It is useful to see what the effect of the filter is on a sinusoidal signal, say x(t)=cos(Ω0t)x(t)=cos(Ω0t). If y(t)y(t) is the output of the filter, then we can write
y
(
t
)
=
∫
-
∞
∞
cos
(
Ω
0
(
t
-
τ
)
)
h
(
τ
)
d
τ
y
(
t
)
=
∫
-
∞
∞
cos
(
Ω
0
(
t
-
τ
)
)
h
(
τ
)
d
τ
(1)Using the Euler formula for cos(Ω0t)cos(Ω0t), right hand side of Equation 1 can be written as:
1
2
∫
-
∞
∞
e
j
(
Ω
0
(
t
-
τ
)
)
+
e
-
j
(
Ω
0
(
t
-
τ
)
)
h
(
τ
)
d
τ
1
2
∫
-
∞
∞
e
j
(
Ω
0
(
t
-
τ
)
)
+
e
-
j
(
Ω
0
(
t
-
τ
)
)
h
(
τ
)
d
τ
(2)This integral can be split into two separate integrals, and written as:
e
j
Ω
0
t
2
∫
-
∞
∞
e
-
j
Ω
0
τ
h
(
τ
)
d
τ
+
e
-
j
Ω
0
t
2
∫
-
∞
∞
e
j
Ω
0
τ
h
(
τ
)
d
τ
e
j
Ω
0
t
2
∫
-
∞
∞
e
-
j
Ω
0
τ
h
(
τ
)
d
τ
+
e
-
j
Ω
0
t
2
∫
-
∞
∞
e
j
Ω
0
τ
h
(
τ
)
d
τ
(3)The first of the two integrals can be recognizes as the Fourier Transform of the impulse response evaluated at Ω=Ω0Ω=Ω0. The second integral is just the complex conjugate of the first integral. Therefore Equation 3 can be written as:
e
j
Ω
0
t
2
H
(
j
Ω
0
)
+
e
-
j
Ω
0
t
2
H
*
(
j
Ω
0
)
e
j
Ω
0
t
2
H
(
j
Ω
0
)
+
e
-
j
Ω
0
t
2
H
*
(
j
Ω
0
)
(4)Since the second term in Equation 4 is the complex conjugate of the first term, we can express Equation 4 as:
R
e
e
j
Ω
0
t
H
(
j
Ω
0
)
R
e
e
j
Ω
0
t
H
(
j
Ω
0
)
(5)or expressing H(jΩ0)H(jΩ0) in terms of polar coordinates:
R
e
e
j
Ω
0
t
|
H
(
j
Ω
0
)
|
e
j
∠
H
(
j
Ω
0
)
=
R
e
|
H
(
j
Ω
0
)
|
e
j
(
Ω
0
t
+
∠
H
(
j
Ω
0
)
)
R
e
e
j
Ω
0
t
|
H
(
j
Ω
0
)
|
e
j
∠
H
(
j
Ω
0
)
=
R
e
|
H
(
j
Ω
0
)
|
e
j
(
Ω
0
t
+
∠
H
(
j
Ω
0
)
)
(6)Therefore, we find that the filter output is given by
y
(
t
)
=
|
H
(
j
Ω
0
)
|
cos
(
Ω
0
t
+
∠
H
(
j
Ω
0
)
)
y
(
t
)
=
|
H
(
j
Ω
0
)
|
cos
(
Ω
0
t
+
∠
H
(
j
Ω
0
)
)
(7)This is called the sinusoidal steady state response. It tells us that when the input to a linear, time-invariant filter is a cosine, the filter output is a cosine whose amplitude has been scaled by |H(jΩ0)||H(jΩ0)| and that has been phase shifted by ∠H(jΩ0)∠H(jΩ0). The same result applies to an input that is an arbitrarily phase shifted cosine (e.g. a sine wave).
Example 3.1
Find the output of a filter whose impulse response is h(t)=e-5tu(t)h(t)=e-5tu(t) and whose input is given by x(t)=cos(2t)x(t)=cos(2t). It can be readily seen that the frequency response of the filter is
H
(
j
Ω
)
=
1
5
+
j
Ω
H
(
j
Ω
)
=
1
5
+
j
Ω
(8)and therefore H(j2)=0.1857H(j2)=0.1857 and ∠H(j2)=-0.3805∠H(j2)=-0.3805. Therefore, using Equation 7:
y
(
t
)
=
0
.
1857
cos
(
2
t
-
0
.
3805
)
y
(
t
)
=
0
.
1857
cos
(
2
t
-
0
.
3805
)
(9)
