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Digital signal processing Digital signal processing (DSP) is concerned with the representation of
signals by a sequence of numbers or symbols and the processing of these
signals. Digital signal processing and analog signal processing are subfields
of signal processing. DSP includes subfields like: audio and speech signal
processing, sonar and radar signal processing, sensor array processing,
spectral estimation, statistical signal processing, digital image processing,
signal processing for communications, control of systems, biomedical signal
processing, seismic data processing, etc
The goal of DSP is usually to measure, filter and/or compress continuous
real-world analog signals. The first step is usually to convert the signal from
an analog to a digital form, by sampling it using an analog-to-digital
converter (ADC), which turns the analog signal into a stream of numbers.
However, often, the required output signal is another analog output signal,
which requires a digital-to-analog converter (DAC).
A signal : is a function of asset of independent variable, with being perhaps the most prevalent signal variation. The signal itself carries some
kind of information available for observation.
Processing: Means operation in some faction on a signal to extract some useful information.
Digital: means that the processing is done with a digital computer or special
purpose digital hardware.
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We deal mostly with discrete time signals and systems, which are analyzed
in both the time and the frequency domains. The analysis and design of
processing structures called filters and spectrum analyzers is one of the most
important aspects of DSP.
signal processing: is an area of electrical engineering and applied mathematics that deals with
operations on or analysis of signals, in either discrete or continuous time, to
perform useful operations on those signals. Signals of interest can include
sound, images, time-varying measurement values and sensor data, for
example biological data such as electrocardiograms, control system signals,
telecommunication transmission signals such as radio signals, and many
others. Signals are analog or digital electrical representations of time-
varying or spatial-varying physical quantities. In the context of signal
processing, arbitrary binary data streams and on-off signalling are not
considered as signals, but only analog and digital signals that are
representations of analog physical quantities.
However, one needs to convert analog signals into a form suitable for digital
hardware. This form of the signal is called a digital signal. It takes one of the
finite number of values at specific instances in time, and hence it can be
represented by binary numbers, or bits. The processing of digital signals is
called DSP; in block diagram form it is represented by
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where the various block elements are discussed below.
PrF: This is a prefilter or an antaliasing filter, which conditions the analog
signal to prevent aliasing.
ADC: This is called an analog-to-digital converter, which produces a stream
of binary numbers from analog signals.
Digital signal processor: This is the heart of DSP and can represent a
neural-purpose computer or a special-purpose processor, or digital hardware,
and so on.
DAC: This is the inverse operation to the ADC, called a digital-to-analog
converter, which produces a staircase waveform from a sequence of binary
numbers, a first step towards producing an analog signal.
PoF: This is a post filter to smooth out staircase waveform into the desired
analog signal.
Advantages of DSP 1. allows more complex processing than is possible with analog circuitry.
2. provides better signal quality and repeatable performance because of the
digital processing.
3. is more flexible since the digital processing can often be easi1y modified
(e.g. software can be up-graded)
4. usually results in a lower cost for equivalent performance.
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The principal disadvantage of DSP 1. limited to signal with relatively low bandwidth.
2. the point at which DSP becomes too expensive will depend on the
application and the current state of conversion and digital processing
technology.
3. the need for an ADC and DAC makes DSP uneconomical for simple
applications (e-g a simple filters)
4. higher power consumption and size of a DSP implementation can make it
unsuitable for simple very low-power or small size applications
Applications 1. digital sound recording such as CD and DAT.
2. speech and compression for telecommunication and storage .
3. implementation of wire line and radio modems (including digital filtering
modulation, echo cancellation and other functions).
4. image enhancement and compression.
The Sampling Theorem also known as Shannon’s Sampling Theorem, states that a signal can be
exactly reconstructed from its samples if the sampling frequency is greater
than twice the highest frequency of the signal; but requires an infinite
number of samples . In practice, the sampling frequency is often
significantly more than twice that required by the signal's limited bandwidth.
A typical digital signal processing system contains a low-pass analog filter,
often called pre-sampling filter, to ensure that the highest frequency allowed
into the system, will be equal or less the sampling rate so that the signal can
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be recovered. The highest frequency allowed by the pre-sampling filter is
referred to as the Nyquist frequency.
If a signal is not band-limited, or if the sampling rate is too low, the spectral
components of the signal will overlap each another and this condition is
called aliasing. To avoid aliasing, we must increase the sampling rate.
we will adopt the following notations:
N = number of samples in time or frequency period.
fs = sampling frequency = samples per second
Tt = period of a periodic discrete time function
tt = interval between the N samples in time period Tt
Tf = period of a periodic discrete frequency function
tf = interval between the N samples in frequency period Tf
we have the relations:
Example :
The period of a periodic discrete time function is 0.125 millisecond, and it is
sampled at 1024 equally spaced points. It is assumed that with this number
of samples, the sampling theorem is satisfied and thus there will be no
aliasing.
a. Compute the period of the frequency spectrum in KHz
b. Compute the interval between frequency components in KHz.
c. Compute the sampling frequency fs
d. Compute the Nyquist frequency
Solution:
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Intervals between samples and periods in discrete time and frequency
domains Tt = 0.125 ms and N = 1024 point . Therefore, the time between
successive time components is
a. the period Tf of the frequency spectrum is
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b. the interval ft between frequency components is
C. the sampling frequency fs is
d. the Nyquist frequency must be equal or less than half the sampling
frequency, that is,
DISCRETE-TIME SIGNALS Signals are broadly classified into analog and discrete signals. An analog
signal will be denoted by x(t), in which the variable t can represent any
physical quantity, but we will assume that it represents time in seconds. A
discrete signal will be denoted by x (n),in which the variable n is integer
valued and represents discrete instances in time. Therefore it is also called a
discrete time signal, which is a number sequence and will be denoted by one
of the following notations:
where the up-arrow indicates the sample at n = 0.
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Example: X(n)=[1 0 2 -1.5 3+j4 -2.5-j1.025 ]
Example:
X(n)=
otherwiseonn
................22..........5.0
n x(n)
-2 0.25
-1 0.5
0 1 x(n)=[0.25 0.5 1 2 4]
1 2
2 4
3 0
4 0
Example:(period function):
X(n)=[…….. 1 -1 2 1 -1 2 1 -1 2 1 -1 2 ……]
P=3 :period :min. integrate that satisfies x(n)=x(n+p)
x(n)=[……. 2 2 1 2 2 1 2 2 1 -1 2 2 1 2 2 1 2 2 1 -1 2 2 1……]
p=10
x(n)=[……1 2 1 1 2 1 3 1 2 1 1 2 1 1 2 1 …….] not period
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Continuous/Discrete-Time Signals In this section, we introduce several elementary signals which not only occur
frequently in nature, but also serve as building blocks for constructing many
other signals.
Some continuous–time and discrete–time signals
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TYPES OF SEQUENCES 1a- The unit step function us(t) 1b- Unit step sequence
1c-A delayed and scaled step function 1d-A delayed and scaled step sequence
• The unit step function offers a convenient method of describing the sudden
application of a voltage or current source.
2a- Unit impulse function 2b-Unit sample or impulse sequence
2c-A delayed and scaled impulse function 2d-A delayed and scaled impulse sequence
• The unit impulse or delta function, denoted as δ(t) , is the derivative of the
unit step u0(t) . It is also defined as
And
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• The shifting property of the delta function states that
• The sampling property of the doublet function δ'(t) states that
*Relationship between δ(t) and us (t) *Relationship between δ[n] and us [n]
3- The Unit Ramp Function u1(t)
we define u1(t) as
Since u1(t) is the integral of u0(t) , then u0(t) must be the derivative of u1(t)
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Higher order functions of t can be generated by repeated integration of the
unit step function. For example, integrating u0(t) twice and multiplying by
2, we define u2(t) as:
The Unit Step Function u0(t)
is defined as: Waveform for uo(t):
(At t = to) Waveform for uo(t-to)
(At t = -to) Waveform for uo(t+to)
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Other forms of the unit step function are shown in Figure:
A rectangular pulse expressed as the sum of two unit step functions
the pulse of Figure (a) is the sum of the unit step functions of Figures (b) and
(c) is represented as u0(t) – u0(t–1)
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example: x(n)=u(n+2)-u(n-3)
n x(n)
-3 0 u(n+2)
-2 1 - u(n-3)
-1 1 = u(n+2)-u(n-3)
0 1
1 1
2 1
3 0
Ex: u(n+5)-u(n+2)+2u(n)-3u(n-3)+u(n-5)
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Example
Express the square waveform of Figure as a sum of unit step
functions. The vertical dotted lines indicate the discontinuities at T,
2T, 3T and so on.
Solution:
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Systems A system is a plant or a process that produces a response called an output in response to an excitation called an input. If a system’s input and output signals are scalars, the system is called a single-input single-output (SISO) system. If a system’s input and output signals are vectors, the system is called a multiple-input multiple-output (MIMO) system. A single-input multiple-output (SIMO) system and a multiple-input single-output (MISO) system can also be defined in a similar way. For example, is an electric circuit whose inputs are voltage/current sources and whose outputs are voltages/currents/charges in the circuit.
A description of continuous–time and discrete–time systems
DISCRETE SYSTEMS Mathematically, a discrete time system (or discrete system for short) is
described as an operator T[.] that takes a sequence x(n) (called excitation)
and transforms it into another sequence y(n) (called response). That is,
In DSP we will say that the system processes an input signal into an output
signal. Discrete systems are broadly classified into linear and nonlinear
systems. We will deal mostly with linear systems.
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LINEAR SYSTEMS A discrete system T[. ] is a linear operator L[. ] if and only if L[.] satisfies
the principle of superposition, namely,
the output y(n) of a linear system to an arbitrary input x(n) is given by
The response L [ (n - k)] can be interpreted as the response of a linear
system at time n due to a unit sample (a well-known sequence) at time k. It
is called an impulse response and is denoted by h(n ,k). The output then is
given by the superposition summation
Linear time-invariant (LTI) system:
A system is said to be time/shift-invariant if a delay/shift in the input causes
only the same amount of delay/shift in the output without causing any
hanged of the characteristics (shape) of the output. Time/shift-invariance can
be expressed as follows:
Then the time-varying function h (n,k ) becomes a time-invariant function h
(n - k), and the output is given by
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The impulse response of a LTI system is given by h(n). The mathematical
operation is called a linear convolution sum and is denoted by:
Stability: This is a very important concept in linear system theory. The primary reason
for considering stability is to avoid building harmful systems or to avoid
burnout or saturation in the system operation. A system is said to be
bounded-input bounded-output (BIBO) stable if every bounded input
produces a bounded output.
An LTI system is BIBO stable if and only if its impulse response is
absolutely summable.
Causality:
This important concept is necessary to make sure that systems can be built.
A system is said to be causal if the output at index 0n depends only on the
input up to and including the index 0n ; that is, the output does not depend on
the future values of the input. An LTI system is causal if and only if the
impulse response:
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Ex: Linearity of the 3-sample averager .The transformation produced by the
3-sample averager is given by
y(n) =1\3[x(n + 1) + x(n) + x(n - 1)]= L{x(n)}
sol:
L{ax1(n) + bx2(n)) }=
=1\3[ax1(n + 1) + bx2(n + 1) + ax1(n) + bx2(n)+ ax1(n - 1) + bx2(n - 1)]
=a/3[x1(n+ 1) + x1(n) + x1(n- 1)]+ b/3[x2(n + 1) + x2(n) + x2(n - 1)]
= a{y1(n)} + b{y2(n)} its linear system
Example: A nonlinear system. Let us consider a square-law device, in which
the output is the square of the input: {y(n)} =L{x(n))= {x^2(n)}
Sol:
L{ax1(n)+bx2(n)}=[ax1(n)+bx2(n)]^2=a^2x1^2(n)+b^2x2^2(n)+2abx1(n)x
2(n)
which is not equal to ax1^2(n)+bx2^2(n), the output required for the system
to be linear, hence , the square-law device is a nonlinear system.
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CONVOLUTION EXAMPLE: Let the rectangular pulse x(n) = u(n) - u(n - 10) be an input to
an LTI system with impulse response u(n) (0.9) h(n)n ,Determine the output
y(n).
Sol.:
There are three different conditions under which u(n - k) can be evaluated.
CASE 1 : n < 0 Then u(n - k) = 0, 90 K .
In this case the nonzero values of x(n) and h(n) do not overlap.
CASE 2 : 90 n : Then u(n - k) = 1, nK 0
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In this case the impulse response h(n) partially overlaps the input x(n).
CASE 3 : 9n : Then u(n - k) = 1, 90 K
In this last case h(n) completely overlaps x ( n ) .
EXAMPLE : Given the following two sequences
determine the convolution
Sol.:
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the beginning point (first nonzero sample) of y(n) is given by
n = -3 + (-1) = -4, while the end point (the last nonzero sample) is given by
n = 3 + 4 = 7. The complete output is given by
Note that the resulting sequence y (n) has a longer length than both the x (n)
and h (n) sequences.
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A digital filter: A digital filter is a numerical procedure, or algorithm that transforms a given
sequence of numbers into a second sequence that has some more desirable
properties, such as less noise or distortion.
n: an index typically, the set of index values consists of consecutive integers,
which in same cases may take values from minus infinity to plus infinity.
A digital filter consists of the interconnection of three simple elements:
1- adders
2-multiplier
3-positive and negative delay
A positive delay : is implemented by a memory register that stores the
current value of a sequence for one sample interval, thus making it available
for future calculations.
A negative delay: (advance) is used to look ahead to the next value in the
sequence.
1Z 1Z 1Z
Z
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A digital filter design involves selecting and interconnecting a finite number
of these elements and determining the multiplier coefficient values.
Example : Digital filter used as a three-sample averager. consider the
relationship between the values of the output sequence at time n, denoted by
y(n), and the values of the input sequence, x(n-1), x(n) and x(n + 1), given
by y(n) =1/3x(n + 1) + 1/3x(n) + 1/3x(n - I) As shown in Fig.
Non recursive digital filter that acts as a three-sample averager (a second-
order).
Order: denoted the minimum number of delay.
Non recursive: when the filter output is a function of only the input
sequence.
Recursive filter: when the out put is also a function of the previous output
values.
Example : First-order recursive filter. Consider the relationship between the input and output sequence values given by Y(n): aY(n - 1) + x(n) where a is a constant. As shown in Fig. the current value of the output is equal to the sum of the input and a times the value of the previous output. Since only one delay is necessary, this is a first-order filter.
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Not : both positive and negative delay can de applied to the input sequence,
only positive delays can be applied to the output sequence.
Example : unit-sample response of the 3-sample averager.
Let y(n)= 1/3x(n + 1) + 1/3x(n)+ 1/3x(n-1)
To find the unit-sample response {h(n,0)}, we set {x(n)} = {d(n)} and solve
for :
For n= -2, y(-2)=1/3d(-1)+1/3d (-2)+1/3d(-3)=0.
For n -2, y(n)=0, since d(n):0 for n 1 , y (n) = 0, since the nonzero value of {d(n)} has moved out of the
memory of this filter. Since {d(n)} is the input, the output is defined to be
{h(n,0)}. Hence
h(n)=
otherwiseonfor
..................11............3/1
The unit-sample response is computed by stating at a point in time, n, for
which all the previous inputs and the contends of the delays and advances
are all equal to zero. This state of the filter is known as the zero initial
condition.
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Example: unit-sample response of a first-order recursive filter. Let
To find {h(n,0)}, we let {x(n)}= {d(n)} and apply the zero initial condition.
For n < 0, y(n)=0, because d(n) is zero, for n
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impulse-response of the system
one:-finite impulse response FIR
Y(n) = 3x(n) + 2x(n-1) - 4x(n-3) x(n) unit-sample
n x(n) y(n) if n ≤ -1 then y(n)=0
0 1 3 h(n)=[3 2 0 -4]
1 0 2 x(n) y(n)
2 0 0 x(n-1) 3
3 0 -4
4 0 0 x(n-2) 2
5 0 0
its characteristic : x(n-3)
1-finte response -4
2-no feed back
3-stable all the time
4-all zero system
Two:- infinite impulse response IIR
Y(n)=x(n)-2x(n-1)+y(n-1) ,y(-1)=0, y(0)=1, y(1)=0
n x(n) y(n) x(n) y(n) 0 1 1 x(n-1) 1 0 -1 2 0 -1 3 0 -1 4 0 -1 -2 y(n-1) 5 0 -1 h(n)=[1 -1 -1 -1 -1 …..]
Element of h(n) is infinite because the feed back y(n-1)
×
× 1Z
1Z
1Z
∑
×
1Z
1Z ×
∑
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its characteristic :
1-infinte response
2-have feed back
3-generaly unstable
4-have no zero system
Example: y(n) = x(n)-x(n-1)-2y(n-1)+y(n-2), y(-1)=1, y(-2)=-1
x(n) y(n) x(n)=[3 1 -1]
y(n-1)
x(n-1) -2 y(n-2)
-1
n x(n) y(n)
0 3 0 y(0)=x(0)-x(-1)-2y(-1)+y(-2)=3-0-2-1=0
1 1 -1 y(1)=x(1)-x(o)-2y(0)+y(-1)=1-3-0+1=1
2 -1 0 y(2)=x(2)-x(1)-2y(1)+y(0)=0
3 0 0
4 0 0
THE STEP RESPONSE:
just as the unit step function u[n]is the running sum of the unit impulse (n)
so the step response of a LTI processor is the running sum of its impulse
response. Therefore if we denote the step response by s[n] , we have:
h[n] is the first-order difference of s[n]:
h[n] = s[n] - s[n -1]
×
1Z
∑
1Z
1Z
×
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example: find and sketch the first few sample value of the impulse and step
responses of the system y[n] = 0.8 y[n -1] + x[n] also determine the final
value of s(n) as - ∞.
Solution:
y[n] =0.8 y[n -1] + x[n]
Its impulse response is therefore given by:
h[n] = 0.8 h[n-1] + (n)
h[0] =1,
h[1] = o.8
h[2] = 0.8^2=0.64
h[3]=0.8^3=0.512
the step response equals the running sum of h[n]
s[0] = h[0] = 1, s[1] =h[0] + h[1]=1.8 ,
s[2] = h[0] + h[1] + h[2]= s[1] + h[2] = 2.44
s[3] =s[2]+h[3]=2.952, s[4] = s[3] + h[4] = 3.3616 and so on.
s[∞] = 1 + 0.8 + 0.8^2 + 0.8^3 + ... = 1/1-0.8=5
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The Discrete Fourier Transform (DFT) we will denote a discrete time signal as x[n] , and its discrete frequency
transform as X[m] .
Where N represents the number of points that are equally spaced in the
interval 0 to 2π
For m = 0,1,2,…….,N-1. the N-point DFT of X[m]
The Inverse DFT is defined as
For n = 0,1,2,….,N-1
In general, the discrete frequency transform X[m] is complex, and thus we
can express it as:
X[m] = Re{X [m]}+ Im{X [m]}
For m = 0,1,2,…,N-1
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For n = 0, X[m] = x[0] ,
Example : A discrete time signal is defined by the sequence x[0] = 1, x[1] =
2, x[2] = 2, and x[3] = 1 and x[n] = 0 for all other . Compute the frequency
components X[m]
Solution:
Since we are given four discrete values of x[n], we will use a -point DFT,
that is, for this example, N=4
N = 0,1, 2, 3
For m = 0,1, 2, 3
m=0 Re{X [0]}=1+2⋅ (1) +2⋅ (1) +1⋅ (1)=6
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m=1 Re{X [1]}=1+2⋅ (0) +2⋅ (–1) +1⋅ (0)=-1
m=2 Re{X [2]}=1+2⋅ (–1) +2⋅ (1) +1⋅ (–1)=0
m=3 Re{X [3]}=1+2⋅ (0) +2⋅ (–1) +1⋅ (0)= -1
M = 1, 2, 3
m=0 Im{X [0]}=–2 ⋅ (0)–2 ⋅ (0)–1 ⋅ (0)=0
m=1 Im{X [1]}=–2 ⋅ (1)–2 ⋅ (0)–1 ⋅ (–1)= –1
m=2 Im{X [2]}=–2 ⋅ (0)–2 ⋅ (0)–1 ⋅ (0)=0
m=3 Im{X [3]}=–2 ⋅ (–1)–2 ⋅ (0)–1 ⋅ (1)=1
The discrete frequency components X[m] for m = 1, 2, 3 are found by
addition of the real and imaginary components XRe[i] and XIm[i]
X[0]=6+j0=6
X[1] = – 1 – j
X[2]=0+j0=0
X[3] = – 1 + j
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Example : Use the Inverse DFT, and the results of Example above to
compute the values of the discrete time sequence x[n].
sol.: we will use a 4 -point DFT, that is, for this example, N=4 for m= 0, 1,
2, and 3 reduces to:
The discrete frequency components x[n] for n = 0, 1, 2, 3 are,
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Even and Odd Properties of the DFT
Even Time Function f[N – n] = f [n]
Odd Time Function f[N – n] = –f [n]
Even Frequency Function F[N – m] = F[m]
Odd Frequency Function F[N – m] = –F[m]
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Properties and Theorems of the DFT
X[m] = D{x[n]}
x[n] 1D = {X[m]}
1. Linearity ax1[n]+bx2[n]+… aX1[m]+bX2[m]+…
where x1[n] X1[m] , x2[n] X2[m] , and a and b are arbitrary
constants.
2. Time Shift
x[n – k] WN kmN X[m] ,
3. Frequency Shift
4. Time Convolution
4. Frequency Convolution
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The Fast Fourier Transform (FFT)
The DFT introduced earlier is the only transform that is discrete in both the
time and the frequency domains, and is defined for finite-duration
sequences. Although it is a computable transform, the straightforward
implementation is very inefficient, especially when the sequence length N is
large.
This led to the explosion of applications of the DFT, including in the digital
signal processing area. Furthermore, it also led to the development of other
efficient algorithms. All these efficient algorithms are collectively known as
fast Fourier transform (FFT) algorithms.
Consider an N-point sequence x(n). Its N-point DFT is given by:
a. Decimation in Time
If the DFT algorithm is developed in terms of the direct DFT , it is referred
to as decimation in time,
,
b. Decimation in Frequency
If the DFT algorithm is developed in terms of the Inverse DFT, it is
referred to as decimation in frequency.
,
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Some additional properties of NW are given below.
Example:
Find the FFT for x(n)=[1 -1 0 2 1 3 -1 1]
Solution:
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n nW8 nW4 nW2
0 1 1 1 n=0 , 1202
2
eW n
1 2
12
1 j -j -1 n=3 , jeW
4323
4
2 -j -1 1 n=6 , 1202
2
eW n
3 2
121 j j -1
4 -1 1 1
5 2
121 j -j -1
6 j -1 1
7 2
12
1 j j -1 X(K)=[6 _ 3+J _ -4 _ 3-J _ ]
x(0) 1 2 1 6 X(0)
02W 0
4W 0
8W
x(4) 1 12W 0 14W -j
18W X(1)
x(2) 0 -1 24W 3 3+j X(2)
22W
x(6) -1 32W 1 3
4W j X(3)
x(1) -1 2 5
42W
x(5) 3 52W -4 -4-j
x(3) 2 3 -1
62W
x(7) 1 72W 1 -4+j
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Decimation-in-tame FFT structure for N = 8
The Inverse Fast Fourier Transform (IFFT)
Ex: x(n)= [1 -1 2 1 ]
FFT
n bin. Inv. Ord. n nW4 nW2
0 00 00 0 0 1 1
1 01 10 2 1 -j -1
2 10 01 1 2 -1 1
3 11 11 3 3 j -1
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x(0) 1 3 1 3
02W
x(2) 2 12W -1 -j -1+j2
x(1) -1 0 -1 3
22W
x(3) 1 32W -2 j -1-j2
X(K)=[3 -1+j2 3 -1-j2 ]
IFFT
n nW 4 nW 2
0 1 1
1 j -1
2 -1 1
3 -j -1
X(0) 3 6 1 4
X(2) 3 0 j -4
X(1) -1+J2 -2 -1 8
X(3) -1-J2 J4 -j 4
x(n)=1/4[4 -4 8 4 ]
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The Z Transform The Z transform performs the transformation from the domain of discrete
time signals, to another domain which we call z – domain, The Z transform
yields a frequency domain description for discrete time signals, and forms
the basis for the design of digital systems, such as digital filters.
The z-transform of a discrete-time signal x(n) is defined by:
Where z = jre is a complex variable. The values of z for which the sum
converges define a region in the z-plane referred to as the region of
convergence (ROC).
If x (n) has a z-transform X (z), we write
Because the z-transform is a function of a complex variable, it is convenient
to describe it using the complex z-plane. With z = Re (z) + j Im (z) = jre
The contour corresponding to z = 1 is a circle of unit radius referred to as
the unit circle.
At z = l (ω = 0),
z = j (ω = π/2),
z = - 1 ( ),
We obtain the values of X (je ) for 0 , the unit circle must be
within the region of convergence of X (z).
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The unit circle in the complex z-plane.
EXAMPLE: Let us find the z-transform of the sequence x (n) = n u (n).
Using the definition of the z-transform and the geometric series given in
Table, we have
43
The regions of convergence and divergence for the sequence are shown in
Figure
Example:
Find the Z transform of the discrete unit step function u0 [n] shown in Figure
Solution:
44
Example: Let us find the z-transform of the sequence x (n) = - n u (-n - 1). Proceeding as in the previous example. We have
1
0
11
11
)()()(
z
zzznxzX nnn
nnn
Example: Find the z-transform of x (n) = n)21( u (n) -
n2 u (-n - l),
We know that the z-transform of xl (n) = n)21( u (n) is
And the z-transform of x2(n) = - n2 u (-n - 1) is
Therefore, the z-transform of x (n) = x1 (n) + x2(n) is
45
Common z-transform Pairs
PROPERTIES
1 – Linearity
The z-transform is a linear operator. Therefore, if x (n) has a z-transform
X (z) , and if y(n) has a z-transform Y (z) .
2- Shifting Property
Shifting a sequence (delaying or advancing) multiplies the z-transform by a
power of z. That is to say, if x (n) has a z-transform X (z),
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3- Time Reversal
If x (n) has a z-transform X (z), the z-transform of the time-reversed
sequence x(-n) is
4-Multiplication by an Exponential
If a sequence x (n) is multiplied by a complex exponential n ,
5-Convolution Theorem
States that convolution in the time domain is mapped into multiplication in
the frequency domain, that is,
EXAMPLE: Consider the two sequences
The z-transform of x (n) is
And the z-transform of h (n) is
The z-transform of the convolution of x (n) with h (n) is
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6- Conjugation
If X (z) is the z-transform of x (n), the z-transform of the complex conjugate
of x (n) is
If x (n) is real-valued, x (n) = x*(n), then
7- Derivative
If X (z) is the z-transform of x (n), the z-transform of n x(n) is
EXAMPLE: find the z- transform of the second order recursive filter, given
eothererwiso
nnrnh
n
.........................0....).........cos(
][ 0
Solution:
0
1
0
1
000
])()([21
2)cos()(
00
00
n
nj
n
nj
n
nnjnj
n
n
nn
zrezre
zeerznrzH
2210
10
11 )cos(21)cos(1
11
11
21)(
00
zrzrzr
zrezrezH jj
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The Inverse Z Transform The Inverse Z transform enables us to extract f [n] from F[z] It can be found
by any of the following three methods:
a. Partial Fraction Expansion
b. The Inversion Integral
c. Long Division of polynomials
Example: Use the partial fraction expansion method to compute the
Inverse Z transform of
Solution:
We multiply both numerator and denominator by 3z to eliminate the negative powers of . Then,
Next, we form f(z)/z, and we expand in partial fractions as
49
and multiplication of both sides by z yields
We have:
50
Example : Use the partial fraction expansion method to compute the Inverse
Z transform of
Solution:
We have:
,
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Example
Use the inversion integral method to find the Inverse Z transform of
Solution:
by Cauchy’s residue theorem, this integral can be expressed as where KP
represents a pole of KPz
nzzF 1)( and Res 1)( nzzF represents a residue at
z= KP
Multiplication of the numerator and denominator by 2z yields
52
Example Use the long division method to determine f(n) for n=0 ,1 ,2 ,
given that
Solution: we multiply by 3z
By definition of the Z transform,
We obtain f 0=1f 1=52 and f 2= 81 16
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The Transfer Function of Discrete - Time Systems. The discrete - time system, can be described by the linear difference
equation
y [n] + b1y[n – 1]+b2y[n – 2]+..........+ bk y [n – k]
= a0x [n]+a1x [n – 1]+a2x [n – 2]+....+ ak x [n – k]
Where ai and bi are constant coefficients.
Assuming that all initial conditions are zero, taking the Z transform of both
sides
We define the discrete - time system transfer function H[z] as
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The discrete impulse response h [n ] is the response to the input x [n] = δ[n]
, and since
we can find the discrete - time impulse response h[n] by taking the Inverse Z
transform of the discrete transfer function H[z], that is,
Example :
A discrete - time system is described by the difference equation
Y [n] + y [n – 1] = x [n ] where y [n] = 0 for n < 0
a. Compute the transfer function H(z)
b. Compute the impulse response h[n]
c. Compute the response when the input is x [n] = 10 for n ≥ 0
Solution:
a. y [n] + y [n – 1] = x [n]
y [n] = 0 for n < 0
Taking the Z transform of both sides we obtain
55
b.
c. x [n] = 10 for n ≥ 0
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Stability determination based z- Transform: A digital signal can always be describe using z – Transform as the ratio
...).........)((...).........)((
)()()(
21
21
pzpzzzzzK
zDzNzX
K : the system gain.
,.........,, 321 zzz :are called the “zeros” of X(z), because they are the values
of (z) for which X(z) is zero.
,......., 32,1 ppp :are the poles of X(z). the poles and zeros are either real, or
occur in complex conjugate pairs.
The digital system is stable if and only if all the poles of the system lie
inside the unit circle in the z –plane.
Example: check the stability of the system given by:
)5.0)(3.0()1()( 2
2
zzzzKzH
Solution:
)5.05.0)(5.05.0)(3.0()1()(
2
jzjzzzKzH
Im z
Re z
The system is stable because all the poles lie inside
57
Example: find the impulse response and the transfer function of the
following system.
X[n] y[n]
B
Solution:
Y[n]=B y[n-1] +x[n]
For the impulse response x[n]=δ [n] → X(z)=1
Y(z)=B 1Z Y(z)+1
h [n]=y[n] when x[n]= δ[n]
h[n]= 1Z y[z]= nB u[n]
the transfer function H[n]=Z h[n] = 111
BZ
Example: implement the second order recursive filter
Y[n]=2r cos( 0 ) y[n-1]-2r y[n-2]+x[n]- r cos ( 0 ) x[n-1]
Solution:
Y[z]=2r cos( 0 ) 1Z y[z] -2r 2Z y[z] + x[z]- r cos ( 0 ) 1Z x[z]
X[z] Y[z]
-rcos(w) 2rcos(w)
2r
1Z
∑
1Z
∑
1Z
1Z
×
∑
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Frequency response of LTI system
Since a sinusoidal signal can be expressed in terms of an exponential signal,
the response of the LTI system to an exponential input is of practical
interest. This leads to the concept of frequency response.
A transform – domain representation of the LTI discrete – time system :
n
eznjj
jzHenheH )(][)(
Example: find and draw the frequency response of the system shown
x[n] y[n] 2 0.9 Solution: Y(z) =2 x(z) + 0.9 1Z y(z) H(z) = y(z)/x(z) = 2/(1-0.9 1Z ) frequency response = H( je ) = H(z) jez
H( je )= sin9.0cos9.012
9.012
je j
1Z
∑
59
Problems: 1-Use the partial fraction expansion to find f [n] = 1Z [F (z)] given that
2-Use the Inversion Integral to compute the Inverse Z transform of
3-Given a causal system y(n) = 0.9y(n-1)n - 1) + x(n)
a. Find H(z) and sketch its pole-zero plot.
b. Plot /H(je )l and ∟H(e'").
c. Determine the impulse response h(n).
4-A causal LTI system is described by the following difference equation:
y(n) = 0.81y(n - 2) + x(n) - x(n - 2)
Determine
a. the system function H(z),
b. the unit impulse response h(n),
c. the unit step response v(n), that is, the response to the unit step u(n),
d. the frequency response function H(je ), and plot its magnitude and phase
over 0 ≤ w ≤ π
5-Determine the zero-state response of the system
y(n) = 1/4y(n - 1) +x(n) +3x(n - 1), n 2≥ 0; y(-1) = 2 . to the input