Signal representation using Fourier series - Continuous and Discrete Time Signals and Systems - page 151

Proof
The proof of Corollary 4.1 follows the same lines as the proof of Theorem 4.1.
The sinusoidal signals can be expressed as real (Re) and imaginary (Im) com-
ponents of a complex exponential function as follows:

cos( ω 1 t ) = Re e j ω 1 t sin( ω 1 t ) = Im e j ω 1 t . and

Because the impulse response function is real-valued, the output y 1 ( t )to k 1
sin( ω 1 t ) is the imaginary component of y ( t ) given in Eq. (4.23). In other words,

A 1 k 1 e j( ω 1 t +φ 1 )

y 1 ( t ) = Im = A 1 k 1 sin( ω 1 t + φ 1 ) .

Likewise, the output y 2 ( t )to k 1 cos( ω 1 t ) is the real component of y ( t ) given in
Eq. (4.23). In other words,

A 1 k 1 e j( ω 1 t +φ 1 )

y 2 ( t ) = Re = A 1 k 1 cos( ω 1 t + φ 1 ) .

Example 4.5
Calculate the output response if signal x ( t ) = 2 sin(5 t ) is applied as an input to
an LTIC system with impulse response h ( t ) = 2e

− 4 t u ( t ).

Solution
Based on Corollary 4.1, we know that output y ( t ) to the sinusoidal input x ( t )=
2 sin(5 t )isgivenby

y ( t ) = 2 A 1 sin(5 t + φ 1 ) ,

where A 1 and φ 1 are the magnitude and phase of the complex constant H ( ω 1 ),
given by

∞ ∞ ∞

2
4 + j ω .

− j ωτ d τ − 4 τ e − j ωτ d τ − (4 + j ω ) τ d τ

H ( ω ) = h ( τ )e = 2 e = 2 e =

−∞

0 0

The magnitude A 1 and phase φ 1 are given by

2
4 + j ω

√

magnitude A 1 A 1 = H ( ω 1 ) = =

41 .

ω= 5

2
4 + j ω

5
4

− 1

=− 51 . 34 o . phase φ 1 φ 1 = < H ( ω 1 ) = < = 0 − tan

ω= 5

The output response of the system is, therefore, given by

√

41 sin(5 t − 51 . 34 o ) .
As shown in Example 3.4, the LTIC system with impulse response h ( t ) =
2e

y ( t ) =

− 4 t u ( t ) can alternatively be represented by the linear, constant-coefficient
differential equation as follows:
d y
d t

+ 4 y ( t ) = 2 x ( t ) .