Unit - 5
Design of IIR filter
Filters are those electrical circuits which perform signal processing functions, to remove unwanted frequency components.
Filters are so named according to the frequency range of signals that they allow to pass.
(a) Low Pass Filter: These filters reject all frequencies above a specified value called the cut off frequency.
(b). High Pass Filter: These filters reject all frequencies below the cut off frequency.
(c). Band Pass Filter: These filters allow signals falling within a certain frequency band and blocks all the frequencies outside this band.
(d). Band Stop Filter: These filters reject transmission of a limited band of frequencies but allows the transmission of all other frequencies.
The figure below gives the ideal gain v/s frequency response of above all mentioned types of filters.
A BPF is usually obtained by series connections of LPF with a HPF, where FL of HPF < FH of LPF.
A BSP may be obtained by parallel connections of LPF with a HPF, where FH of HPF > FL of LPF.
A Basic LPF circuit can be designed using passive elements R, L and C. The circuit is having single ‘ R ’ in series with capacitor ‘C’.
For RC LPF
VO = = Vi x fC = 1 / 2∏RC
But Z = Phase shift φ = -cot(2∏FC)
And XC =
VO – output voltage
Vi – input voltage
Z – Impedance
XC – Capacitive Reactance
FC – cut off Frequency
For RC LPF the reactance of ‘ C ’ is very high at low frequency so, capacitor acts as open circuit and blocks i/p signal Vi until cut-off frequency is reached.
Cut-off frequency for a LPF is that frequency at which the voltage at output equals 70.7% of input voltage( -3 dB point ).
LPF are used to filter noise from circuit.
Q –A LPF circuit consisting of a resistor of 40KΩ in series with capacitor 47nF across a 10V sinusoidal supply. Calculate VO at frequency 1Kz?
Sol:
R = 40KΩ
C = 47nF
Vi = 10V
F = 1000 Hz
For LPF,
Capacitive Reactance is XC
The output voltage is given as
VO = Vi
XC = =
XC = 3.386KΩ
VO =
=10 x
= 10 x
VO = 0.843 V
(b) RL LPF
FC = R / 2nt
VO = Vi
Comparison of RL – RC circuits
- RC circuit occupy small space. It is not possible to fabricate inductors on surface of semiconductor chip.
- The response of charging and discharging of RL circuit is not proper.
- Capacitors are less expensive.
- RC- circuit is used in high time constant application. While RL used in low time constant applications.
Q – An RL LPF consists of 5.6mH coil a 3.3 KΩ resistor. The output voltage is taken across resistor. Calculate the critical frequency?
Soln: Given:
L = 5.6 x 10-3 H
R = 3.3 x 103Ω
For RL LPF
F ==
F = 93.78 KHz
Q – A sinusoidal voltage with peak to peak value of 10 V is applied to an RC LPF. If reactance at input is zero, find output voltage?
Soln :
VO =
XC =
F = XC = 0, F ->∞
Hence, VO = 0V.
Q. An RC – LPF consists of 120 Ω resistor and 0.02 µF capacitor. The output is taken through capacitor. Calculate the critical frequency?
For RC LPF FC is given as
FC==
FC = 66.31 KHz
High pass
(a). The RC – HPF circuits is shown below, which is having capacitor at input and resistor at output.
Reactance of capacitor is very high at low frequency.
XC =
XC when F ( C acts as open circuit )
XC when F ( C acts as short circuit )
FC =
Phase shift = Cot
VO =Vi
VO = Vi
At low F :XC ->∞ VO = 0
At high F :XC -> 0 VO = Vi
Q. An RC HPF with input capacitor 10nf and the output resistor of 10K Ω. Find the cut off frequency.
Soln:
FC = =
FC = 15.915 Hz
Q. Calculate the break point (FC) for a simple passive HPF consisting of 82pF capacitor connected in series with 240 KΩ resistor?
Soln:
FC = =
FC = 8KHz
RL – HPF:
The basic RL HPF is shown below. The inductors pass low frequency signals with very little resistance, and offers great resistance to high frequency signals.
VO =
FC =
Q. For a RL HPF consists of 470Ω resistor and 600 mH coil output is taken across coil. Calculate the cut off frequencies.
Soln:
FC = == 125 Hz
Band Pass and Band reject proto- type section
It can be designed by making a series connection of HPF with an LPF. The Basic circuit is shown below.
FC = 1 / 2∏RC C2 = 1 / 2∏FHR
C1 = 1 / 2∏FLR FR =
FL = lower -3dB cut off frequency
FH= upper -3dB cut off frequency
Band width (BW) = FH - FL
USES
1>. The BPF has its applications in wireless communication at transmitter and receiver.
2>. It is also used in optical communication.
3>. For reducing the signal to noise ratio at the receiving end.
Q. The value of capacitor C1 required to give a cut off frequency fL = 1KHz and value of resistor is 10K for the BPF. The value of capacitor C2 and higher cut off frequency (FH) is 30 KHz with resistor 10KΩ. Also calculate the central resonat frequency?
Soln: Given values for HPF
FL = 1000 Hz, R1 = 10 x 103Ω
C1 = == 15.9nF
For LPF,
FH = 30 x 103 Hz, R2 = 10 x 103Ω
C2 === 530nF
Central resonat frequency,
Fr =
=
Fr = 5.477 KHz
Band Stop Filter (BSF) OR Notch FILTER
A notch filter is a Band Stop filter with a narrow band stop bandwidth. These filters are used to attenuate a narrow range of frequencies.
FC = 1 / 2∏RC or FR =
R = 1 / 2∏FCC
Applications of BSF
(i). In telephone technology it is used as noise reducer.
(ii). Widely used in guitar amplifiers.
(iii). For reducing unwanted harmonics.
(iv). Used to reduce static on radio.
(v). In image signal processing.
(vi). Used in biometric instruments like ECG for removing line noise.
Before a filter can be designed, a set of filter specifications must be defined. For example, suppose that we would like to design a low-pass filter with a cut-off frequency ωK. The frequency response of an ideal low-pass filter with linear phase and a cut-off frequency ωK. Is
Hd(ejω) = e-jαω for |ω|<|ωc|
=0 for ωc<| ω|<π
hd(n) = sin(n-α) ωc/ π (n- α)
Because this filter is unrealizable (noncausal and unstable), it is necessary to relax the ideal constraints on the frequency response and allow some deviation from the ideal response. The specifications for a low-pass filter will typically have the form
1-p < |H(ejω)|<1+ p
|H(ejω)|<1+ s
Thus, the specifications include the passband cut-off frequency, ωS the stopband cutoff frequency, ωP, the passband deviation, δP. And the stopband deviation, δS . The passband and stopband deviations
αp= -20log(1- p)
αs= -20log(s)
The interval [ωP, ωS] is called the transition hand.
Once the filter specifications have been defined, the next step is to design a filter that meets these specifications.
Butterworth filter design
At the expense of steepness in transition medium from pass band to stop band this Butterworth filter will provide a flat response in the output signal. So, it is also referred as a maximally flat magnitude filter. The rate of falloff response of the filter is determined by the number of poles taken in the circuit. The pole number will depend on the number of the reactive elements in the circuit that is the number of inductors or capacitors used in the circuits.
The amplitude response of nth order Butterworth filter is given as follows:
Vout / Vin = 1 / √{1 + (f / fc)2n}
Where ‘n’ is the number of poles in the circuit. As the value of the ‘n’ increases the flatness of the filter response also increases.
'f' = operating frequency of the circuit and 'fc' = centre frequency or cut off frequency of the circuit.
These filters have pre-determined considerations whose applications are mainly at active RC circuits at higher frequencies. Even though it does not provide the sharp cut-off response it is often considered as the all-round filter which is used in many applications.
First Order Low Pass Butterworth Filter
The below circuit shows the low pass Butterworth filter:
Fig: First order LP Butterworth Filter
The required pass band gain of the Butterworth filter will mainly depends on the resistor values of ‘R1’ and ‘Rf’ and the cut off frequency of the filter will depend on R and C elements in the above circuit.
The gain of the filter is given as Amax = 1 + (R1 / Rf)
The impedance of the capacitor ‘C’ is given by the -jXC and the voltage across the capacitor is given as,
Vc = - jXC / (R - jXC) * Vin
Where XC = 1 / (2πfc), capacitive Reactance.
The transfer function of the filter in polar form is given as
H(jω) = |Vout/Vin| ∟ø
Where gain of the filter Vout / Vin = Amax / √{1 + (f/fH)²}
And phase angle Ø = - tan-1 ( f/fH )
At lower frequencies means when the operating frequency is lower than the cut-off frequency, the pass band gain is equal to maximum gain.
Vout / Vin = Amax i.e. constant.
At higher frequencies means when the operating frequency is higher than the cut-off frequency, then the gain is less than the maximum gain.
Vout / Vin < Amax
When operating frequency is equal to the cut-off frequency the transfer function is equal to Amax /√2. The rate of decrease in the gain is 20dB/decade or 6dB/octave and can be represented in the response slope as -20dB/decade.
Second Order Low Pass Butterworth Filter
An additional RC network connected to the first order Butterworth filter gives us a second order low pass filter. This second order low pass filter has an advantage that the gain rolls-off very fast after the cut-off frequency, in the stop band.
Fig: Second Order LP Butterworth Filter
In this second order filter, the cut-off frequency value depends on the resistor and capacitor values of two RC sections. The cut-off frequency is calculated using the below formula.
fc = 1 / (2π√R2C2)
The gain rolls off at a rate of 40dB/decade and this response is shown in slope -40dB/decade. The transfer function of the filter can be given as:
Vout / Vin = Amax / √{1 + (f/fc)4}
The standard form of transfer function of the second order filter is given as
Vout / Vin = Amax /s2 + 2εωns + ωn2
Where ωn = natural frequency of oscillations = 1/R2C2
ε = Damping factor = (3 - Amax ) / 2
For second order Butterworth filter, the middle term required is sqrt(2) = 1.414, from the normalized Butterworth polynomial is
3 - Amax = √2 = 1.414
In order to have secured output filter response, it is necessary that the gain Amax is 1.586.
Higher order Butterworth filters are obtained by cascading first and second order Butterworth filters.
n (order) | Normalized Denominator polynomials in factored form |
1 | |
2 | |
3 |
The transfer function of the nth order Butterworth filter is given as follows:
H(jω) = 1/√{1 + ε² (ω/ωc)2n}
Where n is the order of the filter
ω is the radian frequency and it is equal to 2πf
And ε is the maximum pass band gain, Amax
Examples
Let us consider the Butterworth low pass filter with cut-off frequency 15.9 kHz and with the pass band gain 1.5 and capacitor C = 0.001µF.
fc = 1/2πRC
15.9 * 10³ = 1 / {2πR1 * 0.001 * 10-6}
R = 10kΩ
Amax = 1.5 and assume R1 as 10 kΩ
Amax = 1 + {Rf / R1}
Rf = 5 kΩ
Ideal Frequency Response of the Butterworth Filter
The flatness of the output response increases as the order of the filter increases. The gain and normalized response of the Butterworth filter for different orders are given below:
Fig: Frequency Response of Butterworth Filter
Example. A Butterworth Amplitude Response design
Let,
Solving for N gives
Matching at
The normal from the table is
Frequency seal impliea that we let
Finally, the frequency scaled system function is
Chebyshev filters
A Chebyshev design achieves a more rapid roll-off rate near the cut-off frequency than the Butterworth by allowing ripple in the passband (type I) or stopband (type II). Monotonicity of the stopband or passband is still maintained respectively.
Fig: Chebyshev Filter Type I and II
Chebyshev Type I
The magnitude response is given by
Where,
Nth order Chebyshev polynomial
And specifies the passband ripple
The Chebyshev polynomials are of the form\
With recurrence formula
An alternate form for which will be useful in both analysis and design is
Design a Chebyshev type I lowpass filter to satisfy the following amplitude specification
Using the design formula for N
From the 2dB ripple table (9-20)
Example. Design from a Rational H (s)
Let,
Inverse Laplace transforming yields
Sampling seconds and gain scaling we obtain
Which implies that
Setting requires that
Q. Design a Chebyshev type 1 low pass filter to satisfy the following amplitude specifications:
Using the design formula for N
N = [2.08]=3
Note:
The normalized system function is
With poles at
To frequency scale let
So in the z domain
Note that when the design originates from discrete time specifications the poles of H (z) are independent of .
Q. Design a Chebyshev type 1 low pass filter to satisfy the following amplitude specifications:
Using the design formula for N
N = [2.08]=3
Note:
The normalized system function is
With poles at
To frequency scale let
So in the z domain
Note that when the design originates from discrete time specifications the poles of H (z) are independent of .
The Impulse Invariance Method is used to design a discrete filter that yields a similar frequency response to that of an analog filter. Discrete filters are amazing for two very significant reasons:
- You can separate signals that have been fused and,
- You can use them to retrieve signals that have been distorted.
We can design this filter by finding out one very important piece of information i.e., the impulse response of the analog filter. By sampling the response we will get the time-domain impulse response of the discrete filter.
When observing the impulse responses of the continuous and discrete responses, it is hard to miss that they correspond with each other. The analog filter can be represented by a transfer function, Hc(s).
Zeros are the roots of the numerator and poles are the roots of the denominator.
Mapping from s-plane to z-plane
The transfer function of the analog filter in terms of partial fraction expansion with real coefficients is
Where A are the real coefficients and P are the poles of the function And k can be 1, 2 …N.
h(t) is the impulse response of the same analog filter but in the time domain. Since ‘s’ represents a Laplace function Hc(s) can be converted to h(t), by taking its inverse Laplace transform.
Using this transformation,
We obtain
However, in order to obtain a discrete frequency response, we need to sample this equation. Replace ‘nTS’ in the place of t where TS represents the sampling time. This gives us the sampled response h(n),
Now, to obtain the transfer function of the IIR Digital Filter which is of the ‘z’ operator, we have to perform z-transform with the newly found sampled impulse response, h(n). For a causal system which depends on past(-n) and current inputs (n), we can get H(z) with the formula shown below
We have already obtained the equation for h(n). Hence, substitute eqn (2) into the above equation
Factoring the coefficient and the common power of n
—(3)
Based on the standard summation formula, (3) is modified and written as the required transfer function of the IIR filter.
–(4)
Hence (4) is obtained from (1), by mapping the poles of the analog filter to that of the digital filter.
That is how you map from the s-plane to z-plane
Relationship of S-plane to Z plane
From the equation above, Since, the poles are the denominators we can say .
Comparing (1) and (4), we can derive that
–(5)
And since , substituting into (5) gives us
–(6)
Where
TS is the sampling time
Now, s is taken to be the Laplace operator
–(7)
σ is the attenuation factor
Ω is the analog frequency
Changing Z from rectangular coordinates to the polar coordinates, we get:
–(8)
Where r is magnitude and ω is digital frequency
Replacing (7) in place of s in (6), and replacing that value as Z in (8)
Compare the real and imaginary parts separately. Where the component with ‘j’ is imaginary.
–(9)
And
Hence, we can make the inference that
To understand the relationship between the s-plane and Z-plane, we need to picture how they will be plotted on a graph. If we were to plot (7) in the ‘s’ domain, σ would be the X-coordinates and jΩ would be the Y-coordinate. Now, if we were to plot (8) in the ‘Z’ domain, the real portion would be the X-coordinate, and the imaginary part would be the Y-coordinate.
Let us take a closer look at equation (9),
There are a few conditions that could help us identify where it is going to be mapped on the s-plane.
Case 1
When σ <0, it would translate that r is the reciprocal of ‘e’ raised to a constant. This will limit the range of r from 0 to 1.
Since σ <0, it would be a negative value and would be mapped on the left-hand side of the graph in the ‘s’ domain
Since 0<r<1, this would fall within the unit circle which has a radius of in the ‘z’ domain.
Case 2
When σ =0, this would make r=e0, which gives us 1, which means r=1. When the radius is 1, it is a unit circle.
Since σ =0, which indicates the Y-axis of the ‘s’ domain.
Since r=1, the point would be on the unit circle in the ‘z’ domain.
Case 3
When σ>0, since it is positive, r would be equal to ‘e’ raised to a particular constant, which means r would also be a positive value greater than 1.
Since σ>0, the positive value would be mapped onto the right-hand side of the ‘s’ domain.
Since r>1, the point would be mapped outside the unit circle in the ‘z’ domain.
Here is a pictorial representation of the three cases:
Mapping of poles located at the imaginary axis of the s-plane onto the unit circle of the z-plane. This is an important condition for accurate transformation.
Mapping of the stable poles on the left-hand side of the imaginary s-plane axis into the unit circle on the z-plane. Poles on the right-hand side of the imaginary axis of the s-plane lie outside the unit circle of the z-plane when mapped.
Disadvantages:
- Digital frequency represented by ‘ω,’ and its range lies between – π and π. Analog frequency is represented by ‘Ω,’ and its range lies between – π/TS and π/TS. When mapping from digital to analog, from – π/TS and π/TS , ‘ω’ maps from – π to π. This would make the range of Ω (k-1)π/TS and (k+1)π/TS, where k is an arbitrary constant. However, mapping the other way, from analog to digital, will mean ω maps from – π to π, which makes it many-to-one. Hence, mapping is not one-to-one.
- Analog filters do not have a definite bandwidth because of which when sampling is performed, this would give rise to aliasing. Aliasing is when the signal eats up into the next signal and so on. This would lead to considerable distortion of the signal. Hence, making the frequency response of the converted digital signal very different from the original frequency response of the analog filter.
- Increasing the sampling time will result in a frequency response that is more spaces out hence decreasing the chances of aliasing. However, this is not the case with this method. Increasing the sampling time has no effect on the amount of aliasing that happens.
Given , that has a sampling frequency of 5Hz. Find the transfer function of the IIR digital filter.
Solution:
Step 1:
Step 2:
Applying partial fractions on H(s),
Step 3:
Step 4:
The bilinear transform is the result of a numerical integration of the analog transfer function into the digital domain. We can define the bilinear transform as:
Find the bilinear transformation which maps points z =2,1,0 onto the points w=1,0,i.
Ans. Let,
And,
Since bilinear transformation preserves cross ratios,
Thus we have,
Use the bilinear transformation to convert the analog filtrt with system function
into a digital IIR filter. Select T =0.1
Consider the following system function
Note that the following is the resonant frequency of the analog filter
Consider that the resonant frequency of analog filter must be mapped by selecting the value of parameter
T= 0,1
Use the following mapping for bilinear transformation
Write the system function H(z) of the resultant digital filter
Frequency warping
• The bilinear transformation method has the following important features: A stable analog filter gives a stable digital filter. t The maxima and minima of the amplitude response in the analog filter are preserved in the digital filter. As a consequence, – the pass band ripple, and – the minimum stop band attenuation of the analog filter are preserved in the digital filter frame.
• To determine the frequency response of a continuous-time filter, the transfer function Ha(s)Ha(s) is evaluated at s=jω which is on the jω axis. Likewise, to determine the frequency response of a discrete-time filter, the transfer function Hd(z) is evaluated at z=ejωT which is on the unit circle, |z|=1|z|=1 .
• When the actual frequency of ω is input to the discrete-time filter designed by use of the bilinear transform, it is desired to know at what frequency, ωa , for the continuous-time filter that this ω is mapped to.
• This shows that every point on the unit circle in the discrete-time filter z-plane, z= ejωT is mapped to a point on the jω axis on the continuous-time filter s-plane, s=jω. That is, the discrete-time to continuous-time frequency mapping of the bilinear transform is
ωa=(2/T) tan(ωt/2)
And the inverse mapping is
ω=(2/T) arc tan(ωaT/2)
• The discrete-time filter behaves at frequency the same way that the continuous-time filter behaves at frequency (2/T)tan(ωT/2). Specifically, the gain and phase shift that the discrete-time filter has at frequency ω is the same gain and phase shift that the continuous-time filter has at frequency (2/T)tan(ωT/2). This means that every feature, every "bump" that is visible in the frequency response of the continuous-time filter is also visible in the discrete-time filter, but at a different frequency. For low frequencies (that is, when ω≪2/T or ωa≪2/T),ω≈ωa.
One can see that the entire continuous frequency range
−∞<ωa<+∞
Is mapped onto the fundamental frequency interval
−πT<ω<+πTω=±π/T.ωa=±∞
One can also see that there is a nonlinear relationship between ωa and ω This effect of the bilinear transform is called frequency warping.
Design a discrete time lowpass filter to satisfy the following amplitude specifications:
Assume
The pre-warped critical frequency are
Since both the passband and stopband are required to be monotonic, a Butterworth approximation will be used
From the Butterworth design tables we can immediately write
Now find H (z) by first noting that
Using the pole/ zero mapping formula
We can now write
Find by setting
Finally after multiplying out the numerator and denominator we obtain
Key takeaway
Sr No. | Impulse Invariance | Bilinear Transformation |
1 | In this method IIR filters are designed having a unit sample response h (n) that is sampled version of the impulse response of the analog filter. | This method of IIR filter design is based on the trapezoidal formula for numerical integration. |
2 | The bilinear transformation is a conformal mapping that transforms the j axis into the unit circle in the z plane only once, thus avoiding aliasing of frequency component. | The bilinear transformation is a conformal mapping tjat transforms the axis into the unit circle in the z plane only once, thus avoiding aliasing of frequency components. |
3 | For design of LPF, HPF and almost all types of bandpass and band stop filters this method is used. | For designing of LPF, HPF and almost all types of bandpass and band stop filters this method is used. |
4 | Frequency relationship is non –linear. Frequency warping or frequency compression is due to non – linearity. | Frequency relationship is non linear. Frequency warping or frequency compression is due to non – linearity. |
5 | All poles are mapped from s plane to the z plane by the relationship . But the zeros in two domain does not satisfy the same relationship. | All poles and zeros are mapped. |
Q1) Design an analog Butterworth filter that has a -2dB pass band attenuation at a frequency of 20 rad/sec and atleast -10 dB stop band attenuation at 30 rad/sec.
A1) Given data:
Pass band attenuation αP= 2 dB;
Stop band attenuation αS= 10 dB;
Pass band frequency ΩP= 20 rad/sec.
Stop band frequency ΩS=30 rad/sec.
The order of the filter
Rounding off ‘N’ to the next higher integer, we get
N=4
The normalized transfer function for N=4.
To find cut off frequency
The transfer function for Ωc=21.3868,
Q2) Design a third order Butterworth digital filter using impulse invariant technique. Assume sampling period T=1 sec.
A2) For N=3, the transfer function of a normalized Butterworth filter is given by
Q3) Design a digital Butterworth filter satisfying the constraints
With T=1 sec using bilinear transformation
A3) Given data:
Pass band attenuation; Pass band frequency;
Stop band attenuation; Stops band frequency;
Step 1: Specifying the pass band and stop band attenuation in db
Step2. Choose T and determine the analog frequencies (i.e) Prewarp band edge frequency
Step3. To find order of the filter
Rounding the next higher value N=2
Step 4: The normalized transfer function
Step 5: Cut off frequency
Step 6: To find Transfer function of H(s):
Step 7. Apply Bilinear Transformation with to obtain the digital filter
Q4) Design a digital Butterworth filter satisfying the constraints
With T=1 sec using Impulse invariant method
A4) Pass band attenuation; Pass band frequency;
Stop band attenuation; Stops band frequency;
Step 1: Specifying the pass band and stop band attenuation in dB.
Step2. Choose T and determine the analog frequencies (i.e) Prewarp band edge frequency
Step3. To find order of the filter
Rounding the next higher value N=4
Step 4: The normalized transfer function
Step 5: Cut off frequency
Step 6: To find Transfer function of H(s):
Step 7: Using partial fraction expansion, expand H(s) into
Q5) Prove that
A5)
Direct form I
The difference equation
Specifies the Direct-Form I (DF-I) implementation of a digital filter. The DF-I signal flow graph for the second-order case is shown in Fig.
Figure: Direct-Form-I implementation of a 2nd-order digital filter.
Direct form II
The signal flow graph for the Direct-Form-II (DF-II) realization of the second-order IIR filter section is shown in Fig.
Figure: Direct-Form-II implementation of a 2nd-order digital filter.
The difference equation for the second-order DF-II structure can be written as
Which can be interpreted as a two-pole filter followed in series by a two-zero filter.
Cascade form
The filter section can be seen to be an FIR filter and can be realized as shown below
W[n] = p0x[n] + p1x[n-1] + p2x[n-2] +p3x[n-3]
The time-domain representation of H2(z) is given by
Y[n] = w[n] –d1y[n-1] –d2y[n-2] – d3y[n-3]
A cascade of the two structures realizing and leads to the realization of shown below and is known as the direct form I structure
Direct form II and cascade form realizations of
Direct form II
Cascade form
References:
1. Mitra S., “Digital Signal Processing: A Computer Based Approach”, Tata McGraw-Hill,1998, ISBN 0-07-044705-5
2. A.V. Oppenheim, R. W. Schafer, J. R. Buck, “Discrete Time Signal Processing”, 2nd Edition Prentice Hall, ISBN 978-81-317-0492-9
3. Steven W. Smith, “Digital Signal Processing: A Practical Guide for Engineers and Scientists”,1st Edition Elsevier, ISBN: 9780750674447