Signal Analysis, Systems, Transforms

Transcription

1 Michael J. Corinthios Signal Analysis, Systems, Transforms Engineering Book (English) August 29, 2007 Springer

2 Contents Discrete-Time Signals and Systems Introduction Linear Constant-Coefficient Difference Equations The Z-Transform Convergence of the Z-Transform Inverse Z-Transform Inversion by Long Division Inversion by a Power Series Expansion Some Properties of the Z-Transform Initial Value Right-Sided Sequence Left-Sided Sequence Convolution in Time Convolution in Frequency Tables of Z-Transform Parseval s Relation Final Value Theorem Multiplication by an Exponential Frequency Translation Reflection Property Multiplication by n Unilateral Z-Transform Time Shift Property of Unilateral z-transform Causality and Stability Delayed Response and Group Delay Discrete-Time Convolution Discrete-Time Correlation in D Convolution and Correlation as Multiplications Two-Dimensional Signals Linear Systems, Convolution and Correlation Correlation of Images Correlation in 2D Discrete-Time Allpass Systems Minimum-Phase and Inverse System

3 II Contents References

4 . Discrete-Time Signals and Systems. Introduction A discrete-time signal is a sequence v[n] which is defined for every integer value n. A discrete-time signal has the general appearance of the sequence v [n] shown in Fig... A sequence x[n] may be obtained by discrete-time Fig... Discrete time signal. sampling or discretization of a continuous-time signal x c (t). The process may be represented as shown in Fig..2, where the switch is closed periodically every T seconds producing at the output a discrete-time signal, namely, the sequence x[n] = x c (t) t=nt = x c (nt). (.) The sampling frequency is f s = /T. If the input is a sinusoid such as xc( t) x [ n] T Fig..2. Discretization of a continuous time signal. the resulting sequence is x c (t) = sin(βt) (.2) x[n] = sin(βnt) sin(γt). (.3) In other words, a continuous-time signal of frequency β results in a sequence of frequency γ = βt. This relation is an important one. If, as will be the

5 2 Discrete-Time Signals and Systems case in the following chapter, the frequency in the continuous time domain is denoted by ω and that in the discrete time domain is denoted Ω then the relation between the frequencies in the two domains is Ω = ωt. (.4) Example. Evaluate the frequency of the sequence that is the discrete-time sampling of the continuous-time signal x c (t) = cos(000πt + 0.π), with a sampling frequency of f s = 2000 Hz. Let the frequency of x c (t) be denoted β rad/sec, and equivalently f c = β/(2π) Hz. We have β = 000π rad/sec and f c = 000/pi/(2π) = 500 Hz. The sampling period is T = /f c = sec. The corresponding discrete time sequence is x[n] = x c (nt) = cos(000πn π) = cos(2πn+0.π), of frequency γ = 2π, which is simply γ = βt..2 Linear Constant-Coefficient Difference Equations Linear time-invariant LTI discrete-time systems are characterized by linear constant-coefficient difference equations of the form N M d k y[n k] = c k v[n k] (.5) k=0 where v [n] is the system input and y [n] its response. We can write N M y[n] = (d k /d 0 )y[n k] + (c k /d 0 )v[n k] k= k= k=0 k=0 N M = a k y[n k] + b k v[n k] where a k = d k /d 0 and b k = c k /d 0..3 The Z-Transform k=0 (.6) The bilateral or two-sided z-transform of a sequence v[n] is defined by V (z) = n= v [n] z n (.7) where z is a complex variable and v[n] is assumed to be a general two-sided sequence, that is, a sequence defined over the entire interval < n <. The bilateral z-transform of a sequence v[n] will be denoted V (z). It will

6 .4 Convergence of the Z-Transform 3 alternatively be denoted V II (z) to put into evidence the fact that it is the two-sided transform. For the bilateral transform we may write and V II (z) = Z (v [n]) (.8) v [n] Z V II (z) (.9) meaning that V II (z) is the z-transform of v[n]. As we shall see shortly, similarly to Laplace transform a unilateral or onesided z-transform also exists and is particularly useful for the solution of linear constant-coefficient difference equations with initial conditions. The following remarks parallel those made in the context of Laplace transform. The z-transform on the unit circle is given by V ( e jω) = n= v[n]e jωn (.0) and is in fact, by definition, the Fourier transform of the sequence v[n]..4 Convergence of the Z-Transform The Fourier transform V ( e jω) is given by an infinite sum of a power series. This series converges uniformly if the sequence is absolutely summable. Similarly, the z-transform in polar form is given by an infinite sum that is absolutely convergent if the sequence v[n]r n is absolutely summable, i.e., n= v[n]r n <. (.) In general, this condition is satisfied in an annular region in the z plane, referred to as the region of convergence ROC, namely, the region: r < z < r 2. (.2) The radii r and r 2 are functions of the sequence v[n] and may thus be also denoted r v and r v2. In particular, we have the following four cases. Case : Finite Duration Sequence For a sequence v[n] of finite duration [n, n 2 ] and nil elsewhere which is of finite value over the interval, i.e., v[n] < for n n n 2. We have V (z) = n 2 n=n v [n] z n = v [n ] z n + v [n + ]z n v [n 2 ] z n2. (.3)

7 4 Discrete-Time Signals and Systems Fig..3. Finite duration sequence and its ROC. The region of convergence is therefore the whole z plane except z = for n < 0 and z = 0 for n 2 > 0, so that, the region of convergence is 0 < z < and may include z = 0 or z =. The sequence and its region of convergence are shown in Fig..3. Case 2: Right-Sided Sequence A right-sided sequence v[n] is one that extends to the right on the n axis starting from a finite value n. In other words it is nil for n < n. In this case we have V (z) = v [n]z n = v[n ]z n +v[n +]z (n+) +v[n +2]z (n+2) n=n (.4) The region of convergence is the exterior of a circle, i.e. z > r, except z = if n < 0; see Fig..4. If the sequence is causal, n 0, the region of Fig..4. Right-sided sequence and its ROC. convergence includes z =. Case 3: Left-Sided Sequence A left-sided sequence v[n] is one that extends to the left on the n axis starting from a finite value n 2. In other words it is nil for n > n 2. We have n 2 V (z) = v [n] z n = v [ m] z m n= m= n 2 = v[n 2 ]z n2 + v[n 2 ]z (n2 ) + v[n 2 2]z (n2 2) +... which is the same as the expression of V (z) in the previous case except for the replacement of n by n and z by z. The region of convergence is therefore

8 .5 Inverse Z-Transform 5 Fig..5. Left-sided sequence and its ROC. z < r 2,except z = 0 if n 2 > 0; see Fig..5. We note that the z-transform of an anti-causal sequence, a sequence that is nil for n > 0, converges for z = 0. Case 4: General Two-Sided Sequence Given a general two-sided sequence v[n] we have V (z) = n= v [n] z n = v [n] z n + n=0 n= v [n]z n. (.5) The first term converges for z > r, the second term for z < r 2, wherefrom there is convergence if and only if r < r 2 and the region of convergence is the annular region r < z < r 2. (.6) The z-transform of the sum of left-handed sequences has a region of convergence that is the interior of the circle that passes through the pole(s) of least radius. For illustration purposes some basic one-sided sequences are shown together with their regions of convergence in the Laplace s plane and in the z plane, respectively, in Fig..6. Two-sided sequences are similarly shown with their regions of convergence in the s and z plane, in Fig Inverse Z-Transform The inverse z-transform can be derived as follows. We have by definition V (z) = n= v[n]z n. (.7) Multiplying both sides by z k and integrating we have V (z)z k dz = v[n]z n+k dz (.8) n= where the integration sign denotes a counter clockwise circular contour centered at the origin. Assuming uniform convergence, the order of summation and integration can be reversed, wherefrom

9 6 Discrete-Time Signals and Systems j z plane n a = e n s plane a 0 n a 0 a n j a 0 n 0 j a n a 0 a n n 0 n 0 0 Fig..6. Right- and left-sided sequences and ROC. V (z)z k dz = v[n]z n+k dz. (.9) n= Now, according to Cauchy s integration theorem { 2πj, k = 0 z k dz = 0, k 0 C j a (.20) where C is a counter-clockwise circular contour that is centered at the origin, wherefrom V (z)z k dz = 2πj v[k] (.2) and replacing k by n C

10 .5 Inverse Z-Transform 7 j a 2 n a n a a 2 0 n 0 2 a n j a 2 n a a 2 0 n 0 2 j a 2 n a n a 2 a 0 n 2 0 a 2 n a n j 2 a 2 a a 2 n 0 n a n n 0 2 Fig..7. Two-sided sequences and ROC. v[n] = V (z)z n dz (.22) 2πj C where the contour C is in the region of convergence of V (z) and encircles the origin. This is the inverse z-transform. Replacing z by e jω we obtain the inverse Fourier transform v [n] = ˆ π V ( e jω) e jωn dω. (.23) 2π π If V (z) is rational, the ratio of two polynomials, the residue theorem may be applied to evaluate this last equation. We can write j a 2 a v [n] = [ residues of V (z)z n at its poles inside C ]. (.24) If V (z)z n is a rational function in z we can write

11 8 Discrete-Time Signals and Systems V (z)z n = F (z) (z z 0 ) m (.25) where V (z)z n has a pole of order m at z = z 0 and F (z) has no poles at z = z 0. The residues of V (z)z n at z = z 0 are given by Res [ V (z)z n at z = z 0 ] = (m )! In particular, for the case of a simple pole (m = ) Example.2 Let [ d m ] F (z) dz m z=z 0. (.26) Res [ V (z)z n at z = z 0 ] = F (z0 ). (.27) V (z) =.5z z 2, 0.5 < z < z + The region of convergence implies a two-sided sequence. The poles are shown in Fig..8. Alternatively, we can may write Fig..8. Annular region of convergence. v [n] = 2πj.5z n (z 2)(z 0.5) dz. For n 0 the circle C encloses the pole z = 0.5, as shown in Fig..9. We have [.5z n ].5 (0.5)n v[n] = Res at z = 0.5 = = (0.5) n. (z 2)(z 0.5) For n < 0 the circle C encloses a simple pole at z = 0.5 and a pole of order n at z = 0, as shown in the figure. Writing m = n we have

12 .5 Inverse Z-Transform 9 C ( n) C Fig..9. Contour of integration in ROC. [ ].5 v[n] = Res at z = 0.5 (z 2)(z 0.5)zm [ ].5 + Res (z 2)(z 0.5)z m at z = 0 [ ] = (0.5) m d m.5 + (m )! dz m (z 2)(z 0.5) z=0 (0.5) n + v 2 [n]. Now if m =, i.e. n =, and For m = 2, i.e. n = 2 v 2 [n] =.5 d dz For m = 3 we obtain v 2 [n] =.5 2 ( 0.5) =.5 v[n] = (0.5).5 = 0.5. z 2 2.5z + = z=0 v[n] = (0.5) = (2z 2.5) (z 2 2.5z + ) 2 z=0 v 2 [n] =.5 d2 2 dz 2 z 2 2.5z + =.5 { (z 2 2.5z + ) 2 [ (2)] + (2z 2.5) 2 2 ( z 2 2.5z + ) (2z 2.5) } { (z / 2 2.5z + ) } 4 z=0 = (3/2) (2/4) = 63/8 We note that for n < 0, v[n] = 2 n v[n] = (0.5) 3 63/8 = 8 63/8 = 2 3. so that

13 0 Discrete-Time Signals and Systems { 2 v[n] = n, n 0 2 n, n < 0. The successive differentiation is needed due to the multiple pole at z = 0. We can avoid such complication by using the substitution z = x in obtaining v[n] = V (z)z n dz 2πj C v[n] = fi 2πj C 2 V ( ) x n dx x where the contour of integration C 2 is now clockwise. Reversing the contour direction we have v[n] = ( ) x n dx 2πj x C 2 V where the direction of integration is now anti-clockwise. We note that if the circle C is of radius r, the circle C 2 is of radius /r. Moreover the poles of V (z) that are inside C are moved by this transformation to outside the new contour. Example.3 Evaluate the inverse transform of the last example for n < 0 using the transformation z = /x. We write v[n] = 2πj = 2πj C 2.5x x n (x 2)(x dx = 0.5) 2πj.5x n (x 0.5)(x 2) dx with n 0 [.5x n v[n] = Res (x 0.5) (x 2) C 2.5x n ( 2x) ( 0.5x) dx ].5 (0.5) n at x = 0.5 = = 2 n, n 0..5 The contour C 2 is shown in Fig..0.The contour encloses a pole at z = 0.5 for n 0.

14 .7 Inversion by a Power Series Expansion Fig..0. Circular contour in z plane..6 Inversion by Long Division Another approach to evaluate the inverse z-transform is the use of a long division. Example.4 Evaluate the inverse transform of V (z) = az a, z < a. z The region of convergence implies a left-sided sequence. The result of the division should reflect this fact as increasing powers of z. We write wherefrom az + + a z + a 2 z 2... a z az az a z a z a z a 2 z 2... V (z) = az + + a z + a 2 z = n= a n z n v[n] = a n u[ n]..7 Inversion by a Power Series Expansion If V (z) can be expressed as a power series in powers of z we would be able to identify the sequence v[n].

15 2 Discrete-Time Signals and Systems Example.5 Using a power series expansion, evaluate the inverse z-transform of V (z) = ( + az ) 2. We have the expansion ( + x) 2 = 2x + 3x 2 4x 3 + 5x 4..., < x <. We can therefore write By definition V (z) = ( + az ) 2 = ( ) n (n + )a n z n, z > a. wherefrom v[n] is the sequence n=0 V (z) = n= v[n]z n v[n] = ( ) n (n + )a n u[n]. Example.6 Using a power series expansion evaluate the inverse z-transform of X (z) = log ( + az ), z > a. Using the power series expansion of the log function we can write X (z) = ( ) n+ a n z n n= x[n] = ( ) n+ a n u[n ]. n.8 Some Properties of the Z-Transform.8. Initial Value Let v[n] be a causal sequence. We have i.e. We note that V (z) = n v[n]z n (.28) n=0 V (z) = v[0] + v[]z + v[2]z (.29) This result can be generalized as follows. v[0] = V ( ). (.30)

16 .8 Some Properties of the Z-Transform Right-Sided Sequence Let v[n] be a right-sided sequence that is non-nil for n N, where N is a positive or negative integer and nil for n < N, as shown in Fig... We can Fig... Right-sided sequence and left-sided sequence. write obtaining V (z) = v[n]z N + v[n + ]z (N+) +... (.3) z k V (z) = v[n]z N+k + v[n + ]z (N+) z k +... (.32) lim z zk V (z) = 0, k < N v [N], k = N, k > N. (.33) We conclude that for a right-sided sequence that is non-nil for n N the limit lim z zk V (z) is equal to the initial value v[n] if k = N, is zero if k < N and is infinite if k > N. By evaluating this limit we may determine the sequence s initial value..8.3 Left-Sided Sequence For a left-sided sequence that is non-nil for n N and nil for n > N, as the sequence shown in the figure, we write obtaining Z k V (z) = v[n]z N z k + v[n ]z (N ) z k +... (.34) 0, k > N lim z 0 zk V (z) = v [N], k = N, k < N. (.35) We conclude that for a left-handed sequence that is non-nil for n N the limit limz k V (z) is equal to v[n] if k = N, is zero if k > N and is infinite if z 0 k < N. By evaluating the limit we may thus deduce the sequence s right-most value. Example.7 Evaluate the initial value of V (z) = az 5 a, z < a. z

17 4 Discrete-Time Signals and Systems We note that lim z 0 z5 V (z) = a lim z 0 zk V (z) = { 0, k > 5, k < 5 wherefrom v[5] = a and v[n] = 0 for n > 5. This result can be easily verified by evaluating v[n]. We obtain v[n] = a n 4 u[5 n]..8.4 Convolution in Time The convolution in time property states that if v [n] V (z), a < z < b (.36) and then v 2 [n] V 2 (z), a 2 < z < b 2 (.37) v [n] v 2 [n] V (z)v 2 (z), max(a, a 2 ) < z < min (b, b 2 ) (.38) where v (n) v 2 (n) is the convolution of v [n] and v 2 [n] defined as v [n] v 2 [n] = k= v [k]v 2 [n k] =.8.5 Convolution in Frequency If then x[n]v[n] 2πj k= v [n k]v 2 [k]. (.39) x[n] X (z) (.40) v[n] V (z) (.4) C X ( ) z V (y)y dy (.42) y where C is a contour in the common region of convergence of X (z/y) and V (y), that is, multiplication in the time domain corresponds to convolution in the z domain. The transforms X (z/y) and V (y) have, respectively, the regions of convergence r x < z y < r x 2 and r v < y < r v2 (.43)

18 wherefrom W (z) has the region of convergence.8 Some Properties of the Z-Transform 5 r x r v < z < r x2 r v2. (.44) Equivalently, W (z) = X (y)v 2πj C ( ) z y dy (.45) y Using polar representation we write r x r v < z < r x2 r v2. (.46) z = re jω, y = ρe jφ (.47) W ( re jω) = ˆ π X ( ρe jφ) { } r V 2π π ρ ej(ω φ) dφ. (.48) The right hand side shows the convolution of two spectra. If r and ρ are constants these spectra are z-transforms evaluated on two circles in the z plane, of radii ρ and r/ρ respectively. For the particular case r = we have the Fourier transform W ( e jω) = 2π ˆ π π X ( ρe jφ) V [ ] ρ ej(ω φ) dφ (.49) wherein if ρ is constant the convolution is of two z spectra, namely, those evaluated on a circle of radius ρ and another of radius /ρ, respectively. If ρ = we have the z-transform. W ( re jω) = 2π and the Fourier transform W ( e jω) = 2π ˆ π π ˆ π π X ( e jφ) V X ( e jφ) V [ re j(ω φ)] dφ (.50) [ e j(ω φ)] dφ (.5) which is simply the convolution of the two Fourier transforms X ( e jω) and V ( ) jω on the unit circle. Example.8 Given v [n] = n u[n] v 2 [n] = a n u[n] evaluate the z-transform of v[n] = v [n]v 2 [n]. We have V (z) = nz n. n=0

19 6 Discrete-Time Signals and Systems To evaluate this sum we note that z n = z, z >. n=0 Differentiating we have ( n)z n = wherefrom Since n=0 V (z) = we have V (z) = ( z V 2πj y = 2πj C z 2 ( z ) 2, z > nz n z = ( z 2, z >. ) n=0 V 2 (z) =, z > a az ) V 2 (y)y dy = 2πj zy (y z) 2 (y a) dy. z y ( z y) y ay dy The contour of integration C must be in the region of convergence common to V (z/y) and V 2 (y), that is, z y > and y > a or a < y < z. The integrand has two poles in the y plane, namely, a double pole at y = z, z being a constant through the integration, and a simple one at y = a. The contour of integration is a circle which lies in the region between these two poles, thus enclosing the poles y = a, as shown in Fig..2. We deduce that [ ] zy V (z) = Res of (y z) 2 (y a) at y = a za = (z a) 2 = az ( az, z > a. ) 2.9 Tables of Z-Transform Table. lists basic properties of the z-transform. Table.2 lists z transforms of some basic sequences.

20 Table.. Basic properties of z-transform..9 Tables of Z-Transform 7 a v[n] + b x(n) av (z) + bx (z) n m=0 v[n n 0] z n 0 V (z), n 0 0 v[m]x[n m] v[n]x[n] n v[n] 2πj C V (z)x (z) V (y)x (z/y)y dy dv (z) z dz v [n] V (z ) a n v[n] V a z lim v[n] n 0 lim V (z) z lim v[n] n R {v[n]} I {v[n]} v[ n] v [n] v 2[n] k= n= v [0] v[k] lim ( /z) V (z) z 2 [V (z) + V (z )] 2j [V (z) V (z )] V (/z) V (z) V 2 (z) V (z) z lim V (z), v [n] = 0, n < 0 v [n]v 2[n] z 2πj C V (y)v 2 (/y)y dy

21 8 Discrete-Time Signals and Systems Table.2. Transforms of basic sequences. Sequence Transform R.O.C. δ[n] All z u[n] u[n m] u[ n ] z z > z m z z > z z < δ[n m] z m All z plane except z = 0 (if m > 0) or z = (if m < 0) α n u[n] αz z > α α n u[ n ] n α n u[n] αz αz ( αz ) 2 z < α z > α n 2 u[n] z 2 +z (z ) 3 z > n α n u[ n ] [cos Ω 0n]u[n] [sin Ω 0n] u[n] [r n cos Ω 0n] u [n] [r n sin Ω 0n] u [n] cosh (nα) u[n] sinh (nα) u[n] n a n u[n] αz ( αz ) 2 [cos Ω 0 ]z z < α [2 cos Ω 0 ]z +z 2 z > [sin Ω 0 ]z [2 cos Ω 0 ]z +z 2 z > [r cos Ω 0 ]z [2r cos Ω 0 ]z +r 2 z 2 z > r [r sin Ω 0 ]z [2r cos Ω 0 ]z +r 2 z 2 z > r z[z cosh(α)] z 2 2z cosh(α)+ z sinh(α) z 2 2z cosh(α)+ z (z a) 2 n(n )...(n m+) m! a n m u[n] z (z a) m+

22 .0 Parseval s Relation 9 C y plane y= z a.0 Parseval s Relation Fig..2. Contour of integration. Parseval s relation states that v[n]x [n] = 2πj n= V (z)x (/z )z dz (.52) the contour of integration being in the region of convergence common to V (z) and X (/z ). Proof Let w[n] = v[n]x [n]. (.53) Using the complex convolution theorem we have W (z) = ( ) z V (y)x 2πj y y dy. (.54) Since W (z) = wherefrom n= n= w[n]z n n= we have v[n]x [n] = 2πj w[n] = W (z) z= (.55) V (y)x (/y )y dy. (.56) Replacing y by z completes the proof. We note that if the unit circle is in the region of convergence common to V (z) and X (z) the Fourier transforms V ( e jω) and X ( e jω) exist. Parseval s relation with z = e jω takes the form n= v[n]x [n] = 2π ˆ π π V ( e jω) X ( e jω) dω (.57)

23 20 Discrete-Time Signals and Systems and with x[n] a real sequence equal to v[n] we have n=.0. Final Value Theorem v[n] 2 = ˆ π V ( e jω) 2 dω. (.58) 2π π The final value theorem for a right-sided sequence states that ( lim v[n] = lim z ) V (z). (.59) n z. Multiplication by an Exponential The multiplication by an exponential property states that a n v[n] V ( a z ). (.60) In fact Z [a n v[n]] = a n v[n]z n = v[n] { a z } n. (.6) n=0.. Frequency Translation As a special case of the multiplication by an exponential property we have the frequency translation property, namely, n=0 v[n]e jβn V ( e jβ z ). (.62) This property is also called the modulation by a complex exponential property...2 Reflection Property Let The reflection property states that v[n] V (z), ROC = r v. (.63) v[ n] V ( ), ROC =. (.64) z r v Indeed 0 Z [v[ n]] = v[ n]z n = v[m]z m = V ( z ). (.65) n= m=0

24 .2 Unilateral Z-Transform 2..3 Multiplication by n This property states that Since we have so that as stated. dv (z) dz = n= dv (z) z = dz n v[n] z V (z) = n= v[n] ( n)z n = n= dv (z). (.66) dz v[n]z n (.67) n v[n]z n (.68) n=0 n v[n]z n = Z [n v[n]] (.69).2 Unilateral Z-Transform The unilateral z-transform is a special form of the z-transform that is an important tool for the solution of linear difference equations with non-zero initial conditions. It is applied in the analysis of dynamic discrete-time linear systems in the same way that the unilateral Laplace transform is used in the analysis of continuous-time dynamic LTI systems. Similarly to the unilateral Laplace transform, the unilateral z-transform of a sequence x[n] is the z- transform of the causal part of the sequence. It disregards any value of the sequence that may happen to exist before zero time n = 0. Denoting by X I (z) the unilateral z-transform of a general sequence x[n] we have X I [z] = x[n]z n = Z I [x[n]] (.70) n=0 and we may write x[n] = Z Z I [X I [z]], and x[n] I XI [z]. We note that if the sequence x[n] is causal, its unilateral transform X I (z) is identical to its bilateral z-transform X(z). Example.9 Compare the unilateral and bilateral z-transforms of the sequences a) v[n] = δ[n] + na n u[n]. b) x[n] = a n u[n 2]. c) y[n] = a n u[n + 5].

25 22 Discrete-Time Signals and Systems Fig..3. Three sequences of example. The sequences are shown in Fig..3, assuming a value a = 0.9 as an illustration a) We have V I [z] = Z I [[v[n]] = [δ[n] + na n u[n]] n=0 az = V II [z] = + ( az 2, z > a. ) The unilateral transform V I (z) is equal to the bilateral transform V (z); the sequence v[n] being causal. b) The sequence x[n] is causal. Its unilateral transform X I (z) is therefore equal to the bilateral transform X II (z). Writing x[n] = a 2 a n 2 u[n 2] [ X I [z] = X II [z] = a n z n] = a2 z 2 az, z > a. Alternatively we can write n=2 X I [z] = X II [z] = a 2 z 2 Z [a n u[n]] = a 2 z 2 az, z > a. c) The sequences y[n] = a n u[n + 5] is not causal. Its bilateral transform Y (z) is given by Y (z) = [a n u[n + 5]z n ] = [a n z n ] n=0 = a 5 z 5 az, z > a n= 5 whereas its unilateral transform is equal to [ Y I (z) = a n z n] =, z > a az n=0 wherefrom Y I (z) Y II [z] as expected.

26 .2 Unilateral Z-Transform Time Shift Property of Unilateral z-transform The unilateral z-transform has almost identical properties to those of the bilateral z-transform. An important distinction exists, however, between the time-shift properties of the two transforms. We have seen that the time shift property of the bilateral z-transform is simply given by x[n n 0 ] Z z n0 X II (z). (.7) We now consider the same property as it applies to the unilateral transform X I (z). To develop this property consider the sequence x[n] which extends from n = 3 to n = 3 shown in Fig..4. The sequence v[n], shown in the x [ n] v [ n] y [ n] n n -5 0 Fig..4. A sequence shifted right and left. figure, is a right shift of x[n] by two points. The sequence y[n], in the same figure, is a left shift of x[n] by two points. We now evaluate the unilateral and bilateral z-transforms of v[n] and y[n], and compare them with those of x[n]. Referring to Fig..4 (a) where the values of x[n] for n = 3, 2,, 3 are denoted as x 3, x 2,, x 3, we have x[n] = x[ 3]δ [n + 3] + x[ 2]δ [n + 2] + x[ ]δ [n + ] + x[0] δ [n] +x[]δ [n ] + x[2]δ [n 2] + x[3]δ [n 3] (.72) X I [z] = x[0] + x[]z + x[2]z 2 + x[3]z 3 (.73) v[n] = x[n 2] = x[ 3]δ[n + ] + x[ 2]δ[n] + x[ ]δ[n ] + x[0]δ[n 2] + x[]δ[n 3] + x[2]δ[n 4] + x[3]δ[n 5] V I [z] = x[ 2] + x[ ]z + x[0]z 2 + x[]z 3 + x[2]z 4 + x[3]z 5 = x[ 2] + x[ ]z + z 2 X I [z] n (.74) (.75) y[n] = x[n + 2] (.76) Y I [z] = x[2] + x[3]z = z 2 [ X I [z] x[0] z x[] ]. (.77) More generally, if n 0 > 0 and v[n] = x[n n 0 ] then

27 24 Discrete-Time Signals and Systems V I (z) = x[ n 0 ] + z x[ n 0 + ] + z 2 x[ n 0 + 2] z (n0 ) x[ ] + x[0]z n0 x[]z (n0+) +... n 0 = x[ k]z k n0 + z n0 X I [z]. k= (.78) In other words [ n0 ] Z x[n n 0 ] I z n 0 x[ k]z k + X I [z] k= (.79) and if y[n] = x[n + n 0 ] then [ Y I (z) = z n0 XI [z] x[0] z x[] z 2 x[2]... z (n0 ) x[n 0 ] ] ] n 0 = z [X n0 I [z] z k x[k] k=0 (.80) i.e In particular x[n ] ] n Z x[n + n 0 ] I 0 z n 0 [X I [z] z k x[k]. (.8) k=0 Z I z [x[ ]z + X I [z]] = x[ ] + z X I [z] (.82) x[n 2] x[n + 2] x[n + ] Z I z [X I [z] x[0]] (.83) Z I z 2 [ x[ ]z + x[ 2]z 2 + X I [z] ] (.84) Z I z 2 [ X I [z] x[0] z x[] ]. (.85) Example.0 Evaluate the response y[n] of the system described by the difference equation y[n] 3 4 y[n ] + y[n 2] = x[n] 8 if the input is a unit pulse δ[n] and the initial conditions are y[ ] = and y[ 2] = 2. Since x[n] = δ(n) we have X I (z) =. Applying the unilateral z-transform to both sides of the difference equation we have Y I [z] 3 4 z [y[ ]z + Y I [z]] + 8 z 2 [ y[ ]z + y[ 2]z 2 + Y I [z] ] = Y I [z] [ 34 z + 8 ] z 2 = y[ ] 8 z y[ ] 8 y[ 2]

28 Y I [z] =.3 Causality and Stability 25 [ ] 8 z y[ ] 8 y[ 2] 3 4 z + 8 z 2 Y I [z] z + 3 = z 2 4 y[ ] 8 y[ 2] 8 y[ ]z z z + 8 [ + 34 y[ ] 8 ] y[ 2] z 8 y[ ] /8 = = (z /4)(z /2) (z /4)(z /2). Using a partial fraction expansion we have Y I [z] = [ ] z 2 (z /2) (z /4) Y I [z] = [ ] 2 ( (/2)z ) ( (/4)z ) wherefrom y[n] = 2 [(0.5)n (0.25) n ] u[n]..3 Causality and Stability As we have seen with continuous-time systems a system is causal if its impulse response h [n] is zero for n < 0. It is stable if and only if n= It is therefore stable if and only if h [n] <. (.86) n= h [n] z n < (.87) and z =. In other words a system is stable if the Fourier transform, H ( e jω) of its impulse response, that is, its frequency response, exists. If the system impulse response h [n] is causal the region of convergence of H (z) includes the unit circle and hence, the Fourier transform H ( e jω) exists if and only if all the poles are inside the unit circle. A causal system is therefore stable if and only if all its poles lie inside the z plane unit circle.

29 26 Discrete-Time Signals and Systems Example. For the system described by the linear difference equation y [n] 0.7y [n ] y [n 2].575y [n 3] = x[n] evaluate the system function H (z) and its conditions for causality and stability. Transforming both sides we have H (z) = Y (z) X (z) = 0.7z z 2.575z 3 = ( z 2 )( 0.7z ) = z 3 (z )(z 0.7). The zeros and poles are shown in Fig..5. We note that neither the dif- Fig..5. Poles and zeros in z plane. ference equation nor the system function H (z) implies a particular region of convergence, and hence whether or not the system is causal or stable. In fact there are three distinct possibilities for the region of convergence, namely, z < 0.7, 0.7 < z <.5 and z >.5. These correspond respectively to a left-sided, two-sided and right-sided impulse response. Since the system is stable if and only if the Fourier transform H ( e jω) exists, only the second possibility, namely, the region of convergence 0.7 < z <.5 corresponds to a stable system. In this case the system is stable but not causal. Note that the third possibility, z >.5, corresponds to a causal but unstable system..4 Delayed Response and Group Delay An ideal lowpass filter has a frequency response H ( e jω) defined by H ( e jω) = {, Ω < Ωc 0, Ω c < Ω π (.88) and has zero phase

30 The impulse response of this ideal filter is given by h [n] = 2π ˆ π π.4 Delayed Response and Group Delay 27 arg [ H ( e jω)] = 0. (.89) H ( e jω) e jωn dω = 2π ˆ πc e jωn dω = sin(nω c) π c πn. (.90) The ideal lowpass filter is not realizable since the impulse response h [n] is not causal. To obtain a realizable filter we can apply an approximation. If we shift the impulse response h [n] to the right by M samples obtaining the impulse response h [n M], and if M is sufficiently large, most of the impulse response will be causal, and will be a close approximation of h [n] except for the added delay. An ideal filter with added delay M has the frequency response shown in Fig..6. Its frequency response H ( e jω) is defined by Fig..6. Ideal lowpass filter with linear phase. H ( e jω) { e = jmω, Ω < Ω c (.9) 0, Ω c < Ω π that is, ( H e jω ) {, Ω < Ωc = (.92) 0, Ω c < Ω π arg [ H ( e jω)] = MΩ, Ω < π (.93) and its impulse response is h [n] = sin [(n M)Ω c]. (.94) π (n M) As shown in the figure, the resulting filter has a linear phase ( MΩ) corresponding to the pure delay by M samples. Such a delay does not cause distortion to the signal and constitutes a practical solution to the question of causality and realizability of ideal filters.

31 28 Discrete-Time Signals and Systems Phase linearity is a quality that is often sought in the realization of continuous-time as well as digital filters. A measure of phase linearity is obtained by differentiating it, leading to a constant equal to the delay if the phase is truly linear. The measure is called the group delay and is defined as τ (Ω) = d [ [ ( arg H e jω )]]. (.95) dω Before differentiating the phase any discontinuities are eliminated first by adding integer multiples of π, thus leading to a phase arg [ H ( e jω)] that is continuous without the discontinuities caused by crossing the boundary points ±π on the unit circle. Example.2 For the first-order system of transfer function H (z) = az where a is real, evaluate the amplitude spectrum and group delay of its frequency response. We have H ( ( e jω) ) ae jω = ae jω = ( ae jω )( ae jω ) ae jω = 2a cosω + a 2 = ae jω 2a cosω + a 2 H ( e jω) 2 = H ( e jω) H ( e jω) = = 2a cosω + a 2 H ( e jω) = 2a cosω + a 2 ( ae jω )( ae jω ) arg [ H ( e jω)] = tan a sinω a cosω τ (Ω) = d [ ] tan a sin Ω dω a cosω = { } a [ ( a cosω)cosω a sin 2 Ω ] a sin Ω + ( a cosω) 2 a cosω a cosω a2 = 2a cosω + a 2.

32 .5 Discrete-Time Convolution.5 Discrete-Time Convolution 29 As seen above, the discrete convolution of two sequences v[n] and x[n] is given by y [n] = v [m]x[n m]. (.96) m= The discrete correlation r vx [n] is given by r vx [n] = m= v[n + m]x[m]. (.97) The same analytic and graphic approaches used in the convolution and correlation of continuous time systems can be used in evaluating discrete convolutions and correlations. The approach is best illustrated by examples. Example.3 Let where h[n] = e 0.n u[n] x[n] = 0.n R N [n] R N [n] = u[n] u[n N]. Evaluate the convolution y [n] = h[n] x[n]. Analytic solution y [n] = k= x[k] h [n k] = 0.k {u [k] u [k N]}e 0.(n k) u [n k] k= { n } { n } = 0.ke 0.(n k) u [n] 0.ke 0.(n k) u [n N] k=0 k=n i.e. { } { n n y[n] = 0.e 0.n ke 0.k u[n] 0.e 0.n ke }u[n 0.k N]. k=0 k=n Letting a = e 0. and using the Weighted Geometric Series WGS Sum S(a, n, n 2 ) as evaluated in the Appendix we may write y[n] = { 0.a n S(a, 0, n) } u[n] { 0.a n S(a, N, n) } u[n N]. This expression can be simplified manually, using the Matlab R sum function wtdsum given in the Appendix or using Mathematica R. The sequences

33 30 Discrete-Time Signals and Systems h [ n] x [ n] 0 (a) n 0 N- (b) n h [ n- m] h [ n- m] 0 n N- m 0 N- n m (c) y [ n] (d) (e) Fig..7. Convolution of two sequences. h[n], x[n], h[n m] for 0 n N, h[n m] for 0 n N, h[n m] for n N and y[n], respectively, are shown in Fig..7 for the case N = 50. Graphic Approach For n < 0, y[n] = 0. For 0 n N n y[n] = 0.me 0.(n m). For n N m=0 y[n] = N m=0 0.me 0.(n m)..6 Discrete-Time Correlation in D The following example illustrates discrete correlation for one-dimensional signals followed by a faster approach to for its analytic evaluation.

34 .6 Discrete-Time Correlation in D 3 Example.4 Evaluate the cross-correlation r xh [n] of the two sequences x[n] = u [n] u [n N] h [n] = e αn u [n]. We start with the usual analytic approach. We have r xh [n] = {u [n + m] u [n + m N]} e αm u[m] m= u [m] u [n + m] 0 iff m 0 and m n i.e. m 0 if 0 n or n 0 m n if n 0 u [n + m N] u [m] 0 iff m 0 and m N n i.e. m 0 if 0 N n or n N m N n if N n 0 or n N { { } r xh [n] = e }u αm [n] + e αm u [ n ] m=0 { m= n { e }u αm [n N] e }u αm [N n]. m=0 m=n n The graphic approach proceeds with reference to Fig..8 (a) to (d): For n + N < 0 i.e. n > N, r xh [n] = 0. For n + N 0 and n 0 i.e. 0 n N r xh [n] = For n 0 i.e. n 0 r xh [n] = n+n m=0 n+n e αm = e α(n n) e α. m= n e αm αn e αn = e e α. The sequence r xh [n] is shown in Fig..8 (e) for the case N = 50. A Short-Cut Analytic Approach To avoid the decomposition of the correlation expression into four sums as just seen, a simpler short-cut approach consists of referring to the rectangular sequence x[n] by the rectangle symbol R rather than decomposing it into the sum of two step functions. To this end we define a mobile rectangular window R n0,n [n], which starts at n = n 0 and is of duration N

35 32 Discrete-Time Signals and Systems Using this window we can write r xh [n] = Fig..8. Cross-correlation of two sequences. R n0,n [n] = u [n n 0 ] u [n (n 0 + N)]. = m= m= e αm u [m] {u [n + m] u [n + m N]} e αm u [m] R n,n [m] m= e αm p. Referring to Fig..8 (f) we draw the following conclusions. If n + N < 0, r xh [n] = 0. If n 0 and n + N 0 i.e. 0 n N, the product p 0 iff 0 m n + N. If n 0 i.e. n 0 then p 0 iff n m n + N. r xh [n] = { n+n m=0 e αm } {u [n] u [n N]}+ { n+n m= n e αm } u [ n ]. Example.5 Evaluate the cross-correlation r xv [n] of the two sequences

36 .6 Discrete-Time Correlation in D 33 x[n] = βn {u[n] u[n N]} v[n] = e αn u[n]. The sequences are shown in Fig..9 (a) and (b), respectively. r xv (n) = β[n + m] {u[n + m] u[n + m N]} e αm u[m]. m= Referring to Fig..9 (c) and (d) we may write v [ n] x [ n] 0 n 0 (a) (b) N- n x [ n+ m] x [ n+ m] -n 0 -n+n- m 0 -n (c) r n xv[ ] (e) Fig..9. Discrete cross-correlation. 0 (d) n -n+n- m For n + N < 0 i.e. n > N, r xv (n) = 0. For 0 n + N N i.e. 0 n N r xv [n] = N n m=0 e αm β (n + m). For n 0 i.e. n 0

37 34 Discrete-Time Signals and Systems r xv [n] = N n m= n e αm β (n + m). Letting a = e α we can write this result using the Weighted Geometric Series WGS Sum S(a, n, n 2 ) evaluated in the Appendix. We obtain for 0 n N r xv [n] = βn and for n 0 r xv [n] = βn { N n m=0 e αm } + β { N n m=0 me αm } = βn ( e α(n n)) / ( e α ) + βs (a, 0, N n ) { N n m= n e αm } + β { N n m= n me αm } = βne αn ( e αn) / ( e α ) + βs (a, n, N n ). The cross-correlation sequence r xv [n] is shown in Fig..9 (e). The result can be confirmed using the cross-correlation Matlab R command xcorr(x,v)..7 Convolution and Correlation as Multiplications Given a finite duration sequence its z-transform is a polynomial in z. The convolution in time of two finite duration sequences corresponds to multiplication of the two polynomials in the z domain. As the following examples illustrate it is possible to use this property to evaluate convolutions and correlations as simple spatial multiplications. Example.6 Evaluate the convolution z[n] of the two sequences defined by: x[n] = {2, 3, 4} and y[n] = {, 2, 3}, for n = 0,, 2 and zero elsewhere. The following multiplication structure evaluates the convolution, where x k stands for x[k] and y k stands for y[k]. As in hand multiplication, each value z[k] is deduced by adding the elements above it. The result is z[n] = 2, 7, 6, 7, 2, for n = 0,, 2, 3, 4, respectively. x[2] x[] x[0] y [2] y [] y [0] y 0 x 2 y 0 x y 0 x 0 y x 2 y x y x 0 y 2 x 2 y 2 x y 2 x 0 z [4] z [3] z [2] z [] z [0] Example.7 Evaluate the correlation r vx [n] of the two sequences defined by: v[n] = {, 2, 3} and x[n] = {2, 3, 4}, for n = 0,, 2 and zero elsewhere.

38 .8 Two-Dimensional Signals 35 The following multiplication structure evaluates the correlation, where again v k stands for v[k] and x k stands for x[k]. x[0] x[] x[2] v [2] v [] v [0] v 0 x 0 v 0 x v 0 x 2 v x 0 v x v x 2 v 2 x 0 v 2 x v 2 x 2 r vx [0] r vx [ ] r vx [ 2] r vx [ 3] r vx [ 4] The result is r vx [n] = 4,, 20, 3, 6, for n = 4, 3, 2,, 0, respectively..8 Two-Dimensional Signals Let x[n, n 2 ] be a two-dimensional sequence representing an image, twodimensional data or any other signal. The z-transform of the sequence is given by In polar notation X (z, z 2 ) = X ( r e jω, r 2 e jω2) = n = n 2= n = n 2= x[n, n 2 ] z n z n2 2. (.98) z = r e jω (.99) z 2 = r 2 e jω2 (.00) x[n, n 2 ] r n r n2 2 e jωn e jω2n2. If r = r 2 = we have the two-dimensional Fourier transform X ( e jω, e jω2) = Convergence: The z-transform converges if n = n 2= (.0) x[n, n 2 ] e jωn e jω2n2. (.02) n = n 2= The inverse transform is given by x[n, n 2 ] = x[n, n 2 ] z n z n2 2 <. (.03) ( ) 2 X (z, z 2 )z n z n2 2 dz dz 2 (.04) 2πj C C 2

39 36 Discrete-Time Signals and Systems where the contours C and C 2 are closed contours encircling the origin and lie in the region of convergence of the integrands. The inverse Fourier transform is written ( ) 2ˆ π ˆ π x[n, n 2 ] = X ( e jω, e jω2) e jωn e jω2n2 dω dω 2. (.05) 2π π π If a sequence is separable, i.e. x[n, n 2 ] = x [n ] x 2 [n 2 ] (.06) then X (z, z 2 ) = X (z )X 2 (z 2 ) (.07) since X (z, z 2 ) = x [n ] x 2 [n 2 ] z n z n2 2 = x [n ] z n x 2 [n 2 ] z n2 2. n n 2 n n 2 Properties If then (.08) x[n, n 2 ] X (z, z 2 ) (.09) x[n + m, n 2 + k] z m z2x k (z, z 2 ) (.0) a n b n2 x[n, n 2 ] X ( a z, b ) z 2 (.) n n 2 x[n, n 2 ] d2 X (z, z 2 ) dz dz 2 (.2) x [n, n 2 ] X (z, z 2 ) (.3) x[ n, n 2 ] X ( z ) (.4), z 2 x[n, n 2 ] y [n, n 2 ] X (z, z 2 )Y (z, z 2 ) (.5) ( ) 2 ( ) z z 2 x[n, n 2 ] y [n, n 2 ] X, Y (w )w 2πj C C 2 w w w 2 dw dw 2. 2 (.6) A two-dimensional system having input x[n, n 2 ] and output y [n, n 2 ] may be described by a difference equation of the general form M k=0 m=0 N a km y [n k, n 2 m] = P k=0 m=0 Q b km x[n k, n 2 m]. (.7) The system function H(z) may be evaluated by applying the z-transform, obtaining M N k=0 m=0 a km z k z m 2 Y (z, z 2 ) = P Q k=0 m=0 b km z k z m 2 X (z, z 2 ) (.8)

40 wherefrom H (z, z 2 ) = Y (z, z 2 ) X (z, z 2 ) =.8 Two-Dimensional Signals 37 P Q k=0 m=0 M N k=0 m=0 Examples of basic two-dimensional sequences follow. Impulse The 2D impulse is defined by {, n = n δ [n, n 2 ] = 2 = 0 0, otherwise. b km z k z m 2 a km z k z m 2. (.9) (.20) The impulse is represented graphically in Fig..20 (a). Unit Step Function Fig..20. (a) Two dimensional impulse, (b) Representation of unit step 2D sequence. {, n, n u [n, n 2 ] = 2 0 0, otherwise. (.2) In what follows, the area in the n n 2 plane wherein a sequence is non-nil will be hatched. The unit step function is non-nil, equal to in the first quarter of the plane and may thus graphically represented as shown in Fig..20 (b). Causal Exponential { a n x[n, n 2 ] = an2 2, n, n 2 0 (.22) 0, otherwise. Complex Exponential x[n, n 2 ] = e j(ωn+ω2n2), n, n 2. (.23) Sinusoid x[n, n 2 ] = sin(ω n + Ω 2 n 2 ). (.24)

41 38 Discrete-Time Signals and Systems.9 Linear Systems, Convolution and Correlation Similarly to one-dimensional systems the system impulse response h[n, n 2 ] is the inverse z-transform of the system transfer function H(z, z 2 ). The system response is the convolution of the input x[n, n 2 ] with the impulse response. y [n, n 2 ] = x[n, n 2 ] h [n, n 2 ] = h[m, m 2 ] x[n m, n 2 m 2 ]. m = m2= (.25) The correlation of two 2D sequences x[n, n 2 ] and y[n, n 2 ] is defined by r xy [n, n 2 ] = m = m2= x[n + m, n 2 + m 2 ] y [m, m 2 ]. (.26) The convolution and correlation of images and in general two-dimensional sequences are best illustrated by examples. Example.8 Evaluate the convolution z[n, n 2 ] of the two sequences x[n, n 2 ] = e α(n+n2) u [n, n 2 ] y [n, n 2 ] = e β(n+n2) u [n, n 2 ]. The sequences x[n, n 2 ] and y [n, n 2 ] are represented graphically by hatching the region in the n n 2 plane wherein they are non-nil. The two sequences are thus represented by the hatched regions in Fig..2 (a) and (b). Let Fig..2. Convolution of 2D sequences. p[m, m 2 ] = e α(m+m2) e β(n m+n2 m2). We have z [n, n 2 ] = m = m 2= p [m, m 2 ]u [m, m 2 ] u [n m, n 2 m 2 ]. The analytic solution is obtained by noticing that the product of the step functions is non-nil if and only if

42 wherefrom.9 Linear Systems, Convolution and Correlation 39 m 0, m 2 0, m n, m 2 n 2, i.e. n 0, n 2 0 Simplifying we obtain { z[n, n 2 ] = z [n, n 2 ] = e β(n+n2) n n n 2 m =0 m 2=0 m =0 n 2 e (β α)m p[m, m 2 ]u [n, n 2 ]. m 2=0 e (β α)m2 } u[n, n 2 ] = e β(n+n2) { e (α β)(n +) }{ e (α β)(n2+)} { e (α β) } 2 u[n, n 2 ]. The graphic solution is obtained by referring to Fig..2 (c) and (d). Similarly to the one dimensional sequences case, the sequence x[m, m 2 ] is shown occupying the first quarter of the m m 2 plane, while the sequence y [n m, n 2 m 2 ] is a folding around the point of origin of the sequence y [m, m 2 ] followed by a displacement of the point of origin, referred to as the mobile axis, shown in the figure as an enlarged dot, to the point n, n 2 in the m m 2 plane. The figure shows that if n < 0 or n 2 < 0 then z [n, n 2 ] = 0. If n 0 and n 2 0 then z [n, n 2 ] = n n 2 m =0 m 2=0 p[m, m 2 ] in agreement with the results obtained analytically. Example.9 Let a system impulse response be the causal exponential h [n, n 2 ] = e α(n+n2) u [n, n 2 ] and the input be an L-shaped image of width N, namely, x[n, n 2 ] L N [n, n 2 ] u [n, n 2 ] u [n N, n 2 N]. Evaluate the system output y [n, n 2 ]. The non-nil regions of the sequences, respectively, are shown in Fig..22 (a) and (b). In what follows, for simplifying the expressions, we shall write p p[m, m 2 ] = e α(m+m2) where we use alternatively the symbols p or p[m, m 2 ]. We have y [n, n 2 ] = p u [m, m 2 ] {u [n m, n 2 m 2 ] m = m 2= u [n m N, n 2 m 2 N]}.

43 40 Discrete-Time Signals and Systems Fig..22. Two 2D sequences. Analytic approach: The analytic approach, redefining the limits of summation based on the range of variable values for which the products of step functions are non-nil, shows that where y [n, n 2 ] = { n n 2 m{ =0 m 2=0 n N n 2 N m =0 m 2=0 p u [n, n 2 ] } p u [n N, n 2 N] = S u [n, n 2 ] S 2 u [n N, n 2 N] ( ) ( e α(n +) e α(n2+)) S = ( e α ) 2 { } { e α(n N+) e α(n2 N+)} S 2 = ( e α ) 2. Graphic approach: As mentioned above, in the graphic approach the sequence x[n, n 2 ] is folded about the point of origin and the point of origin becomes a mobile axis that has the coordinates (n, n 2 ), dragging the folded quarter-plane to the point (n, n 2 ) in the m m 2 plane. Referring to Fig..23 (a), (b) and (c) we can write } Fig..23. Convolution of two 2D sequences.

44 .9 Linear Systems, Convolution and Correlation 4 For n < 0 or n 2 < 0, y [n, n 2 ] = 0. The Region of Validity, namely, n < 0 and n 2 < 0 of this result, will be denoted as the area A. It covers all quarters except the first of the n n 2 plane, as shown in Fig..24 (a). Referring again to Fig..23 (a), (b) and (c) Fig..24. Regions of validity A, B, C of convolution expressions. we deduce the following. For {n 0 and 0 n 2 N } or {n 2 0 and 0 n N } we have n n 2 n y [n, n 2 ] = p [m, m 2 ] = m( =0 m 2=0 ) ( m =0 e α(n +) e α(n2+)) = ( e α ) 2. n 2 e αm m 2=0 e αm2 The Region of Validity of this result, namely, {n 0 and 0 n 2 N } or {n 2 0 and 0 n N }, is shown as the Area B in Fig..24 (b). For the case n N and n 2 N, as shown in Fig..23 (d) we have y [n, n 2 ] = or, equivalently, n n 2 m =n N+ m 2=n 2 N+ y [n, n 2 ] = n n 2 m =0 m 2=0 p[m, m 2 ]+2 p[m, m 2 ] n n 2 N m =n N+ m 2=0 n N n 2 N m =0 m 2=0 p[m, m 2 ]. p[m, m 2 ] The Region of Validity of this result, namely, n N and n 2 N, is shown as the Area C in Fig..24 (c). Combining these results we may write, corresponding to the Region of Validity B y B [n, n 2 ] = n n 2 m =0 m 2=0 p[m, m 2 ] {u[n, n 2 ] u[n N, n 2 N]}

45 42 Discrete-Time Signals and Systems and for the Region of Validity C { n y C [n, n 2 ] = n 2 m =0 m 2=0 p[m, m 2 ] u [n N, n 2 N] n N n 2 N m =0 m 2=0 so that the overall result may be written in the form y [n, n 2 ] = y B [n, n 2 ] + y C [n, n 2 ]. p[m, m 2 ] }.20 Correlation of Images The cross-correlation of two continuous-domain images is written r xy (s, t) = ˆ ˆ x(s + σ, t + τ)y (σ, τ)dσ dτ. (.27) Discrete Correlation The cross-correlation of two discrete-domain images is written r xy [n, n 2 ] = m = m 2=.2 Correlation in 2D x[n + m, n 2 + m 2 ] y [m, m 2 ]. (.28) Example.20 Evaluate the cross-correlation r xh [n, n 2 ] of the L-shaped sequence and x[n, n 2 ] = L N [n, n 2 ] u [n, n 2 ] u [n N, n 2 N] h [n, n 2 ] = e α(n+n2) u [n, n 2 ]. The regions of non-zero values of the two sequences, respectively, are shown in Fig..25. Let p p[m, m 2 ] = e α(m+m2). We have r xh [n, n 2 ] = m = m 2= p u [m, m 2 ] {u [n + m, n 2 + m 2 ] u [n + m N, n 2 + m 2 N]}. The graphic approach to the evaluation of the cross-correlation sequence is written with reference to Fig..26 (a), (b), (c), (d), (e) and (f). In these

46 .2 Correlation in 2D 43 Fig..25. Two 2D sequences. Fig..26. Cross-correlation of two 2D sequences. figures the mobile axis is shown as the enlarged dot at ( n, n 2 ) in the m m 2 plane. The inner corner of the L-section has the coordinates ( n + N, n 2 + N ). Referring to Fig..26 (a) we can write: For n +N < 0 and n 2 +N < 0, i.e. n N and n 2 N, that is, the region of validity A shown in Fig..26 (d), denoting the cross-correlation by r xh, [n, n 2 ] we have r xh, [n, n 2 ] = 0. Referring to Fig..26 (b) we can write: For 0 n 2 + N N and n + N < 0, i.e. 0 n 2 N and n N, that is, the region of validity B shown in Fig..26 (e), we have r xh [n, n 2 ] = m =0 n 2+N m 2=0 p[m, m ]. Given the region of validity of this expression we can re-write it in the form

47 44 Discrete-Time Signals and Systems r xh,2 [n, n 2 ] = m =0 n 2+N m 2=0 p[m, m ] {u [n N, n 2 ] u [n N, n 2 N]}. Referring to Fig..26 (c) we can write: For 0 n + N N and 0 n 2 + N N, i.e. 0 n N and 0 n 2 N, that is, the region of validity C shown in Fig..26 (f), we have r xh [n, n 2 ] = or equivalently r xh [n, n 2 ] = m =0 n 2+N m 2=0 m =0 m 2=0 n +N p[m, m 2 ] + p[m, m 2 ] + m =0 m 2= n 2+N m = n +N m 2= n 2+N p[m, m 2 ] p[m, m 2 ] and given the region of validity of this expression we can re-write it in the form { } r xh,3 [n, n 2 ] = p + p m =0 m 2=0 m = n +N m 2= n 2+N {u [n, n 2 ]u [ n + N, n 2 + N ]} where the product of the step functions defines the region of validity as the area C in the n n 2 plane as required. Two subsequent steps are shown in Fig..27 (a), (b), (c) and (d). Referring to Fig..27 (a) we can write: For n 0 and 0 n 2 +N N, Fig..27. Correlation steps and corresponding regions of validity. i.e. n and 0 n 2 N, that is, the region of validity D shown in Fig..27 (b), we have { } n 2+N n +N r xh,4 [n, n 2 ] = p + p m = n m 2=0 m = n m 2= n 2+N {u [ n, n 2 ] u [ n, n 2 N]}.

48 .2 Correlation in 2D 45 Referring to Fig..27 (c) we can write: For n 2 0 and n + N 0, i.e. n N and n 2, the region of validity E shown in Fig..27 (d), we have r xh,5 [n, n 2 ] = { m =0 n 2+N m 2= n 2 p [m, m 2 ] } u [n N, n 2 ]. Proceeding similarly and referring to Fig..28 (a) we obtain for the region of validity F shown in Fig..28 (b): Fig..28. Correlation steps and corresponding regions of validity. { } n 2+N n +N r xh,6 [n, n 2 ] = p + p m =0 m 2= n 2 m =0 m 2= n 2+N {u [n, n 2 ] u [n N, n 2 ]}. Similarly, referring to Fig..28 (c) we obtain { n 2+N r xh,7 [n, n 2 ] = p + m 2= n 2 m = n u [ n, n 2 ] n +N m = n m 2= n 2+N which has the region of validity G in Fig..27 (d). We can write equivalently { } r xh,7 [n, n 2 ] = p + p m 2= n 2 m = n u [ n, n 2 ]. m = n +N m 2= n 2+N p } Referring to Fig..29 (a), (b), (c) and (d) we have For the region of validity H shown in Fig..29 (b): { n+n } r xh,8 [n, n 2 ] = p[m, m 2 ] u [ n, n 2 N] m =0 m 2=0 and over the region I shown in Fig..29 (d)

49 46 Discrete-Time Signals and Systems Fig..29. Correlation steps and regions of validity. { n+n } r xh,9 [n, n 2 ] = p[m, m 2 ] m =0 m 2=0 {u [n, n 2 N] u [n N, n 2 N]}. and the cross-correlation over the whole n n 2 plane is given by r xh [n, n 2 ] = 9 r xh,i [n, n 2 ]. i= The regions of validity A, B,..., I corresponding to the cross correlations r xh, [n, n 2 ], r xh,2 [n, n 2 ],..., r xh,9 [n, n 2 ] over the whole plane are shown in Fig..30. Fig..30. Correlation regions of validity..22 Discrete-Time Allpass Systems As with continuous-time systems an allpass system has a magnitude spectrum that is constant for all frequencies. To be causal and stable the system s poles should be inside the unit circle. Similarly to continuous-time systems every pole has an image reflection which in the present case is reflected into the unit circle producing a zero outside the unit circle. In fact a pole z = p and its conjugate z = p are accompanied by their reflections into the unit circle

50 .22 Discrete-Time Allpass Systems 47 as the two zeros z = /p and z = /p, respectively. A pole z = p 0 where p 0 is real is accompanied by its reflection, simply its reciprocal, as the zero z = /p 0. Such relations are illustrated for the case of a third order system with two complex conjugate poles and a real one in Fig..3, where the poles p, p 2 and p 3 are seen to be accompanied by the three zeros z, z 2 and z 3. To illustrate the evaluation of the system frequency response, the figure also shows vectors u, u 2 and u 3, extending from the zeros to an arbitrary point z = e jω on the z plane unit circle and vectors v, v 2 and v 3, extending from the poles to the same point. The transfer function of a first order allpass z = /p * u p v z=e j v 3 u 3 p 3 v 2 u 2 - -j v * z =e z 3= /p 3 p 2 = p * z 2 =z * = /p Fig..3. Vectors from poles and zeros to a point on unit circle. system, having a single generally complex pole z = p, has the form H (z) = z p pz. (.29) An all-pass system of a higher order is a cascade of such first order systems. The transfer function of the third order system shown in the figure is given by H (z) = z p z p 2 z p 3 p z p 2 z p 3 z (.30) where p 2 = p and p 3 = p 3. We can write H (z) = ( p )( p 2 )( p 3 ) z /p z p z /p 2 z p 2 z /p 3 z p 3 (.3) and the system frequency response is given by H ( e jω) = ( p )( p 2) ( p 3) ejω /p e jω /p e jω p e jω p = p p 2p 3 e jω /p 3 e jω p 3 u u 2 u (.32) 3 v v 2 v 3