Lecture-2.pptx

Digital Signal & Image Processing
Lecture-4

Overview
• Filters
• Common Filter Types
• Analog vs. Digital Filters
• Difference Equation
• Difference Equation Diagram Elements
• Nonrecursive Difference Equation Diagrams
• Recursive Difference Equation Diagrams
• Impulse Response
• Step Response

Filters
• Filters change a signal’s characteristics by selectively removing
some of its frequency elements.
• A low pass filter, for example, removes the high frequency
components of a signal, but passes low frequency
components.
• It does this because the filter’s gain, the amplification factor it
applies to an input, varies with frequency.
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Filters
• For a low pass filter, the gains are highest at low frequency
and much lower at high frequency.
• A high pass filter has exactly the opposite shape: its gains are
highest at high frequency and lowest at low frequency.
• The pass band of a filter determines the range of frequencies
that are passed.
• The stop band of the filter determines the range of
frequencies that are strongly attenuated.
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Filters
Cutoff Frequency
• A filter is considered to pass signals at frequencies where the filter’s
gain exceeds 0.707 of its maximum gain.
• The frequency or frequencies where the gain equals 0.707 of the
maximum gain are called the cut-off frequencies of the filter.
• Since 20log0.707 = -3 dB, they are also called –3 dB frequencies.
5

Filters
Bandwidth
• The bandwidth of a low pass filter is the range of frequencies from
0 to the –3 dB frequency.
• For a high pass filter, the bandwidth is the range of frequencies
from the –3 dB frequency to half the sampling frequency.
• For band pass filters, the bandwidth is the distance in Hz between
the cut-off frequencies.
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Common Filter Types
Low Pass Filter
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_____ Real Filter
….... Ideal Filter

Common Filter Types
High Pass Filter
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_____ Real Filter

Common Filter Types
Band Pass Filter
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_____ Real Filter

Common Filter Types
Band Stop or Notch Filter
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_____ Real Filter

Filters
Roll-Off
• Another important feature of a filter is its roll-off.
• This characteristic determines how quickly the gain drops
outside the pass band.
• The higher the order of a digital filter, defined by the number
of coefficients needed to specify it, the steeper the slope, and
the higher the quality of the filter is said to be.
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Filters
Gain
• A filter’s gain at a certain frequency determines the
amplification factor that the filter applies to an input at this
frequency.
• A gain may have any value.
• In the pass band region, filter’s gain is high.
• In the stop band region, filter’s gain is low.
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Filters
Gain
• The cutoff frequency of the filter occurs when the gain is
1
2
~0.707 𝑜𝑟 70.7%
• A gain in dB is calculated as 𝐺𝑎𝑖𝑛 𝑑𝐵 = 20𝑙𝑜𝑔 𝐺𝑎𝑖𝑛
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Filters
Bandwidth
Example-1: Find the bandwidth of the band pass filter.
The edges of the pass
Band occur where the
Gain equals 0.707.
The bandwidth of the
Filter is
BW = FH - FL
BW = 4000 - 2000
BW = 2000 Hz
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Square Wave
• A square wave can be constructed from multiple sine waves at different
frequencies.
• The sine waves added in addition to the fundamental frequency are
called harmonics.
• A square wave has harmonics at odd multiples of the fundamental
frequency.
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Square Wave
• As higher harmonics are added, the result gets closer to an ideal square
wave, which contains infinite harmonics.
• sin(angle) + sin(3*angle)/3 + sin(5*angle)/5 + sin(7*angle)/7 + ...
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Effects of Low Pass Filters
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• Each filter type has a unique effect on an input signal.
• Low pass filters tend to smooth signals by averaging out sudden changes.

Effects of High Pass Filters
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• Each filter type has a unique effect on an input signal.
• High pass filters tend emphasize sharp transition.

Analog vs. Digital Filters
• Filters may be implemented in either analog or digital form.
• Analog filters are defined in hardware, while digital filters are
defined in software.
• In general, digital filters are much less susceptible to noise
and component variation than are analog filters.
• Furthermore, digital filter re-design is simply a matter of
editing a list of coefficients, while analog re-design requires
building an entirely new circuit.
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Difference Equation
• A difference equation is one way to specify a filter or system.
• It may be presented in equation or diagram form.
• The general form of a difference equation is
a0y[n] + a1y[n-1] + a2y[n-2] + … + aNy[n-N]
= b0x[n] + b1x[n-1] + b2x[n-2] + … + bMx[n-M]
• The ak and bk weightings are called filter coefficients.
• Generally difference equation has N+1 ak coefficients and M+1 bk
coefficients.
• N is the past output required, also referred as a filter order.
• M is the number of past input required.
• The equation defines how each new output y[n] is obtained.
• In general, a0 is assumed to be one.
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Difference Equation
• The general form of a difference equation is
a0y[n] + a1y[n-1] + a2y[n-2] + … + aNy[n-N]
= b0x[n] + b1x[n-1] + b2x[n-2] + … + bMx[n-M]
The compact form of the recursive difference equation is
• In the above equation, both inputs and outputs are needed to
compute a new output, so the difference equation is said to
be recursive.
22

Difference Equation
• When only inputs are needed to compute a new output, the
difference equation is said to be non-recursive, that is,
y[n] = b0x[n] + b1x[n-1] + b2x[n-2] + … + bMx[n-M]
The compact form of the non-recursive difference equation is
23

Difference Equation
Example-2: A filter has a difference equation y[n] = 0.5y[n-1] + x[n]
a. Is this a recursive or non-recursive difference equation?
b. Identify all ak and bk coefficients.
c. If the input x[n] is u[n], find 5 samples of the output y[n].
Solution
a. The equation is recursive as it depends on a past output y[n-1]
b. Since y[n] – 0.5y[n-1] = x[n], therefore a0 = 1, a1 = -0.5, b0 = 1
c. y[n] = 0.5y[n-1] + x[n]
When n = 0, y[0] = 0.5y[0-1] + x[0] = 0.5(0) + 1 = 1
When n = 1, y[1] = 0.5y[1-1] + x[1] = 0.5(1) + 1 = 1.5
When n = 2, y[2] = 0.5y[2-1] + x[2] = 0.5(1.5) + 1 = 1.75
When n = 3, y[3] = 0.5y[3-1] + x[3] = 0.5(1.75) + 1 = 1.875
When n = 4, y[4] = 0.5y[4-1] + x[4] = 0.5(1.875) + 1 = 1.9375
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Y[n] = {1, 1.5, 1.75, 1.875, 1.9375}

Difference Equation
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Y[n] = {1, 1.5, 1.75, 1.875, 1.9375}

Difference Equation Diagram Elements
Delay Element
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Coefficient Multiplier Element
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Summer Element
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Nonrecursive Difference Equation Diagram
• The general form of the nonrecursive difference equation is
y[n] = b0x[n] + b1x[n-1] + b2x[n-2] + … + bMx[n-M]
• It can be schematically presented as
29

Finite Word Length Effects
• When difference equations are implemented by a computer,
quantization means that filter coefficients cannot be represented
perfectly.
• This means that filter behavior will not match the design exactly.
• Furthermore, computer arithmetic is always subject to rounding
and truncation errors, again due to the limited number of bits
available.
• All of these effects are called finite word length effects.
30

• In order to reduce quantization error of filter’s coefficient one
strategy is to break higher order filter into second order chunks
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Figure: Cascaded second order nonrecursive filter sections

Example-3: Draw a diagram for the difference equation
y[n] = 0.5x[n] + 0.4x[n-1] - 0.2x[n-2]
Solution
32

Example-4: Write the difference equation of the following diagram.
Solution
y[n] = x[n] - 0.3x[n-2] + 0.7x[n-3] 33

Recursive Difference Equation Diagram
• The general form of a recursive difference equation is
a0y[n] + a1y[n-1] + a2y[n-2] + … + aNy[n-N]
= b0x[n] + b1x[n-1] + b2x[n-2] + … + bMx[n-M]
34

Example-5: Draw a direct form I diagram for the following
recursive difference equation
y[n] + 0.5y[n-2] = 0.8x[n] + 0.1x[n-1] - 0.3x[n-2]
Solution
35

Solution
y[n] + 0.5y[n-2] = 0.8x[n] + 0.1x[n-1] - 0.3x[n-2]
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• In direct form I realization, calculation for output y[n] require
M + 1 input states
N output states
M + N + 1 coefficient multiplications
M + N additions
• When more than 2 or 3 delays are needed, this realization is
very sensitive to the finite word length effects.
37

• Recursive filters have an alternative representation, called direct form 2,
that has certain advantages for implementation.
• This representation is defined by the pair of equations
w[n] = x[n] – a1w[n-1] – a2w[n-2] – … – aNw[n-N]
y[n] = b0w[n] + b1w[n-1] + b2w[n-2] + … + bMw[n-M]
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• Transpose of the direct form 2 realization is an another
popular implementation model.
39

• The output of the bottom summer is 0.1x[n] – 0.3y[n]
• The output of the middle summer is 0.1x[n-1] – 0.3y[n-1] + 0.2x[n] – 0.2y[n]
• The final out top summer is y[n] = 0.1x[n-2] – 0.3y[n-2] + 0.2x[n-1] – 0.2y[n-1] + 0.8x[n]
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Impulse Response
• The impulse response h[n] is an important way of characterizing a
filter or system.
• By definition, it is the system’s response to an impulse function
input δ[n].
41

Impulse Response
• For a musical instrument like a piano, an impulse response
corresponds to the note obtained by striking a single key.
42

Impulse Response
• The impulse response may be calculated from a system’s
difference equation.
• The impulse function δ[n] substitutes for the input x[n]
• The impulse response h[n] substitutes for the output y[n]
43

Impulse Response
• For a non-recursive system, the impulse response is
h[n] = b0δ[n] + b1δ[n-1] + b2δ[n-2] + … + bMδ[n-M]
• For a recursive system, the impulse response is
h[n] = – a1h[n-1] – a2h[n-2] – … – aNh[n-N]
+ b0δ[n] + b1δ[n-1] + b2δ[n-2] + … + bMδ[n-M]
44

Impulse Response
• Because non-recursive systems do not rely on past outputs,
they have finite impulse responses, which return to zero after
a finite number of samples have elapsed.
• Recursive systems, on the other hand, have infinite impulse
responses, because each new output depends on past
outputs as well as inputs.
• For causal systems, both non-recursive and recursive, the
impulse response h[n] is zero for n < 0.
45

Impulse Response
Example-8: Find the first six samples of the impulse response
h[n] for the difference equation
Y[n] – 0.4y[n-1] = x[n] – x[n-1]
Solution
The impulse response of the difference equation is
h[n] – 0.4h[n-1] = δ[n] – δ[n-1]
So the first six samples of the impulse response are
h[0] = 1 h[3] = -0.096
h[1] = -0.6 h[4] = -0.0384
h[2] = -0.24 h[5] = -0.01536
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Impulse Response
So the first six samples of the impulse response are
h[0] = 1 h[3] = -0.096
h[1] = -0.6 h[4] = -0.0384
h[2] = -0.24 h[5] = -0.01536
h[n] = [1, 0.6, -0.24, -0.096, -0.0384, -0.01536]
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Impulse Response
Example-9: write the difference equation whose impulse
response is shown in the figure.
Solution
The impulse response can be written as a sum of impulse
functions.
h[n] = δ[n] + 0.8δ[n-1] + 0.2δ[n-2]
The difference equation is y[n] = x[n] + 0.8x[n-1] + 0.2x[n-2]
48

Impulse Response
Example-10: The input signal x[n] and impulse response h[n] are
shown in the figures. Find the output y[n] by breaking the input
signal into impulse functions and finding the response to each.
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Impulse Response
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Y[n] = 3h[n] + 2h[n-1] + h[n-2]

Step Response
• The step response s[n] for a filter is its response to a step
input u[n].
• It records the response of the system to a change in level, and
may be calculated in two ways.
• The first way mimics that for the impulse response, but with
s[n] replacing y[n] and u[n] replacing x[n].
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Step Response
• For non-recursive systems,
s[n] = b0u[n] + b1u[n-1] + b2u[n-2] + … + bMu[n-M]
• For recursive systems,
s[n] = – a1s[n-1] – a2s[n-2] – … – aNs[n-N]
+ b0u[n] + b1u[n-1] + b2u[n-2] + … + bMu[n-M]
52

Step Response
• The step function u[n] is equivalent to a sum of impulse
functions:
u[n] = δ[n] + δ[n-1] + δ[n-2] + …
• The step response s[n] is the same sum of impulse responses:
s[n] = h[n] + h[n-1] + h[n-2] + …
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Step Response
• The second way to compute the step response s[n] relies on
the impulse response.
• each step response sample is a cumulative sum of the impulse
response samples.
• For example
s[4] = h[4] + h[3] + h[2] + h[1] + h[0]
54

Step Response
Example-11: Find the first eight samples of the step response
s[n] for the difference equation
Y[n] – 0.2y[n-1] = 0.5x[n] + 0.3x[n-1]
Solution
The step response s[n] of the difference equation is
s[n] – 0.4s[n-1] = u[n] – u[n-1]
So the first eight samples of the step response are
s[0] = 0.5 s[3] = 0.96 s[6] = 0.996
s[1] = 0.8 s[4] = 0.98 s[7] = 0.9984
s[2] = 0.9 s[5] = 0.992
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Step Response
So the first eight samples of the step response are
s[0] = 0.5 s[3] = 0.96 s[6] = 0.996
s[1] = 0.8 s[4] = 0.98 s[7] = 0.9984
s[2] = 0.9 s[5] = 0.992
S[n] = [0.5, 0.8, 0.9, 0.96, 0.98, 0.992, 0.996, 0.9984]
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