ENEL 471 -W2026
More practice problems for Chapter 2
Problem 2.1
Determine the inverse Fourier transform of the frequency function 𝐺(𝑓) defined by the
amplitude and phase spectra shown in Fig. 2.1.
Fig. 2.1.
Solution
%
'
&'
%
𝑔(𝑡) = ( 𝑒 !"/$ ⋅ 𝑒 !$"() 𝑑𝑓 + ( 𝑒 &!"/$ ⋅ 𝑒 !$"() 𝑑𝑓
%
=( 𝑗⋅𝑒
!$"()
'
𝑑𝑓 + ( (−𝑗) ⋅ 𝑒 !$"() 𝑑𝑓
&'
%
%
'
𝑒 !$"()
𝑒 !$"()
= 𝑗0
3
−𝑗0
3
𝑗2𝜋𝑡 (*&'
𝑗2𝜋𝑡 (*%
=
1
1
51 − 𝑒 &!$"') 6 +
51 − 𝑒 !$"') 6
2𝜋𝑡
2𝜋𝑡
=
=
1
72 − 5𝑒 !$"') + 𝑒 &!$"') 68
2𝜋𝑡
1
1
(2 − 2 cos(2𝜋𝑊𝑡)) = (1 − cos(2𝜋𝑊𝑡))
2𝜋𝑡
𝜋𝑡
+
Note: If we let 𝑊 → ∞, 𝐺(𝑓) → −𝑗 sgn(𝑓), the inverse of which "). This result agrees with
the limiting value of the solution for 𝑊 = ∞.
Problem 2.2
Suppose that, for a given signal 𝑥(𝑡), the integrated value of the signal over an interval 𝑇 is
required, as shown by
)
𝑦(𝑡) = ( 𝑥 (𝜏) 𝑑𝜏
)&,
Show that 𝑦(𝑡) can be obtained by processing the signal 𝑥(𝑡) with a filter having the
transfer function
𝐻(𝑓) = 𝑇 sinc(𝑓𝑇)exp(−𝑗𝜋𝑓𝑇)
Solution
The integrator output is
)
𝑦(𝑡) = ( 𝑥 (𝜏) 𝑑𝜏
)&,
Let 𝑥(𝑡) ↔ 𝑋(𝑓); then
-
𝑥(𝑡) = ( 𝑋 (𝑓)exp(𝑗2𝜋𝑓𝑡) 𝑑𝑓
&-
Therefore,
)
𝑦(𝑡) = (
-
0( 𝑋 (𝑓)exp(𝑗2𝜋𝑓𝜏) 𝑑𝑓3 𝑑𝜏
)&,
&-
Interchanging the order of integration:
-
)
𝑦(𝑡) = ( 𝑋 (𝑓) 0( exp (𝑗2𝜋𝑓𝜏) 𝑑𝜏3 𝑑𝑓
&-
)&,
-
= ( [𝑇𝑋(𝑓) sinc(𝑓𝑇)exp(−𝑗𝜋𝑓𝑇)] exp(𝑗2𝜋𝑓𝑡) 𝑑𝑓
&-
The Fourier transform of the integrator output is therefore
𝑌(𝑓) = 𝑇𝑋(𝑓) sinc(𝑓𝑇)exp(−𝑗𝜋𝑓𝑇)
(1)
Equation (1) shows that 𝑦(𝑡) can be obtained by passing the input signal 𝑥(𝑡) through a
linear filter whose transfer function is equal to 𝑇 sinc(𝑓𝑇)exp(−𝑗𝜋𝑓𝑇).