Digital Signal Processing
|
|
- Janis Quinn
- 5 years ago
- Views:
Transcription
1 Digital Signal Processing The -Transform and Its Application to the Analysis of LTI Systems Moslem Amiri, Václav Přenosil Embedded Systems Laboratory Faculty of Informatics, Masaryk University Brno, Cech Republic March, 2012
2 The Direct -Transform -transform of x(n) is defined as power series: X () where is a complex variable For convenience n= x(n) n X () Z{x(n)} x(n) X () Since -transform is an infinite power series, it exists only for those values of for which this series converges Region of convergence (ROC) of X () is set of all values of for which X () attains a finite value Any time we cite a -transform, we should also indicate its ROC ROC of a finite-duration signal is entire -plane except possibly points = 0 and/or = 2 / 63
3 The Direct -Transform Example Determine -transforms of following finite-duration signals x(n) = {1, 2, 5, 7, 0, 1} X () = ROC: entire -plane except = 0 x(n) = {1, 2, 5, 7, 0, 1} X () = ROC: entire -plane except = 0 and = x(n) = δ(n) X () = 1 [i.e., δ(n) 1], ROC: entire -plane x(n) = δ(n k), k > 0 X () = k [i.e., δ(n k) k ], ROC: entire -plane except = 0 x(n) = δ(n + k), k > 0 X () = k [i.e., δ(n + k) k ], ROC: entire -plane except = 3 / 63
4 The Direct -Transform Example Determine -transform of x(n) = ( 1 2 )n u(n) -transform of x(n) X () = ( 1 2 )2 2 + ( 1 2 )n n + = ( 1 2 )n n = n=0 ( ) n n=0 For < 1 or > 1 2, X () converges to X () = , ROC: > / 63
5 The Direct -Transform Expressing complex variable in polar form = re jθ where r = and θ = X () = n= In ROC of X (), X () < X () = x(n)r n e jθn X () n= 1 n= n=0 x(n)r n e jθn n= x(n)r n e jθn = n=1 n= x(n)r n + x(n) r n x( n)r n + n=0 x(n)r n x(n) r n If X () converges in some region of complex plane, both sums must be finite in that region 5 / 63
6 The Direct -Transform Figure 1: Region of convergence for X () and its corresponding causal and anticausal components. 6 / 63
7 The Direct -Transform Example Determine -transform of We have x(n) = α n u(n) x(n) = α n u(n) = X () = α n n = n=0 X () = { α n, n 0 0, n < 0 (α 1 ) n n=0 1, ROC: > α 1 α 1 7 / 63
8 The Direct -Transform Example (continued) Figure 2: The exponential signal x(n) = α n u(n) (a), and the ROC of its -transform (b). 8 / 63
9 The Direct -Transform Example Determine -transform of { x(n) = α n 0, n 0 u( n 1) = α n, n 1 We have where l = n X () = 1 n= ( α n ) n = (α 1 ) l l=1 α 1 x(n) = α n u( n 1) X () = 1 α 1 = 1 1 α 1 ROC: < α 9 / 63
10 The Direct -Transform Example (continued) Figure 3: Anticausal signal x(n) = α n u( n 1) (a), and the ROC of its -transform (b). 10 / 63
11 The Direct -Transform From two preceding examples Z{α n u(n)} = Z{ α n u( n 1)} = 1 1 α 1 This implies that a closed-form expression for -transform does not uniquely specify the signal in time domain Ambiguity can be resolved if ROC is also specified A signal x(n) is uniquely determined by its -transform X () and region of convergence of X () ROC of a causal signal is exterior of a circle of some radius r 2 ROC of an anticausal signal is interior of a circle of some radius r 1 11 / 63
12 The Direct -Transform Example Determine -transform of We have X () = α n n + n=0 x(n) = α n u(n) + b n u( n 1) 1 n= b n n = (α 1 ) n + (b 1 ) l n=0 First sum converges if > α, second sum converges if < b If b < α, X () does not exist If b > α X () = ROC: α < < b l=1 1 1 α b 1 = b α α + b αb 1 12 / 63
13 The Direct -Transform Example (continued) Figure 4: ROC for -transform in the example. 13 / 63
14 The Direct -Transform Table 1: Characteristic families of signals with their corresponding ROCs 14 / 63
15 Properties of the -Transform Combining several -transforms, ROC of overall transform is, at least, intersection of ROCs of individual transforms Linearity If x 1 (n) X 1 () and then x 2 (n) x(n) = a 1 x 1 (n) + a 2 x 2 (n) X 2 () X () = a 1 X 1 () + a 2 X 2 () for any constants a 1 and a 2 To find -transform of a signal, express it as a sum of elementary signals whose -transforms are already known 15 / 63
16 Properties of the -Transform Example Determine -transform and ROC of Defining signals x 1 (n) and x 2 (n) According to linearity property Recall that x(n) = [3(2 n ) 4(3 n )]u(n) x 1 (n) = 2 n u(n) and x 2 (n) = 3 n u(n) x(n) = 3x 1 (n) 4x 2 (n) α n u(n) X () = 3X 1 () 4X 2 () 1 1 α 1, ROC: > α Setting α = 2 and α = 3 X 1 () = 1, ROC: > 2 and X () = 1, ROC: > Intersecting ROCs, overall transform is X () = 3 4, ROC: > / 63
17 Properties of the -Transform Time shifting If x(n) X () then x(n k) k X () ROC of k X () is same as that of X () except for = 0 if k > 0 and = if k < 0 17 / 63
18 Properties of the -Transform Scaling in -domain If x(n) X (), ROC: r 1 < < r 2 then a n x(n) X (a 1 ), ROC: a r 1 < < a r 2 for any constant a, real or complex Proof Z{a n x(n)} = n= a n x(n) n = n= Since ROC of X () is r 1 < < r 2, ROC of X (a 1 ) is r 1 < a 1 < r 2 or a r 1 < < a r 2 x(n)(a 1 ) n = X (a 1 ) 18 / 63
19 Properties of the -Transform Time reversal If then Proof x(n) x( n) Z{x( n)} = X (), ROC: r 1 < < r 2 X ( 1 ), ROC: 1 r 2 < < 1 r 1 n= x( n) n = l= x(l)( 1 ) l = X ( 1 ) where l = n ROC of X ( 1 ) r 1 < 1 < r 2 or 1 r 2 < < 1 r 1 19 / 63
20 Properties of the -Transform Differentiation in -domain If then Proof differentiating both sides dx () d = n= x(n) nx(n) X () = X () dx () d n= x(n) n x(n)( n) n 1 = 1 = 1 Z{nx(n)} Both transforms have same ROC n= [nx(n)] n 20 / 63
21 Properties of the -Transform Convolution of two sequences If x 1 (n) x 2 (n) then X 1 () X 2 () x(n) = x 1 (n) x 2 (n) X () = X 1 ()X 2 () ROC of X () is, at least, intersection of that for X 1 () and X 2 () Proof: convolution of x 1 (n) and x 2 (n) x(n) = x 1 (k)x 2 (n k) -transform of x(n) X () = x(n) n = = n= k= x 1 (k) = X 2 ()X 1 () [ n= k= n= [ k= x 1 (k)x 2 (n k) x 2 (n k) n ] = X 2 () ] k= n x 1 (k) k 21 / 63
22 Properties of the -Transform Example Compute convolution x(n) of signals x 1 (n) = {1, 2, 1} and x 2 (n) = -transforms of these signals X 1 () = { 1, 0 n 5 0, elsewhere X 2 () = Convolution of two signals is equal to multiplication of their transforms Hence X () = X 1 ()X 2 () = x(n) = {1, 1, 0, 0, 0, 0, 1, 1} 22 / 63
23 Properties of the -Transform Convolution property is one of most powerful properties of -transform It converts convolution of two signals (time domain) to multiplication of their transforms Computation of convolution of two signals using -transform 1 Compute -transforms of signals to be convolved X 1 () = Z{x 1 (n)} X 2 () = Z{x 2 (n)} 2 Multiply the two -transforms X () = X 1 ()X 2 () 3 Find inverse -transform of X () x(n) = Z 1 {X ()} 23 / 63
24 Properties of the -Transform Correlation of two sequences If x 1 (n) x 2 (n) then r x1x 2 (l) = n= x 1 (n)x 2 (n l) X 1 () X 2 () R x1x 2 () = X 1 ()X 2 ( 1 ) Proof r x1x 2 (l) = x 1 (l) x 2 ( l) Using convolution and time-reversal properties R x1x 2 () = Z{x 1 (l)}z{x 2 ( l)} = X 1 ()X 2 ( 1 ) ROC of R x1x 2 () is at least intersection of that for X 1 () and X 2 ( 1 ) 24 / 63
25 Properties of the -Transform The initial value theorem If x(n) is causal (x(n) = 0 for n < 0), then Proof: since x(n) is causal X () = x(0) = lim X () x(n) n = x(0) + x(1) 1 + x(2) 2 + n=0 as, n 0 and hence X () = x(0) 25 / 63
26 Properties of the -Transform Table 2: Some common -transform pairs Signal, x(n) -Transform, X () ROC δ(n) 1 All 1 u(n) 1 1 > 1 a n 1 u(n) 1 a 1 > a na n u(n) a 1 (1 a 1 ) 2 a n 1 u( n 1) 1 a 1 na n a u( n 1) 1 (cos ω 0 n)u(n) (sin ω 0 n)u(n) (a n cos ω 0 n)u(n) (a n sin ω 0 n)u(n) (1 a 1 ) cos ω 0 > a < a < a cos ω > 1 1 sin ω cos ω > 1 1 a 1 cos ω 0 1 2a 1 cos ω 0 +a 2 2 a 1 sin ω 0 1 2a 1 cos ω 0 +a 2 2 > a > a 26 / 63
27 Rational -Transforms An important family of -transforms are those for which X () is a rational function X () is a ratio of two polynomials in 1 (or ) Some important issues of rational -transforms are discussed here 27 / 63
28 Poles and Zeros Zeros of a -transform X () are values of for which X () = 0 Poles of a -transform are values of for which X () = If X () is a rational function (and if a 0 0 and b 0 0) X () = B() A() = b 0 + b b M M a 0 + a a N N = = b 0 M M + (b 1 /b 0 ) M b M /b 0 a 0 N N + (a 1 /a 0 ) N a N /a 0 = b 0 M+N ( 1)( 2 ) ( M ) a 0 ( p 1 )( p 2 ) ( p N ) = G N M M k=1 ( k) N k=1 ( p k) M k=0 b k k N k=0 a k k X () has M finite eros at = 1, 2,..., M N finite poles at = p 1, p 2,..., p N N M eros if N > M or poles if N < M at = 0 X () has exactly same number of poles as eros 28 / 63
29 Poles and Zeros Example Determine pole-ero plot for signal x(n) = a n u(n), a > 0 From Table 2 X () = 1 1 a 1 = a, ROC: > a X () has one ero at 1 = 0 and one pole at p 1 = a Figure 5: Pole-ero plot for the causal exponential signal x(n) = a n u(n). 29 / 63
30 Poles and Zeros Example Determine pole-ero plot for signal { a x(n) = n, 0 n M 1 0, elsewhere where a > 0 -transform of x(n) X () = M 1 n=0 M 1 a n n = n=0 Since a > 0, M = a M has M roots at (a 1 ) n = 1 (a 1 ) M 1 a 1 = M a M M 1 ( a) k = ae j2πk/m, k = 0, 1,..., M 1 30 / 63
31 Poles and Zeros Example (continued) Zero 0 = a cancels pole at = a. Thus X () = ( 1)( 2 ) ( M 1 ) M 1 which has M 1 eros and M 1 poles Figure 6: Pole-ero pattern for the finite-duration signal x(n) = a n, 0 n M 1 (a > 0), for M = / 63
32 Poles and Zeros Example Determine -transform and signal corresponding to following pole-ero plot Figure 7: Pole-ero pattern. 32 / 63
33 Poles and Zeros Example (continued) We use X () = G N M M k=1 ( k) N k=1 ( p k) There are two eros (M = 2) at 1 = 0, 2 = r cos ω 0 There are two poles (N = 2) at p 1 = re jω 0, p 2 = re jω 0 X () = G ( 1)( 2 ) ( p 1 )( p 2 ) = G ( r cos ω 0 ) ( re jω 0 )( re jω 0) 1 r 1 cos ω 0 = G 1 2r 1 cos ω 0 + r 2, ROC: > r 2 From Table 2 we find that x(n) = G(r n cos ω 0 n)u(n) 33 / 63
34 Poles and Zeros -transform X () is a complex function of complex variable X () is a real and positive function of Since represents a point in complex plane, X () is a surface -transform 1 2 X () = has one ero at 1 = 1 and two poles at p 1, p 2 = 0.9e ±jπ/4 Figure 8: Graph of X () for the above -transform. 34 / 63
35 Pole Location & Time-Domain Behavior for Causal Signals Characteristic behavior of causal signals depends on whether poles of transform are contained in region < 1 or > 1 or on circle = 1 Circle = 1 is called unit circle If a real signal has a -transform with one pole, this pole has to be real The only such signal is the real exponential x(n) = a n u(n) 1 X () =, ROC: > a 1 a 1 having one ero at 1 = 0 and one pole at p 1 = a on real axis 35 / 63
36 Pole Location & Time-Domain Behavior for Causal Signals Figure 9: Time-domain behavior of a single-real-pole causal signal as a function of the location of the pole with respect to the unit circle. 36 / 63
37 Pole Location & Time-Domain Behavior for Causal Signals A causal real signal with a double real pole has the form x(n) = na n u(n) X () = a 1 (1 a 1, ROC: > a ) 2 In contrast to single-pole signal, a double real pole on unit circle results in an unbounded signal 37 / 63
38 Pole Location & Time-Domain Behavior for Causal Signals Figure 10: Time-domain behavior of causal signals corresponding to a double (m = 2) real pole, as a function of the pole location. 38 / 63
39 Pole Location & Time-Domain Behavior for Causal Signals Configuration of poles as a pair of complex-conjugates results in an exponentially weighted sinusoidal signal x(n) = (a n cos ω 0 n)u(n) x(n) = (a n sin ω 0 n)u(n) X () = ROC: > a X () = ROC: > a 1 a 1 cos ω 0 1 2a 1 cos ω 0 + a 2 2 a 1 sin ω 0 1 2a 1 cos ω 0 + a 2 2 Distance r of poles from origin determines envelope of sinusoidal signal Angle ω 0 with real positive axis determines relative frequency 39 / 63
40 Pole Location & Time-Domain Behavior for Causal Signals Figure 11: A pair of complex-conjugate poles corresponds to causal signals with oscillatory behavior. 40 / 63
41 Pole Location & Time-Domain Behavior for Causal Signals Figure 12: Causal signal corresponding to a double pair of complex-conjugate poles on the unit circle. In summary Causal real signals with simple real poles or simple complex-conjugate pairs of poles, inside or on unit circle, are always bounded in amplitude A signal with a pole, or a complex-conjugate pair of poles, near origin decays more rapidly than one near (but inside) unit circle Thus, time behavior of a signal depends strongly on location of its poles relative to unit circle Zeros also affect behavior of a signal but not as strongly as poles E.g., for sinusoidal signals, presence and location of eros affects only their phase 41 / 63
42 Inversion of the -Transform There are three methods for evaluation of inverse -transform 1 Direct evaluation by contour integration 2 Expansion into a series of terms, in variables and 1 3 Partial-fraction expansion and table lookup Inverse -transform by contour integration x(n) = 1 X () n 1 d 2πj C Integral is a contour integral over a closed path C C encloses origin and lies within ROC of X () Figure 13: Contour C. 42 / 63
43 The Inverse -Transform by Power Series Expansion Given X () with its ROC, expand it into a power series of form X () = c n n n= By uniqueness of -transform, x(n) = c n for all n When X () is rational, expansion can be performed by long division Long division method becomes tedious when n is large Although this method provides a direct evaluation of x(n), a closed-form solution is not possible Hence this method is used only for determining values of first few samples of signal 43 / 63
44 The Inverse -Transform by Power Series Expansion Example Determine inverse -transform of X () = When ROC: > 1 2 When ROC: < 0.5 ROC: > 1 Since ROC is exterior of a circle, x(n) is a causal signal. Thus we seek negative powers of by dividing numerator of X () by its denominator X () = = x(n) = {1, 3 2, 7 4, 15 8, 31 16,...} 44 / 63
45 The Inverse -Transform by Power Series Expansion Example (continued) ROC: < 0.5 Since ROC is interior of a circle, x(n) is anticausal. To obtain positive powers of, write the two polynomials in reverse order and then divide X () = = x(n) = { 62, 30, 14, 6, 2, 0, 0 } 45 / 63
46 The Inverse -Transform by Partial-Fraction Expansion In table lookup method, express X () as a linear combination X () = α 1 X 1 () + α 2 X 2 () + + α K X K () X 1 (),..., X K () are expressions with inverse transforms x 1 (n),..., x K (n) available in a table of -transform pairs Using linearity property x(n) = α 1 x 1 (n) + α 2 x 2 (n) + + α K x K (n) If X () is a rational function X () = B() A() = b 0 + b b M M a 0 + a a N N dividing both numerator and denominator by a 0 X () = B() A() = b 0 + b b M M 1 + a a N N This form of rational function is called proper if a N 0 and M < N 46 / 63
47 The Inverse -Transform by Partial-Fraction Expansion An improper rational function (M N) can always be written as sum of a polynomial and a proper rational function Example Express improper rational function X () = in terms of a polynomial and a proper function Terms 2 and 3 should be eliminated from numerator Do long division with the two polynomials written in reverse order Stop division when order of remainder becomes 1 X () = / 63
48 The Inverse -Transform by Partial-Fraction Expansion Any improper rational function (M N) can be expressed as X () = B() A() = c 0 + c c M N (M N) + B 1() A() Inverse -transform of the polynomial can easily be found by inspection We focus our attention on inversion of proper rational transforms 1 Perform a partial fraction expansion of proper rational function 2 Invert each of the terms 48 / 63
49 The Inverse -Transform by Partial-Fraction Expansion Let X () be a proper rational function (a N 0 and M < N) X () = B() A() = b 0 + b b M M 1 + a a N N Eliminating negative powers of Since N > M, the function X () = b 0 N + b 1 N b M N M N + a 1 N a N X () = b 0 N 1 + b 1 N b M N M 1 N + a 1 N a N is also proper To perform a partial-fraction expansion, this function should be expressed as a sum of simple fractions First factor denominator polynomial into factors that contain poles p 1, p 2,..., p N of X () 49 / 63
50 The Inverse -Transform by Partial-Fraction Expansion Distinct poles Suppose poles p 1, p 2,..., p N are all different. We seek expansion X () = A 1 p 1 + A 2 p A N p N To determine coefficients A 1, A 2,..., A N, multiply both sides by each of terms ( p k ), k = 1, 2,..., N, and evaluate resulting expressions at corresponding pole positions, p 1, p 2,..., p N ( p k )X () = ( p k)a 1 p 1 A k = ( p k)x (), =pk + + A k + + ( p k)a N p N k = 1, 2,..., N 50 / 63
51 The Inverse -Transform by Partial-Fraction Expansion Example Determine partial-fraction expansion of X () = Eliminate negative powers by multiplying by 2 X () = p 1 = j 1 2 and p 2 = 1 2 j 1 2 p 1 p 2 X () A 1 = ( p 1)X () A 2 = ( p 2)X () = + 1 ( p 1 )( p 2 ) = A 1 + A 2 p 1 p p 2 = + j =p1 2 + j j 1 = 1 2 j = + 1 =p1 = + 1 =p2 p 1 = =p2 1 2 j j j 1 2 = j 3 2 Complex-conjugate poles result in complex-conjugate coefficients 51 / 63
52 The Inverse -Transform by Partial-Fraction Expansion Multiple-order poles If X () has a pole of multiplicity m (there is factor ( p k ) m in denominator), partial-fraction expansion must contain the terms A 1k + A 2k p k ( p k ) A mk ( p k ) m Coefficients {A ik } can be evaluated through differentiation 52 / 63
53 The Inverse -Transform by Partial-Fraction Expansion Example Determine partial-fraction expansion of X () = 1 (1 + 1 )(1 1 ) 2 Expressing in terms of positive powers of X () = 2 ( + 1)( 1) 2 X () has a simple pole at p 1 = 1 and a double pole at p 2 = p 3 = 1 X () = 2 ( + 1)( 1) 2 = A A A 3 ( 1) 2 A 1 = ( + 1)X () = 1 = / 63
54 The Inverse -Transform by Partial-Fraction Expansion Example (continued) A 3 = ( 1)2 X () = 1 =1 2 To obtain A 2 ( 1) 2 X () = ( 1)2 + 1 A 1 + ( 1)A 2 + A 3 Differentiating both sides and evaluating at = 1, A 2 is obtained [ ( 1) 2 ] X () A 2 = d d =1 = / 63
55 The Inverse -Transform by Partial-Fraction Expansion Having performed partial-fraction expansion, final step in inversion is as follows If poles are distinct X () = A 1 + A p 1 p 2 p N 1 X () = A 1 1 p A p A 1 N 1 p N 1 x(n) = Z 1 {X ()} is obtained by inverting each term and taking the corresponding linear combination From table 2 { } { Z 1 1 (pk ) 1 p k 1 = n u(n), if ROC: > p k (causal) (p k ) n u( n 1), if ROC: < p k (anticausal) If x(n) is causal, ROC is > p max, where p max = max{ p 1, p 2,..., p N } In this case all terms in X () result in causal signal components A N x(n) = (A 1 p n 1 + A 2 p n A N p n N)u(n) 55 / 63
56 The Inverse -Transform by Partial-Fraction Expansion If all poles are distinct but some of them are complex, and if signal x(n) is real, complex terms can be reduced into real components If p j is a pole, its complex conjugate pj is also a pole If x(n) is real, the polynomials in X () have real coefficients If a polynomial has real coefficients, its roots are either real or occur in complex-conjugate pairs Their corresponding coefficients in partial-fraction expansion are also complex-conjugates Contribution of two complex-conjugate poles is x k (n) = [A k (p k ) n + A k (p k )n )]u(n) Expressing A j and p j in polar form A k = A k e jα k and p k = r k e jβ k which gives x k (n) = A k rk n[ej(β k n+α k ) + e j(β k n+α k ) ]u(n) or x k (n) = 2 A k rk n cos(β kn + α k )u(n) Thus ( Z 1 A k 1 p k 1 + A ) k 1 pk = 2 A k r n 1 k cos(β k n + α k )u(n) if ROC is > p k = r k 56 / 63
57 The Inverse -Transform by Partial-Fraction Expansion In case of multiple poles, either real or complex, inverse transform of terms of the form A/( p k ) n is required In case of a double pole, from table 2 { Z 1 p 1 } (1 p 1 ) 2 = np n u(n) provided that ROC is > p In case of poles with higher multiplicity, multiple differentiation is used 57 / 63
58 The Inverse -Transform by Partial-Fraction Expansion Example Determine inverse -transform of X () = 1 If ROC: > 1 2 If ROC: < If ROC: 0.5 < < 1 Partial-fraction expansion for X () p 1=1 X () = p 2=0.5 X () = ( 1)X () A 1 = = 2 =1 ( 0.5)X () A 2 = = 1 =0.5 ( 1)( 0.5) = A A / 63
59 The Inverse -Transform by Partial-Fraction Expansion Example (continued) X () = When ROC is > 1, x(n) is causal and both terms in X () are causal 1 1 p k 1 (p k ) n u(n) x(n) = 2(1) n u(n) (0.5) n u(n) = (2 0.5 n )u(n) When ROC is < 0.5, x(n) is anticausal and both terms in X () are anticausal 1 1 p k 1 (p k ) n u( n 1) x(n) = [ 2 + (0.5) n ]u( n 1) 59 / 63
60 The Inverse -Transform by Partial-Fraction Expansion Example (continued) When ROC is 0.5 < < 1 (ring), signal x(n) is two-sided One of the terms corresponds to a causal signal and the other to an anticausal signal Since the ROC is overlapping of > 0.5 and < 1, pole p 2 = 0.5 provides causal part and pole p 1 = 1 anticausal x(n) = 2(1) n u( n 1) (0.5) n u(n) 60 / 63
61 The Inverse -Transform by Partial-Fraction Expansion Example Determine causal signal x(n) whose -transform is X () = We have already obtained partial-fraction expansion as X () = A 1 1 p A 2 1 p 2 1 A 1 = A 2 = 1 2 j 3 2 and p 1 = p 2 = 1 2 +j 1 2 For a pair of complex-conjugate poles (A k = A k e jα k and p k = r k e jβ k ) ( Z 1 A k 1 p k 1 + A ) k 1 pk = 2 A k r n 1 k cos(β k n + α k )u(n) A 1 = ( 10/2)e j and p 1 = (1/ 2)e jπ/4 x(n) = 10(1/ 2) n cos(πn/ )u(n) 61 / 63
62 The Inverse -Transform by Partial-Fraction Expansion Example Determine causal signal x(n) having -transform X () = 1 (1 + 1 )(1 1 ) 2 We have already obtained partial-fraction expansion as For causal signals X () = (1 1 ) p 1 (p) n u(n) and 1 p 1 (1 p 1 ) 2 np n u(n) x(n) = 1 4 ( 1)n u(n) u(n) nu(n) = [ 1 4 ( 1)n n 2 ] u(n) 62 / 63
63 References John G. Proakis, Dimitris G. Manolakis, Digital Signal Processing: Principles, Algorithms, and Applications, Prentice Hall, / 63
Discrete-Time Signals and Systems. The z-transform and Its Application. The Direct z-transform. Region of Convergence. Reference: Sections
Discrete-Time Signals and Systems The z-transform and Its Application Dr. Deepa Kundur University of Toronto Reference: Sections 3. - 3.4 of John G. Proakis and Dimitris G. Manolakis, Digital Signal Processing:
More informationELEG 305: Digital Signal Processing
ELEG 305: Digital Signal Processing Lecture 4: Inverse z Transforms & z Domain Analysis Kenneth E. Barner Department of Electrical and Computer Engineering University of Delaware Fall 008 K. E. Barner
More informationECE-S Introduction to Digital Signal Processing Lecture 4 Part A The Z-Transform and LTI Systems
ECE-S352-70 Introduction to Digital Signal Processing Lecture 4 Part A The Z-Transform and LTI Systems Transform techniques are an important tool in the analysis of signals and linear time invariant (LTI)
More informationChapter 7: The z-transform
Chapter 7: The -Transform ECE352 1 The -Transform - definition Continuous-time systems: e st H(s) y(t) = e st H(s) e st is an eigenfunction of the LTI system h(t), and H(s) is the corresponding eigenvalue.
More informationGeneralizing the DTFT!
The Transform Generaliing the DTFT! The forward DTFT is defined by X e jω ( ) = x n e jωn in which n= Ω is discrete-time radian frequency, a real variable. The quantity e jωn is then a complex sinusoid
More informationNeed for transformation?
Z-TRANSFORM In today s class Z-transform Unilateral Z-transform Bilateral Z-transform Region of Convergence Inverse Z-transform Power Series method Partial Fraction method Solution of difference equations
More informationUNIT-II Z-TRANSFORM. This expression is also called a one sided z-transform. This non causal sequence produces positive powers of z in X (z).
Page no: 1 UNIT-II Z-TRANSFORM The Z-Transform The direct -transform, properties of the -transform, rational -transforms, inversion of the transform, analysis of linear time-invariant systems in the -
More informationDIGITAL SIGNAL PROCESSING. Chapter 3 z-transform
DIGITAL SIGNAL PROCESSING Chapter 3 z-transform by Dr. Norizam Sulaiman Faculty of Electrical & Electronics Engineering norizam@ump.edu.my OER Digital Signal Processing by Dr. Norizam Sulaiman work is
More informationDigital Signal Processing
Digital Signal Processing Discrete-Time Signals and Systems (2) Moslem Amiri, Václav Přenosil Embedded Systems Laboratory Faculty of Informatics, Masaryk University Brno, Czech Republic amiri@mail.muni.cz
More information(i) Represent discrete-time signals using transform. (ii) Understand the relationship between transform and discrete-time Fourier transform
z Transform Chapter Intended Learning Outcomes: (i) Represent discrete-time signals using transform (ii) Understand the relationship between transform and discrete-time Fourier transform (iii) Understand
More informationSEL4223 Digital Signal Processing. Inverse Z-Transform. Musa Mohd Mokji
SEL4223 Digital Signal Processing Inverse Z-Transform Musa Mohd Mokji Inverse Z-Transform Transform from z-domain to time-domain x n = 1 2πj c X z z n 1 dz Note that the mathematical operation for the
More informationy[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)
7. The Z-transform 7. Definition of the Z-transform We saw earlier that complex exponential of the from {e jwn } is an eigen function of for a LTI System. We can generalize this for signals of the form
More informationChapter Intended Learning Outcomes: (i) Understanding the relationship between transform and the Fourier transform for discrete-time signals
z Transform Chapter Intended Learning Outcomes: (i) Understanding the relationship between transform and the Fourier transform for discrete-time signals (ii) Understanding the characteristics and properties
More informationChapter 13 Z Transform
Chapter 13 Z Transform 1. -transform 2. Inverse -transform 3. Properties of -transform 4. Solution to Difference Equation 5. Calculating output using -transform 6. DTFT and -transform 7. Stability Analysis
More informationSignals and Systems Lecture 8: Z Transform
Signals and Systems Lecture 8: Z Transform Farzaneh Abdollahi Department of Electrical Engineering Amirkabir University of Technology Winter 2012 Farzaneh Abdollahi Signal and Systems Lecture 8 1/29 Introduction
More information8. z-domain Analysis of Discrete-Time Signals and Systems
8. z-domain Analysis of Discrete-Time Signals and Systems 8.. Definition of z-transform (0.0-0.3) 8.2. Properties of z-transform (0.5) 8.3. System Function (0.7) 8.4. Classification of a Linear Time-Invariant
More informationThe Z transform (2) 1
The Z transform (2) 1 Today Properties of the region of convergence (3.2) Read examples 3.7, 3.8 Announcements: ELEC 310 FINAL EXAM: April 14 2010, 14:00 pm ECS 123 Assignment 2 due tomorrow by 4:00 pm
More informationESE 531: Digital Signal Processing
ESE 531: Digital Signal Processing Lec 6: January 30, 2018 Inverse z-transform Lecture Outline! z-transform " Tie up loose ends " Regions of convergence properties! Inverse z-transform " Inspection " Partial
More informationLet H(z) = P(z)/Q(z) be the system function of a rational form. Let us represent both P(z) and Q(z) as polynomials of z (not z -1 )
Review: Poles and Zeros of Fractional Form Let H() = P()/Q() be the system function of a rational form. Let us represent both P() and Q() as polynomials of (not - ) Then Poles: the roots of Q()=0 Zeros:
More informationX (z) = n= 1. Ã! X (z) x [n] X (z) = Z fx [n]g x [n] = Z 1 fx (z)g. r n x [n] ª e jnω
3 The z-transform ² Two advantages with the z-transform:. The z-transform is a generalization of the Fourier transform for discrete-time signals; which encompasses a broader class of sequences. The z-transform
More informationDigital Signal Processing
Digital Signal Processing Introduction Moslem Amiri, Václav Přenosil Embedded Systems Laboratory Faculty of Informatics, Masaryk University Brno, Czech Republic amiri@mail.muni.cz prenosil@fi.muni.cz February
More informationThe Laplace Transform
The Laplace Transform Introduction There are two common approaches to the developing and understanding the Laplace transform It can be viewed as a generalization of the CTFT to include some signals with
More informationESE 531: Digital Signal Processing
ESE 531: Digital Signal Processing Lec 6: January 31, 2017 Inverse z-transform Lecture Outline! z-transform " Tie up loose ends " Regions of convergence properties! Inverse z-transform " Inspection " Partial
More informationThe Z-Transform. Fall 2012, EE123 Digital Signal Processing. Eigen Functions of LTI System. Eigen Functions of LTI System
The Z-Transform Fall 202, EE2 Digital Signal Processing Lecture 4 September 4, 202 Used for: Analysis of LTI systems Solving di erence equations Determining system stability Finding frequency response
More information7.17. Determine the z-transform and ROC for the following time signals: Sketch the ROC, poles, and zeros in the z-plane. X(z) = x[n]z n.
Solutions to Additional Problems 7.7. Determine the -transform and ROC for the following time signals: Sketch the ROC, poles, and eros in the -plane. (a) x[n] δ[n k], k > 0 X() x[n] n n k, 0 Im k multiple
More information6.003: Signals and Systems
6.003: Signals and Systems Z Transform September 22, 2011 1 2 Concept Map: Discrete-Time Systems Multiple representations of DT systems. Delay R Block Diagram System Functional X + + Y Y Delay Delay X
More informationDiscrete-Time Signals and Systems. Frequency Domain Analysis of LTI Systems. The Frequency Response Function. The Frequency Response Function
Discrete-Time Signals and s Frequency Domain Analysis of LTI s Dr. Deepa Kundur University of Toronto Reference: Sections 5., 5.2-5.5 of John G. Proakis and Dimitris G. Manolakis, Digital Signal Processing:
More informationSignals & Systems Handout #4
Signals & Systems Handout #4 H-4. Elementary Discrete-Domain Functions (Sequences): Discrete-domain functions are defined for n Z. H-4.. Sequence Notation: We use the following notation to indicate the
More informationDiscrete-Time Fourier Transform (DTFT)
Discrete-Time Fourier Transform (DTFT) 1 Preliminaries Definition: The Discrete-Time Fourier Transform (DTFT) of a signal x[n] is defined to be X(e jω ) x[n]e jωn. (1) In other words, the DTFT of x[n]
More information! z-transform. " Tie up loose ends. " Regions of convergence properties. ! Inverse z-transform. " Inspection. " Partial fraction
Lecture Outline ESE 53: Digital Signal Processing Lec 6: January 3, 207 Inverse z-transform! z-transform " Tie up loose ends " gions of convergence properties! Inverse z-transform " Inspection " Partial
More informationECE503: Digital Signal Processing Lecture 4
ECE503: Digital Signal Processing Lecture 4 D. Richard Brown III WPI 06-February-2012 WPI D. Richard Brown III 06-February-2012 1 / 29 Lecture 4 Topics 1. Motivation for the z-transform. 2. Definition
More informationSIGNALS AND SYSTEMS. Unit IV. Analysis of DT signals
SIGNALS AND SYSTEMS Unit IV Analysis of DT signals Contents: 4.1 Discrete Time Fourier Transform 4.2 Discrete Fourier Transform 4.3 Z Transform 4.4 Properties of Z Transform 4.5 Relationship between Z
More informationThe Z transform (2) Alexandra Branzan Albu ELEC 310-Spring 2009-Lecture 28 1
The Z transform (2) Alexandra Branzan Albu ELEC 310-Spring 2009-Lecture 28 1 Outline Properties of the region of convergence (10.2) The inverse Z-transform (10.3) Definition Computational techniques Alexandra
More informationA system that is both linear and time-invariant is called linear time-invariant (LTI).
The Cooper Union Department of Electrical Engineering ECE111 Signal Processing & Systems Analysis Lecture Notes: Time, Frequency & Transform Domains February 28, 2012 Signals & Systems Signals are mapped
More informationEE123 Digital Signal Processing. M. Lustig, EECS UC Berkeley
EE123 Digital Signal Processing Today Last time: DTFT - Ch 2 Today: Continue DTFT Z-Transform Ch. 3 Properties of the DTFT cont. Time-Freq Shifting/modulation: M. Lustig, EE123 UCB M. Lustig, EE123 UCB
More informationSignals and Systems. Spring Room 324, Geology Palace, ,
Signals and Systems Spring 2013 Room 324, Geology Palace, 13756569051, zhukaiguang@jlu.edu.cn Chapter 10 The Z-Transform 1) Z-Transform 2) Properties of the ROC of the z-transform 3) Inverse z-transform
More informationZ-Transform. The Z-transform is the Discrete-Time counterpart of the Laplace Transform. Laplace : G(s) = g(t)e st dt. Z : G(z) =
Z-Transform The Z-transform is the Discrete-Time counterpart of the Laplace Transform. Laplace : G(s) = Z : G(z) = It is Used in Digital Signal Processing n= g(t)e st dt g[n]z n Used to Define Frequency
More informationDSP-I DSP-I DSP-I DSP-I
DSP-I DSP-I DSP-I DSP-I Digital Signal Processing I (8-79) Fall Semester, 005 OTES FOR 8-79 LECTURE 9: PROPERTIES AD EXAPLES OF Z-TRASFORS Distributed: September 7, 005 otes: This handout contains in outline
More informationEE Homework 5 - Solutions
EE054 - Homework 5 - Solutions 1. We know the general result that the -transform of α n 1 u[n] is with 1 α 1 ROC α < < and the -transform of α n 1 u[ n 1] is 1 α 1 with ROC 0 < α. Using this result, the
More informationHow to manipulate Frequencies in Discrete-time Domain? Two Main Approaches
How to manipulate Frequencies in Discrete-time Domain? Two Main Approaches Difference Equations (an LTI system) x[n]: input, y[n]: output That is, building a system that maes use of the current and previous
More informationTopic 4: The Z Transform
ELEN E480: Digital Signal Processing Topic 4: The Z Transform. The Z Transform 2. Inverse Z Transform . The Z Transform Powerful tool for analyzing & designing DT systems Generalization of the DTFT: G(z)
More informationThe Laplace Transform
The Laplace Transform Generalizing the Fourier Transform The CTFT expresses a time-domain signal as a linear combination of complex sinusoids of the form e jωt. In the generalization of the CTFT to the
More informationSignal Analysis, Systems, Transforms
Michael J. Corinthios Signal Analysis, Systems, Transforms Engineering Book (English) August 29, 2007 Springer Contents Discrete-Time Signals and Systems......................... Introduction.............................................2
More informationDIGITAL SIGNAL PROCESSING UNIT 1 SIGNALS AND SYSTEMS 1. What is a continuous and discrete time signal? Continuous time signal: A signal x(t) is said to be continuous if it is defined for all time t. Continuous
More informationZ-Transform. 清大電機系林嘉文 Original PowerPoint slides prepared by S. K. Mitra 4-1-1
Chapter 6 Z-Transform 清大電機系林嘉文 cwlin@ee.nthu.edu.tw 03-5731152 Original PowerPoint slides prepared by S. K. Mitra 4-1-1 z-transform The DTFT provides a frequency-domain representation of discrete-time
More informationEE102B Signal Processing and Linear Systems II. Solutions to Problem Set Nine Spring Quarter
EE02B Signal Processing and Linear Systems II Solutions to Problem Set Nine 202-203 Spring Quarter Problem 9. (25 points) (a) 0.5( + 4z + 6z 2 + 4z 3 + z 4 ) + 0.2z 0.4z 2 + 0.8z 3 x[n] 0.5 y[n] -0.2 Z
More informationz-transform Chapter 6
z-transform Chapter 6 Dr. Iyad djafar Outline 2 Definition Relation Between z-transform and DTFT Region of Convergence Common z-transform Pairs The Rational z-transform The Inverse z-transform z-transform
More informationAS PURE MATHS REVISION NOTES
AS PURE MATHS REVISION NOTES 1 SURDS A root such as 3 that cannot be written exactly as a fraction is IRRATIONAL An expression that involves irrational roots is in SURD FORM e.g. 2 3 3 + 2 and 3-2 are
More informationIf every Bounded Input produces Bounded Output, system is externally stable equivalently, system is BIBO stable
1. External (BIBO) Stability of LTI Systems If every Bounded Input produces Bounded Output, system is externally stable equivalently, system is BIBO stable g(n) < BIBO Stability Don t care about what unbounded
More information# 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.
[ ] = 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 ( )
More informationEE 521: Instrumentation and Measurements
Aly El-Osery Electrical Engineering Department, New Mexico Tech Socorro, New Mexico, USA November 1, 2009 1 / 27 1 The z-transform 2 Linear Time-Invariant System 3 Filter Design IIR Filters FIR Filters
More informationVI. Z Transform and DT System Analysis
Summer 2008 Signals & Systems S.F. Hsieh VI. Z Transform and DT System Analysis Introduction Why Z transform? a DT counterpart of the Laplace transform in CT. Generalization of DT Fourier transform: z
More informationVery useful for designing and analyzing signal processing systems
z-transform z-transform The z-transform generalizes the Discrete-Time Fourier Transform (DTFT) for analyzing infinite-length signals and systems Very useful for designing and analyzing signal processing
More informationZ-TRANSFORMS. Solution: Using the definition (5.1.2), we find: for case (b). y(n)= h(n) x(n) Y(z)= H(z)X(z) (convolution) (5.1.
84 5. Z-TRANSFORMS 5 z-transforms Solution: Using the definition (5..2), we find: for case (a), and H(z) h 0 + h z + h 2 z 2 + h 3 z 3 2 + 3z + 5z 2 + 2z 3 H(z) h 0 + h z + h 2 z 2 + h 3 z 3 + h 4 z 4
More informationThe Laplace Transform
The Laplace Transform Syllabus ECE 316, Spring 2015 Final Grades Homework (6 problems per week): 25% Exams (midterm and final): 50% (25:25) Random Quiz: 25% Textbook M. Roberts, Signals and Systems, 2nd
More informationELEN E4810: Digital Signal Processing Topic 4: The Z Transform. 1. The Z Transform. 2. Inverse Z Transform
ELEN E480: Digital Signal Processing Topic 4: The Z Transform. The Z Transform 2. Inverse Z Transform . The Z Transform Powerful tool for analyzing & designing DT systems Generalization of the DTFT: G(z)
More informationAdvanced Training Course on FPGA Design and VHDL for Hardware Simulation and Synthesis
065-3 Advanced Training Course on FPGA Design and VHDL for Hardware Simulation and Synthesis 6 October - 0 November, 009 Digital Signal Processing The z-transform Massimiliano Nolich DEEI Facolta' di Ingegneria
More informationThe z-transform Part 2
http://faculty.kfupm.edu.sa/ee/muqaibel/ The z-transform Part 2 Dr. Ali Hussein Muqaibel The material to be covered in this lecture is as follows: Properties of the z-transform Linearity Initial and final
More informationLecture 7 Discrete Systems
Lecture 7 Discrete Systems EE 52: Instrumentation and Measurements Lecture Notes Update on November, 29 Aly El-Osery, Electrical Engineering Dept., New Mexico Tech 7. Contents The z-transform 2 Linear
More informationModule 4 : Laplace and Z Transform Problem Set 4
Module 4 : Laplace and Z Transform Problem Set 4 Problem 1 The input x(t) and output y(t) of a causal LTI system are related to the block diagram representation shown in the figure. (a) Determine a differential
More informationUse: Analysis of systems, simple convolution, shorthand for e jw, stability. Motivation easier to write. Or X(z) = Z {x(n)}
1 VI. Z Transform Ch 24 Use: Analysis of systems, simple convolution, shorthand for e jw, stability. A. Definition: X(z) = x(n) z z - transforms Motivation easier to write Or Note if X(z) = Z {x(n)} z
More informationReview of Fundamentals of Digital Signal Processing
Chapter 2 Review of Fundamentals of Digital Signal Processing 2.1 (a) This system is not linear (the constant term makes it non linear) but is shift-invariant (b) This system is linear but not shift-invariant
More informationDigital Signal Processing. Lecture Notes and Exam Questions DRAFT
Digital Signal Processing Lecture Notes and Exam Questions Convolution Sum January 31, 2006 Convolution Sum of Two Finite Sequences Consider convolution of h(n) and g(n) (M>N); y(n) = h(n), n =0... M 1
More informationChapter 6: The Laplace Transform. Chih-Wei Liu
Chapter 6: The Laplace Transform Chih-Wei Liu Outline Introduction The Laplace Transform The Unilateral Laplace Transform Properties of the Unilateral Laplace Transform Inversion of the Unilateral Laplace
More informationChap 2. Discrete-Time Signals and Systems
Digital Signal Processing Chap 2. Discrete-Time Signals and Systems Chang-Su Kim Discrete-Time Signals CT Signal DT Signal Representation 0 4 1 1 1 2 3 Functional representation 1, n 1,3 x[ n] 4, n 2 0,
More information2.161 Signal Processing: Continuous and Discrete Fall 2008
IT OpenCourseWare http://ocw.mit.edu 2.161 Signal Processing: Continuous and Discrete all 2008 or information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. assachusetts
More informationLecture 04: Discrete Frequency Domain Analysis (z-transform)
Lecture 04: Discrete Frequency Domain Analysis (z-transform) John Chiverton School of Information Technology Mae Fah Luang University 1st Semester 2009/ 2552 Outline Overview Lecture Contents Introduction
More informationUniversity of Illinois at Urbana-Champaign ECE 310: Digital Signal Processing
University of Illinois at Urbana-Champaign ECE 0: Digital Signal Processing Chandra Radhakrishnan PROBLEM SET : SOLUTIONS Peter Kairouz Problem. Hz z 7 z +/9, causal ROC z > contains the unit circle BIBO
More informationSolutions: Homework Set # 5
Signal Processing for Communications EPFL Winter Semester 2007/2008 Prof. Suhas Diggavi Handout # 22, Tuesday, November, 2007 Solutions: Homework Set # 5 Problem (a) Since h [n] = 0, we have (b) We can
More informationLike bilateral Laplace transforms, ROC must be used to determine a unique inverse z-transform.
Inversion of the z-transform Focus on rational z-transform of z 1. Apply partial fraction expansion. Like bilateral Laplace transforms, ROC must be used to determine a unique inverse z-transform. Let X(z)
More information5 Z-Transform and difference equations
5 Z-Transform and difference equations Z-transform - Elementary properties - Inverse Z-transform - Convolution theorem - Formation of difference equations - Solution of difference equations using Z-transform.
More informationECE-314 Fall 2012 Review Questions for Midterm Examination II
ECE-314 Fall 2012 Review Questions for Midterm Examination II First, make sure you study all the problems and their solutions from homework sets 4-7. Then work on the following additional problems. Problem
More information1 + lim. n n+1. f(x) = x + 1, x 1. and we check that f is increasing, instead. Using the quotient rule, we easily find that. 1 (x + 1) 1 x (x + 1) 2 =
Chapter 5 Sequences and series 5. Sequences Definition 5. (Sequence). A sequence is a function which is defined on the set N of natural numbers. Since such a function is uniquely determined by its values
More informationThe z-transform and Discrete-Time LTI Systems
Chapter 4 The z-transform and Discrete-Time LTI Systems 4.1 INTRODUCTION In Chap. 3 we introduced the Laplace transform. In this chapter we present the z-transform, which is the discrete-time counterpart
More informationUnit 2: Modeling in the Frequency Domain Part 2: The Laplace Transform. The Laplace Transform. The need for Laplace
Unit : Modeling in the Frequency Domain Part : Engineering 81: Control Systems I Faculty of Engineering & Applied Science Memorial University of Newfoundland January 1, 010 1 Pair Table Unit, Part : Unit,
More informationEC Signals and Systems
UNIT I CLASSIFICATION OF SIGNALS AND SYSTEMS Continuous time signals (CT signals), discrete time signals (DT signals) Step, Ramp, Pulse, Impulse, Exponential 1. Define Unit Impulse Signal [M/J 1], [M/J
More informationDigital Signal Processing:
Digital Signal Processing: Mathematical and algorithmic manipulation of discretized and quantized or naturally digital signals in order to extract the most relevant and pertinent information that is carried
More information1. Z-transform: Initial value theorem for causal signal. = u(0) + u(1)z 1 + u(2)z 2 +
1. Z-transform: Initial value theorem for causal signal u(0) lim U(z) if the limit exists z U(z) u(k)z k u(k)z k k lim U(z) u(0) z k0 u(0) + u(1)z 1 + u(2)z 2 + CL 692 Digital Control, IIT Bombay 1 c Kannan
More informationSection Taylor and Maclaurin Series
Section.0 Taylor and Maclaurin Series Ruipeng Shen Feb 5 Taylor and Maclaurin Series Main Goal: How to find a power series representation for a smooth function us assume that a smooth function has a power
More informationDiscrete Systems. Step response and pole locations. Mark Cannon. Hilary Term Lecture
Discrete Systems Mark Cannon Hilary Term 22 - Lecture 4 Step response and pole locations 4 - Review Definition of -transform: U() = Z{u k } = u k k k= Discrete transfer function: Y () U() = G() = Z{g k},
More informationTransform Analysis of Linear Time-Invariant Systems
Transform Analysis of Linear Time-Invariant Systems Discrete-Time Signal Processing Chia-Ping Chen Department of Computer Science and Engineering National Sun Yat-Sen University Kaohsiung, Taiwan ROC Transform
More informationDiscrete Time Systems
1 Discrete Time Systems {x[0], x[1], x[2], } H {y[0], y[1], y[2], } Example: y[n] = 2x[n] + 3x[n-1] + 4x[n-2] 2 FIR and IIR Systems FIR: Finite Impulse Response -- non-recursive y[n] = 2x[n] + 3x[n-1]
More informationLecture 2 OKAN UNIVERSITY FACULTY OF ENGINEERING AND ARCHITECTURE
OKAN UNIVERSITY FACULTY OF ENGINEERING AND ARCHITECTURE EEE 43 DIGITAL SIGNAL PROCESSING (DSP) 2 DIFFERENCE EQUATIONS AND THE Z- TRANSFORM FALL 22 Yrd. Doç. Dr. Didem Kivanc Tureli didemk@ieee.org didem.kivanc@okan.edu.tr
More informationLecture 18: Stability
Lecture 18: Stability ECE 401: Signal and Image Analysis University of Illinois 4/18/2017 1 Stability 2 Impulse Response 3 Z Transform Outline 1 Stability 2 Impulse Response 3 Z Transform BIBO Stability
More informationECE 421 Introduction to Signal Processing
ECE 421 Introduction to Signal Processing Dror Baron Assistant Professor Dept. of Electrical and Computer Engr. North Carolina State University, NC, USA The z-transform [Reading material: Sections 3.1-3.3]
More informationDigital Signal Processing. Outline. Hani Mehrpouyan, The Z-Transform. The Inverse Z-Transform
Digital Signal Processing Hani Mehrpouyan, Department of Electrical and Computer Engineering, California State University, Bakersfield Lecture (Digital Signal Processing) Feb 9 th, 206 The following references
More informationAnalog vs. discrete signals
Analog vs. discrete signals Continuous-time signals are also known as analog signals because their amplitude is analogous (i.e., proportional) to the physical quantity they represent. Discrete-time signals
More informationCh. 7: Z-transform Reading
c J. Fessler, June 9, 3, 6:3 (student version) 7. Ch. 7: Z-transform Definition Properties linearity / superposition time shift convolution: y[n] =h[n] x[n] Y (z) =H(z) X(z) Inverse z-transform by coefficient
More informationReview of Fundamentals of Digital Signal Processing
Solution Manual for Theory and Applications of Digital Speech Processing by Lawrence Rabiner and Ronald Schafer Click here to Purchase full Solution Manual at http://solutionmanuals.info Link download
More informationQUESTION BANK SIGNALS AND SYSTEMS (4 th SEM ECE)
QUESTION BANK SIGNALS AND SYSTEMS (4 th SEM ECE) 1. For the signal shown in Fig. 1, find x(2t + 3). i. Fig. 1 2. What is the classification of the systems? 3. What are the Dirichlet s conditions of Fourier
More informationZ - Transform. It offers the techniques for digital filter design and frequency analysis of digital signals.
Z - Transform The z-transform is a very important tool in describing and analyzing digital systems. It offers the techniques for digital filter design and frequency analysis of digital signals. Definition
More informationLecture 16: Zeros and Poles
Lecture 16: Zeros and Poles ECE 401: Signal and Image Analysis University of Illinois 4/6/2017 1 Poles and Zeros at DC 2 Poles and Zeros with Nonzero Bandwidth 3 Poles and Zeros with Nonzero Center Frequency
More informationDiscrete Fourier Transform
Discrete Fourier Transform Virtually all practical signals have finite length (e.g., sensor data, audio records, digital images, stock values, etc). Rather than considering such signals to be zero-padded
More informationTransform Representation of Signals
C H A P T E R 3 Transform Representation of Signals and LTI Systems As you have seen in your prior studies of signals and systems, and as emphasized in the review in Chapter 2, transforms play a central
More informationDiscrete-time Signals and Systems in
Discrete-time Signals and Systems in the Frequency Domain Chapter 3, Sections 3.1-39 3.9 Chapter 4, Sections 4.8-4.9 Dr. Iyad Jafar Outline Introduction The Continuous-Time FourierTransform (CTFT) The
More information( ) John A. Quinn Lecture. ESE 531: Digital Signal Processing. Lecture Outline. Frequency Response of LTI System. Example: Zero on Real Axis
John A. Quinn Lecture ESE 531: Digital Signal Processing Lec 15: March 21, 2017 Review, Generalized Linear Phase Systems Penn ESE 531 Spring 2017 Khanna Lecture Outline!!! 2 Frequency Response of LTI System
More informationEE 225a Digital Signal Processing Supplementary Material
EE 225A DIGITAL SIGAL PROCESSIG SUPPLEMETARY MATERIAL EE 225a Digital Signal Processing Supplementary Material. Allpass Sequences A sequence h a ( n) ote that this is true iff which in turn is true iff
More informationTable 1: Properties of the Continuous-Time Fourier Series. Property Periodic Signal Fourier Series Coefficients
able : Properties of the Continuous-ime Fourier Series x(t = a k e jkω0t = a k = x(te jkω0t dt = a k e jk(/t x(te jk(/t dt Property Periodic Signal Fourier Series Coefficients x(t y(t } Periodic with period
More information1. Use the properties of exponents to simplify the following expression, writing your answer with only positive exponents.
Math120 - Precalculus. Final Review. Fall, 2011 Prepared by Dr. P. Babaali 1 Algebra 1. Use the properties of exponents to simplify the following expression, writing your answer with only positive exponents.
More informationDigital Signal Processing Lecture 9 - Design of Digital Filters - FIR
Digital Signal Processing - Design of Digital Filters - FIR Electrical Engineering and Computer Science University of Tennessee, Knoxville November 3, 2015 Overview 1 2 3 4 Roadmap Introduction Discrete-time
More information