Continuous-time Fourier transform - Continuous and Discrete Time Signals and Systems - page 197

Table 5.1. Magnitude X (ω) and phase < X (ω) for the CTFT of x ( t ) = exp(−3 t ) u ( t ) in Example 5.1

ω (radians / s) −∞ − 1000 − 100 − 10 − 1 1 0 1 0 0 ∞

Magnitude: X ( ω ) 0 0.001 0.01 0.096 0.316 0.333 0.316 0.096 0.01 0

Phase: < X ( ω ) π/ 2 1.57 1.54 1.28 0.32 0 − 0 . 32 − 1 . 28 − 1 . 54 −π/ 2

R +

The notation a ∈ implies that a is real-valued within the range −∞ <

a < ∞ .

Solution
(i) Based on the definition of the CTFT, Eq. (5.10), we obtain

∞ ∞

− at u ( t ) = − at u ( t )e − j ω t d t − ( a + j ω ) t d t

X 1 ( ω ) =ℑ e e = e

−∞

0

1
( a + j ω )

1
( a + j ω )

∞
0

− ( a + j ω ) t − ( a + j ω ) t − 1

=− e =−

t →∞ e

lim ,

where the term

− ( a + j ω ) t − at lim − j ω t − j ω t

t →∞ e

lim =

t →∞ e

lim

t →∞ e

= 0 lim

t →∞ e

= 0 .

Therefore,

1
a + j ω .

X 1 ( ω ) =

The magnitude and phase of X 1 ( ω ) are given by

1
a + j ω

1

magnitude X 1 ( ω ) = = √

a 2 + ω 2 ;

1
a + j ω

ω
a

− 1

phase < X 1 ( ω ) = < = < 1 − < ( a + j ω ) =− tan .

Table 5.1 lists the amplitude and phase of X ( ω ) for several values of ω with
a = 3. The exponentially decaying function x 1 ( t ) and its magnitude and phase
spectra are plotted in Fig. 5.3.
(ii) Based on the definition of the CTFT, Eq. (5.10), we obtain

∞

− a t = − a t − j ω t d t

X 2 ( ω ) =ℑ e e e

−∞

∞ ∞

− a t − a t

= e co s( ω t ) d t − j e si n( ω t ) d t .

−∞ even function −∞ odd function

Since the integral of an odd function with limits [ − L , L ] is zero, the above