1.2 SOME USEFUL SIGNAL [OPERATIONS](#page-7-0)

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1.2 SOME USEFUL SIGNAL OPERATIONS

We discuss here three useful signal operations: shifting, scaling, and inversion. Since the independent variable in our signal description is time, these operations are discussed as time shifting, time scaling, and time reversal (inversion). However, this discussion is valid for functions having independent variables other than time (e.g., frequency or distance).

1.2-1 Time Shifting

Consider a signal x(t) (Fig. 1.4a) and the same signal delayed by T seconds (Fig. 1.4b), which we shall denote by φ(t). Whatever happens in x(t) (Fig. 1.4a) at some instant t also happens in φ(t) (Fig. 1.4b) T seconds later at the instant t +T. Therefore

$$\phi(t+T) = x(t) \qquad \text{and} \qquad \phi(t) = x(t-T)$$

Therefore, to time-shift a signal by T, we replace t with t − T. Thus x(t − T) represents x(t) time-shifted by T seconds. If T is positive, the shift is to the right (delay), as in Fig. 1.4b. If T is negative, the shift is to the left (advance), as in Fig. 1.4c. Clearly, x(t − 2) is x(t) delayed (right-shifted) by 2 seconds, and x(t + 2) is x(t) advanced (left-shifted) by 2 seconds.

Figure 1.4 Time-shifting a signal.

EXAMPLE 1.3 Time Shifting

An exponential function x(t) = e−2*^t* shown in Fig. 1.5a is delayed by 1 second. Sketch and mathematically describe the delayed function. Repeat the problem with x(t) advanced by 1 second.

Figure 1.5 (a) Signal x(t). (b) Signal x(t) delayed by 1 second. (c) Signal x(t) advanced by 1 second.

The function x(t) can be described mathematically as

$$x(t) = \begin{cases} e^{-2t} & t \ge 0\\ 0 & t < 0 \end{cases} \tag{1.5}$$

Let xd(t) represent the function x(t) delayed (right-shifted) by 1 second, as illustrated in Fig. 1.5b. This function is x(t − 1); its mathematical description can be obtained from x(t) by replacing t with t −1 in Eq. (1.5). Thus,

$$x_d(t) = x(t-1) = \begin{cases} e^{-2(t-1)} & t-1 \ge 0 \text{ or } t \ge 1\\ 0 & t-1 < 0 \text{ or } t < 1 \end{cases}$$

Let xa(t) represent the function x(t) advanced (left-shifted) by 1 second, as depicted in Fig. 1.5c. This function is x(t + 1); its mathematical description can be obtained from x(t) by replacing t with t +1 in Eq. (1.5). Thus,

$$x_a(t) = x(t+1) = \begin{cases} e^{-2(t+1)} & t+1 \ge 0 \text{ or } t \ge -1\\ 0 & t+1 < 0 \text{ or } t < -1 \end{cases}$$

DR ILL 1.4 Working with Time Delay and Time Advance

Write a mathematical description of the signal x3(t) in Fig. 1.3c. Next, delay this signal by 2 seconds. Sketch the delayed signal. Show that this delayed signal xd(t) can be described mathematically as xd(t) = 2(t − 2) for 2 ≤ t ≤ 3, and equal to 0 otherwise. Now repeat the procedure with the signal advanced (left-shifted) by 1 second. Show that this advanced signal xa(t) can be described as xa(t) = 2(t +1) for −1 ≤ t ≤ 0, and 0 otherwise.

1.2-2 Time Scaling

The compression or expansion of a signal in time is known as time scaling. Consider the signal x(t) of Fig. 1.6a. The signal φ(t) in Fig. 1.6b is x(t) compressed in time by a factor of 2. Therefore, whatever happens in x(t) at some instant t also happens to φ(t) at the instant t/2 so that

$$\phi\left(\frac{t}{2}\right) = x(t)$$

and $\phi(t) = x(2t)$

Observe that because x(t) = 0 at t = T¹ and T2, we must have φ(t) = 0 at t = T1/2 and T2/2, as shown in Fig. 1.6b. If x(t) were recorded on a tape and played back at twice the normal recording speed, we would obtain x(2t). In general, if x(t) is compressed in time by a factor a (a > 1), the resulting signal φ(t) is given by

$$\phi(t) = x(at)$$

Using a similar argument, we can show that x(t) expanded (slowed down) in time by a factor a (a > 1) is given by

$$\phi(t) = x \left(\frac{t}{a}\right)$$

Figure 1.6c shows x(t/2), which is x(t) expanded in time by a factor of 2. Observe that in a time-scaling operation, the origin t = 0 is the anchor point, which remains unchanged under the scaling operation because at t = 0, x(t) = x(at) = x(0).

In summary, to time-scale a signal by a factor a, we replace t with at. If a > 1, the scaling results in compression, and if a < 1, the scaling results in expansion.

EXAMPLE 1.4 Continuous Time-Scaling Operation

Figure 1.7a shows a signal x(t). Sketch and describe mathematically this signal time-compressed by factor 3. Repeat the problem for the same signal time-expanded by factor 2.

The signal x(t) can be described as

$$x(t) = \begin{cases} 2 & -1.5 \le t < 0 \\ 2e^{-t/2} & 0 \le t < 3 \\ 0 & \text{otherwise} \end{cases}$$

(1.6)

Figure 1.7b shows xc(t), which is x(t) time-compressed by factor 3; consequently, it can be described mathematically as x(3t), which is obtained by replacing t with 3t in the right-hand side of Eq. (1.6). Thus,

$$x_c(t) = x(3t) = \begin{cases} 2 & -1.5 \le 3t < 0 \text{ or } -0.5 \le t < 0 \\ 2e^{-3t/2} & 0 \le 3t < 3 \text{ or } 0 \le t < 1 \\ 0 & \text{otherwise} \end{cases}$$

Observe that the instants t = −1.5 and 3 in x(t) correspond to the instants t = −0.5, and 1 in the compressed signal x(3t).

Figure 1.7c shows xe(t), which is x(t) time-expanded by factor 2; consequently, it can be described mathematically as x(t/2), which is obtained by replacing t with t/2 in x(t). Thus,

$$x_e(t) = x\left(\frac{t}{2}\right) = \begin{cases} 2 & -1.5 \le \frac{t}{2} < 0 \text{ or } -3 \le t < 0 \\ 2e^{-t/4} & 0 \le \frac{t}{2} < 3 \text{ or } 0 \le t < 6 \\ 0 & \text{otherwise} \end{cases}$$

Observe that the instants t = −1.5 and 3 in x(t) correspond to the instants t = −3 and 6 in the expanded signal x(t/2).

DR ILL 1.5 Compression and Expansion of Sinusoids

Show that the time compression by an integer factor n (n > 1) of a sinusoid results in a sinusoid of the same amplitude and phase, but with the frequency increased n-fold. Similarly, the time expansion by an integer factor n (n > 1) of a sinusoid results in a sinusoid of the same amplitude and phase, but with the frequency reduced by a factor n. Verify your conclusion by sketching a sinusoid sin 2t and the same sinusoid compressed by a factor 3 and expanded by a factor 2.

1.2-3 Time Reversal

Consider the signal x(t) in Fig. 1.8a. We can view x(t) as a rigid wire frame hinged at the vertical axis. To time-reverse x(t), we rotate this frame 180◦ about the vertical axis. This time reversal [the reflection of x(t) about the vertical axis] gives us the signal φ(t) (Fig. 1.8b). Observe that whatever happens in Fig. 1.8a at some instant t also happens in Fig. 1.8b at the instant −t, and vice versa. Therefore,

φ(t) = x(−t)

Thus, to time-reverse a signal we replace t with −t, and the time reversal of signal x(t) results in a signal x(−t). We must remember that the reversal is performed about the vertical axis, which acts as an anchor or a hinge. Recall also that the reversal of x(t) about the horizontal axis results in −x(t).

Figure 1.8 Time reversal of a signal.

EXAMPLE 1.5 Time Reversal of a Signal

For the signal x(t) illustrated in Fig. 1.9a, sketch x(−t), which is time-reversed x(t).

The instants −1 and −5 in x(t) are mapped into instants 1 and 5 in x(−t). Because x(t) = et/2, we have x(−t) = e−t/2. The signal x(−t) is depicted in Fig. 1.9b. We can describe x(t) and x(−t) as

$$x(t) = \begin{cases} e^{t/2} & -1 \ge t > -5\\ 0 & \text{otherwise} \end{cases}$$

and its time-reversed version x(−t) is obtained by replacing t with −t in x(t) as

x(−t) = e−t/² −1 ≥ −t > −5 or 1 ≤ t < 5 0 otherwise

1.2-4 Combined Operations

Certain complex operations require simultaneous use of more than one of the operations just described. The most general operation involving all the three operations is x(at − b), which is realized in two possible sequences of operation:

• 1. Time-shift x(t) by b to obtain x(t−b). Now time-scale the shifted signal x(t−b) by a [i.e., replace t with at] to obtain x(at −b).
• 2. Time-scale x(t) by a to obtain x(at). Now time-shift x(at) by b/a [i.e., replace t with t − (b/a)] to obtain x[a(t − b/a)] = x(at − b). In either case, if a is negative, time scaling involves time reversal.

For example, the signal x(2t−6) can be obtained in two ways. We can delay x(t) by 6 to obtain x(t − 6), and then time-compress this signal by factor 2 (replace t with 2t) to obtain x(2t − 6). Alternately, we can first time-compress x(t) by factor 2 to obtain x(2t), then delay this signal by 3 (replace t with t −3) to obtain x(2t −6).