# EE513 Audio Signals and Systems Digital Signal Processing (Systems) Kevin D. Donohue

```EE513
Audio Signals and Systems
Digital Signal Processing (Systems)
Kevin D. Donohue
Electrical and Computer Engineering
University of Kentucky
Laplace to begin with
The Laplace transform is used to characterize and
analyze signal and system interactions. If x(t) is the time
domain signal, its Laplace transform is defined as:
~
X (s) 


x(t ) exp(  st )dt

where s belongs to the set of complex numbers over
which the integral converges. What is the Laplace
transform of a system impulse response? What does the
Laplace transform become if it is evaluated along the
imaginary axis?
Z-Transform
If the Laplace transform is made discrete by sampling the
time axis with interval Ts, it becomes:
~
X ( s) 

 x(nT ) exp( snT )T
n  
s
s
s
Now let:
z  exp( sTs )
s  RE[ s]  j IM[ s]    j
z  exp( Ts  jTs )  exp( Ts ) exp( jTs )
z n  exp( Ts ) exp( jTs ) n  exp( nTs ) exp( jnTs )
Z-Transform
Substitute z to obtain:
 
~ 1
n


X  ln( z )    x(nTs ) z Ts
 Ts
 n
Normalize sampling rate and define Z-transform as:
Xˆ z  


x( n ) z  n
n  
where z belongs to the set of complex numbers for which
summation converges. Describe mapping from S-plane to Zplane.
j Axis in s to z Mapping
j axis in s corresponds to a zero real part,  = 0 for all
imaginary values, . In z this results in:
z  exp(  jTs )  exp( 0Ts ) exp( jTs )  exp( jTs )
z  1, z  Ts   2fTs  2
S-plane
f
fs
Z-plane
IM
IM
2f s
2
fs
 f s
2
j
RE
 f s
 2f s
1
1
j
RE
Z-plane S-plane Relationships
Imaginary axis (j) in S-plane, maps into the unit circle in Zplane, where segments on the j axis described by:
(2k  1)
s
2
   (2k  1)
s
2
2
for k  I and  s 
 2f s
Ts
map into a complete unit circle for every integer k.
Since the mapping from S to Z is not unique, let the range for
k=0 be referred to as the root frequency range and all other
values of k with the aliased frequency ranges.
Negative Real Axis s to z Mapping
Negative real axis in s corresponds to  &lt; 0. In z this
results in:
z  exp( Ts  jTs )  exp( Ts ) exp( jTs )
z  exp( Ts )  1, z  Ts 
S-plane
Z-plane
IM
3f s
IM
2f s
j
f s
RE
 f s
 2f s
 3f s
1
1
j
RE
Z-plane S-plane Relationships
The negative real axis of the S-plane ( &lt; 0 ) maps into
the area inside the unit circle of Z (| z | &lt; 1) . Therefore the
stable region of the S-plane (left-half plane) corresponds to
the area inside the unit circle of the Z-plane. Similarly, the
unstable region of the S-plane (right-half plane)
corresponds to the area outside the unit circle of the Zplane.
Because of aliasing from the sampling and scaling from the
exponential transformation, there is no simple (linear)
scaling between the Z and S planes. A warping or
distortion occurs when matching domain points:
Warping
z  exp(Ts ) or
z  Ts
Possible Aliasing
ln z  Ts
Z-Transform One-Sided
Many applications assume the input starts at t = 0
(n=0 for discrete) and no response exists before t = 0.
So the Z-transform is often written as:

Xˆ  z    xnz  n
n 0
Examples: Find the z-transforms of x[n] = u[n] and
x[n] = anu[n]; Assuming the z-transform of x[n] is
X(z), find the Z-transform of x[n-k] for k&gt;0.
Homework(1)
Find the z-Transforms of:
a) xn  cosbnu[n]
Use definition
b) xn  un  un  k 
for k &gt; 0. Use definition
c) Find Y ( z )
X ( z)
given yn  .5 yn  1  xn  .5 xn  1
Use ZT properties (delay)
Convolution
Given the impulse response of a discrete linear system h[n]
the input-output relationship is described by discrete
convolution:


m 0
m 0
y[n]  x[n] * h[n]   x[m]h[n  m]   x[n  m]h[m]
For x[n] and h=[n] below, graphically demonstrate their
convolution.
x(n)
h(n)
2
1
0.8
1
0.6
0
0.4
-1
-2
-6
0.2
-4
-2
0
2
4
n
6
8
10
12
14
0
-6
-4
-2
0
2
4
n
6
8
10
12
14
Convolution output up to n = -1
4
2










  

0
-2
-4
-4
-2
0
2
4
6
8
n
System response h(-1-m) in black X, Signal x(m) in red O
10
12
2
1
0
-1
-2
-6
-4
-2
0
2
4
m
6
8
10
12
14
Convolution output up to n = 0
4
2










  

0
-2
-4
-4
-2
0
2
4
6
8
n
System response h(0-m) in black X, Signal x(m) in red O
10
12
2
1
0
-1
-2
-6
-4
-2
0
2
4
m
6
8
10
12
14
Convolution output up to n = 1
4
2










  

0
-2
-4
-4
-2
0
2
4
6
8
n
System response h(1-m) in black X, Signal x(m) in red O
10
12
2
1
0
-1
-2
-6
-4
-2
0
2
4
m
6
8
10
12
14
Convolution output up to n = 2
4
2










  

0
-2
-4
-4
-2
0
2
4
6
8
n
System response h(2-m) in black X, Signal x(m) in red O
10
12
2
1
0
-1
-2
-6
-4
-2
0
2
4
m
6
8
10
12
14
Convolution output up to n = 3
4
2










  

0
-2
-4
-4
-2
0
2
4
6
8
n
System response h(3-m) in black X, Signal x(m) in red O
10
12
2
1
0
-1
-2
-6
-4
-2
0
2
4
m
6
8
10
12
14
Convolution output up to n = 4
4
2










  

0
-2
-4
-4
-2
0
2
4
6
8
n
System response h(4-m) in black X, Signal x(m) in red O
10
12
2
1
0
-1
-2
-6
-4
-2
0
2
4
m
6
8
10
12
14
Convolution output up to n = 5
4
2










  

0
-2
-4
-4
-2
0
2
4
6
8
n
System response h(5-m) in black X, Signal x(m) in red O
10
12
2
1
0
-1
-2
-6
-4
-2
0
2
4
m
6
8
10
12
14
Convolution output up to n = 6
4
2










  

0
-2
-4
-4
-2
0
2
4
6
8
n
System response h(6-m) in black X, Signal x(m) in red O
10
12
2
1
0
-1
-2
-6
-4
-2
0
2
4
m
6
8
10
12
14
Convolution output up to n = 7
4
2










  

0
-2
-4
-4
-2
0
2
4
6
8
n
System response h(7-m) in black X, Signal x(m) in red O
10
12
2
1
0
-1
-2
-6
-4
-2
0
2
4
m
6
8
10
12
14
Convolution output up to n = 8
4
2










  

0
-2
-4
-4
-2
0
2
4
6
8
n
System response h(8-m) in black X, Signal x(m) in red O
10
12
2
1
0
-1
-2
-6
-4
-2
0
2
4
m
6
8
10
12
14
Convolution output up to n = 9
4
2










  

0
-2
-4
-4
-2
0
2
4
6
8
n
System response h(9-m) in black X, Signal x(m) in red O
10
12
2
1
0
-1
-2
-6
-4
-2
0
2
4
m
6
8
10
12
14
Convolution output up to n = 10
4
2










  

0
-2
-4
-4
-2
0
2
4
6
8
n
System response h(10-m) in black X, Signal x(m) in red O
10
12
2
1
0
-1
-2
-6
-4
-2
0
2
4
m
6
8
10
12
14
Convolution output up to n = 11
4
2










  

0
-2
-4
-4
-2
0
2
4
6
8
n
System response h(11-m) in black X, Signal x(m) in red O
10
12
2
1
0
-1
-2
-6
-4
-2
0
2
4
m
6
8
10
12
14
Convolution output up to n = 12
4
2










  

0
-2
-4
-4
-2
0
2
4
6
8
n
System response h(12-m) in black X, Signal x(m) in red O
10
12
2
1
0
-1
-2
-6
-4
-2
0
2
4
m
6
8
10
12
14
Convolution output up to n = 13
4
2










  

0
-2
-4
-4
-2
0
2
4
6
8
n
System response h(13-m) in black X, Signal x(m) in red O
10
12
2
1
0
-1
-2
-6
-4
-2
0
2
4
m
6
8
10
12
14
Sinusoidal Response
Consider sinusoidal input: x[n]  A exp( j0 n)
Note this input is always on (steady-state). Show that for
impulse response h(n), the convolution sum for evaluating the
output becomes:
y[n] 


 x[n  m]h[m]  A exp j n Hˆ ( z)
0
m  
z exp( j0 )

Input
Important Concepts:
Complex Coefficient
 The response to a sinusoidal input is a phasor multiplication between input
phasor and transfer function value at the excitation frequency.
 The frequency response (Transfer Function) of a discrete system is the ztransform of its impulse response evaluated on the unit circle!
 Convolution in time domain is equivalent to multiplication in the frequency
domain.
Sinusoidal Response Example
For sinusoidal input: x[n]  A exp   jn 
 4 
And system described by:
y[n]  x[n]  .5 y[n  1]
Derive an expression for and plot the frequency response
(phase and magnitude) of the system output.
Homework(2)
For sinusoidal inputs:
a)
b)
c)
 n 
x[n]  0.5 exp   j 
2 

x[n]  5 exp  jn
 j 2n 
x[n]  exp 

 3 
d)
x[n]  4
e)
x[n]  3 cos(0.2216n)
And system described by:
y[n]  x[n  1]  .5 y[n  1]  .25 y[n  2]
Plot the frequency response (phase and magnitude) and
compute the corresponding outputs.
Sampling
Sampling rate determines the highest signal frequency that can be
reconstructed from the signal samples without error. At least 3 samples
(2 complete sampling intervals) must fall within a period for digitization
without aliasing. In other words the sampling rate must be greater than
twice the highest signal frequency for a band limited signal.
Red: 180 Hz, Blue Aliased to: 20 Hz
1
0.5
Amplitude
Fs =200 Hz
Ts = 5 ms
0
-0.5
-1
0
0.02
0.04
0.06
Seconds
0.08
0.1
Sampling
Aliasing I – The Movie (FS=200, Range 50-150 Hz) Run following mfile at:
http://www.engr.uky.edu/~donohue/ee513/mfiles/aliasexm.m
Sampling
Aliasing II – The Sequel (FS=200, Range 350-450 Hz). Change
beginning and ending frequency parameters in following mfile to run:
http://www.engr.uky.edu/~donohue/ee513/mfiles/aliasexm.m
The Sampling Theorem
A band-limited continuous signal s(t) can be reconstructed
without error from its samples provided:
f s  2 fb
where fs is the sampling frequency in samples per second, and
fb is the frequency above which s(t) has no energy.
Aliased Signal Spectra
Sampling in time with sampling frequency fs creates an
infinite pattern of a shifted analog spectra so that frequency
domain has a periodicity of fs. Let Sa(f) be the spectra of the
original analog signal, the spectrum of the sampled signal
becomes:
Sd ( f )  f s

S
n  
a
( f  nf s )
Aliased Signal Spectra
Spectral periodicity of a low-pass signal (not really bandlimited) resulting from an 8 kHz sampling
Original Analog Spectrum
0.2
Amplitude
0.15
0.1
0.05
0
-20
-15
-10
-5
0
kHz
5
10
15
20
Spectrum of Discrete Signal for Sampling Rate 8 kHz
0.025
Folding
Frequency
Amplitude
0.02
0.015
0.01
0.5fs
fs
0.005
2fs
0
-20
-15
-10
-5
0
kHz
5
10
15
20
Restoring Sampled Signals
A sampled signal is reconstructed by low-pass filtering the
samples with a cut-off near the folding frequency
Spectrum of Discrete Signal for Sampling Rate 8 kHz
0.025
Amplitude
0.02
0.015
0.01
0.005
0
-20
-15
-10
-5
0
kHz
5
10
15
20
Aliased Signal Example
Before sampling at a given rate, signals are often low-pass
filtered (anti-aliasing filter) to limit distortions from aliasing.
Tell Me Ma - Average Spectrum CD quality
-20
 Original Sound
-40
dB
-60
-80
 Limited Bandwidth (LPF
-100
with 900 Hz cutoff) and
sampled at 2 kHz
-120
2
3
10
4
10
Hertz
10
Tell Me Ma - Average Spectrum 900 Hz cutoff
0
 Original Sound sampled at
-50
dB
2 kHz (aliasing)
-100
-150
1
10
2
10
3
10
Hertz
4
10
5
10
Homework(3)
Determine the aliased frequencies in the range of
fs
fs

 f 
2
2
For the following sampling frequency and signal pairs:
a)
f s  44.1 kHz
s (t )  cos(( 2 32000)t )
b)
f s  8 kHz
s (t )  cos(( 2 8000)t )
c)
f s  11.025 kHz
d)
f s  8 kHz
s (t )  cos(( 2 50000)t )
e)
f s  4 kHz
s (t )  exp( j (2 25000)t )
s (t )  cos(( 2 10000)t )
Linear Difference Equations
Discrete systems are described by Z-transforms in the
frequency domain and difference equations in the time
domain. Digital filters can be designed in either domain.
Find the impulse response of the following filters.
(FIR)
y[n]  x[n]  2 x[n  2]  x[n  4]
(IIR)
y[n]  x[n]  .4 y[n  1]  .08 y[n  2]
a)
b)
c)
d)
Compute impulse response directly by hand
Use Matlab function “filter”
Take inverse of Z transform
Examine poles and zeros of filters
Linear Difference Equations
(FIR)
Input
Coeff:
n=-4
n=-3
n=-2
n=-1
n=0
n=1
n=2
n=3
n=4
n=5
n=6
y[n]  x[n]  2 x[n  2]  x[n  4]
x[n]
x[n-1]
1
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
x[n-2]
0
0
0
0
0
1
0
0
0
0
0
x[n-3]
-2
0
0
0
0
1
0
0
0
0
x[n-4]
0
0
0
0
0
1
0
0
0
y{n]
1
0
0
0
0
1
0
0
1
0
-2
0
1
0
0
Linear Difference Equations
(IIR)
y[n]  x[n]  .08 y[n  2]  .4 y[n  1]
Input
Coeff:
n=-2
n=-1
n=0
n=1
n=2
n=3
n=4
n=5
n=6
n=7
n=8
n=9
1
0
0
0
0
0
0
0
0
0
x[n]
1
0
0
0
0
0
0
0
0
0
y[n-2]
y[n-1]
1
-0.08
-0.4
0
0
1
-0.4
0.08
0
-0.0064
0.00256
-0.00051
0
0
1
-0.4
0.08
0
-0.0064
0.00256
-0.00051
0
4.1E-05
y{n]
0
0
1
-0.4
0.08
0
-0.0064
0.00256
-0.00051
0
4.1E-05
-1.6E-05
Homework (4)
Find the impulse response of the following filters.
a) y (n)  x(n)  x(n  1)
b) y(n)  2 x(n)  y (n  1)  .5 y(n  2)
c) y (n)  .5 x(n)  4 y (n  1)  8 y (n  2)
1) Use Matlab function “filter”
2) Take inverse of Z transform
3) Examine poles of filter and comment on expected
stability
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