UNIT - 7: FIR Filter Design

Transcription

1 UNIT - 7: FIR Filter Design Dr. Manjunatha. P manjup.jnnce@gmail.com Professor Dept. of ECE J.N.N. College of Engineering, Shimoga October 5, 06

2 Unit 7 Syllabus Introduction FIR Filter Design:[,, 3, 4] Slides are prepared to use in class room purpose, may be used as a reference material All the slides are prepared based on the reference material Most of the figures/content used in this material are redrawn, some of the figures/pictures are downloaded from the Internet. This material is not for commercial purpose. This material is prepared based on Digital Signal Processing for ECE/TCE course as per Visvesvaraya Technological University (VTU) syllabus (Karnataka State, India). Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 / 94

3 Unit 7 Syllabus Unit 7: FIR Filter Design: PART-B-Unit 7: FIR Filter Design: Introduction to FIR Filters Design of FIR Filters using Rectangular window Hamming window 3 Hanning window 4 Bartlet window 5 Kaiser window Design of FIR Filter using frequency sampling technique. Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 3 / 94

4 Introduction Advantages of the FIR digital filter Relatively easy to design and computationally more efficient. FIR filters are implemented in hardware or software. The phase response is linear. Linear phase property implies that the phase is a linear function of the frequency. FIR filter output is delayed by the same amount of time for all frequencies, thereby eliminating the phase distortion (Group delay). FIR filters are always stable i.e. for a finite input, the output is always finite. In linear phase, for the filter of length N the number of operations are of the order of N/. Disadvantages of the FIR digital filter (compared to IIR filters) They require more memory and/or calculation to achieve a given filter response characteristic. Also, certain responses are not practical to implement with FIR filters. For a desired frequency response, with tight constraints on the passband, transition band and the stopband, a FIR filter may have large number of coefficients, thereby have more arithmetic operations and hardware components. An LTI system is causal iff Input/output relationship: y[n] depends only on current and past input signal values. Impulse response: h[n] 0 for n < 0 System function: number of finite zeros number of finite poles. Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 4 / 94

5 An ideal lowpass filter is given by FIR Filter Design { ω ωc H(ω) 0 ω c < ω π Introduction Hd ( e jω ) The impulse response is given by ω c h(n) H(ω)e jωn dω π ω c { ωc n 0 π ω c sin(ω c n) n 0 π ω c n π ω c 0 ω c ω Figure : Ideal low pass filter π Paley-Wiener Theorem: If h(n) has finite energy and h(n) 0 for n < 0 then π ln H(ω) dω < π Figure : Unit sample response H(ω) can be zero at some frequencies. but it cannot be zero over any finite of frequencies, since the integral then becomes infinite. H(ω) cannot be exactly zero over any band of frequencies. (Except in the trivial case where h[n] 0.) Furthermore, H(ω) cannot be flat (constant) over any finite band. Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 5 / 94

6 Introduction Magnitude Characteristic of FIR filter The magnitude response can be expressed as { δ H(ω) + δ Magnitude for 0 ω ω p 0 H(ω)δ for ω s ω π Approximate formula for order N is N 0log 0(δ δ ) 5 4 f ωs ωp where f f π s f p Approximate formula for order N is π N k ω s ω p The width of the main lobe is N k π M δ Passband ripple δ Stopband ripple ωp Passband edge frequency ωs Stopband edge frequency Figure 3: Magnitude Specification of FIR Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 6 / 94

7 Introduction Ideal filters are noncausal, hence physically unrealizable for real time signal processing applications. Causality implies that the frequency response characteristic H(ω) of the filter cannot be zero, except at finite set of points in the frequency range. And also H(ω) cannot have an infinitely sharp cutoff from passband to stopband, that is H(ω) cannot drop from unity to zero abruptly. It is not necessary to insist that the magnitude be constant in the entire passband of the filter. A small amount of ripple in the passband is usually tolerable. The filter response may not be zero in the stopband, it may have small nonzero value or ripple. The transition of the frequency response from passband to stopband defines transition band. The passband is usually called bandwidth of the filter. The width of transition band is ω s ω p where ω p defines passband edge frequency and ω s defines stopband edge frequency. The magnitude of passband ripple is varies between the limits ± δ where δ is the ripple in the passband The ripple in the stopand of the filter is denoted as δ Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 7 / 94

8 FIR Filter Design FIR Filter Design Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 8 / 94

9 FIR Filter Design An FIR system does not have feedback. Hence y(n k) term is absent in the system. FIR output is expressed as y(n) M b k x(n k) k0 If there are M coefficients then M y(n) b k x(n k) k0 The coefficients are related to unit sample response as { bn for 0 n M h(n) 0 otherwise Expanding the summation y(n) b 0 x(n) + b x(n ) + b x(n ) +... b (M ) x(n M + ) Since h(n) b n then y(n) is M y(n) h(k)x(n k) k0 Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 9 / 94

10 Symmetric and Antisymmetric FIR Filters Symmetric and Antisymmetric FIR Filters Linear Phase FIR structure Linear phase is a property of a filter, where the phase response of the filter is a linear function of frequency. The result is that all frequency components of the input signal are shifted in time (usually delayed) by the same constant amount, which is referred to as the phase delay. And consequently, there is no phase distortion due to the time delay of frequencies relative to one another. Linear-phase filters have a symmetric impulse response. The FIR filter has linear phase if its unit sample response satisfies the following condition: h(n) h(m n) n 0,,,..., N The Z transform of the unit sample response is given as M H(z) h(n)z n n0 Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 0 / 94

11 Symmetric and Antisymmetric FIR Filters Symmetry: h(n)h(m--n) Odd M h[n] Symmetry: h(n)h(m--n) Even M h[n] n n Center of Symmetry Center of Symmetry Antisymmetry: h(n)-h(m--n) Odd M h[n] Antisymmetry: h(n)-h(m--n) Even M h[n] n n Center of Symmetry Center of Symmetry Figure 4: Symmetric and antisymmetric responses The unit sample response of FIR filter is symmetric if h(n) h(m n) The unit sample response of FIR filter is antisymmetric if h(n) h(m n) Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 / 94

12 H(z) FIR Filter Design Frequency response of Linear Phase FIR Filter: M n0 h(n)z n Symmetric and Antisymmetric FIR Filters Symmetric with Modd Symmetry: h(n)h(m--n) Odd M h[n] Sy h Symmetric impulse response with Modd Then h(n) h(m n) and (z e jω ) ( M H(z) h ) z ( ) (M 3)/ M + n0 h(n) [z n + z (M n)] n M 3 ( ) ( ) M H(e jω ) h e jω M + h(n) [e jωn + e jω(m n)] n0 e jωn e jωn e M jω( ) e e jω(m n) e jω(m ) e jωn e M jω( ) e M jω( n).e M jω( ) M jω(.e ( ) e jωn + e jω(m n) e jω M [ ( e jω M Center of Symmetry Antisymmetry: h(n)-h(m--n) Odd M h[n] M jω( ) ) e jωn e ) n M jω( ) M jω( n) 5.e6 7 8 n + e jω ( M ( ) e jω M ( ) M cosω n )] Center n of Symmetry An h[ Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 / 94

13 Symmetric and Antisymmetric FIR Filters H(e jω ) h( M )e ( M h ( e jω M M jω( ) + M 3 n0 M 3 ) ( ) e jω M + ) h ( M h(n)[e jωn + e jω(m n) ] n0 M 3 ) + ( ) h(n)e jω M ( ) M cosω n n0 ( n) M h(n)cos ω ( M H(ω) h H(ω) H(ω) e j H(ω) M 3 ) + n0 ( ) M h(n)cos ω n H(ω) ( ) ω M ( ω M for H(ω) > 0 ) + π for H(ω) < 0 Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 3 / 94

14 Symmetric and Antisymmetric FIR Filters Frequency response of Linear Phase FIR Filter: ( H(ω) e jω M M ) n0 ( M h(n)cos ω Symmetric with MEven n) Symmetry: h(n)h(m--n) Odd M h[n] Symmetry: h(n)h(m--n) Even M h[n] H(ω) H(ω) e j H(ω) M ( ) n M H(ω) h(n)cos ω n n0 Center of Symmetry ( ) ω M for H(ω) > 0 Antisymmetry: h(n)-h(m--n) Odd M H(ω) ( ) ω M h[n] + π for H(ω) < n Center of Symmetry Antisymmetry: h(n)-h(m--n) Even M h[n] n n Center of Symmetry Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Center Design of Symmetry October 5, 06 4 / 94

15 Design of linear-phase FIR filters using windows Design of linear-phase FIR filters using windows Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 5 / 94

16 Design steps for Linear Phase FIR Filter (Fourier Series method) Based on the desired frequency response specification H d (e jω ) determine the corresponding unit sample response h d (n). H d (e jω ) h d (n)e jωn n0 Obtain the impulse response h d (n) for the desired frequency response H d (ω) by evaluating the inverse Fourier transform. h d (n) π H d (e jω )e jωn dω π π 3 In general the sample response h d (n) is infinite in duration and must be truncated at some point to get an FIR filter of length M. Truncation is achieved by multiplying h d (n) by window function. h(n) h d (n)w(n) where w(n) is window function 4 Obtain the H(z) for h(n) by taking z transform 5 Obtain the magnitude response H(e jω ) and phase response θ(ω) Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 6 / 94

17 Figure 5: Frequency response characteristic of different types of filters Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 7 / 94 Low-pass filter is used to eliminate high-frequency fluctuations (eg. noise filtering, demodulation, etc.) High-pass filter is used to follow small-amplitude high-frequency perturbations in presence of much larger slowly-varying component (e.g. recording the electrocardiogram in the presence of a strong breathing signal) Band-pass is used to select a required modulated carrier signal (e.g. radio) Band-stop filter is used to eliminate single-frequency (e.g. mains) interference (also known as notch filtering) H d ( ω) H d ( ω) 0 ω p π Low pass Filter ω 0 ω p π High pass Filter ω H d ( ω) H d ( ω) 0 ω ω π ω ω ω 3 4 Band pass Filter ω 0 ω π Band stop pass Filter ω

18 Different Types of Windows Rectangular: Hanning Hamming: Blackman: Bartlett (Triangular) Window Kaiser window Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 8 / 94

19 Window Design Techniques Rectangular window w ( ) R n This is the simplest window function but provides the worst performance from the viewpoint of stopband attenuation. The width of main lobe is 4π/N M- Figure 6: Rectangular window n { for n 0,, M ω R (n) 0 otherwise Magnitude response of rectangular window is W R (ω) sin( ωm ) sin( ω ) Figure 7: Rectangular window Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 9 / 94

20 Window Design Techniques Bartlett (Triangular) Window wt ( n) M- n Figure 8: Bartlett window Bartlett Window is also Triangular window. The width of main lobe is 8π/M ω T (n) n M M Figure 9: Bartlett window Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 0 / 94

21 Hanning window FIR Filter Design Window Design Techniques This is a raised cosine window function given by: w(n) [ ( )] πn cos M W (ω) 0.5W R (ω) [W R (ω πm ) + W R(ω + πm ] ) The width of main lobe is: 8π M w ( ) H n M- n Figure 0: Hanning window Figure : Hanning window Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 / 94

22 Hamming window FIR Filter Design Window Design Techniques This is a modified version of the raised cosine window [ ( )] πn w(n) cos M W (ω) 0.54W R (ω) [W R (ω πm ) + W R(ω + πm ] ) The width of main lobe is: 8π M w ( ) H n M- n Figure : Hamming window Figure 3: Hamming window Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 / 94

23 Blackman window FIR Filter Design Window Design Techniques This is a nd -order raised cosine window. The width of main lobe is: π M ( n) wt [ ( ) ( )] πn 4πn w(n) cos cos M M W (ω) 0.4W R (ω) [ W R (ω π M ) + W R(ω + π M )] [ W R (ω 4π M ) + W R(ω + 4π M )] M- n Figure 4: Blackman window Figure 5: Blackman window Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 3 / 94

24 Kaiser window FIR Filter Design Window Design Techniques This is one of the most useful and optimum windows. ) ( ) I 0 (β n M w(n) I 0 (β) Where I 0 (X ) is the modified zero-order Bessel function, and is a parameter that can be chosen to yield various transition widths and stop band attenuation. This window can provide different transition widths for the same N. w ( ) H n β 0 rectangularwindow β 5.44 Hammingwindow β 8.5 Blackmanwindow M- n Figure 6: Kaiser window Figure 7: Kaiser window Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 4 / 94

25 Window Design Techniques Table : Window and its functions Window name Window Function { for 0 n M Rectangular ω R (n) 0 otherwise M n Triangular ω T (n) M (Bartlet) [ ( )] Hamming w(n) cos πn [ ( N )] Hanning w(n) cos πn [ ( N ) ( )] Blackman w(n) cos πn cos 4πn N N Table : Summary of window function characteristics Window name Rectangular Hanning Hamming Bartlett Blackman Transition Min. stopband Peak value width of main attenuation of side lobe lobe 4π - db - db M+ 8π -44 db -3 db M 8π -53 db -4 db M 8π -5 db -5 db M π -74 db -57 db M Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 5 / 94

26 Window Design Techniques Gibbs Phenomenon The magnitude of the frequency response H(ω) is as shown in Figure. Large oscillations or ripples occur near the band edge of the filter. The oscillations increase in frequency as M increases, but they do not dimmish in amplitude. These large oscillations are due to the result of large sidelobes existing in the frequency characteristic W (ω) of the rectangular window. The truncation of the Fourier series is known to introduce ripples in the frequency response characteristic H(ω) due to the nonuniform convergence of the Fourier series at a discontinuity. The oscillatory behavior near the band edge of the filter is called the Gibbs Phenomenon. To alleviate the presence of large oscillations in both the passband and the stopband window function is used that contains a taper and decays toward zero gradually. Figure 8: LPF designed with rectangular window M6 and 0 Figure 9: LPF designed with Hamming, Hanning and Blackman window M6 Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 6 / 94

27 Low Pass FIR Filter Design Design a LPF using rectangular window for the desired frequency response of a low pass filter given by ω c π rad/sec, and take M. Find the values of h(n). Also plot the magnitude response. Solution: e jωτ ω c ω ω c H d (e jω ) 0 π ω ω c 0 ω c ω π τ M 5 { H d (e jω e jωτ ω ) c ω ω c 0 Otherwise π π 0 Hd ( e jω ) π π ω Figure 0: Frequency response of LPF By taking inverse Fourier transform h d (n) π H d (e jω )e jωn dω π π ωc e jωτ e jωn dω π ω c ωc e jω(n τ) dω π ω c h d (n) [ ] e jω(n τ) ωc π j(n τ) ω c [e jωc (n τ) jωc e (n τ)] jπ(n τ) π(n τ) [ e jωc (n τ) e j ] jωc (n τ) Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 7 / 94

28 Low Pass FIR Filter Design h d (n) sin [ωc(n τ)] π(n τ) for n 5 and ω c π, τ M 5 h d (n) sin [ωc(n 5)] π(n 5) [ ] π(n 5) sin π(n 5) for n5 h d (n) 0. Using L Hospital s Rule 0 sin Bθ lim B θ 0 θ sin π (n 5) lim π/ n 5 π(n 5) π 0.5 where π 3.46 h d (0) h d () sin π(0 5) π(0 5) sin π( 5) π( 5) 0 h d () h d (3) h d (4) h d (5) h d (6) h d (7) h d (8) h d (9) h d (0) sin π( 5) π( 5) 0.06 sin π(3 5) π(3 5) 0 sin π(4 5) π(4 5).38 sin π(5 5) π(5 5).5 sin π(6 5) π(6 5).38 sin π(7 5) π(7 5) 0.0 sin π(8 5) π(8 5).06 sin π(9 5) π(9 5) 0 sin π(0 5) π(0 5).063 Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 8 / 94

29 Low Pass FIR Filter Design The given window is rectangular window { for 0 n 0 ω(n) 0 Otherwise This is rectangular window of length M. h(n) h d (n)ω(n) h d (n) for 0 n 0 M H(z) h(n)z n n0 The impulse response is symmetric with Modd 0 h(n)z n n0 M 3 ( ) M H(z) h z M 3 + h(n)[z n + z (M n) ] n0 h(5)z 5 + h(0)[z 0 + z 0 ] + h()[z + z 9 ] + h()[z + z 8 ] + +h(3)[z 3 + z 7 ] + h(4)[z 4 + z 6 ] H(e jω ) M 3 ( ) M ( ) M h + h(n)cos ω n n0 4 h(5) + h(n)cos ω(5 n) n0 h(5) + h(0)cos 5ω + h()cos 4ω + h()cos 3ω + h(3)cos ω + h(4)cos ω cos 5ω 0.cos 3ω cos ω Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 9 / 94

30 Low Pass FIR Filter Design H(e jω ) cos 5ω 0.cos 3ω cos ω H(e jω ) db 0log H(e jω ) ω H(e jω ) H(e jω ) db π π π π π π π π π π H(e jw ) db ω, 0 to π in radians Figure : Frequency response of LPF Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

31 The desired frequency response of low pass filter is given by Low Pass FIR Filter Design { H d (e jω e j3ω 3π ) 4 ω 3π 4 3π 0 4 ω π Determine the frequency response of the FIR if Hamming window is used with N7 June-05, Dec-04, June-0 Solution: τ M 3 h d (n) π H d (e jω )e jωn dω π π ωc e jωτ e jωn dω π ω c ωc e jω(n τ) dω π ω c π [ e jω(n τ) j(n τ) π(n τ) ] ωc ω c [ e jωc (n τ) e j ] jωc (n τ) π 3π 4 0 h d (n) Hd ( e jω ) 3π 4 sinωc(n τ) π(n τ) ω π Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 3 / 94

32 Low Pass FIR Filter Design n 3 ω c 3π 4 τ M 3 [ ] 3π(n 3) sin 4 h d (n) π(n 3) for n3 h d (n) 0. Using L Hospital s Rule 0 lim n 3 sin [ 3π (n 3)] 4 3π/ π(n 3) π h d (0) h d () h d () h d (3) h d (4) h d (5) h d (6) ( ) 3π(0 3) sin π(0 3) ( ) 3π( 3) sin π( 3) ( ) 3π( 3) sin π( 3) ( ) 3π(3 3) sin π(3 3) ( ) 3π(4 3) sin π(4 3) ( ) 3π(5 3) sin π(5 3) ( ) 3π(6 3) sin π(6 3) Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 3 / 94

33 Low Pass FIR Filter Design The given window is Hamming window ( ) πn w(n) cos M ( ) 0 ω(0) cos ( π ω() cos 6 ( 4π ω() cos 6 ( 6π ω(3) cos 6 ( 8π ω(4) cos 6 ( 0π ω(5) cos 6 ( π ω(6) cos 6 ).3 ).77 ) ) 0.77 ).3 ).08 To calculate the value of h(n) h(n) h d (n)w(n) h(0) h d (0)w(0) h() h d ()w() h() h d ()w() h(3) h d (3)w(3) h(4) h d (4)w(4) h(5) h d (5)w(5) h(6) h d (6)w(6) Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

34 The frequency response is symmetric with Modd7 Low Pass FIR Filter Design H(e jω ) M 3 ( ) M ( ) M h + h(n)cos ω n n0 h(3) + h(n)cos ω(3 n) h(3) + h(0)cos 3ω + h()cos ω + h()cos ω n cos 3ω 0.098cos ω cos ω H(e jω ) db 0log H(e jω ) ω H(e jω ) H(e jω ) db π π π π π π π π π π H(e jw ) db ω, 0 to π in radians Figure : Frequency response of LPF Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

35 Low Pass FIR Filter Design Matlab code clc; clear all; close all; M input( enter the value of M: ); omega input( enter the value of omega: ); tau(m-)/ ; for n0:m-; % c(n+).5-.5*cos((*pi*n)/(m-)); c(n+) *cos((*pi*n)/(m-)); if ntau h(n+)omega/pi; else h(n+)sin(omega*(n-tau))/(pi*(n-tau)); end end h c for n:m yh(n)*c(n) end Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

36 Low Pass FIR Filter Design clc; clear all; close all; range0; %M input( enter the value of M: ); for omega0:.*pi:pi rangerange+; H_omegaabs( *cos(3*omega)-.098*cos(*omega)+.346*cos(omega)); %H_omegaabs( *cos(omega)+.38*cos(*omega)); H_indB(range)0*log0(H_omega) end omega0:.*pi:pi; i:range; plot(omega, H_indB(i), linewidth, ) xlabel( \omega, 0 to \pi in radians, fontsize,3) ylabel( H(e^{jw}) _{db}, fontsize,3) Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

37 Low Pass FIR Filter Design Determine the filter coefficients h d (n) for the desired frequency response of a low pass filter given by { H d (e jω e jω for π ) 4 ω π 4 0 for π ω π 4 If we define the new filter coefficients by h d (n) h d (n)ω(n) where { for 0 n 4 ω(n) 0 otherwise Determine h(n) and also the frequency response H(e jω ) July-03, July-0 Solution: h d (n) π π π π π π H d (e jω )e jωn dω π/4 e jω e jωn dω π/4 π/4 e jω(n ) dω π/4 [ ] π/4 e jω(n ) j(n ) π/4 h d (n) π π 4 0 Hd ( e jω ) π 4 π ω [e j π 4 (n ) e j π (n )] 4 jπ(n ) [ ] e j π 4 (n ) e j π 4 (n ) π(n ) j Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

38 Low Pass FIR Filter Design n h d (n) sin π(n ) 4 π(n ) for n h d (n) 0. Using L Hospital s Rule 0 sin π (n ) 4 lim π/4 n π(n ) π 0.5 The given window function is { for 0 n 4 ω(n) 0 Otherwise This is rectangular window of length M5. In this case h(n) h d (n) for 0 n 4 n h d (n) n h d (n) Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

39 The frequency response is symmetric with Modd5 Low Pass FIR Filter Design H(e jω ) M 3 ( ) M ( ) M h + h(n)cos ω n n0 h() + h ( n) cos ωn h() + h(0)cos ω + h()cos ω n cos ω cos ω H(e jω ) db 0log H(e jω ) ω H(e jω ) H(e jω ) db π π π π π π π π π π H(e jw ) db ω, 0 to π in radians Figure 3: Frequency response of LPF Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

40 Low Pass FIR Filter Design Design the symmetric FIR lowpass filter whose desired frequency response is given as { e jωτ for ω ω H d (ω) c 0 Otherwise The length of the filter should be 7 and ω c radians/sample. Use rectangular window. Solution: Desired frequency response H d (ω) Length of the filter M7 Cut-off frequency ω c radians/sample. Unit sample response is defined as h d (n) π H d (e jω )e jωn dω π π Given H d (ω) is { e ωτ for ω H d (ω) 0 Otherwise π 0 Hd ( e jω ) Figure 4: Frequency response of LPF π ω Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

41 Low Pass FIR Filter Design h d (n) for n τ h d (n) 0 0. Rule e ωτ e jωn dω π e jω(n τ) dω π [ ] e ω(n τ) π j(n τ) sin(n τ) for n τ π(n τ) sin(n τ) lim n τ π(n τ) π Using L Hospital s Thus h d (n) is { sin(n τ) for n τ h d (n) π(n τ) for n τ π Determine the value of τ τ M 3 { sin(n 3) for n τ h d (n) π(n 3) for n τ π This is rectangular window of length M7. In this case h(n) h d (n).w(n) h d (n) n h(n) n h(n) This is the unit sample response of required FIR filter. The filter is symmetric and satisfies h(n) h(m n) Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 4 / 94

42 Design the FIR filter using Hanning window Solution:For M7 Low Pass FIR Filter Design ω(n) 0.5( cos πn M ) ω(n) 0.5( cos πn 6 ) To calculate the value of h(n) h(n) h d (n)w(n) ω(0) 0.0 ω() 0.5( cos π 6 ).5 ω() 0.5( cos 4π 6 ).75 ω(3) 0.5( cos 6π 6 ) ω(4) 0.5( cos 8π 6 ).75 ω(5) 0.5( cos 0π 6 ).5 ω(6) 0.5( cos π 6 ) 0 h(0) h d (0)w(0) h() h d ()w() h() h d ()w() h(3) h d (3)w(3) h(4) h d (4)w(4) h(5) h d (5)w(5) h(6) h d (6)w(6) Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 4 / 94

43 Low Pass FIR Filter Design Design a lowpass digital filter to be used in an A/D Hz D/A structure that will have a -3 db cut-off at 30 π rad/sec and an attenuation of 50 db at 45 π rad/sec. The filter is required to have linear phase and the system will use sampling rate of 00 samples/second. Solution: 3 db cut-off at 30 π rad/sec ω c 30πrad/sec Sampling frequency F SF 00 Hz Stopband attenuation of 50 db at 45 π rad/sec A s 50 db for ω s 45πrad/sec ω Ω F sf Hd ( e jω ) ω Ω F sf ω Ω F sf 30π 0.3π rad/sample 00 45π 0.45π rad/sample 00 π 0.3π 0 0.3π π ω Figure 5: Frequency response of LPF 3 db attenuation at ω 0.3π rad/sample 50 db attenuation at ω 0.45π rad/sample Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

44 Low Pass FIR Filter Design Type of window is The stopband attenuation of 50 db is provided by the Hamming window which of -53 db. Hence Hamming window is selected for the given specifications. To determine the order of the filter The width of the main lobe in Hamming window is 8π M k π M 8π M M 8π ω ω The order of the filter M is: M 8π 0.45π 0.3π Assume linear phase FIR filter of odd length Hence select next odd integer length of 55. { e jωτ for ω H d (ω) c ω ω c 0 Otherwise τ (M )/ 55 / 7 ω c 0.3π h d (n) when n 7 ωc e jωτ e jωn dω π ω c 0.3π e jω(n 7) dω π 0.3π π [ e ω(n 7) j(n 7) sin[ωc(n 7)] π(n 7) h d (n) ωc π 0.3 ] 0.3π 0.3π for n 7 Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

45 Low Pass FIR Filter Design The selected window is Hamming M7 ( ) πn w(n) cos M ( πn ) cos 8 The value of h(n) for M 7 h(n) for n 7 h(n) h d (n)w(n) sin[0.3π(n 7)] π(n 7) [ cos [ h(n) cos ( πn )] 8 ( πn )] 8 n h(n) n h(n) Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

46 Low Pass FIR Filter Design An analog signal contains frequencies upto 0 KHz. The signal is sampled at 50 KHz. Design an FIR filter having linear phase characteristic and transition band of 5 KHz. The filter should provide minimum 50 db attenuation at the end of transition band. Solution: 3 db cut-off at 30 π rad/sec Ω p π rad/sec Ω s π (0 + 5) 0 3 rad/sec Sampling frequency F SF 00 Hz Stopband attenuation of 50 db at 45 π rad/sec A s 50 db for ω s 45πrad/sec ω Ω F sf Hd ( e jω ) ω p Ωp π 0 03 F sf π ω s Ωs π (0 + 5) 03 F sf π π 0.5π 0 0.5π π ω Figure 6: Frequency response of LPF ω p 0.4π rad/sample ω s 0.6π rad/sample Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

47 Low Pass FIR Filter Design Type of window is The stopband attenuation of 50 db is provided by the Hamming window which of -53 db. Hence Hamming window is selected for the given specifications. To determine the order of the filter The width of the main lobe in Hamming window is 8π M k π M 8π M M 8π ω s ω p The order of the filter M is: M 8π 0.6π 0.4π 40 Assume linear phase FIR filter of odd length Hence select next odd integer length of 4. { e jωτ for ω H d (ω) c ω ω c 0 Otherwise τ (M )/ 4 / 0 ω c ω p + ω 0.4π + 0.π ω c 0.5π h d (n) when n 0 ωc e jωτ e jωn dω π ω c 0.5π e jω(n 0) dω π 0.5π π [ e ω(n 7) j(n 0) sin[ωc(n 0)] π(n 0) ] 0.5π 0.5π h d (n) ωc π 0.5π π 0.5 for n 0 Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

48 Low Pass FIR Filter Design The selected window is Hamming M4 ( ) πn w(n) cos M ( ) πn cos 40 The value of h(n) for M 4 n 0 h(n) for n 0 sin[0.5π(n 0)] π(n 0) h(n) h d (n)w(n) [ cos [ h(n) cos ( )] πn 0 ( )] πn 0 n h(n) n h(n) Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

49 Low Pass FIR Filter Design Design an FIR filter (lowpass) using rectangular window with passband gain of 0 db, cutoff frequency of 00 Hz, sampling frequency of khz. Assume the length of the impulse response as 7. Solution: F c 00 Hz, F s 000 Hz, f c Fc 00 F s cycles/sample ω c π f c π πrad M7 Hd ( e jω ) { e jωτ for ω H d (ω) c ω ω c 0 Otherwise τ (M )/ 7 / 3 ω c 0.4π h d (n) ωc e jωτ e jωn dω π ω c 0.4π π π e jω(n 3) dω 0.4π [ ] 0.4π e ω(n 3) j(n 3) 0.4π π 0.4π 0 0.4π π ω Figure 7: Frequency response of LPF when n 3 h d (n) when n 3 sin[0.4π(n 3)] π(n 3) h d (n) 0.4π π 0.4 Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

50 Low Pass FIR Filter Design Determine the value of h(n) Since it is rectangular window h(n) w(n) h d (n) h(n) For M7 n h(n) n h(n) Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

51 Low Pass FIR Filter Design Using rectangular window design a lowpass filter with passband gain of unity, cutoff frequency of 000 Hz, sampling frequency of 5 khz. The length of the impulse response should be 7. DEC:03,DEC:0 Solution: F c 000 Hz, F s 5000 Hz, f c Fc 000 F s cycles/sample ω c πf c π πrad M7 The filter specifications (ω c and M7) are similar to the previous example. Hence same filter coefficients are obtained. h(0) , h()0.0935, h() h(3)0.4, h(4) , h(5)0.0935, h(6) Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 5 / 94

52 Low Pass FIR Filter Design Design a normalized linear phase FIR low pass filter having phase delay of τ 4 and at least 40 db attenuation in the stopband. Also obtain the magnitude/frequency response of the filter. Solution: The linear phase FIR filter is normalized means its cut-off frequency is of ω c rad/sample The length of the filter with given τ is related by τ M For τ 4 M9 Desired unit sample response h d (n) is π 0 Hd ( e jω ) Figure 8: Frequency response of LPF π ω h d (n) ωc e jωτ e jωn dω π ω c π π e jω(n 4) dω [ ] e ω(n 4) j(n 4) when n 4 h d (n) when n 4 sin[(n 4)] π(n 4) h d (n) dω ω π π π Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 5 / 94

53 Low Pass FIR Filter Design ( ) πn w(n) cos M for n 4 and M 9 h(n) for n 4 h(n) π sin(n 4) π(n 4) [ cos [ cos ( π4 4 ( πn )] 4 )] π for n 0 to 8 h(n) h d (n)w(n) The stopband attenuation required for h(0) this filter is 40 db. From the table the h() Hanning window satisfies this requirement. The Hanning window function given by: h() h(3) h(4) h(5) h(6) h(7) h(8) sin(0 4) π(0 4) sin( 4) π( 4) sin( 4) π( 4) sin(3 4) π(3 4) sin(4 4) π(4 4) sin(5 4) π(5 4) sin(6 4) π(6 4) sin(7 4) π(7 4) sin(8 4) π(8 4) [ cos [ cos [ cos [ cos [ cos [ cos [ cos [ cos [ cos ( π0 4 ( π 4 ( π 4 ( 3π 4 ( 4π 4 ( 5π 4 ( 6π 4 ( 7π 4 ( 8π 4 )] )] 0.00 )] )] 0.86 )] )] 0.86 )] )] 0.00 )] Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

54 The frequency response is symmetric with Modd9 Low Pass FIR Filter Design H(e jω ) M 3 ( ) M ( ) M h + h n cos ω n0 3 h(4) + h(n)cos ω (4 n) H(e jω ) db 0log H(e jω ) n0 h(4) + h(0)cos 4ω + h()cos 3ω + h()cos ω + h(3)cos ω cos 3ω cos ω cos ω ω H(e jω ) H(e jω ) db π π π π π π π π π π H(e jw ) db ω, 0 to π in radians Figure 9: Frequency response of LPF Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

55 High Pass FIR Filter Design Design a HPF using Hamming window. Given that cutoff frequency the filter coefficients h d (n) for the desired frequency response of a low pass filter given by ω c rad/sec, and take M7. Also plot the magnitude response. Solution: e jωτ π ω ω c H d (e jω ) e jωτ ω c ω π 0 ω c ω ω c τ M 3 h d (n) { H d (e jω e jωτ π ω ω ) c 0 Otherwise π π π π [ π ωc H d (e jω )e jωn dω π e jω(n τ) dω + π [ ] e jω(n τ) ωc j(n τ) π(n τ) + π e jω(n τ) dω ω c [ ] π e jω(n τ) j(n τ) π ω c [ e jωc (n τ) e jπ(n τ) + e jπ(n τ) e j ] jωc (n τ) Hd ( e jω ) ω c π ω c 0 π ω Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

56 High Pass FIR Filter Design h d (n) π(n τ) [ e jπ(n τ) e jπ(n τ) [ ] e jωc (n τ) jωc e (n τ)] [sinπ(n τ) sinωc(n τ] π(n τ) τ 3 ω c h d (n) [sinπ(n 3) sin(n 3)] π(n 3) when n τ using L Hospital rule h d (n) [ ] sinπ(n 3) sinωc(n 3) π (n 3) (n 3) π [π ωc] [π ] π The given window function is Hamming window. In this case h(n) h d (n)ω(n)) for 0 n 6 ( ) πn w(n) cos M [ h(n) cos [sinπ(n 3) sin(n 3)] π(n 3) j n h(n) n h(n) ( )] πn M Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

57 High Pass FIR Filter Design The magnitude response of a symmetric FIR filter with Modd is For M7 ( M H(e jω ) h (M 3) ) + n0 ( ) M h(n)cosω n H(e jω ) h (3) + h(n)cosω(3 n) n0 h (3) + h(0)cos3ω + h()cosω + h()cosω cos3ω cosω 0.44cosω Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

58 Band Pass FIR Filter Design Design the bandpass linear phase FIR filter having cut off frequencies of ω c rad/sample and ω c rad/sample. Obtain the unit sample response through following window. { for 0 n 6 ω(n) 0 Otherwise Solution: H d (ω) { e jωτ ω c ω c ω c 0 Otherwise h d (n) π π π π π π H d (ω)e jωn dω π 0 Hd ( e jω ) ω c ω c ω c ω c [ ωc ωc ] ω c [ ωc ωc ω c ] e jωτ e jωn dω + e jωτ e jωn dω ω c ω c [ ] e jω(n τ) ωc [ ] e jω(n τ) ωc e jω(n τ) + e jω(n τ) dω (n τ) (n τ) ω c + sinωc (n τ) sinωc (n τ) π(n τ) ω c for n τ Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94 π ω

59 Band Pass FIR Filter Design h d (n) sinωc (n τ) sinωc (n τ) π(n τ) ωc ωc π for n τ h d (n) { sinωc (n τ) sinω c (n τ) ω c ω c π π(n τ) for n τ for n τ The linear phase FIR filter is normalized means its cut-off frequency is of ω c rad/sample The length of the filter with given τ is related by τ M 7 3 n h(n) n h(n) and ω c rad/sample ω c rad/sample { sin(n 3) sin(n 3) for n 3 h d (n) π(n τ) for n 3 for π The given window is rectangular hence h(n) h d (n)w(n) h d (n) Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

60 Band Pass FIR Filter Design For n0,,..6 estimate the FIR filter coefficients h(n). For M7 The magnitude response of the FIR filter is given by ( M H(ω) h M 3 ) + n0 ( ) M h(n)cos ω n H(ω) h(3) + h(n)cos ω (n 3) n0 Estimate the H(ω) by substituting the required values in the above equation. Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

61 Design an ideal bandpass filter having frequency response Band Pass FIR Filter Design H d e (jω) for π 4 ω 3π 4 Use rectangular window with N in your design Solution: H d (ω) { e jωτ ω c ω c ω c 0 Otherwise h d (n) π π π π π π H d (ω)e jωn dω π 0 Hd ( e jω ) ω c ω c ω c ω c [ ωc ωc ] ω c [ ωc ωc ω c ] e jωτ e jωn dω + e jωτ e jωn dω ω c ω c [ ] e jω(n τ) ωc [ ] e jω(n τ) ωc e jω(n τ) + e jω(n τ) dω (n τ) (n τ) ω c + sinωc (n τ) sinωc (n τ) π(n τ) ω c for n τ Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 6 / 94 π ω

62 Band Pass FIR Filter Design h d (n) sinωc (n τ) sinωc (n τ) π(n τ) ωc ωc π for n τ { sinωc (n τ) sinω c (n τ) for n τ h d (n) π(n τ) ω c ω c for n τ for π The length of the filter with given τ is related by and ω c π 4 τ M rad/sample ωc 3π 4 rad/sample The given window is rectangular hence 5 [ ] [ ] 3π(n 5) π(n 5) sin sin 4 4 for n 5 h d (n) π(n 5) 3π 4 π 4 for n 5 for π h(n) h d (n)w(n) h d (n) For n0,,..0 estimate the FIR filter coefficients h(n). Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 6 / 94

63 Band Pass FIR Filter Design Design a BPF using Hanning window with M7. Given that lower cutoff frequency ω c rad/sec and ω c 3rad/sec. Solution: e jωτ for ω c ω ω c H d (e jω ) e jωτ for ω c ω ω c 0 for ω c ω c ω ω c τ M 3 The inverse transform of the H d (e ω ) is Hd ( e jω ) h d (n) π π π π [ π ωc H d (e jω )e jωn dω e jω(n τ) dω + ω c [ ] e jω(n τ) ωc j(n τ) π ω c + ωc ω c [ e jω(n τ) dω e jω(n τ) j(n τ) ] ωc π(n τ) [sinω c(n τ) sinω c (n τ] ω c π ω c ω c ω c ω c 0 π ω Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

64 Band Pass FIR Filter Design τ M 7 3 ω c rad/sec ω c 3 rad/sec for n 3 for n τ h d (n) h d (n) π [sin3(n 3) sin(n 3)] π(n 3) [ ] sinω c (n τ) sinω c (n τ) lim lim n τ (n τ) n τ (n τ) n h(n) n h(n) h d (n) π [ω c ω c ] π The given window function is Hanning window ω(n) cos πn 0 n M M This is rectangular window of length M. In this case h(n) h d (n)ω(n) h d (n) for 0 n 6 [ ] [ sin3(n τ) sin(n τ) h(n) cos πn ] π(n 3) M Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

65 Band Pass FIR Filter Design H(z) M n0 h(n)z n 6 h(n)z n n0 h(0) + h()z + h()z + h(3)z 3 + h(4)z 4 + h(5)z 5 + h(6)z z 0.843z z z z The magnitude response of a symmetric FIR filter with Modd is For M7 ( M H(e jω ) h H(e jω ) h (3) + H(e jω ) h (3) + ) (M )/ + n ( ) M h n cosωn 3 h (3 n) cosωn n 3 h (3 n) cosωn n 5 h(3) + h (5 n) cosωn n h(3) + h()cosω + h()cosω + h(0)cos3ω cosω cosω Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

66 Bandstop FIR Filter Design Design a bandstop filter to reject the frequencies from to 3 rad/sec using rectangular window with M5. Find the frequency response. Solution: e jωτ for π ω ω c H d (e jω e ) jωτ for ω c ω ω c e jωτ for ω c ω π 0 for ω c ω ω c Hd ( e jω ) π 0 ω ω c ω c ω c ω c π h d (n) π π π π [ π H d (e jω )e jωn dω ωc ωc π e jω(n τ) dω + e jω(n τ) dω + e jω(n τ) dω π ω c ω c [ ] e jω(n τ) ωc [ ] + e jω(n τ) ωc [ ] π e jω(n τ) j(n τ) π j(n τ) π j(n τ) π π(n τ) [sinω c(n τ) + sinπ(n τ sinω c (n τ] ω c + ω c Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

67 Bandstop FIR Filter Design The inverse transform of the H d (e ω ) is τ M 5 ω c rad/sec ω c 3 rad/sec h d (n) [sin(n ) + sinπ(n ) sin3(n )] for n π(n ) for n τ h d (n) π [ ] sinω c (n τ) sinπ(n τ) sinω c (n τ) lim + lim lim n τ (n τ) n τ (n τ) n τ (n τ) h d (n) π [ω c + π ω c ] [π ] π The given window function is Rectangular window ω(n) 0 n M This is rectangular window of length M5. In this case h(n) h d (n)ω(n) h d (n) for 0 n 4 h d (n) [sin(n ) + sinπ(n ) sin3(n )] for n π(n ) Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

68 Bandstop FIR Filter Design n 0 h(0) n h() [ sin(0 ) + sinπ(0 ) sin3(0 ) π(0 ) [ sin( ) + sinπ( ) sin3( ) π( ) ] ] n h() [π ] π [ sin(3 ) + sinπ(3 ) sin3(3 ) n 3 h(3) π(3 ) [ sin(4 ) + sinπ(4 ) sin3(4 ) n 4 h(4) π(4 ) ] ] H(z) M n0 h(n)z n 4 h(n)z n n0 h(0) + h()z + h()z + h(3)z 3 + h(4)z z z z z 4 Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

69 Bandstop FIR Filter Design The magnitude response of a symmetric FIR filter with Modd is For M5 ( M H(e jω ) h H(e jω ) h () + ) (M )/ + n ( ) M h n cosωn h ( n) cosωn n 3 H(e jω ) h (3) + h (3 n) cosωn n h() + h()cosω + h(0)cosω (0.445)cosω + h( )cosω cosω 0.58cosω Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

70 FIR Filter Design Using Kaiser Window FIR Filter Design Using Kaiser Window The Kaiser window is parametric and its bandwidth as well as its sidelobe energy can be designed. Mainlobe bandwidth controls the transition characteristics and sidelobe energy affects the ripple characteristics. The Kaiser window function is given by ] ( ) I 0 [α n M w k (n) I 0 (α) where M is the order of the filter, I 0 (x) is a zeroth Bessel function of the first kind [ ( x ) ] k I 0 (x) + k! k + 0.5x (!) + (0.5x ) (!) + (0.5x ) 3 (3!) + α 0 if A < 0.584(A ) (A ) if A 50 db 0.0(A 8.7)) if A > 50 db Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

71 FIR Filter Design Using Kaiser Window + δ δ H ( ω) A P δ δ Passband ripple Stopband ripple Ideal LPF A S δ Gain Passband ω P ω C Transition band ω S Stopband ω Figure 30: Frequency response of LPF Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 7 / 94

72 Kaiser Window Design Equations FIR Filter Design Determine ideal frequency response FIR Filter Design Using Kaiser Window { H d (e jω for ω ωc ) 0 for ω c ω π where ω c (ωp + ωs) Chose δ such that the actual passband ripple, A p is equal to or less than the specified passband ripple Ãp, and the actual minimum stopband attenuation A is equal or greater than the specified minimum stop attenuation Ãs δ min(δ p, δ s) where δ p 00.05Ãp and Ãp δs à s + 3 The actual stopband attenuation is 4 The parameter α is α A 0log 0 δ 0 for A 0.584(A ) (A a ) for < A (A 8.7) for A > 50 Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 7 / 94

73 FIR Filter Design Using Kaiser Window 5 The value of M is found by M A f where f ω ωs ωp and ω is the width of transition band π π 6 Obtain impulse response by multiplying Kaiser window function 7 Obtain the causal finite impulse response 8 The system function is given by h(n) h d (n)w k (n) M H(z) h(n)z n n0 Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

74 FIR Filter Design Using Kaiser Window Design a lowpass filter with a cutoff frequencies from w c π ω 0.0π and a stopband 4 ripple δ s 0.0. Use Kaiser window Solution: { H d (e jω for ω ωc ) 0 for ω c ω π A 0logδ s 0log(0.0) 40 db The inverse transform of the H d (e jω ) is h d (n) π H d (e jω )e jωn dω π π ωc e jωn dω π ω c [ e jωn ] ωc π jn ω c [e jωc n jωc e n] jπn [ e jω c n e jωc n ] πn j sinωcn πn α 0.584(A ) (A ) 0.584(40 ) (40 ) 3.4 f 0.0π π 0.0 M A f τ Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

75 FIR Filter Design Using Kaiser Window where w k (n) w k (n) I 0 [α I 0 (α) ] ( ) n M I 0 [3.4 ( ] ) n 4 I 0 (3.4) 0 n M 0 n M h(n) h d w k (n) I 0 [3.4 ( ] ) n πn [sinωcn] 4 I 0 (3.4) ω c ω c + ω 0.5π + 0.0π 0.6π Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

76 FIR Filter Design Using Kaiser Window Find an expression for the impulse response h(n) of a linear phase lowpass FIR filter using Kaiser window to satisfy the following magnitude response specifications for the equivalent analog filter. Stopband attenuation: 40 db Passband ripple: 0.0 db Transition width: 000 π rad/sec Ideal cutoff frequency: 400 π rad/sec Sampling frequency: 0 KHz Solution: x() t x( n ) yn ( ) yt () A/D Filter D/A A 0logδ s 40 db logδ s δ s 0.0 0log( + δ p) 0.0 log( + δ p) δ p δmin(δ p, δ s) A 0log(0.005) 58.8 db ω Ω 000π 0.π rad f s 0 03 M A f f 0.π π τ 7 35 Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

77 FIR Filter Design Using Kaiser Window The attenuation is 58.8 db then the parameter α is α 0.0(A 8.7) 0.0( ) 5.5 w k (n) I 0 [5.5 ( ] ) n 70 I 0 (5.5) ω c Ω c T 400π 0.4π ω c ω c + ω 0.4π + 0.π 0.9π h(n) h d w k (n) { H d (e jω for ω ωc ) 0 for ω c ω π The inverse transform of the H d (e jω ) is h d (n) π H d (e jω )e jωn dω π π ωc e jωn dω π ω c [ e jωn ] ωc π I 0 [5.48 [sinωc(n τ)] π(n τ) jn ω c [e jωc n jωc e n] jπn [ e jω c n e jωc n ] πn j sinωcn πn ( n 4 I 0 (5.5) Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94 ) ]

78 FIR Filter Design Using Kaiser Window Find an expression for the impulse response h(n) of a linear phase Design a lowpass FIR filter satisfying the following specifications using Kaiser window Solution: α p 0. db α s 44 db ω p 0 rad/sec ω s 30 rad/sec ω sf 00 rad/sec H d (e jω ) { for ω ωc 0 for ω c ω π The inverse transform of the H d (e jω ) is h d (n) π H d (e jω )e jωn dω π π ωc e jωn dω π ω c [ e jωn ] ωc π jπn sinωcn πn jn ω c [e jωc n jωc e n] ω ω s ω p 0rad/sec ω c (ωp + ωs) 5rad/sec ω c(in discrete and radian) 5 00 (π) π rad δ s As δ p 00.05Ap Ap δ min(δ p, δ s) A 0log 0 (δ) dB Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

79 FIR Filter Design Using Kaiser Window α 0.584(A ) (A ) 0.584( ) ( ) M A f f ω ω sf τ 7 3 ] ( ) I 0 [α n M w k (n) I 0 (α) 0 n M w k (n) I 0 [3.954 ( ] ) n 7 I 0 (3.954) 0 n M h(n) h d w k (n) I 0 [3.954 ( ] ) n πn [sinωcn] 7 I 0 (3.954) Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

80 FIR Filter Design Using Kaiser Window Design a high pass digital satisfying the following specifications using Kaiser window Solution: Passband cut-off frequency f p 300Hz Passband ripple α p 0.dB Sampling frequency F 0000Hz H d (e jω ) { 0 for ω ωc for ω c ω π The inverse transform of the H d (e jω ) is h d (n) [ ωc π π [ ( e jωn π Stopband cut-off frequency f s 600Hz Stopband ripple α s 40dB π ] e jωn dω + e jωn dω ω c ) ωc ( e jωn ) π ] + jn π jn ω c [e jωc n e jπn + e jπn jωc e n] jπn sinπn sinωcn πn ω p πf p 6400π rad/sec ω s πf s 300π rad/sec ω sf πf 0000π rad/sec ω ω p ω s 300π rad/sec ω c (ωp + ωs) 4800π rad/sec ω c(discrete, radian) 4800 (π) 0.48πrad 0000 δ s As δ p 00.05Ap Ap δ min(δ p, δ s) A 0log 0 (δ) dB Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

81 FIR Filter Design Using Kaiser Window α 0.584(A ) (A ) 0.584( ) ( ) f ω ω sf M A f h(n) h d w k (n) 300π 0000π sinπn sinωcn πn τ 7 8 ] ( ) I 0 [α n M w k (n) I 0 (α) 0 n M w k (n) I 0 [3.954 ( ] ) n 7 I 0 (3.954) 0 n M I 0 [3.954 ( n 7 I 0 (3.954) Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 8 / 94 ) ]

82 Design of FIR system Frequency Sampling for FIR Filters Design of FIR filter using Frequency Sampling With necessary mathematical analysis explain the frequency sampling technique of FIR filter design Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, 06 8 / 94

83 Design of FIR system Frequency Sampling for FIR Filters In this method a set of M equally spaced samples in the interval (0, π)are taken in the desired frequency response H d (ω). The continuous frequency ω is replaced by ω ω k π k k 0,,... M M The discrete time Fourier transform (DTFT) is H(k) H d (ω) ωωk ( ) π H d M k k 0,,... M The inverse M point DFT (IDFT) h(n) is h(n) M M M n0 M n0 H(k)e jωn H(k)e j πkn M n 0,,... M H d ( ω) π k π k ω k M 7 K θ ( ω) 4π 8π π 7 7 6π k θk 8ωk 7 π 0π 3π 30π π ( k 7) θk 7 ω π Magnitude frequency response is symmetric about π, while ideal phase response is antisymmetric about π Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94 ω

84 Design of FIR system Frequency Sampling for FIR Filters For the FIR filter to be realizable the coefficients h(n) must be real. This is possible if all complex terms appear in complex conjugate pairs. Consider the term H(M k)e jπn(m k)/m H(M k)e jπn(m k)/m H(M k)e jπn e jπkn/m H(M k)e jπn(m k)/m H(M k)e jπkn/m e jπn cos(πn) + jsin(πn) substituting the H(M k) H(k) H(M k)e jπn(m k)/m H(k)e jπkn/m The term H(k)e jπkn/m is complex conjugate of H(k)e jπkn/m. Hence H(M k)e jπn(m k)/m is complex conjugate of H(k)e jπkn/m H(M k) H (k) Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

85 Design of FIR system Frequency Sampling for FIR Filters If H(M k) H (k) then h(n) h(n) M ( H(0) + P k [ Re H(k)e jπkn/m]) where P is { M if M is odd P M if M is even This equation is used to compute the coefficients of FIR filter. H(z) is M H(z) h(n)z n n0 M H(ω) h(n)e jωn n0 Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

86 Design of FIR system Frequency Sampling for FIR Filters Design a lowpass FIR filter using frequency sampling technique having cut-off frequency of π/ rad/sample. The filter should have linear phase and length of 7. Solution: The Ideal LPF frequency response H d (ω) for the linear phase is H d (ω) { ( ) e jω M 0 0 ω π π ω π H d ( ω) π k π k ω k M 7 K { e j8ω 0 ω π H d (ω) π 0 ω π To sample put ω πk M H d (ω) H d (ω) { { j πk e 0 πk πk 7 π π πk 7 π j 6πk e k k 7 θ ( ω) 4π 8π π 7 7 6π k θk 8ωk 7 The range of k is πk 7 π k πk 7 π k 7 8 π 0π 3π 30π π ( k 7) θk 7 ω π ω Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

87 Design of FIR system Frequency Sampling for FIR Filters The range of k is 0 k 7 4 k is an integer. Hence the range is 0 k 4 Similarly 7 4 k k 8.5 The range 5 k 8 H(k) 0 k k 8 3 k 6 The value of h(n)is given by h(n) M H(0) + ( + 7 H(k) 0 k 4 h(n) ( + 7 ( k 4 k 4 M Re [H(k)e jπkn/m] k [ Re H(k)e jπkn/7]) j 6πk Re [e 7 e jπkn/7]) [ Re e jπk(n 8)/7]) k ( 4 [ ] ) πk(n 8) + cos 7 k Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94

88 Design of FIR system Frequency Sampling for FIR Filters Determine the impulse response h(n) of a filter having desired frequency response H d (w) { ( ) e j (M )ω M7 use frequency sampling approach. Solution: The Ideal LPF frequency response H d (ω) is H d (ω) { ( ) e jω M ω π π ω π { e j3ω 0 ω π H d (ω) π 0 ω π To sample put ω πk M H d (ω) { H d (ω) { j πk e 0 πk πk π 7 π πk π 7 j 6πk e k k 7 0 ω π π ω π π k π k H d ( ω) ω k M 7 K θ ( ω) π π 4π 6π π 8π 0π π k θk 3ωk 7 The range of k is πk 7 π k 7 4 πk 7 π k 7 3 3π π 7 6 π ( k 7) θk 7 ω π Dr. Manjunatha. P (JNNCE) UNIT - 7: FIR Filter Design October 5, / 94 ω