Intermediate test
Luigi Palopoli
January 6, 2014
1
Question
Consider a signal whose spectrum is given by:
 2
4ω
if ω ∈ [−1, 1]



ω + 5
if ω ∈ [−5, −1]
S(ω) =

−ω
+
5
if
ω ∈ [1, 5]



0
Otherwise.
1. Compute the Energy of the signal
2. Compute the time domain signal corresponding having spectrum S(ω)
3. Suppose you want to transmit the signal using sampling. Show the block
scheme for transmitting and reconstructing the signal and say what is the
minimum frequency that should be used.
4. Suppose you first sample with sampling frequency ωs = 6 rad/s, you
transmit it and then you use a low pass filter with bandwidth 5rad/s to
recover it. What is the resulting signal?
1.1
Response
1. Applying Parseval Theorem, the Energy of the signal is given by:
Z ∞
1
S 2 (ω)dω
E=
2π −∞
Z ∞
1
=2
S 2 (ω)dω
2π 0
1 3
5 !
10ω 2 1 16 5 ω
ω +
+ 25ω −
=
π
5 0
3
2
1
1
1 16 125
+
+ 125 − 125 − − 25 + 5
=
π 5
3
3
≈ 7.81
1
2. We apply duality. Let us first consider the FT of the signal S(t). It can
be seen that:
Z t
S(t) =
S1 (τ )dτ
−∞
Z
t
S1 (t) = G4 (t + 3) − G4 (t − 3) +
S2 (τ )dτ
−∞
S2 (t) = 8 (−δ(t + 1) + G2 (τ ) − δ(t − 1)) .
Hence,
F (S2 (τ )) = 8 −ejω − e−jω + 2Sinc (ω)
= 16 (Sinc (ω) − cos ω)
16
(Sinc (ω) − cos ω)
F (S1 (τ )) = 4Sinc(2ω) ejω3 − e−jω3 +
jω
16j
= 8jSinc(2ω) sin 3ω −
(Sinc (ω) − cos ω)
ω
16
8
F (S(τ )) = Sinc(2ω) sin 3ω − 2 (Sinc (ω) − cos ω)
ω
ω
16
= 24Sinc(2ω)Sinc(3ω) − 2 (Sinc (ω) − cos ω)
ω
By applying duality, the continuous time signal has to be
4
2
s(t) =
3Sinc(2t)Sinc(3t) − 2 (Sinc (t) − cos t)
π
t
3. The transmission scheme is based on a sampler, which has to operate with
a minimum frequency of 10 rad/s.
4. If we sample at 6rad/s, and then we
 2
4ω



4
S 0 (ω) =

4



0
filter we get
if ω ∈ [−1, 1]
if ω ∈ [−5, −1]
if ω ∈ [1, 5]
Otherwise.
Applying the same line of reasoning as in the first question, we can write:
Z t
S 0 (t) = 4G4 (t + 3) + 4G4 (t − 3) +
S3 (τ )dτ
−∞
Z
t
S3 (t) = 4δ(t + 1) − 4δ(t − 1) +
S2 (τ )dτ
−∞
F (S3 (t)) = 8j sin ω −
16j
(Sinc (ω) − cos ω)
ω
F (S 0 (t)) = 32Sinc (2ω) cos 3ω + 8Sinc (ω) −
2
16
(Sinc (ω) − cos ω)
ω2
Applying duality, we obtain:
1
16
s0 (t) =
32Sinc (2t) cos 3t + 8Sinc (t) − 2 (Sinc (t) − cos t)
2π
t
3
2
Question
Consider a system associated with the following differential equation:
ÿ + 5ẏ + 4y = u̇ + 3u
Show the Bode plot with amplitude and phase. Given an input signal cos ω0 t;
say if the system has a resonance peak and, in the affirmative case, compute
the frequency.
2.1
Response
1. The Transfer Function is given by:
Y (s) =
s+3
U (s).
s2 + 5s + 4
We can write it in Bode Form as follows:
Y (jω) =
1 + jω
3
3
4 (1 + jω) 1 +
jω
4
The asymptotic plot is in Figure 1. In the top part we show the amplitude
plot. The first singularity that operates is the pole at 1, which determines
a decrease of 20db/decade, followed by the zero in 3 that introduces an
increase of 20db/decade flattening the plot. When the pole at 4 kicks in,
the decrease is once again 20dB/decade. The bottom figure reports the
phase plot. The zero introduces an increase of 90 deg, while both poles
introduces a decrease of 90 deg. The plot asymptotically reaches -90 deg.
Conventionally, each singularity starts to be active one decade before and
its operation ends one decade after.
2. Since the transfer function has two real poles, it has no resonance peak,
as clearly visible by the constant sign of the derivative of |H(jω)|2 . So the
maximum response of the system is from ω0 = 0.
4
Amplitude Plot (dB)
0
−20
−40
−60
−2
10
−1
10
0
10
1
10
1
10
10
2
10
3
2
10
w (rad/s)
phase plot (degrees)
0
−50
−100
−2
10
−1
10
0
10
10
Figure 1: Asymptotic bode plot
5
3