SIGNALS AND SYSTEMS

Code-128 Scanner In NoneUsing Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications.

Code 128C Encoder In NoneUsing Barcode generator for Software Control to generate, create USS Code 128 image in Software applications.

[CHAP. 1

Code 128 Recognizer In NoneUsing Barcode decoder for Software Control to read, scan read, scan image in Software applications.

Encode Code128 In Visual C#Using Barcode encoder for .NET framework Control to generate, create Code 128 image in .NET applications.

Case 3 6 5 n 5 10. For 6 5 n 5 10, all of the nonzero values of x ( n sum. and

Code128 Encoder In VS .NETUsing Barcode maker for ASP.NET Control to generate, create Code-128 image in ASP.NET applications.

USS Code 128 Drawer In Visual Studio .NETUsing Barcode encoder for .NET Control to generate, create Code 128C image in .NET framework applications.

- k ) are within the limits of the

Making Code 128C In VB.NETUsing Barcode creation for .NET framework Control to generate, create Code 128C image in VS .NET applications.

Printing Code 3/9 In NoneUsing Barcode printer for Software Control to generate, create Code 39 image in Software applications.

Case 4 1 I 5 n 5 15. When n is in the range I 1 5 n 5 15, the sequences h ( k ) and x ( n - k ) overlap for n - 5 5 k 5 10. Therefore,

Printing EAN / UCC - 14 In NoneUsing Barcode generator for Software Control to generate, create GS1 128 image in Software applications.

Create Barcode In NoneUsing Barcode drawer for Software Control to generate, create bar code image in Software applications.

Case 5 n > 15. Finally, For n > 15, there is again no overlap between h ( k ) and x ( n - k ) , and the product h ( k ) x ( n - k ) is equal to zero for all k. Therefore, y ( n ) = 0 for n 15.

Barcode Creation In NoneUsing Barcode generator for Software Control to generate, create bar code image in Software applications.

Code-128 Creator In NoneUsing Barcode printer for Software Control to generate, create Code 128 Code Set C image in Software applications.

In summary, for the convolution we have

Identcode Encoder In NoneUsing Barcode encoder for Software Control to generate, create Identcode image in Software applications.

2D Barcode Generator In C#Using Barcode creator for VS .NET Control to generate, create Matrix Barcode image in Visual Studio .NET applications.

The correlation of two sequences is an operation defined by the relation

Creating Code 128 Code Set C In NoneUsing Barcode drawer for Online Control to generate, create ANSI/AIM Code 128 image in Online applications.

Print 1D In .NETUsing Barcode generation for .NET framework Control to generate, create 1D image in VS .NET applications.

Note that we use a star * to denote correlation and an asterisk to denote convolution.

GS1 - 13 Printer In C#.NETUsing Barcode creation for Visual Studio .NET Control to generate, create EAN-13 image in .NET applications.

Make Bar Code In JavaUsing Barcode encoder for Java Control to generate, create bar code image in Java applications.

( a ) Find the correlation between the sequence x ( n ) = u ( n ) - u(n - 6 ) and h(n) = u(n - 2 ) - u(n - 5).

Universal Product Code Version A Creation In NoneUsing Barcode generator for Font Control to generate, create UPC A image in Font applications.

Scanning Bar Code In Visual C#.NETUsing Barcode Control SDK for .NET Control to generate, create, read, scan barcode image in .NET applications.

( b ) Find the correlation of x ( n ) = crnu(n)with itself (i.e., h(n) = x(n)). This is known as the autocora relation of x ( n ) . Assume that l1 < 1.

(a) If we compare the expression for the correlation of x ( n ) and h ( n ) with the convolution

we see that the only difference is that, in the case of convolution, h ( k ) is time-reversed prior to shifting by n , whereas for correlation h ( k ) is shifted without time-reversal. Therefore, with a graphical approach to compute the correlation, we simply need to plot x ( k ) and h ( k ) , shift h ( k ) by n (to the left if n > 0 and to the right if n < 0), multiply the two sequences x ( k ) and h ( n k ) , and sum the products. Shown in the figure below is a plot o F x ( k ) and h ( k ) .

CHAP. I]

SIGNALS AND SYSTEMS

Denoting the correlation by r,h(n), it is clear that for n = O the correlation is equal to 3. In fact, this will be the value of r r h ( n )for - I 5 n 5 2. For n = 3, s ( k ) and h(3 k ) only overlap at two points, and r r h ( 3 = 2. ) Similarly, because x ( k ) and h(4 k ) only overlap at one point, r,y,1(4) I . Finally, r r h ( n = 0 for n > 4. = ) Proceeding in a similar fashion for n < 0 , we find that r,,,(-2) = 2, and r-,,,(-3) = I . The correlation is shown in the figure below.

~.vII(~)

(h) Let r,(n) denote the autocorrelation of .x(n),and note that the autocorrelation is the convolution of x ( n ) with x(-n):

In addition observe that r , ( n ) is an even function of n:

Therefore, it is only necessary to find the values of r , ( 1 1 ) for n 1 0.For n 2 0 , we have

Using the symmetry of r , ( n ) , we have, for n < 0,

Combining these two results together, we finally have

Difference Equations 1.37

Consider a system described by the difference equation

Find the response of this system to the input

with initial conditions y ( - I ) = 0.75 and y(-2) = 0.25.

The first step in solving this difference equation is to find the particular solution. With x ( n ) = (OS)"u(n), assume we a solution of the form y,(n) = C l ( 0 . 5 ) " n 20 Substituting this solution into the difference equation, we have

SIGNALS AND SYSTEMS Dividing by (0.5)". CI = 2 c I - 4 C I + o . 5 + 1 which gives

[CHAP. 1

c, = ;

The next step is to find the homogeneous solution. The characteristic equation is z2-z+l=O

Therefore, the form of the homogeneous solution is yh(n) = and the total solution becomes y(n) = (0.5)"+' ein"l> ~~~-1nnl3

+ A , e ~ n n l > +2e-in+

n z 0

The constants A1 and A2 must now be found so that the total solution satisfies the given initial conditions, y(-1) = 0.75 and y(-2) = 0.25. Because the solution given in Eq. (1.25) is only applicable for n > 0, we must derive an equivalent set of initial conditions for y(0) and y(l). Evaluating the difference equation for n = 0 and n = I, we have y(0) = y(- 1) - y(-2) and y(l) = y(0) - y(-1)