# FIR. [ ] = b k. # [ ]x[ n " k] [ ] = h k. x[ n] = Ae j" e j# ˆ n Complex exponential input. [ ]Ae j" e j ˆ. ˆ )Ae j# e j ˆ. y n. y n.
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- Basil Shepherd
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1 [ ] = h k M [ ] = b k x[ n " k] FIR k= M [ ]x[ n " k] convolution k= x[ n] = Ae j" e j ˆ n Complex exponential input [ ] = h k M % k= [ ]Ae j" e j ˆ % M = ' h[ k]e " j ˆ & k= k = H (" ˆ )Ae j e j ˆ ( ) nk ( * Ae j+ e j ˆ ) n let H " ˆ " n M ( ) = h[k]e j ˆ " k H (" ˆ ) frequency response
2 Ex. y[ n] = x n 3 [ ] + x n " 3 [ ] + x n " 3 [ ] h[ n] = " n 3 [ ] + " n 3 [ ] + " n 3 [ ] FIR H (" ˆ ) = h[k]e j ˆ " k = h[]e j ˆ " + h[]e j ˆ " + h[]e j ˆ " = e j ˆ " + 3 e j ˆ " = ( 3 + e j " ˆ + e j " ˆ ) = 3 e j ˆ " = 3 e j ˆ " ( e j" ˆ ++ e j " ˆ ) ( + cos" ˆ )
3 H (" ˆ ) = 3 e j ˆ " ( + cos" ˆ ) ( )( cos" ˆ j sin" ˆ ) = 3 + cos ˆ " H (" ˆ ) = 3 + cos" ˆ Note: "H ( ˆ ) = ˆ ( ) H ( " 3 ) = ' ˆ % 3 & ˆ < = ( ) ˆ + % % & ˆ < % 3 "H ˆ ( ) = "H ( ˆ ) linear phase principal value of phase fn H( " ˆ ). ( ) "H ˆ linear phase low pass filter ˆ " zero
4 Normalized vs. actual frequency H (" ˆ ) = 3 + cos" ˆ Note: ( ) H ( " 3 ) = ˆ f = f f s ˆ " = 3 f =? " ˆ = ˆ f = f f s " f = ˆ f s H( " ˆ ) normalized frequency actual frequency (Hz) linear phase low pass filter " < ˆ < ˆ " f " ˆ = = f s = f s ˆ " ˆ " = 3, f s = 8Hz f = " 3 8 " = 8 3 zero Nyquist Hz =667Hz
5 Convolution / solving the difference equation x[ n] = 3+ 3cos( "n) input y[ n] = x n 3 [ ] + x n " 3 [ ] + x n " 3 [ ] FIR filter sample domain y[ n] =+ cos( "n) ++ cos " ( n ) = 3+ cos "n ( ) = cos "n " ( ) ( ) ( ) ++ cos( " n ) + cos " ( )cos "n ( ) + sin " ( )sin "n ( ) + cos." ( )cos "n ( ) + sin." ( )sin "n ( ) M
6 Multiplication in frequency domain [ ] = 3+ 3cos( "n) x n H( " ˆ ) closer " ˆ to a zero, the smaller the output. "H( ˆ ) H (" ˆ ) = 3 + cos" ˆ ( ) " ˆ H ( ) = H ( " ) =.7 "H ( ˆ ) = ˆ "H ( ) = "H ( ) = y[n] = 3 ( ) ( )cos "n " ( ) = 3+.38cos "n " ( )
7 ! x[ n] h[n]= [/3,/3,/3] h[n]= [/3,-/3,/3] y[ n]! h[n]=[/3,/3,/3] lowpass h[n]=[/3,-/3,/3]! highpass.9.9 H (" ˆ ).7.5 H (" ˆ ) ! ˆ "! ˆ "! x[ n] h[n]* h[n]! y[ n]!! H (" ˆ )H (" ˆ ) bandstop!! " ˆ h[n]* h[n]=[/9,,/9,,/9] H " ˆ " " ( ) = H ( ˆ ) H ( ˆ ) sample domain convolution = freq domain multiplication (and visa versa)
8 [ ] Low pass High pass y[ n] x n Ideal!c=.8 Ideal!c=.6 lowpass highpass H (" ˆ ) H (" ˆ ) ˆ " ˆ " [ ] h[n]* h[n] y[ n] x n null.9 H (" ˆ )H (" ˆ ) passbands don t overlap. " ˆ
9 x n Ideal [ ] Low pass High pass y[ n]!c=..6 Ideal!c= lowpass highpass H (" ˆ ) H (" ˆ ) ˆ " ˆ " [ ] Ideal Bandpass y[ n] x n h[n]* h[n]!c= ->.6 bandpass H (" ˆ )H (" ˆ ) passbands overlap. " ˆ
10 x[ n] lowpass Ideal Low pass!c=.8 Ideal High pass!c=.6 y[ n] highpass H (" ˆ ) H (" ˆ ) ˆ " ˆ " [ ] Ideal Bandstop y[ n] x n h[n]+ h[n]!c= ->.6 bandstop H (" ˆ ) + H (" ˆ ) passbands don t overlap. " ˆ
11 x[ n] Ideal All - pass "[n] Ideal Low pass!c=.8 - y[ n] all-pass lowpass H (" ˆ ) H (" ˆ ) ˆ " ˆ " [ ] Highpass y[ n] x n "[n]- h[n]!c= highpass " H ( ˆ ). " ˆ
12 h[n]=[/3,/3,/3] h[n]="[n]-h[n]=[/3,-/3,-/3].4 lowpass highpass.4. H (" ˆ ) H (" ˆ ) = H (" ˆ )...! ˆ " h3[n] =h[n]* h[n]! =[/9, /9,, -/9, -/9].4 bandpass! ˆ " h4[n] =h[n]+ [/3 -/3 /3]! =[/3 /3]! bandstop h 5 [n].4 H 3 (" ˆ ) = H (" ˆ )H (" ˆ ). H 4 (" ˆ ) = H 4 ˆ " ( ) + H 5 (" ˆ ) ˆ "! ˆ "!!
13 FIR low-pass filter response vs. number of taps N N=5 N=. b =[, -.49,.549,.549, -.49, ] N=5 N= More coefficients ( taps ), can achieve more ideal response; Longer to compute, more memory, greater phase shift Note: These aren t L-point averagers, fir(), Hammingwindow, Wn=
14 Causal FIR Filter x[ n] Causal FIR filter y[ n] In an FIR filter, the output y at each sample n is a weighted sum of the present input, x[n], and past inputs, x[n-], x[n-],, x[n-m]. y[ n] = b x[ n] + b x[ n "] +K+ b M x[ n " M] [ ] = b k x[ n " k] M k=
15 x[ n] Causal IIR filter Infinite Impulse Response Causal IIR filter y[ n] For an IIR filter, the output y at each sample n is a weighted sum of the present input, x[n], past inputs, x[n-], x[n-],, x[n-m], and past outputs, y[n-], y[n-],, y[n-n].. [ ] = b x[ n] + b x[ n "] +K+ b M x[ n " M ] + a y[ n "] + a y[ n " ] +K+ a M y[ n " N] N [ ] = a l y[ n " l] + b k x n " k l= M [ ] N " M proper system
16 [ ] =" [ n] [ ] = y[ n "] + x[ n] x n IIR Impulse Response To calculate output at step n, need previous outputs. At step n=, assume initial rest conditions. x[n]=, y[n]= for n< y[ ] = y[ "] + x[ ] y = + = [ ] = y[ "] + 4 x[ ] = y[ ] + 4 x[ ] = + 4 = for n> [ ] = y[ n "] + x[ n] = y[ n "]
17 [ ] [ ] n = " " 3 4 K x n [ ] = K [ ] = y[ n "] + x[ n] [ ] = [ 4 8 K " n ] = h[n] impulse response y[n] TextEnd n decays exponentially, goes on forever infinite impulse response
18 N [ ] = a k x[ n " l] x n l= IIR impulse response M [ ] + b k x n " k IIR filter [ ] = "[n] = n = % otherwise M [ ] x="[ n] = h[n] b k "[ n k] Delta function % Can t read filter coefficients off of impulse response impulse response [ ] = h[n]x[ n " k] k=" Convolution sum: LTI systems: FIR/IIR
19 [ ] = x[k]h[ n " k] k=" IIR convolution x[ n] = x[k]" [ n k] We can decompose x[n] into scaled delayed impulses. Here assume finite length x[n]. Convolution sum: LTI: FIR / IIR y[ n] = x[]h[ n] + x[]h[n "]+ x[]h[n " ]+K Each impulse has a scaled impulse response. The final output is a sum of those scaled and delayed impulse responses. Now the impulse responses are Infinite in length.
20 [ ] = x[k]h[ n " k] k=" IIR convolution x[ n] = " [ n] 3" [ n ] + " [ n ] [ ] = h n y[ n] = " h n & = & & & % [ 4 8 K " n ] n= [ ] 3" h[n ]+ " h[n ] Each impulse has a scaled impulse response. The final output is a sum of those scaled and delayed impulse responses K " n 3 8 K 3" 4 K " n K n> n [ ] n ' ) ) ) ) ( K [ ] n"3 [ " 3 + ]
21 IIR convolution x[n] h[n] 5 TextEnd TextEnd *h[n] -3*h[n-] *h[n-] TextEnd TextEnd TextEnd y[n] TextEnd n [ K [ 4 ] n" ] y[n] = 4 "4
22 x n IIR convolution [ ] = " [ n] 3" [ n ] + " [ n 3] [ ] = 4 h n [ 8 K " n ] y[n] = h[n]* x[n] [ 4 K [ ] n" ] y[n] = 4 "4 4 3 y[n] TextEnd n
23 Frequency response [ ] = h k [ ]x n " k [ ] convolution x[ n] = Ae j" e j ˆ n Complex exponential input % [ ] = & h k [ ]Ae j" e j ˆ nk ( ) ( ) Ae j& e j ˆ = h[ k]e " j ˆ % k = H (" ˆ )Ae j e j " ˆ n frequency response n H let (" ˆ ) = h[k]e j ˆ " k This sum must converge for H (" ˆ ) to exist. H (" ˆ ) is the Discrete Time Fourier Transform of the impulse function h[n]
24 Frequency response H H FIR M k= [ ] = b k x[ n " k] M [ ] = b k "[ n k] h k M (" ˆ ) = h[k]e j ˆ M (" ˆ ) = b k e j ˆ Easy to go from difference equation to frequency response because h[n] finite length and h[n] = [b, b, ]. " k " k h k H IIR % [ ] " & b k [ n k] (" ˆ ) = h[k]e j ˆ " k Tough to go from diff.eqn. to freq. response because h[n] infinite length, h[n]=f(al,bk) is complicated, and N [ ] = a l y[ n " l] H (" ˆ ) + b k x n " k l= x Argh! M may be unbounded. [ ]
25 temporal space - n N [ ] = a l y[ n " l] % [ ] " & b k [ n k] h k Fourier transform H l= (" ˆ ) = h[k]e j ˆ frequency space -! + b k x n " " k. Get around road block by using z-plane and z-transforms. Compute system function from diff.eq. coefficients, then evaluate on the unit circle to find the frequency response.. z-plane (pole/zeros) will tell us if system stable and frequency response exists. 3. By using z-transforms, solution to diff.eq goes from solving convolution in n-space to solving algebraic equations in z-domain (easier). And lots more! M Argh! road block [ ] z-transform Hurray! H( z) = H " Benefits of z-plane and z-transforms: complex frequency space- z M N b k z "k " a k z "k k= ( ) = H e j" The frequency response is H(z) evaluated on unit circle = M ( i= z " zzi ) N z " zpi i= ( ) = H z ( ) ( ) z=e j"
26 Frequency response [ ] = h k [ ]x n " k [ ] convolution x[ n] = Ae j" e j ˆ n Complex exponential input % [ ] = & h k [ ]Ae j" e j ˆ nk ( ) ( ) Ae j& e j ˆ = h[ k]e " j ˆ % k = H (" ˆ )Ae j e j " ˆ n frequency response n H let (" ˆ ) = h[k]e j ˆ " k This sum must converge for H (" ˆ ) to exist. H (" ˆ ) is the Discrete Time Fourier Transform of the impulse function h[n]
27 System function [ ] = h k [ ]x[ n " k] convolution k=" x[ n] = z n power sequence of complex numbers [ ] = h k k=" [ ]z n"k ( ) = h[ k]z "k k=" system function ( ) z n = H( z)z n scaled and shifted = H( z)e j"h ( z) z n power sequence H(z) = k=" h[ k]z "k let H(z) = k=" h[ k]z "k LTI
28 z n characteristic functions of LTI systems Re( z n ) z n +(z * ) n z = e j " 6 sinusoid Re( z n ) z n +(z * ) n z =.9e j " 6 damped sinusoid.9.7 z n z = exponentials
29 z-plane and sample responses z n - Im(z) Re(z)
30 z-plane z n high freq Im(z) low freq z = - 5 DC,!= Re(z) z = e ± j" ˆ - 5 unit circle sinusoids z = " - 5 Nyquist sampled sinusoid
31 z-plane z n z =.5e ± j " 4 high freq Im(z) low freq z =.5e ± j " Re(z) - 5 z < Damped sinusoids z =.5e ± j 3" 4-5 z = ".5-5 Nyquist sampled damped sinusoid
32 z-plane Im(z) z = z n - 5 Re(z) z = z = - 5 +Re(z) axis exponential decay z = - 5 Im(z) - 5 Re(z) z = ".5-5 -Re(z) axis " s samples/cycle z = " - 5 Nyquist sampled signals
33 z-plane z n Im(z) z = ±j Re(z) z = ± j.5 z = Im(z) axis every other sample Im(z) - 5 Re(z) z = ± j. - 5 z > unstable z < stable
34 System function [ ] = h k [ ]x[ n " k] convolution k=" x[ n] = z n power sequence of complex numbers [ ] = h k k=" [ ]z n"k ( ) = h[ k]z "k k=" system function ( ) z n = H( z)z n scaled and shifted = H( z)e j"h ( z) z n power sequence H(z) = k=" h[ k]z "k let H(z) = k=" h[ k]z "k LTI
35 z-transform Ex. Ex. impulse x[n] = " n z % k= x[n] " X(z) = x[k]z k z [ ] " y[n] = h[n]" x[n] = h[n]"[n] c Y(z) = H(z) " = H(z) c z z y[n] = h[n] X(z) = % k= = z = "[ k]z k H(z) is the z-transform of the impulse response h[n] H(z) = definition k=" h[ k]z "k LTI
36 Ex. n sample delayed impulse n th power of z - x[n] = "[ n n ] " % k= X(z) = "[ k n ]z k Ex. finite response 8 6 = z n = ( z ) n 4 x[n] TextEnd n x[n] =" [ n] " [ n ] + 5" [ n ] 7" [ n 3] cz X(z) =" z " + 5z " " 7z "3
37 Ex. y[ n] = x n 3 [ ] + x n " 3 [ ] + x n " 3 [ ] h[ n] = " n 3 [ ] + " n 3 [ ] + " n 3 [ ] H z k= ( ) = h[k]z "k = h[]z + h[]z " + h[]z " = z" + 3 z" FIR = ( 3 z" ) + ( 3 z" ) + ( 3 z" ) polynomial in z - h[n] = [ ] b k " n k % H z k= k= ( ) = b k z k FIR only! sequence n-domain (sample space) " polynomial z-domain (complex freq space)
38 Ex. num= denom= H( z) = z" + 3 z" = z + z + 3z H z y[n] = H( z)z n ( ) = y[n] = z + z += z = ("± j 3) = e ± j 3 ( ) = " y[n] = " z = H z z =, roots of denominator zeros roots of numerator poles 5 H(z) 4 3 Im(z) Re(z) FIR filters only have zeros on unit circle, and poles are either at or. poles=zeros extra zero/poles are at z=.
39 system response H(z) pz plot H(z) Imaginary part TextEnd Im(z) Re(z) Real part H(e j! ) ! frequency response H " ( ) = H e j" The frequency response is H(z) evaluated on unit circle ( ) = H z ( ) z=e j"!
40 Ex. N [ ] = a l y[ n " l] + b k x n " k l= N [ ] = a l h[ n " l] h n M [ ] + b k n " k l= M [ ] IIR H( z) = N H(z) = a l H (z)z "l + b k z "k l= M N b k z "k " a k z "k k= M ( ) = b kz "k N H(z) " a l z "l l= M freqz(b,a) sequence n-domain (sample space) " polynomial z-domain (complex freq space)
41 Equivalent ways to represent the system x[n]="[n] N [ ] = a l y[ n " l] c + b k x n " k l= M difference equation [ ] c " inspection x[n] b b a unit delay + + block diagram y[n] 3 h[n] = y[n] x[n ]="[ n] H z impulse response sequence 6 H " z z = e j" ( ) = H e j" ( ) = c 4 ( ) = H z frequency response M % N b k z k % a k z k k= system function polynomial ( ) z=e j" = & M ( z zzi ) i= N & z z pi i= ( ) pole-zero locations 5 All poles must be inside unit circle for H (") to converge and the system to be stable. (FIR filter always stable)
42 difference equation IIR filter [ ] = x[ n] + y[ n "] ".5y[ n " ] Imaginary part. -. TextEnd H( z) = M N b k z "k " a k z "k k= Real part frequency response H " ( ) = H e j" ( ) = H z ( ) z=e j" H( z) = = " z " +.5z " = z z " z +.5 = M ( i= z " zzi ) N z " zpi i= ( ) z z z ".5 " j.5 ( )( z ".5 + j.5) zeros:, poles:.5±j.5
43 zeros:, poles:.5±j.5 IIR filter Imaginary part TextEnd Real part ( ) = " M ( z zzi ) i= N H z = " i= ( z zpi ) z z z.5 j.5 ( )( z.5 + j.5) = z z z +.5 = z +.5z " z " +.5z = " = z z " z +.5 " z " +.5z frequency response H " ( ) = H e j" ( ) = H z " z z ( ) z=e j" [ ] = x[ n] + y[ n "] ".5y[ n " ] H( z) = M N b k z "k " a k z "k k= difference equation
44 zeros:, poles:.5±j.5 IIR filter Imaginary part TextEnd H( z) = = z ( z ".5 " j.5) z ".5 + j.5 ( ) z z " z +.5 = " z " +.5z " Real part H( " ˆ ) = H( z) z= e j" ˆ y[ n] = x[ n] + y[ n "] ".5y[ n " ] Magnitude Response (db) Normalized frequency (Nyquist == ) TextEnd y[n]. Phase (degrees) 5-5 TextEnd Normalized frequency (Nyquist == ) n
45 zeros:, poles:.5±j.5 IIR filter Imaginary part TextEnd H( z) = = z ( z ".5 " j.5) z ".5 + j.5 ( ) z z " z +.5 = " z " +.5z " Real part H( " ˆ ) = H( z) z= e j" ˆ y[ n] = x[ n] + y[ n "] ".5y[ n " ] To go from p-z map to impulse response: Look at locations of poles. Closer poles are to zeros, the weaker their response. y[n] n
46 zeros:, poles:.5±j.5 IIR filter Imaginary part TextEnd H( z) = = z ( z ".5 " j.5) z ".5 + j.5 ( ) z z " z +.5 = " z " +.5z " Real part H( " ˆ ) = H( z) z= e j" ˆ y[ n] = x[ n] + y[ n "] ".5y[ n " ] Magnitude Response (db) Phase (degrees) Normalized frequency (Nyquist == ) 5-5 TextEnd TextEnd From zp map to freq response: Go around unit circle from DC to Nyquist. Response goes down near zeros, goes up near poles. Closer poles are to zeros, the weaker their effect Normalized frequency (Nyquist == )
47 x[n] = x[k] % n " k k=" z-transform k=" [ ] & X(z) = x[k]z "k sequence n-domain (sample space) " polynomial z-domain (complex freq space) impulse response sequence h[n] " H( z) = M k= N b k z k a k z k system function k= polynomial H(z) is the z-transform of the impulse response h[n]. LTI
48 z-transform x[n] = x[k] % n " k k=" k=" [ ] & X(z) = x[k]z "k sequence n-domain (sample space) " polynomial z-domain (complex freq space) [ ] = h k M Why re we interested? [ ]x[ n " k] convolution sum k= Y(z) = H(z)X(z) polynomial multiplication
49 Ex. impulse x[n] = "[ n] " z-transform % k= x[n] " X(z) = x[k]z k y[n] = h[n]" x[n] = h[n]"[n] c Y(z) = H(z) " = H(z) c z z y[n] = h[n] X(z) = % k= = z = "[ k]z k H(z) is the z-transform of the impulse response h[n] LTI
50 n sample delayed impulse n th power of z - x[n] = "[ n n ] " % k= X(z) = "[ k n ]z k Ex. finite response 8 6 = z n = ( z ) n 4 x[n] TextEnd n x[n] =" [ n] " [ n ] + 5" [ n ] 7" [ n 3] cz X(z) =" z " + 5z " " 7z "3
51 Infinite signals % k= x[n]u[n] " X(z) = x[k]z k right-sided signals k= x[n] = a n u[n] " X(z) = a k z "k ( ) k k= ( ) + ( az " ) 3 K = az " =+ az " + az " geometric series
52 Infinite signals k= x[n] = a n u[n] " X(z) = a k z "k =+ az " + az " geometric series ( ) + ( az " ) 3 K - ( ) =+ az " + az " ( ) = " az " " az " X z az " X z ( ) + ( az " ) 3 K ( ) " ( az " ) 3 " ( az " ) 4 K ( " az " )X( z) = X( z) = az " < " az " or z > a region of convergence
53 Infinite series: x[n] = a n u[n] " X(z) = a k z "k x[n]= n< right sided X( z) = k= " az " IIR filter z > a region of convergence Finite series: x[n] = a n ( u[n " M] " u[n " N] ) ( ) = az" X z ( ) M " ( az " ) N " az " lim z"a X(z) = N M " N" k= M X(z) = a k z "k all z region of convergence FIR filter
54 Ex. x[n] =u[ n] X(z) =? k=" X(z) = x[k]z "k k= = z "k ( ) k k= = z " let a = We know: ( az " ) k k= = " az " z > a roc = " z " z >
55 Ex. x[n] = cos( ˆ " n)u n [ ] X(z) =? X(z) = x[k]z "k = cos( " ˆ k)z k k=" e = % j" ˆ k +e j ˆ k= " k z k % k= ( " k )z k = % e j ˆ + e j ˆ k= ( ) k % k= " k z k = % e j" ˆ z + e j ˆ = k= let a = e j ˆ " " e j ˆ z + " %( " z ) k k= let a = e " j ˆ " e " j ˆ z " We know: ( az " ) k k= = intersection of roc s z > e j ˆ " e j ˆ " z > " az " z > a roc
56 Ex. x[n] = cos( " ˆ n)u[ n] X(z) =? Cont. X(z) = x[k]z "k = cos( " ˆ k)z k k=" % k= = = " e j ˆ z + " z z " e + j ˆ " e " j ˆ z " z z " e " j ˆ = z z " e " j ˆ " e j ˆ z " ( e " j ˆ + e j ˆ )z + = z z " cos( ˆ ) z " cos( ˆ )z + z >
57 Table of z-transforms signal z-transform ROC "[n k] z "k u[n] n u[n] " n cos (n)u[n] " n sin (n)u[n] z z " z( z +) ( z ") 3 ( ) z z " cos z " ( cos )z + z" sin z ( " cos )z + " all z z > z > z > " z > "
58 z-transform properties a x [ n] + a x [ n] " a X ( z) + a X ( z) x[ n " m] z "m X( z) linearity/superposition sample shift h[ n] " x[ n] H( z)x( z) sample domain convolution z domain multiplication nx[ n] " z dx(z) dz a n x[ n] " X( z ) a multiply by a ramp z domain differentiation multiply by an exponential z domain scaling
59 Ex. x[n] = na n u[n] X( z) =? We already know a n u[n] " " az " and ny[ n] " z dy(z) dz d dz & % ( = d z & % ( " az " ' dz z " a' na n u[ n] " z " = " "a ( z " a) = "az " az " az ( az ) az ( az ) ( )
60 Next: Solve difference equation using z-transforms. Convolution problem in the temporal domain becomes an algebra problem in the z-domain. Need to know how to convert from z-domain back to temporal domain (inverse z-transform).
61 [ ] = x n 3 [ ] + x n " 3 [ ] + x n " 3 [ ] [ ] = " 3 [ n ] + " 3 [ n ] + " 3 [ n ] " n)u[ n] h n x[n] = cos( ˆ y[ n] = h[ n] " x[ n] Y(z) = H( z)x( z) H z ( ) = z" + 3 z" = z + z + X(z) = z Y(z) = z + z + 3z z " cos( ˆ ) z " cos( ˆ )z + & z % * z + z + y[n] = Z ", & z + 3z % 3z z " cos( ˆ ) ' ) z " cos( ˆ )z +( z " cos( ˆ ) '- )/ z " cos( ˆ )z +(.
Z-Transform. x (n) Sampler
Chapter Two A- Discrete Time Signals: The discrete time signal x(n) is obtained by taking samples of the analog signal xa (t) every Ts seconds as shown in Figure below. Analog signal Discrete time signal
Responses of Digital Filters Chapter Intended Learning Outcomes:
Responses of Digital Filters Chapter Intended Learning Outcomes: (i) Understanding the relationships between impulse response, frequency response, difference equation and transfer function in characterizing
Z Transform (Part - II)
Z Transform (Part - II). The Z Transform of the following real exponential sequence x(nt) = a n, nt 0 = 0, nt < 0, a > 0 (a) ; z > (c) for all z z (b) ; z (d) ; z < a > a az az Soln. The given sequence
y[n] = = h[k]x[n k] h[k]z n k k= 0 h[k]z k ) = H(z)z n h[k]z h (7.1)
![y[n] = = h[k]x[n k] h[k]z n k k= 0 h[k]z k ) = H(z)z n h[k]z h (7.1)](/thumbs/71/66089461.jpg "y[n] = = h[k]x[n k] h[k]z n k k= 0 h[k]z k ) = H(z)z n h[k]z h (7.1)")
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