DISCRETE SIGNAL OPERATIONS

EDITOR: B. SOMANATHAN NAIR

1. INTRODUCTION

In the previous two blogs, we had discussed operations of scaling and shifting on continuous time-domain signals. In this blog we discuss how these operations are performed on discrete time-domain signals.

Example 1: Consider the discrete-time function x(n) shown in Fig. 1. Obtain the plot of the function x(n–2).    

Solution: The given discrete-time function is shifted to the right by two units; the resulting plot is shown in Fig. 2.

Example 2: Obtain the plot of the function x(n–2) using the plot shown in Fig. 3.                                                                   

                                                                       

Solution: The desired function is obtained by shifting x(n) to the right by two units as shown in Fig. 4.

Example 3: Using the plot shown in Fig. 3, obtain the function x(2n–3).    

Solution: The desired function x(2n–3) is a time-shifted and compressed version of the given function x(n). For the discrete version, we shift the given function by three units to the right as shown in Fig. 5.

            This function is compressed by two units of time. In discrete-time domain, compression by 2 units of time means dividing the sample time by the factor of 2 and then choosing those samples represented by whole numbers (or integers), and neglecting those samples represented by fractions.  To illustrate this idea further, consider the sample of amplitude –1 at time unit n = 1 in Fig. 5. When this is divided by the compression factor of 2, we get the fractional number 1/2; since this is not an integer, this sample is neglected.

            Next, we consider the sample of amplitude –1 at time n = 2 in Fig. 5. When this is divided by the compression factor of 2, we get 2/2 = 1; since this is a whole number, it is selected and drawn in Fig. 6 as the sample of amplitude –1 at n =1.

            In the same manner as described above, we divide the sample of amplitude +1 at time n = 4 by factor 2 to get a valid whole number at 4/2 = 2. This is then drawn as the sample of amplitude +1 at n = 2, as shown in Fig. 6. Extending the same argument to the sample of +1 at n = 5, we find that this is invalid as 5/2 = 2.5, which is a fraction and hence neglected. The desired function x(2n–3)  is thus obtained as shown in Fig. 6.

Example 4: Figure 7 shows a discrete-time function x(n). Get the function x(2n).

Solution: The function x(2n)  represents discrete-time compression of the function x(n) shown in Fig. 7. As explained in Example 3, we divide each value of n by 2 to get the samples of x(2n). Thus dividing sample time n of values equal to -3, -1, 1, and 3 by the compression factor 2 yields fractions of -3/2, -1/2, 1/2, and 3/2, respectively. However dividing n = -2 and 2 by factor 2 yields whole numbers of -1, and 1, respectively. Hence the sample amplitudes of 2 each at these two points (i.e., -2 and 2) are selected as valid sample amplitudes at instants of -1, and 1, respectively and marked in Fig. 8 to yield the function x(2n).