Signals and Systems - Electrical Engineering

74 C H A P T E R 1: Continuous-Time Signals

Likewise, we have that

x(t+ 2 )=

{

1 0≤t+ 2 ≤1 or− 2 ≤t≤− 1

0 otherwise

The signalx(t+ 2 )can be seen to be the advanced version ofx(t), as it is this signal shifted to the

left by two units of time. The valuex( 0 )forx(t+ 2 )now occurs att=−2, which is ahead oft=0.

Finally, the signalx(−t)is given by

x(−t)=

{

1 0≤−t≤1 or− 1 ≤t≤ 0

0 otherwise

This signal is a mirror image of the original: the valuex( 0 )still occurs at the same time, butx( 1 )

occurs whent=−1. n

nExample 1.5

When the shifting and reflecting operations are considered together the best approach to visual-

ize the operation is to make a table computing several values of the new signal and comparing

these with those from the original signal. Consider the pulse in Example 1.4, and plot the signal

x(−t+ 2 ).

Solution

Although one can see that this signal is reflected, it is not clear whether it is advanced or delayed

by 2. By computing a few values:

t x(−t+ 2 )

2 x( 0 )= 1

1.5 x(0.5)= 1

1 x( 1 )= 1

0 x( 2 )= 0

− 1 x( 3 )= 0

it becomes clear thatx(−t+ 2 )is reflected and “delayed” by 2. In fact, as indicated above, whenever

the signal is a function of−t(i.e., reflected), the−t+τoperation becomes reflection and “delay,”

and−t−τbecomes reflection and “advancing.” n

RemarksWhen computing the convolution integral later on, we will consider the signal x(t−τ)as a

function ofτfor different values of t. As indicated from Example 1.5, this signal is a reflected version of

x(τ)being shifted to the right t seconds. To see this, consider t= 0 then x(t−τ)|t= 0 =x(−τ), the reflected

version, and x( 0 )occurs atτ= 0. When t= 1 , then x(t−τ)|t= 1 =x( 1 −τ)and x( 0 )occurs atτ= 1 , so

that x( 1 −τ)is x(−τ)shifted to the right by 1 , and so on.