequency domain
Time-domain Input, Output
h(n) ˆ( )jH e
, k ka b
?
DSP-8-IIR 25
1 2 0 1 2
1 20 1 2
1 21 2
ˆ ˆ2ˆ 0 1 2
ˆ ˆ1 2
[ ] [ 1] [ 2] [ ] [ 1] [ 2]
( )1
( )1
j jj
j j
y n a y n a y n b x n b x n b x n
b b z b zH za z a z
b b e b eH ea e a e
Solution difference equation?
DSP-8-IIR 26
Inverse z-Transform
10 1
11
We consider a first order system
( ) ( ) ( ) ( )1
1. Determine transform X( ) of input [ ]2. Muliply ( ) ( ) to get Y(z)3. Determine inverse transform of ( )
b b zH z Y z H z X za zz z x n
H z X zz Y z
DSP-8-IIR 27
1
1
0
1
0
1
1
1
Step response of a first order system[ ] [ ]
( )
this sum is finite for 1
1( ) ... für 1
U
1[ ] 1
nit step for 1
n
nn n
n n
na
h n a n
H z a z az
az
H z a zaz
a
naz
0
1 für 11
k
k
x xx
DSP-8-IIR 28
1
11 1
11
1 21 1
11
11
1
1
1
1
1
( ) ( ) ( )1 (1 )
(determine and by comparing coefficients)1
or faster using residue meth
1
1
(1(1 )(1
1
ode
)
1oo
o
b b zY z H z X za z a
b b za zz
A B A Bz
b b
a z
z BY z Az
a z
z
11 1
1
1
11
1 1
1 11 1 1
11
11 1
1
1 11
1)
(1 )( )1 1
(1 )
1 1
oz a
z a z a
o o
z a
zb b z BY z A A
z z
a
b b z b b aA
aa
a
z
zz
z
Partial fraction expansion
DSP-8-IIR 29
1 0 11
1
11 1 0 1
1 1 111 1
( ) (1 )1
[ ] [ ] [ ]1 1
z
no
b bB Y z za
b b a b by n a n na a
11 [ ] 1
nA Aa naz
DSP-8-IIR 30
Inverse transform (M<N)
1
11
1
11
1. Factoring numerator polynomial of ( ) (1 ) für 1,2,...,2. Partial fraction expansion
( )1
( )(1 )
3. Inverse transform
[ ] [ ]
k
k
Nk
k k
k k z p
Nn
k kk
H zp z k N
AH zp z
A H z p z
h n A p n
DSP-8-IIR 31
Important transform pairs
0
1 2 1 2
0
0
[ ] [ ] ( ) ( )
[ ] z ( )[ ] [ ] [ ] ( ) ( ) ( )[ ] 1[ - ]
n
ax n bx n aX z bX z
x n n X zy n x n h n Y z X z H z
nn n
0
1 1
1[ ]
1
n
n
z
a naz
DSP-8-IIR 32
1 1
1 2 1 1
1 1
1 2.1 1 2.1( )1 0.3 0.4 (1 0.5 )(1 0.8 )
( )1 0.5 1 0.8
z zX zz z z z
A BX zz z
1 1 11
1 1 1
0.50.5
1 2.1 1 0.5 (1 0.5 )( )(1 0.
0
5 ) 21 0.8 1 0.5 1
10.8
zz
z z B zX z z Az z z
11
10.8
1 2.1( )(1 0.5 ) 11 0.8 z
zB X z zz
1 1[ ] 2( 0.5) [ ] (0.8) [ ]n nx n n n
DSP-8-IIR 33
A = [1 -0.3 -0.4]; B = [1 -2.1]; [R,P,K] = residuez(B,A) R' = -1 2 P' = 0.8 -0.5 K = [] impz(B,A)
0 5 10 15 20 25 30 35 40-2
-1.5
-1
-0.5
0
0.5
1
n (samples)
Am
plitu
de
Impulse Response
DSP-8-IIR 34
ˆ
ˆ ˆ
0 0ˆ
ˆ
0
Frequency response of FIR-filters
[ ] [ ]
ˆ = ( ) Frequency response with z-transform
ˆ( ) ( ) ( )
e.g.: ( )1
j
M Mj k j n
k kk k
j n
jz e
y n b x n k b e Ae
Ae
H e H z
bH z
H
H
0 0 0
0
ˆ 0ˆ1
1 1
ˆ ˆ ˆ0ˆ
1
( )1
[ ] ( ) 1
jj
j j n j nj
bH ea z a e
by n H e e e na e
Frequency response of IIR-Filter
DSP-8-IIR 35
0
0
0
0 0
ˆ1 ˆ 1
0ˆ1 1
1
0 1 0ˆ ˆ
1 1ˆ1
1
Suddenly applied complex exponential input sequence:1[ ] [ ] ( )
11( ) ( ) ( )
1 1After partial fraction expansion:
1( )1 1
j nj
j
j j
j
x n e n X ze z
bY z H z X za z e z
b a ba e a eY z
a z e
0 1z
Steady state - transient response
DSP-8-IIR 36
0 0
0 0
0
0
1 1
1 1
decays if stable steady st
0 1 0ˆ ˆ
1 1ˆ
1
ˆ1
ate
0 1 0ˆ ˆ
1 1
( )1 1
[ ] [ ] [ ]
1[ ]
1
1-
1j
nn
n
j
j
j jj
Y zz z
y n n n
u nz
a e
b a ba e a e
b a ba
a
aa
e ee
a
1
DSP-8-IIR 37
( )
( )
1 10.2
0.3
0.
3131
0.2 0.29712 1
1
1
0.22
1
0
0.9[ ] [ ]0.9
0.4980 [ ]
1[ ] [ ] 0.5533 [ ]1 0.9
0.5533 ( 0.2971) [ ]0.55
0.
cos 0.2sin 033 ( 0.2
1;
9
0.
71)
9
.2
9
0.9
j
j
j n j j
n
jn
n
y n ne
e n
y n e n e ne
nj
b
e
nn
a
π
π ππ
δ
δ
δ
ππ
δ
δ
−
−−
− −−
−
− = = − −
= = = +
−
−
= ++ +
= = −
1
1 2
[ ][ ] [ ] [ ]
ny n y n y n
δ−
= +
DSP-8-IIR 38
0 5 10 15 20-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
b1 = 1 a1 = -0.9
Am
plitu
de
Impulse Response
DSP-8-IIR 39
1Instable system: Pole at 1.1a =
DSP-8-IIR 40
IIR-system second order
1 2 1 2
1 2 1 21 2 1 2
1 21 2
1 21 2
[ ] [ 1] [ 2] [ ] [ 1] [ 2]
( ) ( ) ( ) ( ) ( ) ( )
( )1
o
o
o
y n a y n a y n b x n b x n b x n
Y z a z Y z a z Y z b X z b z X z b z X z
b b z b zH za z a z
DSP-8-IIR 41
Poles and Zeros 1 2 2
0 1 2 0 1 21 2 2
1 2 1 2
( )( )( ) 1
Y z b b z b z b z b z bH zX z a z a z z a z a
A polynomial of degree N has N roots. For real coefficients of the polynomial the roots are either real or complex conjugate
DSP-8-IIR 42
Impulse response 1 2
0 1 21 2
1 2
2 1 21 1
2 1 2
20 1 1 1 2 2 1
2
( )( )( ) 1
( )1 1
[ ] [ ] [ ] [ ]n n
b b z b zY zH zX z a z a z
b A AH za p z p z
bh n n A p n A p na
DSP-8-IIR 43
Real poles
1 2 1 15 1 1 16 6 2 3
1 11 12 3
1 11 12 3
1 1( )1 (1 )(1 )
3 2( )(1 ) (1 )
[ ] 3 [ ] 2 [ ]nn
H zz z z z
H zz z
h n n n
1 2n
k
p pp
For real und the impulse responseconsist of two functions in the form
DSP-8-IIR 44
5 16 6B =[1]; A = 1,- ,
impz(B,A)
0 2 4 6 8 10 120
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
n (samples)
Am
plitu
deImpulse Response
DSP-8-IIR 45
Complex conjugate poles
DSP-8-IIR 46
Complex poles at the unit circle 1 1
/ 4 1 / 4 1 2 2
1 1( )(1 )(1 ) 1 1.4142j j
z zH ze z e z z zπ π
− −
− − − − −
+ += =
− − − +
1.1781 1.1781
/ 4 1 / 4 1
1.3066 1.3066( )(1 ) (1 )
j j
j j
e eH ze z e zπ π
−
− − −= +− −
( ) ( )/ 4 / 41.1781 1.17811 1[ ] 1.3066 [ ] 1.3066 [ ]j n j nj jh n e e n e e nπ πδ δ−−− −= +
1[ ] 2 1.3066cos 1.1781 [ ]4
h n n nπ δ− = ∗ −
DSP-8-IIR 47
Complex conjugate poles at unit circle Second-order oscillator
DSP-8-IIR 48
Complex poles inside the unit circle 1 1
2 / 4 1 / 4 11 12 2
1.2490 1.2490
/ 4 1 / 4 11 12 2
1 1( )1 1 0.5 (1 )(1 )
1.5811 1.5811(1 ) (1 )
j j
j j
j j
z zH zz z e z e z
e ee z e z
π π
π π
− −
− − − −
−
− − −
+ += =
− + − −
= +− −
-1 -0.5 0 0.5 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Real Part
Imag
inar
y Pa
rt B = [1 1]; A = [1 -1 0.5] zplane(B,A)
DSP-8-IIR 49
1[ ] 2 1.5811 cos 1.24942
n
h n nπ = × −
0 5 10 15 20 25-0.5
0
0.5
1
1.5
2
n (samples)
Am
plitu
deImpulse Response
1
2
1( )1 1 0.5
zH zz z
−
−
+=
− +
DSP-8-IIR 50
10
0 1 1 20
10
0 1 1 20
10
0 1 1 2 20
10
0 1
(sin )sin [ ] 1 2(cos )
1 (cos )cos [ ] 1 2(cos )
1 ( cos )cos [ ] 1 (2 cos )
( sin )sin [ ] 1 (2
n
n
zn nz z
zn nz z
r zr n nr z r z
r zr n n
0 0
1 2 20
0 1 1 1
10 1
0 1 1 2 2
cos )
0.5 0.5cos( ) [ ] 1 1
cos( ) [ ] 1 2
j jn
j j
n
r z r zKe KeK r k n
re z re zb b zK r k naz r z
2 2 21 0 0 1 0 1
02 2 2 2
2 arccos arctanb r b ab b ab baKr a r r a
DSP-8-IIR 51
PZ-Video
Running average FIR filter
SigProc-7-IIR 52 Direct form I Direct form II
1 1( ) ( )
( ) ( )B z B z
A z A z
SigProc-7-IIR 53 Transposed direct form II
Attention!
SigProc-7-IIR 54
IIR filters are often represented in the form
In such cases the coefficients a in the block diagrams are negative!
1
01
0
( )( )( ) 1
Lk
kk
Mm
mm
b zY zH zX z a z
DSP-8-IIR 55
Filterdesign
SPtool
DSP-8-IIR 56
• Sine wave 2 kHz + noise (randn) (average = 0 , variance = 0.01)
• Band pass ripple 1.5 dB, stop band 35 dB
SigProc-6-FIR 57
SigProc-6-FIR 58
SigProc-6-FIR 59
No evidence of noise in filtered signal spectrum
SigProc-6-FIR 60
Spectrum filter input
Spectrum filter output
Sampling Theorem
DSP_9-Sampling Theorem 2
Sampling
ˆ[ ] ( ) cos( ) cos( )
ˆ
s s
s
x n x nT A nT A n
T
Normalized angular frequency
DSP_9-Sampling Theorem 3
Normalized angular frequency
ˆˆ
ω has the dimension rad / sec, ω = ωT has the dimension rad,i.e. !The time axis is lost after sampling,the discrete signal is merely a sequence of numbers anddoesn't c
ω is dimen
ontain inf
sionl
orm
ess
ation about the sampling period.To reconstruct the original signal the sampling frequency must be known.I An infinite number of continuous sine signals is transformed into the n other
same diswords :
crete sine representation.
DSP_9-Sampling Theorem 4
Number of samples
0 1 2 3 4 5
-2
0
2
Zeit t
Am
plitu
de
0 1 2 3 4 5
-2
0
2
Index
Am
plitu
de
0 1 2 3 4 5
-2
0
2
Index
Am
plitu
de
DSP_9-Sampling Theorem 5
Signal f(t) multiplied by a sequence of impulses δT(t) with time-step of T seconds (sampling time).
( ) ( ) ( ) ( ) ( )S Tk
f t f t t f kT t kT
The impulse sequence is a periodic function and can be represented as Fourier series.
1( ) [1 2 cos 2 cos2 2 cos 3 ...
2 cos ] ( )
T s s s
s
t t t tT
k t k
Representation of the sampling process
DSP_9-Sampling Theorem 6
/ 2
/ 2
11 ( )
1 2( )
ss
s
s
s
T jk tk TT
s
jk tT s
ks s
D t e dtT
t eT T
Tω
ω
δ
πδ ω
−
−
∞
=−∞
= =
= =
∫
∑
1( ) [1 2cos 2cos2 2cos3 ... 2 cos ]
( )
T s s s ss
t t t t k tT
k
kδ ω ω ω ω= +
∞
+ + + +
→ (real representation)
-5 0 5 100
0.5
1
x 1/Ts
complex representation
DSP_9-Sampling Theorem 7
1
( ) ( ) ( ) [ ( ) 2 ( )cos 2 ( )cos2 ...]S T s sf t f t t f t f t t f t tT
Sampled cosine (ω0) signal
0 0 0
1cos( )cos( ) cos( ) cos( )
2s s st t t t
Sampling generates for spectral component
0( )s
0
DSP_9-Sampling Theorem 8
1( ) ( ) ( ) [ ( )cos 0 2 ( )cos 2 ( )cos2 ...]S T s sf t f t t f t t f t t f t t
T
e.g. :
0 0
0
0
( )
1cos( ) cos( ) cos(
co
)2
s
cos( )s s s
f t
t
t
t t t
ω
ω
DSP_9-Sampling Theorem 9
Spectrum of sampled signal
( ) ( )1( ) cos [ ( ) ( )]2
( ); ( 2 ), ( 2 ); ( 3 ), ( 3 );...1( ) ( )
s s s
s s s s
s sns
f t F
f t t F F
F F F F F
F F nT
ω
ω ω ω ω ω
ω ω ω ω ω ω ω ω ω
ω ω ω∞
=−∞
⇔
⇔ − + +
− + − +
= −∑
The spectrum of the sampled signal is periodic with period ωs.
DSP_9-Sampling Theorem 10
The spectrum of the original signal f(t) is contained in the spectrum of the sampled signal fs(t) and can be recovered from Fs(ω) by „cutting out" using an (ideal) low pass filter.
DSP_9-Sampling Theorem 11
Distance of spectra depends on sampling frequency
DSP_9-Sampling Theorem 12
Shannon Sampling Theorem
A continuous-time signal with frequencies not higher than fmax can be reconstructed exactly from its samples, x[n] = x(nTs), if the samples are taken at a rate fs = 1/Ts that is greater than 2fmax.
DSP_9-Sampling Theorem 13
Examples
• Scale 14 kHz [44,1 kHz] • Scale 14 kHz [11 kHz]
• Phasor • Helicopter
DSP_9-Sampling Theorem 14
Spectrum of sampled sine function Sampling frequency = 90 Hz
100 Hz -100 Hz
f = 10 Hz ! Aliasing
DSP_9-Sampling Theorem 15
Sampling frequency 110 Hz 100 Hz
-100 Hz
f = – 10 Hz ! Folding
DSP_9-Sampling Theorem 16
Sampling frequency = 250 Hz 100 Hz
f = 100 Hz
-100 Hz
Selectivity of filter
DSP_9-Sampling Theorem 17
Sampling frequency = 350 Hz
Filter can be less selective
100 Hz
-100 Hz
DSP_9-Sampling Theorem 18
Spectrum of a continuous-time signal
Spectrum of the sampled continuous-time signal (continued periodically)
Aliasing (Folding), if sampling theorem violated.
DSP_9-Sampling Theorem 19
Reconstruction/Interpolation
Conversion discret => continous(Interpolation with pulses)
( ) [ ] ( )
( ) characteristic pulse shape of the converter
sn
y t y n p t nT
p t
DSP_9-Sampling Theorem 20
Interpolation in the time domain Zero-Order Hold Linear Interpolation
DSP_9-Sampling Theorem 21
1 x
4 x
DSP_9-Sampling Theorem 22
Interpolation in the frequency domain
DSP_9-Sampling Theorem 23
Ideal Filter
• To reconstruct the original signal from the (periodic) spectrum, a filter with rectangular frequency response would be required (ideal filter)
• The input signal of the filter is the sequence of impulses of the sampled signal
• The output signal is the superposition of weighted and time-delayed pulse responses
DSP_9-Sampling Theorem 24
Impulse response of the ideal low pass
( )2
2
1 | | 2( )
0 | | 21( ) 2 sinc 2
2B j t
B
BH
B
h t e d BT Btπ ω
π
ω πω
ω π
ω ππ −
<= ≥
= =∫
00
H(ω
)
ω
0
0
1
h(t)
Zeit t
-2 πB
1/2B
2πB
T
4/2B3/2B2/2B-4/2B -3/2B -2/2B -1/2B
DSP_9-Sampling Theorem 25
The output signal is the superposition of time-delayed, weighted impulse responses
( )(( ) ( )) ( )sinc 2k k
kTf kT f kTf t h t Bt kππ= − = −∑ ∑
For clarity only every second sinc-function drawn.
DSP_9-Sampling Theorem 26
The interpolation with sinc-pulses reconstructs the continuous-time signal exactly.
But the impulse response of the ideal low pass is non-causal, i.e. the filter responds before the pulse is applied.
It is not possible to realize non-causal filters!
DSP_9-Sampling Theorem 27
Interpolation filters (1) Ideal (analog) low pass filters are non-causal and are therefore not realizable.
A practical solution is to use sampling frequencies higher than the Nyquist-Frequency. In this case the gaps in the periodic spectrum are wider and the selectivity of the interpolation-low-pass-filter can be less than ideal. (The continuous-time signal is not reconstructed exactly.)
DSP_9-Sampling Theorem 28
Interpolation filters (2) It is possible to design sharp (near ideal) analog filters with high attenuation in the stop band. But it is not possible to design filters which fully repress signals in the whole stop band.
Every practical signal is of finite duration. We know from Fourier-Transform, that a signal with finite duration has spectral components in the frequency range from −∞ to ∞.
DSP_9-Sampling Theorem 29
No signal can be time-limited and frequency-limited at the same time:
Time-limited (finite duration) means that the spectrum reaches from – ∞ bis ∞ (is not band-limited).
Band-limited (finite bandwidth) signals have infinite signal duration (not time-limited).
DSP_9-Sampling Theorem 30
Overlapping Spectra
DSP_9-Sampling Theorem 31
Antialiasing-Filter To avoid overlapping of spectra Antialiasing-Filter limit the bandwidth of signals.
DSP_9-Sampling Theorem 32
Digitization
The resolution of the Analog-/Digital-Converter depends on quality requirements.
For voice signals a resolution of 8 bit is sufficient, music signals on Audio–CD use 16 bit resolution.
The lower the resolution of the A/D converter the more the digital signal deviates from the analog signal.
DSP_9-Sampling Theorem 33
Quantization error
0 20 40 60-1
0
1
Zeit t
f(t)
0 20 40 60-1
0
1
Zeit t
f(t)
0 20 40 60-1
0
1
Zeit t
F(t)
0 20 40 60-1
0
1
Zeit t
F(t)
0 20 40 60-0.5
0
0.5Fehler: ∆ = f (t) - F(t)
Zeit t
∆
0 20 40 60-5
0
5x 10
-10 Fehler: ∆ = f (t) - F(t)
Zeit t
∆ Quantization noise
DSP_9-Sampling Theorem 34
Quantization noise (1) Quantization error
For equal probability of all signal amplitudes we get the average power of the error signal (noise)
/ 22 2
/ 2
112
~ 0.2912 12
eff
eff
e s ds
LSBe LSB
∆
−∆
∆= =
∆∆
= =
∫
8 bit: 1/900 12 bit: 1/14.000 16 bit: 1/226.000
of value range
DSP_9-Sampling Theorem 35
Quantization noise (2)
[y,fs,nbits]=wavread('file.wav')
sound(y,fs,6) % Play back with 6 bits
wavwrite(a,fs,nbits,'wavefile.wav')
nbits = 8, 16, 32 or 64
DSP_9-Sampling Theorem 36
Sampling in the frequency domain Sampling in the time domain: band-limited signals Sampling in the frequency domain: time-limited signals
DSP_9-Sampling Theorem 37
DSP_9-Sampling Theorem 38
Discrete Fourier-Transform (DFT)
We calculate the DFT of a time-limited signal f(t) with duration (a). The spectrum F(ω) is not band-limited (b).
From the continuous time-limited signal f(t) we get the
discrete signal fs(t) by sampling with the interval T=1/Fs (c). Due to sampling the spectrum becomes periodic with period Fs=1/T and we get spectrum Fs(ω) (d).
Sampling the spectrum with F0=1/T0 (f) leads to the periodic
time-signal width period T0 (e).
DSP_9-Sampling Theorem 39
00
00
1
001
0
1[ ] [ ]
[ ] [ ]
Njk n
k
Njk n
n
f n F k eN
F k f n e
−Ω
=
−− Ω
=
=
=
∑
∑
DFT
IDFT
Diskrete Fourier-Transformation
DSP_9-Sampling Theorem 40
0 0
0
0
1 1ˆ
0 0
1( ) [ ] [ ] ( )
1( ) ( ) ( ) ( ) 2
1[ ] [ ] [ ] [ ]
Tj j
j j
N Nj
k t t
t t
n
n
k
k
k
k j
f F e F f e dtT
f F e d F f e dt
t
f F e F
t
t t
n nk e
k
k fN
kω ω
ω
ω ωωπ
ω ω
+∞−
=−∞
∞ ∞−
−∞ −∞
− −−
= =
= =
= =
= =
∑ ∫
∫ ∫
∑ ∑
⇔
⇔
⇔ 0ˆ
0
2ˆ
k n
TN
ω
πω ω= =
Time domain Frequency domain
FR:
DFT:
FT:
DSP_9-Sampling Theorem 41
Spectral lines of DFT • The spectral lines of the DFT lie on a raster
• There are no spectral lines outside the raster, e.g. 15 Hz. The DFT algorithm must compose a 15 Hz signal from other frequency components!
0
0
1000; e.g.: 10 Samples: 0, 10, 20, ..., 90 Hz100
s
s
T fNT f
= = =
DSP_9-Sampling Theorem 42
( )
( ) ( )
( )
( )
( )( ) ( )
0
0
0
0
0
ˆ
N-1ˆ 2 /
n=0
ˆ2 /
ˆ2 /
Phase onl
ˆ 02
y
2 / 1 /
[ ] 0,1,2,..., 1 [ ]
X[k]= ...
1 1
ˆsin 2 / / 2
j n
j n j N kn
j k N Nj
j k N
j k N Nj
x n e n Nx n
e e
eee
k N Ne e
ω ϕ
ω ϕ π
π ωϕ
π ω
π ωϕ π ω
+
+ −
− −
− −
− − −
= = −
= =
−= =
−− =
∑
The N- point DFT of is
( )0
Diriclet Function
ˆsin 2 / / 2k Nπ ω−
X[k] contains samples of the Dirichlet Function.
2 2 2
2 sin
1j j jj
j
e e e e
0 1 2 3 4 5 6-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
DSP_9-Sampling Theorem 43
0 0.5 1 1.5 2-1
-0.5
0
0.5
1Time-Domain (f = 3, T = 1, fs = 8, N = 8)
t
x(t)
0 2 4 6 8 10 12 14 160
1
2
3
4Frequency-Domain
f
|X(f)
|
Sampling of f = 3 delivers a non-zero value only at this frequency. (Cosine function continued periodically.)
DSP_9-Sampling Theorem 44
The frequency f = 2.2 doesn‘t lie on the raster and must be composed of other frequencies! The Dirichlet Function delivers non-zero values at the raster frequencies. Note the periodic continuation of the cosine function!
0 0.5 1 1.5 2-1
-0.5
0
0.5
1Time-Domain (f = 2.2, T = 1, fs = 8, N = 8)
t
x(t)
0 2 4 6 8 10 12 14 160
1
2
3
4Frequency-Domain
f
|X(f)
|
DSP_9-Sampling Theorem 45
In the previous representation we calculated the spectrum of ( )0ˆ[ ] j nx n e ω ϕ+=
For the real-valued signal x[n] we must compute the spectrum of
( ) ( )( )0 0ˆ ˆ[ ] j n j nx n e eω ϕ ω ϕ+ − += +
(by adding X [k] for .) 0ω−
DSP_9-Sampling Theorem 46
0 0.5 1 1.5 2-1
-0.5
0
0.5
1Time-Domain (f = 2.2, T = 1, fs = 8, N = 8)
t
x(t)
0 2 4 6 8 10 12 14 160
1
2
3
4
Frequency-Domain
f
|X(f)
|
DSP_9-Sampling Theorem 47
We always compute the spectrum of the periodic continuation! The periodic continuation delivers as sine-(cosine-)function, and therefore single spectral line, only if the period fits exactly in the raster, i.e. only for spectral lines on the raster.