Embed Size (px)
Citation preview
Part A: Signal Processing
Chapter 3: Digital Signal Processing
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Chapter 3: Digital Signal Processing
3.1 Introduction to Digital Signal Processing3.2 Analogue to Digital Conversion Process3.3 Quantisation and Encoding3.4 Sampling of Analogue Signals3.5 Aliasing3.6 Digital to Analogue Conversion3.7 Introduction to digital filters
3.7.1 Non-Recursive Digital filters3.7.2 Recursive Digital filters
3.8 Digital filter Realisation3.8.1 Parallel Realisation 3.8.2 Cascade Realisation
3.9 Magnitude and Phase Responses3.10 Minimum/Maximum/Mixed Phase Systems3.11 All-pass Filters3.12 Second Order Resonant Filter3.13 Stability of a Second Order Filter3.14 Digital Oscillators3.15 Notch FiltersProblem Sheet A3
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Chapter 3: Digital Signal Processing (DSP)
3.0 Introductions
Digital Signal Processing is a rapidly developing technology for scientists and engineers. In the 1990s the digital signal processing revolution started, both in terms of the consumer boom in digital audio, digital telecommunications and the wide used of technology in industry.
Due to the availability of low cost digital signal processors, manufacturers are producing plug-in DSP boards for PCs, together with high-level tools to control these boards. Prof
esso
r E. A
mbikair
ajah
UNSW, A
ustra
lia
There are many areas where DSP technology is now being used and the current proliferation of such technology will open up further applications.
Audiologists and speech therapists are exposed to DSP systems for both testing a person’s level of hearing and subsequently DSP hearing aid filtering.
The professional music industry uses spectrum analysers, digital filtering, sampling conversion filters etc and is one of the biggest users and exploiters of DSP technology.Prof
esso
r E. A
mbikair
ajah
UNSW, A
ustra
lia
In summary, DSP is applied in the area of control and power systems, biomedical engineering, instrumentation (test and measurement), automotive engineering, telecommunications, mobile communication, speech analysis and synthesis, audio and video processing, seismic, radar and sonar processing and neural computing.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
There are many advantages to using DSP techniques for variety of applications, these include:
high reliability and reproducibilityflexibility and programmabilitythe absence of component drift problemcompressed storage facility
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
DSP hardware allows for programmable operations. Through software, one can easily modify the signal processing functions to be performed by the hardware. For all these reasons, there has been vast growth in DSP theory & applications over the past decade.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
3.1 Introductions to DSPAn analogue signal processing system is shown in Fig 3.1, in which both the input signal and output signal are in analogue form
Analogue signal processor (e.g. low-pass filter)Analogue
Input SignalAnalogueOutput Signal
s(t)x(t) = s(t) + n(t)
signal noise
Figure 3.1. A general description of analogue systems whose input and output are in analogue form.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
A digital signal processing system Figure 3.2 provides an alternative method for processing the analogue signal.
s(n)
xa(t)
dB 3 dBA/D
Converter
x(n)Digital Signal
Processor
analogue prefilteror antialiasing filter analogue to digital
converter
Lowpass filtered signal
x(t)
Sampling frequency
discrete-time signal
D/A converter
Digital to analogue converter
dBs(t)
reconstruction filter (analogue filter) same as the pre-filter
Figure 3.2. A general process of converting analogue signals into digital signals and back to analogue form.Prof
esso
r E. A
mbikair
ajah
UNSW, A
ustra
lia
Note:
The A/D converter converts the analogue input signal into a digital form.
The D/A converter converts the processed signal back into analogue form.
The reconstruction filter smooths out the outputs of the D/A and removes unwanted high frequency components.
The analogue input filter is used to band-limit the analogue input signal prior to digitisation to reduce aliasing (see later)
The heart of the system in Figure 3.2 is the digital signal processor which may be based on a DSP chip such as Texas instruments TMS 320C60Prof
esso
r E. A
mbikair
ajah
UNSW, A
ustra
lia
The digital signal processor may implement one of the several DSP algorithms, for example digital filtering
(low-pass filter) mapping the input x[n]into the output s[n].
Digital signal processor implies that the input signal must be in a digital form before it can be processed.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
3.2 Analogue to digital conversion Process
Before any DSP algorithm can be performed, the signal must be in a digital form. The A/D conversion process involves the following steps:
The signal (Band-limited) is first sampled, converting the analogue signal into a discrete-time signalThe amplitude of each sample is quantised into one of 2B levels (where B is the number of bits used to represent a sample in the A/D converter)The discrete amplitude levels are represented or encoded into distinct binary words each of length B bits.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
A practical representation of the A/D conversion process is shown in Figure 3.3.
A/D converter
2B
1
close & open the switch at fs Hz
Analogue Signal(bandlimited)
xa(t)
T
Logic circuitB bits
x[n]
digitaloutput
Sample & Hold Quantiser encoder
Figure 3.3. Analogue to digital conversion process.Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Sample and hold (S/H) takes a snapshot of the
analogue signal every T sec and then holds that
value constant for T secs until the next snapshot is obtained.
Tfs
1=
S/H output
t
xa(t)
tT
Input signal
Figure 3.4. An example of “sample and hold” process to convert analogue signals into digital signals.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Example :
Sampled Signaln
0V
-6V
-12V
12V x[n]
6V
T = sampling period Analogue Signal
Samples
Figure 3.5. An example of sampling analogue signals to discrete-time signals. The sampling period is T.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
54321
digital
-1 1111-2 1110
-3 1101-4 1100-5 1011
01010100
00110010
0001
analogue
Example: 4-bit (B = 4) A/D converter (bipolar)
Input-output characteristic of 4-bit quantiser (linear) (two’s complement notation)Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
3.3 Quantisation and encoding [1]
Before conversion to digital, the analogue
sample is assigned one of 2B values (see Fig 3.6). This process, termed quantization, introduces an error, which cannot be removed.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
123456
sampling instants
Quantisation Level
123456
100
001
101
010
110
3 bits code output
Quantisation Level 3-bit A/D Converter(Unipolar)
encoder output
Figure 3.6. Quantisation of discrete-time signals.
LSB
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
A 12 bit A/D converter (bipolar) with an input voltage
range of ±10V will have a least significant bit (LSB) of
(resolution)mVmVV 9.412
2012 =−
mVVV 9.412
2012 =−
=Δ
1000 0000 000012 bits
0111 1111 111112 bits
+10V
-10V
212 levels = 4096
Resolution (step-size)
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Quantisation error = (one half of an
LSB) = 4.9 mV / 2 = 2.45 mV
Note:
2VΔ
level n+1
level n
level n-1
sampling instant
v∆V
∆V/2
∆V/2∆V
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
For an A/D converter with Binary digits the number
of quantisation level is 2B, and the interval between levels, that is the quantisation step size (ΔV) –resolution is given by
V-full scale range of the A/D converter with bipolar signal inputs. The maximum quantisation error, for the case where the values are rounded up or down .
BB
VVV212
≈−
=Δ
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
For a sine wave input of amplitude A, the quantisation step size becomes
The quantisation error (e) for each sample, is normally assumed to be random and uniformly distributed in the interval with zero mean.
In this case, the quantisation noise power or variance is given by
BAV
22
≈Δ2AA
-A
2VΔ
±
e = actual amplitude - quantised amplitude Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
∫∫Δ
Δ−
Δ
Δ−
Δ==
2
2
22
2
22 1)(
V
V
V
Ve dee
VdeePeσ
constant = VΔ1
Hence,
12
22 Ve
Δ=σ for uniform quantisation
(Note : Uniform quantisation - all steps are of equal size)Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
For the sine wave input, the average
signal power is , ie. rms value
The signal-to-quantisation noise power ratio (SQNR) in decibels is
2
2⎟⎠⎞
⎜⎝⎛ A
2
2A
( )
⎟⎟⎠
⎞⎜⎜⎝
⎛ ×=
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
=
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
Δ=
223log10
122/2
2log10
12
2log10
2
2
2
2
2
B
BA
A
V
A
SQNR
SQNR = 6.02B + 1.76 dB
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
The SQNR increases with the number of bits, B. In many DSP applications, an A/D converter resolution between 12 and 16 bits is adequate.
Thus, the signal-to-quantisation noise ratio increases approximately 6dB for each bit.
43.8 dB1287
37.7 dB646
31.6 dB325
25.3 dB164
18.7 dB83
SQNRLevelsNumber of Bits
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Summary: Analogue to Digital Converter
12-bitA/D
unipolar
12-bit
Example: Sampling frequency
x(t) x[n]
Conversion time (say) = 35 μs
0 to 5 volts
Step size or Resolution = volts12
512 −
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
In practice, the A/D is proceeded by a sample and hold (S/H) which freezes the signal during conversion.
Two parameters related to S/H:
Then the maximum frequency that can be converted becomes
maximum sampling frequency for the above A/D converter.
aperture time ≈ 25 ns (for example)acquisition time ≈ 2 μs (for example)
kHzfs 2710)025.0235(
16max=
++= −
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
3.4 Sampling of Analogue Signal
Suppose that an analogue signal x(t) is sampled
every T seconds, then at the output of the sampler
we obtain a discrete-time signal x(n) = x(t)|t = nT.
A/D
Sampling
freq.( )
x[n]
X(θ)
x(t)
x(ω)
Tfs
1=
s
a
ff
T
πθ
ωθ
2=
=
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
∫∞
∞−= ωω
πω deXtx tj)(
21)(
[ ] ∫∞
∞−= ωω
πω deXnx nTj)(
21
t=nT
Inverse Fourier Transform
(3.1)
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Now ejnωT is a periodic function of period 2π. Equation (3.1) becomes
[ ] ∫∞
∞−= ωω
πω deXnx nTj)(
21
(3.1)
[ ]
TdeT
kXT
TdeT
kXT
TTdeXnx
nTj
nTT
kj
k
k
k
nTj
k
ωπωπ
ωπωπ
ωωπ
ωπ
π
π
π
πω
ππ
ππ
ω
⎥⎦⎤
⎢⎣⎡ +=
+=
=
∫
∫∑
∫∑
−
−
+∞
−∞=
+
−
∞
−∞=
)2(121
)2(121
1)(21
)2(
2
2
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Digital spectrum Analogue spectrum
Let ωT = θ
We have
By comparing (3.2) and (3.3), We obtain
[ ] (3.2) )2(121 θπωπ
θπ
πde
TkX
Tnx nj
k⎥⎦
⎤⎢⎣
⎡+= ∑∫
∞
−∞=−
X(θ)
[ ] (3.3) )(21 θθπ
θπ
πdeXnx nj∫−=
Inverse Fourier Transform for discrete signal
(3.4) )2(1)( πθππωθ <<−+= ∑∞
−∞=k TkX
TX
Repeats itself
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
It is seen that X(θ) is periodic with
period 2π. The digital spectrum is a repetition of the analogue spectrum.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
3.5 Aliasing
Figure 3.7 illustrates the relationship between the digital spectrum X(θ) and the analogue spectrum X(ω) for the case X(ω) = 0, or .
Tπω >
2sff >
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Figure 3.7. Above : Frequency response of an analogue signal. Below : Frequency response of the sampled analogue signal.
X(ω)A
-π/T π/T ω
X(θ)A/T
-3π -2π -π 0 π 2π 3π ω
fs
Analogue Spectrum
Digital Spectrum
2sf
Tπω >Case 1: X(ω) = 0, (sampling theorem holds)
2sf
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Note: corresponds to θ = π (or )
The digital spectrum is the same as the original analogue spectrum and repeats at
multiples of the sampling frequency fs.
Tπω =
2sff =
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Case 2: X(ω) ≠ 0, , but X(ω) = 0,Tπω >
T23πω >
X(ω)A
Figure 3.8. Above: Frequency response of an analogue signal whose highest frequency component is larger than the sampling frequency. Below: Frequency response of the sampled analogue signal. The overlapped region represents aliasing.
-3π -2π -π π 2π 3π θ
aliasing
X(θ)A/T
23π
−2
3π
Tπ3
−Tπ
−Tπ
Tπ3 ω
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
If the sampling frequency, fs is not sufficiently high, the spectrum centred on fs will fold over or alias into the base band frequencies (Fig 3.8). Equation (3.4) tells us that aliasing can only be avoided if the analogue signal is band limited such that X(ω) = 0, ⇒ .
This results in the familiar sampling theorem. The minimum sampling frequency for which equation (3.4) holds is called the Nyquistfrequency.
Tπω >
22 sff
Tf ≥⇒≥
ππ
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Note: It should be noted that even if X(ω) is not strictly band limited, that it has some negligible energy outside , a small enough T can be chosen so that the overlap of the components of the summation in equation (3.4) is below a prescribed level. This is important when a sampling frequency is to be selected for a particular signal.
Tπ
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Digital-to-Analogue Conversion (D/A) – Signal recovery
The D/A conversion process is employed to convert the digital signal into an analogue form after it has been digitally processed. The reason for such conversion may be for example, to generate an audio signal to drive a loudspeaker or to sound an alarm.
The D/A process is shown in Figure 3.9. A register is used to buffer the D/A’s input to ensure that its output remains the same until the D/A is fed the next digital input.Prof
esso
r E. A
mbikair
ajah
UNSW, A
ustra
lia
Note: The inputs to the D/A are series of impulses, while the output of the DAC has a staircase shape as
each impulse is held for a time T sec.
)(ˆ ty
)(ˆ ty
n t
T-Sampling period T – Sampling period
DigitalSignal
ProcessorD/A
Low pass filter
y[n]
8 or 12 bits
y(t)
y[n]reconstruction filteror smoothing filter
Figure 3.9. Conversion process from digital signals to analogue signals.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
The D/A shown in Figure 3.9 is referred to as a zero-order hold.
By comparing its output and its input
y[n], it is evident that for each digital code fed into the D/A, its output is held for a time
T. The result is the characteristic staircase shape a the D/A output.
The D/A output approximates the analogue signal by a series of rectangular pulses whose height is equal to the corresponding value of the signal pulse.
)(ˆ ty
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
⎩⎨⎧ ≤≤
=otherwise
Ttth
001
)(
h(t)1
T t
Just consider one pulse.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
The corresponding frequency response is
2
2sin
22
2
]1[1
)()(
222
2
222
00
T
T
eTTjeeeT
jeeee
jw
jedte
dtethH
TjTjTj
Tj
TjTjwTjjwT
TtjTtj
tj
ω
ω
ω
ω
ω
ω
ωωω
ω
ωω
ωω
ω
−−
−
−−−
−−
∞
∞−
−
=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡−
=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−
=−−
=
⎥⎦
⎤⎢⎣
⎡−
==
=
∫
∫
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
The magnitude of H(ω) is plotted in Figure 3.10.
|H(ω)|
0 ω
Figure 3.10. Magnitude response of a rectangular pulse.
76π−
74π−
72π−
72π
74π
76π
2T
xxsin
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
In the frequency domain, the staircase action of the DAC introduces a type of distortion known as the or aperture distortion, where . x
xsin
2Tx ω
=
Y(θ)input to the D/A
-4 π -3 π -2 π - π 0 π 2 π 3 π 4 π θ
D/A output
ω
xxsin
)(ˆ ωy
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
The amplitude of the output signal spectrum is
multiplied by the function, which acts like a
lowpass filter, with the high frequencies heavily
attenuated.
The effect is due to the holding action of the
DAC and, in signal recovery, introduces an amplitude
distortion.
xxsin
xxsin
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
For a zero-order hold, the function falls to
about 4dB at half the sampling frequency
giving an average error of about 36.4%.
Aperture error can be eliminated by equalization. In
practice this can be achieved by first applying the
signal, before converting it to analogue, through a
digital filter whose amplitude-frequency response
has a shape.
xxsin
⎟⎠⎞
⎜⎝⎛
2sf
xx
sin
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Reconstruction Filter [4]
The output of the D/A converter contains unwanted high frequency at multiples of the sampling frequency as well as the desired frequency components.
The role of the output filter is to smooth out the steps in the D/A output thereby removing the unwanted high frequency components. In general, the requirements of the anti-imaging filter are similar to those of the anti-aliasing filter.Prof
esso
r E. A
mbikair
ajah
UNSW, A
ustra
lia
Ideal D/A Converter
Ideal D/A
T
y[n] y(t)
Ideal Lowpass filter
y(t)^ T
Tπ
−Tπ
)(ωH
Impulse not square pulses as in the case of an non-ideal D/A
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Consider Just an impulse
1h(t)h(t)T)(tδ
)(ωH
ω
TtTt
thπ
πsin)( =
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
If we apply the signal to the input of the filter, we obtain
y[n] can be written in this form.
)(ˆ ty
∑∞
−∞=
−=
=
nnTtny
TtTttythty
)()(*)/(
)/sin()(ˆ*)()(
δππ
[ ]∑∞
−∞= ⎭⎬⎫
⎩⎨⎧
−=n Tt
TtnTtnyty)/(
)/sin(*)()(ππδ
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Using the property
we can obtain
)()(*)( 00 ttxtttx −=−δ
[ ] (3.5) )(
)(sin)( ∑
∞
−∞= −
⎥⎦⎤
⎢⎣⎡ −
=n
TnTtT
nTt
nyty π
π
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
The original signal can be obtained by adding
together an infinite number of pulses. The nth
pulse here is shifted through a distance nT with
respect to the origin and multiplied (weighted) by a
factor y[n].
This recovery process is called interpolation. Figure
3.11 shows the implementation of equation (3.5).
xxsin
xxsin
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
t2TT0-T
y(1)
y(2)y(0)
y(-1)
y(-2)
y(x) = x(t)original signal
Figure 3.11. Each discrete-time sample is multiplied by a shifted sinc function. Summing these sinc functions will produce the original analogue signal. Prof
esso
r E. A
mbikair
ajah
UNSW, A
ustra
lia
The signal x(t) is reconstructed from the
samples of by summation of
weighted and shifted pulse.
[ ] [ ]nynx ≡
xxsin
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Consider the analogue signal [1]
x(t) = 3 cos50πt + 10 sin 300πt – cos 100πt
What is the Nyquist rate for this signal?
The frequencies present in the signal above are
f1 = 25Hz; f2 = 150 Hz; f3 = 50 Hz
Hence, fmax = 150 Hz
fsampling > 2 fmax = 300 Hz
The Nyquist rate is fN = 2 fmax = 300 Hz.
Example:
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Note: Consider x(t) = 10 sin 300πtfs ≥ 2 × f = 300 Hz
[ ] ( )
( )n
nf
nTnx
s
π
π
π
sin10
300sin10
300sin10
=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
= x[n]
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
We are sampling the analogue sinusoid at its zero-crossing points and hence we miss the signal completely. The situation will not occur if the sinusoid is offset by some phase (here).
In such case we have
( ) ( )φπ += ttx 300sin10 and sf
T 1= , where fs = 300Hz.
[ ] ( )( ) ( ) ( ) ( )[ ]( ) ( )φπ
φπφπφπ
sincos10sincoscossin10
sin10
nnn
nnx
=+=
+=for n = 0,1,2,..Prof
esso
r E. A
mbikair
ajah
UNSW, A
ustra
lia
Since cos(πn) = (-1)n , [ ] ( ) ( )φsin101 nnx −=
If θ ≠ 0 or θ ≠ π, the samples of the sinusoid taken at the Nyquist rate are not all zero.
Note: x(t) = A cos(2πf0t) is a continuous-time sinusoidal signal
22
(3.6) 2cos)(
0
0
ss
s
fff
nffAnx
≤≤−
⎟⎟⎠
⎞⎜⎜⎝
⎛= π
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
On the other hand, if the sinusoids,
[where fk = f0 + kfs , k = ±1, ±2, ±3, ….] (3.7)
are sampled at a rate fs, it is clear that the frequency fk is outside the fundamental frequency range
; consequently the sampled signal is
( ) ( )tfAtx kπ2cos=
22 0ss fff
≤≤−
[ ]
⎟⎟⎠
⎞⎜⎜⎝
⎛+=
⎟⎟⎠
⎞⎜⎜⎝
⎛ +=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
knnffA
nf
kffA
nffAnx
s
s
s
s
k
ππ
π
π
22cos
)(2cos
2cos
0
0 [ ] (3.8) 2cos 0 nffAnx
s⎟⎟⎠
⎞⎜⎜⎝
⎛=⇒ π
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Which is identical to the discrete-time signal in equation (3.6). If we are given a
sequence x[n] there is an ambiguity as to
which continuous-time signal x(t) these values represent. We can say the
frequencies fk = f0+kfs are indistinguishable
from the frequency f0 after sampling and
hence they are aliases of f0.
[ ] nffAnx
s⎟⎟⎠
⎞⎜⎜⎝
⎛= 02cos π
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Note:fs – sampling frequency
corresponds to θ = π
is the highest frequency that can be
represented uniquely with a sampling rate fs
is called half the sampling frequency or folding frequency.
2sf
sffT πωθ 2==
2sf
2sf
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Example :
Consider the analogue signal
(a) What is the Nyquist rate for this signal?
The frequencies existing in the analogue signal are:
f1 = 1 kHz; f2 = 3 kHz; f3 = 6 kHz
Thus fmax = 6 kHz and according to the sampling theorem,
fs > 2 fmax = 12 kHzThe Nyguist rate is = 12 kHz.
( ) ( ) ( ) ( )ttttx πππ 12000cos106000sin52000cos3 ++=
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(b) Assume now that we sample this signal x(t) using a
sampling rate fs = 5 KHz (samples/sec). What is the discrete-time signal obtained after sampling?
fs = 5000Hz ⇒ 25002=sf
x(t) = 3 cos (2π × 1000t) + 5 sin(2π × 3000t) + 10 cos (2π×6000t)
[ ]
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛−+⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛=
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ ++⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ −+⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛=
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛=
⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=
nnn
nnn
nnn
nnnnx
512cos10
522sin5
512cos3
5112cos10
5212sin5
512cos3
562cos10
532sin5
512cos3
50006000cos10
500030002sin5
500010002cos3
πππ
πππ
πππ
ππ
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
The same result can be obtained using equation (3.7).
We have fk = f0 + kfs ;
f0 = fk – kfs can be obtained by
subtracting from fk an integer multiple
of fs such that .
[ ] ⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛= nnnx
522sin5
512cos13 ππ
Second Method:
kHzf
kHzf ss 5.2
25 =⇒=
22 0ss f
ff
≤≤−Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
The frequency f1 = 1000 Hz is (= 2500 Hz)
and thus it is not affected by aliasing.
However, the other two frequencies f2 & f3 are above the folding frequency and they will be changed by the aliasing effect.
f2' = f2 – 1 fs = 3000 – 5000 = -2 kHzf3' = f3 – 1 fs = 6000 – 5000 = 1 kHz
2sf
<
[ ] ⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛−+⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛= nnnnx
500010002cos10
500020002sin5
500010002cos3 πππ
This is agreement with the result obtained before.Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(c) What is the analogue signal y(t) we can reconstruct from the samples if we use ideal interpolation.
Since only frequency components at 1 kHz and 2 kHz are present in the sampled signal, the analogue signal we can recover is,
y(t)=13cos(2000πt)-5sin(4000πt)
which is obviously different from the original signal x(t).The distortion of the original analogue signal was caused by the aliasing effect, due to the low sampling rate used.Prof
esso
r E. A
mbikair
ajah
UNSW, A
ustra
lia
Example:An analogue signal x(t) =sin(480πt)+3sin(720πt)is sampled 600 times per second. [1](a) Determine the Nyguist sampling rate for x(t)
(b) Determine the folding frequency (or half the sampling frequency)
(c) What are the frequencies, in radians, in the resulting discrete time signal x[n]?
(d) If x[n] is passed through an ideal D/A converter what is the reconstructed signal y(t)Prof
esso
r E. A
mbikair
ajah
UNSW, A
ustra
lia
(a) x(t) = sin(2π 240t) + 3sin(2π 360t)
f1 = 240 Hz f2 = 360 Hz
∴ fmax = 360 Hz ∴FNyquist = 2 × fmax = 720 Hz
(b) fs = 600 Hz ∴ffold or = 300 Hz2
sf
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(c)
[ ]
⎟⎠⎞
⎜⎝⎛−=
⎟⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛=
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ −+⎟
⎠⎞
⎜⎝⎛=
⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=
⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛==
=
n
nn
nn
nn
nntxnxnTt
54sin2
54sin3
54sin
542sin3
54sin
56sin3
54sin
6003602sin3
6002402sin)(
π
ππ
πππ
ππ
ππ
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
We have fk = f0 + kfs and therefore f0 = fk – k fs
f1 = 240 Hz and is (= 300 Hz)
f2 = 360 Hz and is (= 300 Hz)
Aliased frequency f0 = fk – k fs =360 – 1 × 600= -240 Hz
Second Method
22 0ss f
ff
≤≤−
2sf
<
2sf
<
-not affected by aliasing
-affected by aliasing
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
[ ]
⎟⎠⎞
⎜⎝⎛−=
⎟⎠⎞
⎜⎝⎛ −
+⎟⎠⎞
⎜⎝⎛=
n
nnnx
54sin2
6002402sin3
6002402sin
π
ππ
( )tty
nfnnTtnny
s
π
π
480sin2)(
600
6002402sin2)(
−=
===⎟⎠⎞
⎜⎝⎛−=
(d)
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Note :
(a) bit rate = fs × no of bits
= 8000 samples/sec × 12
=96,000 samples/sec
x[n]
12
x(t) 12-bits A/D(fs = 8,000
kHz)
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(b) In the case of PCM, speech signals are filtered to remove effectively all frequency components above 3.4 kHz and the sampling rate is 8000 samples per sec
Bit rate (bits per second)= sampling frequency × bits/sample= 8000 samples/second × 8-bits/sample= 64,000 bits/sec
0-3.4 kHz
8-bit persample
fs = 8000 Hz(8000 samples/sec)
speech signal x(t) 8-bits(compressed PCM)A/D
converter
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
bit rate=16×44100 bits/sec=0.7056 Mbits/sec
CD16 bit
fs= 44.1 kHz
CDReader
16 bitD/A
lowpassfilter
AMP
16 bit fs = 44.1kHz
(c)
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
3.7 Introduction to Digital Filters
There are two types of digital filters:
Recursive (there is at least one feedback path in the filter)Non-recursive (no feedback paths)
A linear time invariant discrete (LTD) system described by the following equation is commonly called a digital filter:
[ ] [ ] [ ] (3.9) 10∑∑==
−−−=L
kk
M
kk knybknxany
Feed forward Feedback
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
where x[n] is the input signal, y[n] is the
output signal. a0, a1, a2, ...., aM; b1, b2,
b3, ..., bL are constants (filter coefficients). These coefficients determine the characteristics of the system.
when bk = 0 the filter is said to be non-recursive type
when bk ≠ 0 recursive type.
[ ] [ ] [ ] (3.9) 10∑∑==
−−−=L
kk
M
kk knybknxany
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
3.7.1 Non Recursive Digital Filters (FIR)
If bk = 0, then the calculation of y[n] does not require the use of previously calculated samples of the output (see equation (3.9)).
This is recognised as a convolution sum.
[ ] [ ] [ ] [ ] [ ]knxanxanxaknxany M
M
kk −++−+=−= ∑
=
L1100
[ ] [ ] [ ]∑=
−=M
kknxkhny
0Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Therefore the impulse response is identical to the coefficients, that is,
Any filter that has an impulse response of finite duration is called Finite Impulse Response (FIR) filter.
[ ]⎩⎨⎧ ≤≤
=otherwise
Mnanh n
00
[ ] [ ] [ ] [ ] [ ]knxanxanxaknxany M
M
kk −++−+=−= ∑
=
L1100
h(0) h(1) h(M)
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Example :
[ ] [ ] [ ] [ ]2
21
10
210
)()()(
21
−− ++==
−+−+=
zazaazXzYzH
nxanxanxany
This is a (non-recursive) second order FIR filter
Property: A property of the FIR filter is that it will always be stable.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(a) Stability requires that there should be no poles outside the unit circle. This condition is automatically satisfied since there are no poles at all outside the origin. (In fact, all poles are located at the origin.)
(b) Another property of non-recursive filter is that we make filters with exactly linear phase characteristics [4]
Note: The ability to have an exactly linear phase response is one of the most important properties of a LTD system (filter). When a signal passes through a filter, it is modified in amplitude and/or phase. The nature and extent of the modification of the signal is dependent on the amplitude and phase characteristics of the filter.Prof
esso
r E. A
mbikair
ajah
UNSW, A
ustra
lia
The phase delay or group delay of the filter provides a useful measure of how the filter modified the phase characteristic of the signal. If we consider a signal that consists of several frequency components (eg. speech waveform) the phase delay of the filter is the amount of time delay each frequency component of the signal suffers in going through the filter.
(3.10) )()(_θθφ
−=pTdelayphase
[the negative of the phase angle divided by frequency]
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
The group delay on the other hand is the average time delay the composite signal suffers at each frequency as it passes from the input to the output of the filter.
(3.11) )()(_θθφdTdelaygroup g −=
[the negative of the derivative of the phase with respect to frequency]Prof
esso
r E. A
mbikair
ajah
UNSW, A
ustra
lia
A constant group delay means that signal components at different frequencies receive the same delay in the filter.
A linear phase filter gives same time delay to all frequency components of the input signal. A filter with a nonlinear phase characteristic will cause a phase distortion in the signal that passes through it.
φ(θ)
φ(θ) = -θ
- π π θ
Figure 3.12. Phase response of a linear phase filter
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
This is because the frequency components in the signal will each be delayed by an amount not proportional to frequency, thereby altering their harmonic relationship. Such a distortion is undesirable in many applications, for example music, video etc.
A filter is said to have a linear phase response if its phase response satisfies one of the following relationships:
( )( ) (3.12) θθφ
θθφab
a−=
−=
where ‘a’ and ‘b’ are constants.Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Example
Two filter structures are shown below. Show that both filters have linear phase.
z-1 z-1
+1 +1 +1
+
x[n]z-1 z-1
+1 -1
+
x[n]
y[n]y[n]
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
z-1 z-1
+1 -1
+
x[n]
y[n]
[ ] [ ] [ ] [ ]( )( )
( )( )θ
θ
θ
θθθ
θθ
cos211
11
21
2
21
+=
++=
++=
++=
−+−+=
−
−−
−−
−−
j
jjj
jj
eeee
eeHzzzH
nxnxnxny
z-1 z-1
+1 +1 +1
+
x[n]
y[n]
[ ] [ ] [ ]( )( )
( )( )θθ
θ
θ
θ
θθθ
θ
π
π
sin2
sin2
22
11
2
2
2
2
2
−
−
−−
−
−
=
=
⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
−=
−=
−−=
j
jj
jjj
j
e
ee
jeeje
eHzzH
nxnxny
phase: φ(θ) = π/2 - θ ,linear phase
phase: φ(θ) = -θ ,linear phaseProf
esso
r E. A
mbikair
ajah
UNSW, A
ustra
lia
3.7.2 Recursive Digital filter (IIR)Every recursive digital filter must contain at least one closed loop. Each closed loop contains at least one delay element.
For Recursive digital filters bk ≠ 0. Let a0 = a0 ,ak = 0 for k > 0, b1 = b1 & bk = 0 for k > 1
[ ] [ ] [ ]∑∑==
−−−=L
kk
M
kk knybknxany
10
[ ] [ ] [ ]
( ) 11
0
10
1
1
−+=
−−=
zbazH
nybnxany
IIR filter
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
A recursive filter is an infinite impulse response filter (IIR).
Examples :2
21
10)( −− ++= zazaazH
22
111
1)( −− ++=
zbzbzH
22
11
22
11
11)( −−
−−
++++
=zbzbzazazH
Zeros only
Poles and Zeros
(2nd order FIR filter)
2nd order IIR filter (all pole filter)
IIR filter
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Example :
The difference equation is: y[n] = x[n] + ay[n-1].
The DC gain of H(z) can be obtained by substituting θ = 0.
If dc gain is undesirable, introduce a constant gain
factor of (1-a), so that H(z) becomes
∴
Note: Poles and zeros can be real or imaginary
101
1)( 1 <<−
= − aaz
zH
aaeH j −
=−
= −= 11
11|)( )0(0θθ
111)( −−−
=az
azH
dc gain = 1]1[][)1(][ −+−= naynxanyProf
esso
r E. A
mbikair
ajah
UNSW, A
ustra
lia
Example:
Consider a lowpass filter
(i) Determine ‘b’ so that |H(0)| = 1
(ii) Determine the 3dB bandwidth (here) for the normalised filter in part (i)
10 1 <<+= a bx[n],] ay[n-y[n]
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(i) Y(z) = aY(z) z-1 + b X(z) ⇒ 11)( −−=
azbzH
θjaebzH −−
=1
)(
11
)0( =−
=a
bH
θθθθ
θθθ θ
cos211
)sin()cos1(1)(
sin)cos1(1
11)(
222 aaa
aaaH
jaaa
aeaH j
−+
−=
+−
−=
+−−
=−−
= −
we have |H(0)| =1
⇒ b = 1 - a
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Second Method:
⎭⎬⎫
⎩⎨⎧
+−−
=−−
⋅−−
=⋅= − 2
2*2
cos21)1(
11
11)()(|)(|
aaa
aea
aeaHHH jj θ
θθθ θθ
222 |)0(|21|)(|1|)0(| H
cHH =
=∴=
θθθ
(half-power point)θ
21
1
|H(θ)|0 dB
3 dB
θc
2cos211|)(|
aaaH
+−
−=
θθ∴
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
θ
21
1
|H(θ)|2
θc
⎥⎦
⎤⎢⎣
⎡ −+−=
−+−
=
−
aaa
aaa
c
c
214cos
cos21)1(
21
21
2
2
θ
θ
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Example: Consider a filter described by
2
2
)(azcczazH
++
= where a & c are constants.
Show that the magnitude response |H(θ)| is unity for all θ.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
1][][
)()(|)(|
2222
2222
2
2
2
2*2
=++++++
=
++
⋅++
=⋅=
−
−
−
−
θθ
θθ
θ
θ
θ
θ
θθθ
jj
jj
j
j
j
j
eeaccaeeacac
ceaaec
ceaaecHHH
|H(θ)|
-π π θ
This is an all-pass filter.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
3.8 Digital filter Realisation [11][ ] [ ] [ ]
1
1
1)()()(
)()()(
1
1
0
1
0
10
10
444 3444 2143421
43421
structure
poles
L
k
kk
zeros
M
k
kkL
k
kk
M
k
kk
L
k
kk
M
k
kk
L
kk
M
kk
zbza
zb
za
zXzYzH
zzYbzzXazY
knybkxany
∑∑
∑
∑
∑∑
∑∑
=
−=
−
=
−
=
−
=
−
=
−
==
+⋅=
+==∴
−=
−−−=
444 3444 21
43421
43421
2
0
11
1
structure
zeros
M
k
kk
poles
L
k
kk
zazb
∑∑ =
−
=
−⋅
+=
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
-b1
-b2
-bL
X(z)+
z-1
z-1
z-1
a1
a2
aM
a0 Y(z)
z-1
z-1
z-1
Figure 3.13. Structure 1 or Direct Form 1
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
a1
a2
aM
-b1
-b2
-bL
a0X(z) Y(z)
+ +
z-1
z-1
z-1
z-1
z-1
z-1
Figure 3.14. Structure 2 or Direct Form IIProfes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
In the case when L = M, we have Canonic form realisation.
A discrete-time filter is said to be canonic if it contains the minimum numbers of delay elements necessary to realise the associated frequency response.
a0
a1
a2
aM
-b1
-b2
-bL
X(z)+
z-1
z-1
z-1
Y(z)+
Figure 3.15. Canonic form
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
3.8.1 Parallel Realisation [11]
44444444 344444444 21
L
L
structureparallel
k
k
ii
LL
MM
zHzHzHzHzH
zbzbzbzazazaazH
_
3211
22
11
22
110
)(...)()()()(
1)(
++++==
++++++++
=
∑=
−−−
−−−
(use partial fraction to obtain Hi(z))
Y(z)
…
H1(z)
H2(z)
Hk(z)
X(z)+
Figure 3.16. Parallel structure
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
3.8.2 Cascade Realisation [11]
44444 344444 21
L
L
structurecascade
k
k
ii
LL
MM
zHzHzHzHzHzH
zbzbzbzazazaazH
_
3211
22
11
22
110
)(ˆ)...(ˆ)(ˆ)(ˆ)(ˆ)(
1)(
⋅⋅==
++++++++
=
∏=
−−−
−−−
(Product of lower order transfer function ie. 1st or 2nd order sections)The cascade structure is the most popular form
Y(z)X(z)
Fig 3.17. Cascade structure
)(ˆ1 zH )(ˆ
2 zH )(ˆ zH k
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Example:
A parallel realisation of a third order system
H(z) is given by
21
1
1
211
321
3252
)325)(2(19364023)(
−−
−
−
−−−
−−−
++−
++
+=
++++++
=
zzDzC
zBA
zzzzzzzH
H0(z) H1(z) H2(z)Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
21
1
1
21
1
1
21
1
1
6.04.01
)151
1523(
5.01)5.2(
319
)6.04.01(5)23(
31
5.011
25
319)(
325)23(
31
25
319)(
−−
−
−
−−
−
−
−−
−
−
++
+−+
+−
+=
++−
−+
−=
++−
−+
−=
zz
z
z
zzz
zzH
zzz
zzH
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
-0.5
-0.4
-0.6
x[n]
+
z-1
z-1
+
-2.5
z-1
y[n]+ +
319
1523
−
151
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Example: A cascade realisation of a third-order system is given by
⎟⎟⎠
⎞⎜⎜⎝
⎛++++
⎟⎟⎠
⎞⎜⎜⎝
⎛++
=
++++
++
=
++++
++
=
++++
++
=
++++++
=
−−
−−
−
−
−−
−−
−
−
−−
−−
−
−
−−
−−
−
−
−−−
−−−
21
21
1
1
21
21
1
1
21
21
1
1
21
21
1
1
321
321
6.04.018.34.36.4
5.01)5.05.0(
6.04.018.34.36.4
5.01)5.05.0(
)6.04.01(5191723
)5.01(2)1(
325191723
)2()1(
3891019364023)(
zzzz
zz
zzzz
zz
zzzz
zz
zzzz
zz
zzzzzzzH
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
x[n] y[n]
0.5-0.5
0.5
z-1
+ +
3.4-0.4
4.6
z-1
+
3.8-0.6 z-1
+
Cascade
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Example:
Implement the following system in the cascade, direct form II and parallel structures. All coefficients are real.
)1)(1(1)( ).( 11 −− −+
=bzaz
zHa
cascade structure
x[n] y[n]
-az-1
+
bz-1
+
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
21)(11)( −− −−+
=abzzba
zH
y[n]x[n]+
-(a-b)z-1
+ab z-1
Direct form II
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
1111 1111)( −−−− −
++++=
−+
+=
bzba
b
azba
a
bzB
azAzH
Parallel structure
-a
b
x[n]
z-1
z-1
y[n]baa+
bab+
+
+
+
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
31)1(1)( )( −−
=az
zHb
a
x[n] y[n]
az-1
+
az-1
+
z-1
+
cascade structureS
No parallel structure exists because partial fraction expansion cannot be performed.Prof
esso
r E. A
mbikair
ajah
UNSW, A
ustra
lia
33221 3311)( −−− −+−
=zazaaz
zH
Direct Form II
a3
x[n] y[n]
3az-1
+
-3a2z-1
z-1
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
cascadezbbzzaaz
zbazba
parallelbzza
zHc
←+−+−
+++−=
←−
+−
=
−−−−
−−
−−
)21)(21()()(21
)1(1
)1(1)( ).(
221221
1221
2121
)1)(1(1
)1)(1(1)( 1111 −−−− −−
+−−
=bzbzzaza
zH
y[n]x[n]a
z-1+
az-1
+
parallel structure
+
bz-1
+
bz-1
+
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
-2(a+b)
a2+b2
y[n]+
2bz-1
-b2 z-1
+
2a z-1
-a2
x[n]
z-1
+
221221
2221
211
)21()()(21)( −−−−
−−
+−⋅
+−+++−
=zbbzzaaz
zbazbazH
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
3.9 Magnitude and Phase Response [11]
We can show that the magnitude response is an even function of frequency
The phase response is an odd function of frequency
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Example:Calculate the magnitude and phase response of the
3-sample averager given by
[ ]⎪⎪⎩
⎪⎪⎨
⎧ ≤≤−=
otherwise
nnh
0
1131
∑ ∑∞
−∞=
−
−=
−− ++===n n
nn zzzznhznhzH 1011
1 31
31
31)()()(
[ ]11 131)( zzzH ++= −Prof
esso
r E. A
mbikair
ajah
UNSW, A
ustra
lia
[ ] [ ]3111
31)()( ⋅++=++== −−
=θθθθ
θθ jjjjez
eeeezHH j
[ ]θθ cos2131)( +=H
Precautions must be taken when determining the phase response of a filter having a real-valuedtransfer function, because negative real values produce an additional phase of π radians.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
For example, let us consider the following linear-phaseform of the transfer function
H(θ) = e-jkθB(θ)
real-valued function of θ that can take positive and negative values.
Let phase angle is φ :
( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )θθθθθ
θθθθθkjBkBH
kjBkBHsincossincos
−−=−+−=
)tan()cos()()sin()(tan θ
θθθθφ k
kBkB
−=−=
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
The phase function φ(θ) includes linear phase term and also accommodates for the sign changes in B(θ). Since -1 can be expressed as , phase jumps of ±π will occur at frequencies where B(θ) changes sign.
If B(θ) > 0, then φ(θ) = -kθ.
If B(θ) < 0, then φ(θ) = -kθ ± π..
tanφ = tan(-kθ) ∴φ = -kθ or φ(θ) = -kθ
phase angle
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Let us get back to our example
πωωπ
ω
θπθφ
θθφ
θθθθ
ππ
ππ
<<−≤≤−
<<−
⎪⎩
⎪⎨⎧
<±=
>=
+=⇒+=
32
32
32
32
and0)(0)(
0)(0)(
]cos21[31|)(|]cos21[
31)(
H
H
HH
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
The appropriate sign of π must be chosen to
make φ(θ) an odd function of frequency.
φ(θ)
-2π -π -2π/3 2π/3 π 2π θ
|H(θ)|
-2π -π -2π/3 2π/3 π 2π θ
π
-π
Odd function
Even function
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Example:Find the magnitude and phase response of the following:
0)(,21)2(,1)1(,
21)0( ==== nhhhh
θθφ
θθ
θ
θ
θ
θθθ
θθ
−=
+=
⎥⎦⎤
⎢⎣⎡ ++=
++=
++=++=
−
−−
−−
−−−−
)(
]cos1[)(211
21
21
21)(
21
21
211
21)(
)(
2
21210
43421B
j
jjj
jj
eH
eee
eeH
zzzzzzH
The amplitude function is never negative (therefore there is no phase jumps of ±π)Prof
esso
r E. A
mbikair
ajah
UNSW, A
ustra
lia
-π π θ
Even function2
|H(θ)|
-π-π π
Odd function
πφ(θ)
θ
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Example :
10)(
01)(case
otherwisennn
⎭⎬⎫
===
δδ
φ(θ)|H(θ)|
1
-π π θ -π π θ
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
|H(θ)|1
-π πθ
-π
φ(θ)
-π πθ
π
k
δ(n-k)
n
h[n] = δ[n-k]H(z) = 1z-k
H(θ) = 1e-jθk
B(θ) = 1 φ(θ) = -kθ
Note: When phase exceeds ±πrange a jump of ±2π is needed to bring the phase back into ±πrange.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Phase Jumps: From the previous examples, we note that there are two occasions for which the phase function experiences discontinuities or jumps.
(1) A jump of ±2π occurs to maintain the phase
function within the principal value range of [-π and π]
(2) A jump of ± π occurs when B(θ) undergoes a change of sign
The sign of the phase jump is chosen such that the resulting phase function is odd and, after the jump, lies in
the range [-π and π].Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Example: Magnitude and phase response of causal 3-sample average.
[ ]
[ ]
πθπ
πθπθπθ
πθπθθθφ
θθθθθφ
θθ
θ
θ
θ
θθθ
θθ
<<
−<<−<±−=
<<−>−=
+==−=
+=∴
++=
++=⇒++=
⎩⎨⎧ ≤≤
=
−
−−
−−−−
32
320)(
32
320)()(
]cos21[31|)(||)(|;)(
]cos21[31)(
131
31
31
31)(
31
31
31)(
otherwise020for
)(
221
31
B
B
BH
eH
eee
eeHzzzH
nnh
B
j
jjj
jj
4434421
-π
φ(θ)
-π π θ
π
1|H(θ)|
-π -π/2 π/2 π θ
π
2π/3
-2π/3
Phase is undefined at points |H(θ)| = 0.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Example:
Determine and sketch the magnitude and phase response of the following filters:
[ ] [ ] [ ]( )
[ ] [ ] [ ][ ] [ ]4 )(
8 )(
121 )(
−=−−=
−−=
nxnyiiinxnxnyii
nxnxnyi
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
2sin
2sin
222
1
21]1[
21)(
]1[21)()]()([
21)(
22
222
222
222
11
θ
θ
θ
θπ
ππθ
θθθ
θθθθ
⎟⎠⎞
⎜⎝⎛ −
−
−−
−−−
−−
=
⎥⎦
⎤⎢⎣
⎡==
⋅⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡−
=
⎥⎦
⎤⎢⎣
⎡−=−=
−=⇒−=
j
jjj
jjj
jjjj
e
ejee
jjeee
eeeeH
zzHzXzzXzY
|H(θ)|
-π πθ
π/2
-π/2
φ(θ)
-π π θ
(i)
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
θπθφ
θ
θθ
θ
θπ
πθ
θθθθ
42
)(
4sin2
4sin222
1)(
1)()()()()(
)(
42
2444
48
88
−=
=
⋅=×⎥⎦
⎤⎢⎣
⎡ −=−=
−=⇒−=
⎟⎠⎞
⎜⎝⎛ −
−−
−−
−−
43421B
j
jjjj
jj
e
eejjeeeeH
zzXzYzXzzXzY
φ(θ)
|H(θ)|
π/4 π/2 3π/2 π θ
-π/2
π/2
π/4 π/2 3π/2 π θ
(ii)
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
[ ] [ ]( ) ( ) ( )( )( ) 1
4
4
44
=
=
=⇒=
−=
−
−−
θθ θ
HeH
zzHzXzzYnxny
j
|H(θ)|1
-π π θ
-π
π
φ(θ) = -4θ
π/4 π θ
(iii)
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Example:Determine and sketch the magnitude and phase response of 1st order recursive filter (IIR filter)
[ ] [ ] [ ]1−+= naynxny
phaseaa
aa
aaa
aaa
aaja
aaa
aeae
aeH
aeH
azzH
zXzY
j
j
j
j
⎥⎦⎤
⎢⎣⎡−−
=
−−
=
+−−
+−−
=
+−−
+−−
=−−
⋅−
=
−=
−==
−
−
−−
θθθφ
θθ
θθθθ
φ
θθ
θθθ
θ
θ
θ
θ
θ
cos1sintan)(
cos1sin
cos21cos1
cos21sin
tan
cos21sin
cos21cos1
11
11)(
11)(
11)(
)()(
1
2
2
22
1
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
θθ jj aeae −⋅
−= − 1
11
1
2cos211|)(|
aaH
+−=
θθ
Even Symmetry|H(θ)|
-π πθ
a= 0.5
φ(θ)Odd Symmetry
-π π θ
Non-linear phase
Magnitude: |H(θ)|2 = H(θ)⋅H*(θ) [H*(θ) is the complex conjugate]
Assuming 0 < a < 1
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Examples :
The gain k0 can be selected as 1 – a, so that the
filter has unity gain at θ = 0.
In this case, (for unity gain at θ = 0).
Filter Pass-Low 1
11
)( ).( 110
1 −− −−
⇒−
=az
aazkzHa
1
1
02 11)( ).( −
−
−+
=azzkzHb
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
The addition of a zero at z = -1 further attenuates the response of the filter at high frequencies
|H2(θ)|
|H(θ)|
-π π θ
Low-Pass Filter|H1(θ)|
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(c) We can obtain simple high-pass filters by reflecting (folding) the pole-zero locations of the low-pass filters about the imaginary axis in the z-plane.
1
1
3 11
21)( −
−
+−
⋅−
=azzazH
π θ-π
1
|H3(θ)|High pass filter
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
( )[ ] [ ] [ ]1
1 14
−+=+= −
nxnxnyzzH
( )[ ] [ ] [ ]1
1 15
−−=−= −
nxnxnyzzH
(d)
(e)
High-Pass Filter
Low-Pass Filter
π θ-π
2|H5(θ)|2
|H5(θ)|2 = 2(1-cosθ)
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
|H6(θ)|2 = 4(1-cosθ)2
π θ-π
4|H6(θ)|2
(f) ( ) ( )21
6 1 −−= zzH
( ) ( )317 1 −−= zzH
(g) |H7(θ)|2 = 8(1-cosθ)3
π θ-π
8|H7(θ)|2
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
3.10 Minimum-phase, Maximum-phase and Mixed phase systems [11]
Let us consider two FIR filters:
12
11
21)(
211)(
−
−
+=
+=
zzH
zzH
21−=
ρ
|z|=1
ρ = -0.5
|z|=1Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
H2(z) is the reverse of the system H1(z). This is due to the reciprocal relationship between the zeros of
H1(z) & H2(z).
The magnitude characteristics for the two filters are
identical because the roots of H1(z) & H2(z) are reciprocal.
θθθ
θθ θθ
cos45|)(||)(|
21)(&
211)(
21
21
+==
+=+= −−
HH
eHeH jj
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Phase:
θθθφ
θ
θθφ
cos2sintan)(
cos21
sintan)(
11
12
+=
+=
−
−
-π
π
φ2 (θ)
-π π θ
-π
πφ1(θ)
-π π θ
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Note: If we reflect a zero z = ρ that is inside
the unit circle into a zero outside the unit
circle the magnitude characteristic of the system is
unaltered, but the phase response changes.
We observe that the phase characters φ1(θ) begins at zero phase at frequency θ = 0 and terminates at zero phase at the frequency ω = π. Hence the net phase change.
Minimum phase filter
ρz 1=
( ) ( ) 0011 =−φπφProfes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
On the other hand, the phase characteristic for the filter with the zero outside the unit circle undergoes a net phase change
As a consequence of these different phase characteristics, we call the first filter a minimum-phase system and the second system is called a maximum-phase system.
If a filter with M zeros has some of its zeros inside the unit circle and the remaining outside the unit circle, it is called a mixed-phase system.
( ) ( ) radians 012 πφπφ −=−
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
A minimum-phase property of FIR filter carries over to IIR filter.
Let us consider
is called minimum phase if all its poles and zeros are inside the unit circle.
)()()(
zAzBzH =
|z|=1
Re(z)Minimum phase
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
If all the zeros lie outside the unit circle, the system is called maximum phase.
|z|=1
Re(z)Maximum phase
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
If zeros lie both inside and outside the unit circle, the system is called mixed-phase.
|z|=1
Re(z) Mixed phase
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Note: For a given magnitude response, the minimum-phase system is the causal system that has the smallest magnitude phase at every frequency
(θ). That is, in the set of causal and stable filters having the same magnitude response, the minimum-phase response exhibits the smallest deviation from zero phase.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Example:
Consider a fourth-order all-zero filter containing a double complex conjugate set of zeros located at
.The minimum-phase, mixed phase and maximum phase system pole-zero patterns having identical magnitude response are shown below.
47.0πj
ez±
=
|z|=1
ρ
4
mixed-phase
ρ=0.72
24
|z|=1
Minimum-phase
|z|=1
4
maximum-phase
2
21/ρProfes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
The magnitude response and the phase response of the three systems are shown below: The minimum-phase system seems to have the phase with the smallest deviation from zero at each frequency.
|H(θ)|
θπ
φ(θ)
θ
minimum phase
mixed-phase (In the case linear phase)
maximum phase
-π
-2π
-3π
-4πProfes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Example:A third order FIR filter has a transfer function
G(z) given by
From G(z), determine the transfer function of an FIR filter whose magnitude response is identical
to that of G(z) and has a minimum phase response.
)25)(21216()( 1−+−−−−= zzzzG
)251)(
341)(
231(12)( 111 −−− ++−= zzzzG
>1
)52)(43)(32()( 111 −−− ++−= zzzzG
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
23
32
34
−25
−
)521)(
431)(
321()(filter phase Minimum The 111 −−− ++−= zzzkzP
lm(z)
Re(z)
|z|=1
-
- 43
−52
−
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
%Exercise 1. Minimum and maximum phase filter%We examine the properties of minimum and maximum phase filter%We start with a maximum phase filter from the example in page 64.zeros_maxphase = [3/2 -4/3 -5/2]' % zeros position%Note the transpose operation since the inputs are zeros and poles.%If the inputs are row vectors, they will be treated as polynomial coefficents.poles_maxphase = [0 0 0]‘%poles are located at origin%After specifying the poles and zeros position, we now plot their position.figure(1)%create a new figure (No.1)zplane(zeros_maxphase, poles_maxphase)%We also find the numerator and denominator coefficients of the transfer function
-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5
-1.5
-1
-0.5
0
0.5
1
1.5
Real Part
Imag
inar
y P
art
3
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
% in order to plot the magnitude and phase responses.
numerator_maxphase = poly(zeros_maxphase)
denominator_maxphase = 1;
% since all poles are at origin
figure(2)
% create a new figure (No.2)
freqz(numerator_maxphase,denominator_maxphase)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
5
10
15
20
Normalized Angular Frequency (×π rads/sample)
Mag
nitu
de (d
B)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-800
-600
-400
-200
0
Normalized Angular Frequency (×π rads/sample)
Pha
se (d
egre
es)
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
% We now consider the minimum phase filter, which is derived from the maximum% phase filter above.% Recall that the minimum phase filter has all zeros inside the unit circle,% that is, their values are less or equal to 1.zeros_minphase = 1 ./ zeros_maxphase% take the reciprocalpoles_minphase = poles_maxphase% no change to the poles % After specifying the poles and zeros position, we now plot their position.figure(3)% create a new figure (No.3)zplane(zeros_minphase, poles_minphase)%% We also find the numerator and denominator coefficients of the transfer function
-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 P
art
3
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
% in order to plot the magnitude and phase responses.numerator_minphase = poly(zeros_minphase)denominator_minphase= 1;% since all poles are at originfigure(4) % create a new figure (No.4)freqz(numerator_minphase, denominator_minphase)%
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-15
-10
-5
0
5
Normalized Angular Frequency (×π rads/sample)
Mag
nitu
de (d
B)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-60
-40
-20
0
20
Normalized Angular Frequency (×π rads/sample)
Pha
se (d
egre
es)
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
3.11 All-Pass Filters [11]
An all-pass filter is one whose magnitude response is constant for all frequencies, but whose phase response is not identically zero.
[The simplest example of an all-pass filter is a pure
delay system with system function H(z) = z-k]
A more interesting all-pass filter is one that is described by
where a0 = 1 and all coefficients are real.
LL
LLLL
zazazazazaa
zH−−
−+−−−
+++
++++=
L
L1
1
01
11
1
1)(
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
If we define the polynomial A(z) as
1|)()(|)(|)()()(
1)(
121
00
=⋅=⇒=∴
==
=−
−−
=
−∑
θθ jezL
L
k
kk
zHzHHzA
zAzzH
azazA
i.e. all pass filter,
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Furthermore, if z0 is a pole of H(z), then is a
zero of H(z) {ie. the poles and zeros are reciprocals of one another}. The figure shown below illustrates typical pole-zero patterns for a single-pole, single-zero filter and a two-pole, two-zero filter.
0
1z
a1
|z|=1
0
All-pass filter
a
|z|=1
r
0
All pass filter
θ0
(1/r, θ0)
(1/r, -θ0)(r, -θ0)
1
1
1
11)( −
−
−
−=
az
zazH |a| < 1 for stabilityProf
esso
r E. A
mbikair
ajah
UNSW, A
ustra
lia
We can easily show that the magnitude response is constant.
Phase response:
22
2
1*2
cos21
1cos21
1
11
1
11
|)()()()(|)(
aaθaa
θa
ae
ea
ae
ea
zHzHθHθHθH
θj
θj
θj
θj
ez θj
=+−
+−=
−
−⋅
−
−=
⋅=⋅=
−
−
=−
−
⎥⎦
⎤⎢⎣
⎡+−−−
=∴
+−−−+−
=−−
⋅−
−=
−
−−
−−
−
−
θθθφ
θθθθ θ
θ
θ
θ
cos)(2sin)(tan)(
cos21sin)(cos)(2
11
1
11)(
1
11
2
11
aaaa
aaaajaa
aeae
ae
eaH j
j
j
j
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
φ(θ)
π
-π
a = 0.5π
θa = -0.5 a= -0.8
When 0 < a < 1, the zero lies on the positive real axis. The phase over 0 ≤ θ ≤ π is positive, at θ = 0 it is equal to π and decreases until ω = π, where it is zero.
When -1< a < 0, the zero lies on the negative real axis. The phase over 0 ≤ θ ≤ π is negative, starting at 0 for θ = 0 and decreases to -π at ω = π.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
3.12 A second Order Resonant Filter
002
001
sincos
sincos0
0
θθ
θθθ
θ
jrrrep
jrrrepj
j
−==
+==−
(A) 1
1)(21
2
2
22
11 bzbz
zzbzb
zH++
=++
= −−
z-1
z-1
+y[n]x[n]
-b1
-b2
θ0r p1
p2
All pole system has poles only (without counting the zeros at the origin)
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(B) cos2)(
)(
))(())(()(
20
2
2
22
2
2
21
2
21
12
2
00
00
rzrzz
rzeerzzzH
rezrezz
pzpzz
bzbzzzH
jj
jj
+−=
++−=
−−=
−−=
++=
−
−−
θθθ
θθ
2201 , cos2 rbrb =−= θ
2
10 2 b
bCos −=θ
sff0
02πθ =
θ0 = resonant frequency
Comparing (A) and (B), we obtain
∴
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
3.13 Stability of a second-order filter
Consider a two-pole resonant filter given by
b1 & b2 are coefficients
This system has two zeros at the origin and poles at
22
1121
2
2
11)( −− ++
=++
=zbzbbzbz
zzH
24
2, 211
21
bbbpp
−±−=Prof
esso
r E. A
mbikair
ajah
UNSW, A
ustra
lia
The filter is stable if the poles lies inside the unit circle i.e. |p1| < 1 & |p2| < 1
For stability b2 < 1. If b2 = 1 then the system is an oscillator (Marginally stable)
Assume that the poles are complex
i.e. b12 – 4b2 < 0 ⇒ b1
2 < 4b2 and
If b12 – 4b2 ≥ 0 then we get real roots.
0 , 2 221 >±< bbb
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
The stability conditions define a region in the
coefficient plane (b1, b2) which is in the form of a triangle (see below)
The system is only stable if and only if the point
(b1, b2) lie inside the stability triangle.
b1
b2
-2 -1 0 1 2
1
-1
Real Poles
Complex Conjugate Poles
112 −= bb112 −−= bb
4
21
2bbparabola =
12 =b
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Stability Triangle
If the two poles are real then they must have a value between -1 and 1 for the system to be stable.
01 and 01)2(4 and 4)2(
24 and 42
242
12
41
2121
212
212
21
21
12212
211
12211
2211
>++<−−+<−−>+−
+<−−−<+−∴
+<−±<+−
<−±−
<−
bbbbbbbbbb
bbbbbb
bbbb
bbb
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
The region below the parabola (b12 > 4b2)
corresponds to real and distinct poles.
The points on the parabola (b12 = 4b2) result in real
and equal (double) poles.
The points above the parabola correspond to complex-conjugate poles.
b1
b2
-2 -1 0 1 2
1
-1
Real Poles
Complex Conjugate Poles
112 −= bb112 −−= bb
4
21
2bbparabola =
12 =b
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
3.14 Digital Oscillators
A digital oscillator can be made using a second order discrete-time system, by using appropriate coefficients. A difference equation for an oscillating system is given by
From the table of z-transforms we know that the
z-transform of p[n] above is
[ ] ( )θnAnp cos=
21
1
cos21cos1)( −−
−
+−−
=zzθ
zθzP
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Let21
1
cos21cos1
)()()(
−−
−
+−
−==
zzθzθ
zXzYzP
Taking inverse z-transform on both sides, we obtain
[ ] [ ] [ ] [ ] [ ]1cos21cos2 −−=−+−− nxnxnynyny θθ
No Input term for an oscillator x[n] = 0, x[n-1] = 0
So the equation of the digital oscillator becomes
[ ] [ ] [ ]21cos2 −−−= nynyny θProfes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
% Exercise 2. Digital oscillator% We now implement a digital oscillator which uses some initial conditions.theta = pi/8 % angular frequency of the oscillatoramplitude = 2 % amplitude of the oscillatornumber_of_samples = 100% The oscillation depends on the initial conditions.% That is, we can make a sine or cosine wave by specifying the right initial conditions.% These are initial conditions for creating a sine wave.% Remove the `%' symbol to use themy1 = -amplitude * sin(theta), y2 = -amplitude * sin(2*theta)% These are initial conditions for creating a cosine wave.% Remove the `%' symbol to use themy1 = amplitude * cos(theta), y2 = amplitude * cos(2*theta)y = zeros(1,number_of_samples); % initializing oscillation samplesweight = 2 * cos(theta)% Notice the oscillation samples are evaluated without calling the commands `cos' or `sin'.% It only requires multiplication and addition(subtraction).for n = 1:number_of_samples
y(n) = weight * y1 - y2;y2 = y1;
% updating previous samplesy1 = y(n);
end Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
figure(4)stem(y)hold onplot(amplitude*cos(theta*(0:number_of_samples-1)),'r-.')hold off%% Notice that if the first set of initial conditions are selected, the oscillation starts% at zero level. If the second set is selected, it starts at the amplitude level.% Other initial conditions can be selected which will make the oscillation to start at % certain phase.%% Experiment with different values of `theta', `amplitude'
0 10 20 30 40 50 60 70 80 90 100-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
So the equation of the digital oscillator becomes
[ ] [ ] [ ]21cos2 −−−= nynyny θ
and its structure is shown below.
b2 = -1
b1 = 2cosθ
y[n-2]y[n-1]+ z-1 z-1
y[n] = A cos(nθ)
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
To obtain y[n] =Acos(nθ), use the following initial conditions:
y[0] = A cos(0.θ) = Ay[-1] = A cos(-1.θ) = A cosθ
The frequency can be tuned by changing the coefficient
b1 (b2 is a constant). The resonant frequencyθ of the oscillator is,
22cos 1
2
1 bbb
−=−
=θ (For an oscillator b2 = 1)
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Example:A digital sinusoidal oscillator is shown below.
(a) Assuming θ0 is the resonant frequency of the digital oscillator, find the values of b1 and b2 for sustaining the oscillation.
x[n]
-b1
+z-1
z-1
y[n] = A sin(n+1)θ
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
20
2
22
22
11
cos2
)(
))((1)(
00
00
rzrzK
rzeerzK
rezrezK
zbzbKzH
jj
jj
+−=
++−=
−−=
++=
−
−−−
θ
θθ
θθ
∴b1 = -2 r cosθ0 ; b2 = r2
For oscillation b2 = 1 ⇒ r = 1 ∴ b1 = -2 cosθ0Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(b). Write the difference equation for the above figure. Assuming
x[n] = (Asinθ0)δ[n], and y(-1) = y(-2) = 0.
Show, by analysing the difference equation, that the application of an impulse at n = 0 serves the purpose of beginning the sinusoidal oscillation, and prove that the oscillation is self-sustaining thereafter.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
[ ] [ ] [ ] [ ][ ] [ ] [ ] [ ]nAnynyny
nxnynybnyδθθ 00
1
sin21cos221
+−−−=+−−−−=
y[0] = A sinθ0
y[1] = 2cosθ0 , Asinθ0 = A sin2θ0
0 0 1
0 0
n = 0y[0] = 2cosθ0 y[-1] – y[-2] + A sinθ0 δ[0]
y[1] = 2cosθ0 y[0] –y[-1] + A sinθ0 δ[1]
n = 1
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
y[2] = 2 cosθ0 y[1] – y[0] + A sinθ0δ[2]= 2 cosθ0 Asin2θ0 – A sinθ0= 2A cosθ0 [2 sinθ0cosθ0] - sinθ0= A sinθ0 [4 cos2θ0 –1 ] = A[3sinθ0 – 4 sin3θ0]
n = 2
where sin3θ0 = 3sinθ0 – 4 sin3θ0
y[2] = A sin3θ0 and so forth.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
By setting the input to zero and under certain initial conditions, sinusoidal oscillation can be obtained using the structure shown above. Find these initial conditions.
(x[n] = 0 for an oscillator)
n = 0 y[0] = 2cosθ0 y[-1] – y[-2]for oscillation, y[-1] = 0 ⇒ no cosine terms
y[0] = -y[-2] ⇒ y[-2] = -Asinθ0 (sine term is equired)
y[0] = 0-(-A sinθ0) = A sinθ0Initial conditions:
[ ] [ ] [ ] [ ]nxnynyny +−−−= 21cos2 θ
y[-1] =0; y[-2] = -Asinθ0Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Sine and cosine oscillators [1]Sinusoidal oscillators can be used to deliver the carrier in modulators. In modulation schemes, both sines and cosines oscillators and needed. A structure that delivers sines and cosines simultaneously is shown below:
y[n]= cos(nθ)
x[n]= sin(nθ)
-sinθ
sinθ
cosθ
cosθ
z-1
+
z-1
+
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Proof:Trigonmetric equation for cos(n+1)θ is:cos(n+1)θ = cos(nθ)cosθ - sin(nθ) sinθLet y[n] = cos(nθ) and x[n] = sin(nθ)
∴ y[n+1] = cosθ y[n] – sinθ x[n]Replace n by n-1
y[n] = cosθ y[n-1] – sinθ x[n-1] (A)
Similarlysin(n+1)θ = sinθ cos(nθ) + sin(nθ) cosθ
∴ x[n+1] = sinθ y[n] + x[n] cosθReplace n → n-1
x[n] = sinθ y[n-1] + x[n-1] cosθ (B)
Using equations A & B above, the structure shown above can be obtained.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Exercise:An oscillator is given by the following coupled difference equations expressed in matrix form.
Draw the structure for the realisation of this oscillator, where θ0 is the oscillation frequency. If the initial conditions yc[-1] = Acosθ0 and ys[-1] = -Asinθ0, obtain the outputs yc[n] & ys[n] using the above difference equations.
[ ][ ]
[ ][ ]⎥⎦
⎤⎢⎣
⎡−−
⎥⎦
⎤⎢⎣
⎡ −=⎥
⎦
⎤⎢⎣
⎡11
cossinsincos
00
00
nyny
nyny
s
c
s
c
θθθθ
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
[ ] [ ] [ ][ ] [ ] [ ]1cos1sin
1sin1cos
00
00
−+−=−−−=
nynynynynyny
scs
scc
θθθθ
yc[n]
ys[n]
-sinθ0
sinθ0
cosθ0
cosθ0
z-1
z-1
+
+
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
n=0 ys[0] = sinθ0 (A cosθ0) + cosθ0 (-Asinθ0) = 0
n=0 yc[0] = cosθ0 (Acosθ0) - sinθ0(-Asinθ0) = A
n=1 yc[1] = cosθ0.A - sinθ0 .0 = Acosθ0
n=1 ys[1] = A sinθ0 + 0 = A sinθ0
n=2 yc[2] = cosθ0 yc[1] - sinθ0 ys[1]= cosθ0 A cosθ0 - sinθ0 A sinθ0 = A cos2θ0
n = n yc[n] = A cos (nθ0)
similarly ys[n] = A sin(nθ0)Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Exercise:For the structure shown below, write down the appropriate difference equations and hence state the function of this structure.
θsin21
z-1
y1[n]
2cosθ
z-1
-1
-1y2[n]
+
+
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
3.15 Notch filters [4]When a zero is placed at a given point on the z-plane, the frequency response will be zero at the corresponding point. A pole on the other hand produces a peak at the corresponding frequency point.
Poles that are close to the unit circle give rise large peaks, where as zeros close to or on the unit circle produces troughs or minima. Thus, by strategically placing poles and zeros on the z-plane, we can obtain sample low pass or other frequency selective filters (notch filters).Prof
esso
r E. A
mbikair
ajah
UNSW, A
ustra
lia
Example:
Obtain, by the pole-zero placement method, the transfer function of a sample digital notch filter (see figure below) that meets the following specifications: [4]
Notch Frequency: 50Hz3db width of the Notch: ±5HzSampling frequency: 500 Hz
π⎟⎟⎠
⎞⎜⎜⎝
⎛ Δ−=
sffr 1
|H(f)|
50 250 f (Hz)0
The radius , r of the poles is determined by :
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
To reject the component at 50Hz , place a pair of complex zeros at points on the unit circle corresponds to 50Hz. i.e. at angles of
To achieve a sharp notch filter and improved amplitude response on either side of the notch frequency , a pair of complex conjugate zeros are placed at a radius r < 1.
937.05001011 =⎟
⎠⎞
⎜⎝⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛ Δ−= ππ
ff
rs
360
360
0.937
|z| =1
π2.03650050360 00 ±=±=×
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
( )( )( )( )
21
21
2
2
2.02.02
2.02.02
2.02.0
2.02.0
878.05161.116180.11
)2.0cos(937.02878.0)2.0cos(21
)(937.0878.0)(1
937.0937.0)(
−−
−−
−
−
−
−
+−+−
=
×−+−+
=
+−++−+
=
−−−−
=
zzzz
zz
zeezeez
ezezezezzH
jj
jj
jj
jj
ππ
ππ
ππ
ππ
ππ
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Problem Sheet A3 [5,1]
Q1. The frequency response of an ideal differentiator is given by
This response is periodic with period 2π. The
quantity τ is the delay of the system in samples.
πθπθθ θτ ≤≤−= − jejH )(
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
PhaseH
H
magnitudeH
←−−=<
+−=>
←
⎩⎨⎧
≤≤−−≤≤
=
2)(arg0
2)(arg0
00
|)(|
πθτθθ
πθτθθ
θπθπθθ
θ
(a). Sketch the magnitude and phase responses of the
system for -π ≤ θ ≤ π.
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(b) Find the impulse response h[n] of the system
as a sum of a sine and cosine function.
[ ]ττπ
τπτπ
−−
+−
−−=
nn
nnnh )(cos
)()(sin)( 2
)1(2 θθθθ
θ jkk
edejk
jk −=∫
Ans:
Note:
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Q2. For the system in Figure 18, sketch the
output y[n] when the input x[n] is δ[n] and
H(θ) is an ideal lowpass filter as follows:
⎪⎩
⎪⎨
⎧
≤<
≤≤=
πθπ
πθθ
||2
02
||01)(H
H(θ)x[n] ω[n] (-1)n
y[n]
Figure 18
+×
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
=n
n
n 2sin1)(
π
πω
[ ]2
2sin
1)1(21)(
π
π
n
n
ny n +−=
y[n]
1
0 1 2 3 4 n
Ans:
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Q3 (a) Show that both digital filters given below have the same magnitude response :
y[n] = output; x[n] = input; ci = coefficients
[ ] [ ]∑−=
−=m
mii inxcnyi )(
[ ] [ ]∑−=
−−=m
mii imnxcnyii )(
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(a) Compute the 3dB bandwidth of the following filters
Which filter has a smaller 3dB bandwidth?
10,11
21)(;
11)( 1
1
211 <<−+−
=−−
= −
−
− aazzazH
azazH
Ans:
.2
12cos
214cos :
12
2
1
21
21
filter
aa
aaaAns
ndcc
c
c
⇒<
⎥⎦⎤
⎢⎣⎡+
=
⎥⎦
⎤⎢⎣
⎡ −−=
∴−
−
θθ
θ
θ
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(c) Find the magnitude response for the system
function H(z) and comment on your result
Ans: |H(θ)| = 1 allpass filter
Draw the canonic realization of the system H(z).
2
2
331)(zzzH
++
=
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Q4.(a) Determine the magnitude and phase response of the multi-path channel y[n] = x[n] + x[n-M]. At what frequencies H(θ) = 0?
(b) Determine and sketch the magnitude and phase response of the system shown below.
z-1 z-1 z-1x[n] y[n]
81
+ + +
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(c) Assuming that the digital filter G(z) is to be realized using the cascade structure, draw a suitable block diagram and develop the difference equation(s).
)3
1)(3
1(6
)1()(3
jzjzz
zzG+−
+=
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(d) Determine the frequency response H(θ) of the ladder filter shown below:
z-1z-1-1
b
a
Y(z)X(z)
θjθj
θj
abebeabeθHAns
2
2
1)(:
−−
−
−+=
+
+
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(e) Determine the frequency response H(θ) of the lattice filter
T
a
-a
T
b
-b
x[n] y[n]+ +
+ +
θjθj aeeabθHAns
2)1(11)(:
−− +++=
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(f) Show that the filter structure shown below has a linear phase characteristic equation
given by: φ(θ) = -2θ.
z-1 z-1
1
z-1
1
z-1
1 1
y[n]
x[n]
1
+
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(g) Determine the transfer function of the system shown below. Check the stability of
the system when r = 0.9 and .40πθ =
+ + z-1 +z-1
r cosθ0
- r sinθ0
y[n]
r cosθ0
x[n]
2210
10
cos21sin)( : −−
−
+−=
zrzrzrzHAns
θθ
filterstableb
b_
1312.164.0
1
2
⎭⎬⎫
−==
,
r sinθ0
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Q5. (a) Consider the following causal IIR transfer function:
Is H(z) a stable function? If it is not stable, find
a stable transfer function G(z) such that
|G(θ)| = |H(θ)|. Is there any transfer function
having the same magnitude response as H(z)?
)5.0)(3(942)( 2
23
++−+−
=zzz
zzzH
{ }filterAllpassz
zzAAHzG
zAzHzzz
zzzGUnstableAns
331)(|,)(||)(||)(|
)()()5.0)(31(
942)( , : 2
23
−−
==
=++−+−
=
θθProfes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(c) Analyse the digital structure given below and determine its transfer function.
(i) Is this a canonic structure?(ii) What should be the value of the multiplier coefficient K so that H(z) has a unity gain at θ = 0?(iii) What should be the value of the multiplier coefficient K so that H(z) has a unity gain at θ = π?(iv) Is there a difference between these two values of K? If not, why not?
)()()(
zXzYzH =
Kz-1 +
z-1
z-1
-1
β
z-1
α
Y(z)X(z)
-1
+ +
+Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
z-1 +
z-1
z-1
-1
β
z-1
α
Y(z)X(z)
-1
+ +
+
21
21
1)( −−
−−
+−+−
=zzzzKzH
βααβ
11
1=
+−+−
=βα
αβK
Ans:
(i) 4 delays ⇒ noncanonic(ii)&(iii)
(iv) no (all pass filter)Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Q6: (a) A first-order digital filter is described
by
where, 0 < a, b < 1 assume
(i) Determine k, so that the maximum value of |H(θ)| is equal to 1.
(ii) Compute the 3-dB bandwidth of the filter H(z).
1
1
11)( −
−
−+
=azbzkzH
21
== ba
⎟⎠⎞
⎜⎝⎛=
=+−
=
−
4435cos
31
11 :
1c
bakAns
θProfes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(iii) Draw a canonic realization of the system function H(z).
(b) A two-pole low-pass filter has the system function
Determine the values of k0 and b1 such that the frequency response H(θ) satisfy the conditions
Ans: k=0.46, b1 = 0.32
211
0
)1()(
−−=
zbk
zH
21)(;1)0( 2
4
===πθ
θHH
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(c) A third order FIR filter has a transfer function G(z) given by
from G(z), determine the transfer function of an FIR filter whose magnitude response is
identical to that of G(z) and has a maximum phase response.
)251)(
34)(
231(30)( 111 −−− ++−= zzzzG
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Q7. For the system shown in the next slide,
(a) Express y1[n] in terms of y1[n-1],y2[n-1], and x[n]; do the same for y2[n].(b) Assume A = cos(θ0); B = sin(θ0);y1(-1) = cos(-θ0); y2(-1) = sin(-θ0).
If x[n] = 0, show that :
y1[n] = cos(nθ0) and y2[n] = sin(nθ0)(c) Calculate (for arbitrary A and B) the
system function and)()(
)( 11 zX
zYzH = )(
)()( 2
2 zXzY
zH =
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(d) Take and draw the poles-zeros plot
for H1(z) and H2(z).
(e) Take and x[n]=δ[n].
Calculate the impulse response h1[n] for -2 ≤ n ≤ 10.
221
== BA
221
== BA
y2[n]
z-1+
z-1 +
x[n]
y1[n]
A
A
B
-B
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Q8. (a) Obtain a parallel realization for the
following H(z)
Implement the parallel realization of H(z)which you have obtained.
(b) Obtain a parallel realization for the following transfer function.
2
2
)5.0(34)(
++−
=zz
zzzH
)41()
21(
1)(2 −−
+=
zz
zzH
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(c) A digital filter is represented by
Does this transfer function represent an FIR or an IIR filter?
Write a difference equation for H(z)using the direct form.
Implement a parallel realization of H(z).
)1)(211(
1
211
21
)(111 −−− −−
+−
=zzz
zH
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Q9. (a) The transfer function of a discrete-time system has poles at z = 0.5, z = 0.1 ±j0.2and zeros at z = -1 and z = 1.
(i) Sketch the pole-zero diagram for the system
(ii) Derive the system transfer function H(z) from the pole-zero diagram.
(iii) Develop the difference equation.
(iv) Draw the block diagram of the discrete system.
321
31
15.07.07.01)( : −−−
−−
−−+−
=zzz
zzzHAns
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
(b) A notch filter is given by
Determine the frequency response at dc,
and . fs – sampling frequency.
Sketch the frequency response in the interval
2
2
8.011)( −
−
++
=z
zzH
4sf
2sf
20 sf
f ≤≤ |H(θ)|
π θ
81.12
81.12
2π
Ans:
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
Q10. A digital filter is shown below.
(i) Determine the system function H(z) for the above structure.
(ii) With a0 = a2 = 1; a1 = 2; b1 = 1.5 and b2 = -0.75,determine the pole-zero pattern of H(z) and indicate if the system is stable or not.
z-1+ z-1+ +y[n]
a0a1a2
b1b2
x[n]
212
212
0)(bzbzazaza
zH−−++
=
Profes
sor E
. Ambik
airaja
h
UNSW, A
ustra
lia
![[Unix Programming] Signal and Signal Processing](https://img.dokumen.tips/doc/110x75/56813a6c550346895da26644/unix-programming-signal-and-signal-processing.jpg)
