Lösningar till Tentamen i Signaler och kommunikation, ETT080 ... - EIT

Lösningar till Tentamen i Signaler och kommunikation, ETT080, 2 juni 2006, kl 14–19 1
Uppgift 1
(a) In the z-transform domain the transfer function becomes
H(z) = 1 − 0.8z−1 + 0.6z −2
1 − 0.2z −1 − 0.9z −2 = z2 − 0.8z + 0.6
z 2 − 0.2z − 0.9z 2 = z2 − 0.8z + 0.6
(z − 1.05)(z + 0.85)
Hence, the system is non-stable.
Answer:False
(b) Counter example: The system with impulse response h[n] = δ[n] is clearly stable. The unit step
response is ys[n] = µ[n]. The energy is
∞
n=−∞
y 2 s[n] =
∞
1 = ∞
n=0
Answer:False
(c) As the unit circle is followed from 0 rad to π rad, there are two zeros both in the beginning and
in the end. In between the zeros there are a couple of poles affecting the frequency response. We
conclude that the amplitude response should be more attenuated in the beginning and the end of
the frequency interval, and less attenuated in between. Therefore, it should be a band pass filter.
Answer:True
(d) Firstly, assume that d0 = 0. The series expansion of H(z) is then
H(z) = p0
+ h1z
d0
−1 + h2z −2 + · · ·
Z−1 → p0
d0
δ[n] + h1δ[n − 1] + h2δ[n − 2] + · · ·
and we see that the system is causal. Secondly, assume that d0 = 0. Then the transfer function
becomes
H(z) = po + p1z−1 + · · · + pMz −M
d1z−1 + · · · + dNz −N = z po + p1z−1 + · · · + pMz −M
d1 + · · · + dNz −N+1
= z ˜h0 + ˜ h1z −1 + ˜ h2z −2
+ · · ·
Z −1
→ ˜ h0δ[n + 1] + ˜ h1δ[n] + ˜ h2δ[n − 1] + · · ·
which is non-causal.
Answer:True
(e) Consider the 8 point DFT of x = (10000000). This becomes x = (11111111). Clearly, the
relation is not true for this example.
Answer:False
Uppgift 2
(a) The frequency response becomes (from the figure)
H(e jω ) = 1 − 0.1ejω − 0.9e jω2
1 − 0.8e jω + 0.64e jω2
The transfer function is
H(z) = 1 − 0.1z−1 − 0.9z −2
1 − 0.8z −1 + 0.64z −2 = z2 − 0.1z − 0.9
z 2 − 0.8z + 0.64
Hence, the zeros are 1 and −0.9, and the poles are 0.8e ±j π
3 . From a pole/zero plot we can sketch
the amplitude response as below.

Lösningar till Tentamen i Signaler och kommunikation, ETT080, 2 juni 2006, kl 14–19 2
Imaginary Part
1
0.5
0
−0.5
−1
−1 −0.5 0
Real Part
0.5 1
6
5
4
3
2
1
0
0 pi
we can easily derive that the ending point is |H(e jπ )| = 0.08 and that |H(e jπ/3 )| = 5.3.
(b) The Z-transform of the step response is
Y (z) =
=
=
1
H(z) =
1 − z−1 1 + 0.9z −1
π
j (1 − 0.8e 3 z−1 π
−j )(1 − 0.8e 3 z−1 )
1 + 0.9z−1 1 − 2 · 0.8 cos( π
3 )z−1 + 0.82z−2 1 − 0.8 cos( π
3 )z−1
1 − 2 · 0.8 cos( π
3 )z−1 + 0.82 2.6
+
z−2 from the Z-transform table we get
0.8 √ 3
y[n] = 0.8 n cos( π 2.6
n)µ[n] +
3 0.8 √ 3 0.8n sin( π
3 n)µ[n]
= 0.8 n
cos( π
n) + 1.9 sin(π
3 3 n)
µ[n]
0.8 sin( π
3 )z−1
1 − 2 · 0.8 cos( π
3 )z−1 + 0.82z−2 (c) We know that if the input is x[n] = A cos( π
π
n + φ), the output is x[n] = A|H(ej 2 )| cos( 2 πn
+ 2
) + φ). To get the desired result we put
Θ( π
2
Uppgift 3
A =
1
π
j |H(e 2 )|
≈ 0.46
φ = −Θ( π
) ≈ 1.095
2
Sampling with sampling rate Fs gives a periodic repetition of the spectra with period Fs. Therefore, a
signal frequency f0 will occur at the frequencies
f = ±f0 + kFs, k ∈ Z
The reconstructed signal is the part of the spectra that lies in the interval −Fs/2 < f < Fs/2.
(a) If Fs = 320Hz the reconstructed signal is in −160Hz < f < 160Hz. The recorded sounds in the
problem becomes reconstructed at
Insekt f [Hz]
Ollonborre 50
Blomfluga 120
Humla 130
Husfluga 120
Mygga 40
Knott 86

Lösningar till Tentamen i Signaler och kommunikation, ETT080, 2 juni 2006, kl 14–19 3
(b) Use a sapling frequency such that the bumble bees are reconstructed at 20Hz. Then everything
below cannot be heard, and the frequencies above can be attenuated by the analog filter after
reconstruction. Use, e.g., Fs = 150Hz to do this. In the following table we check that it is only
the bumble bees that are aliased to 20Hz.
Insekt f [Hz]
Ollonborre 50
Blomfluga 30
Humla 20
Husfluga 50
Mygga 0
Knott 4
Use Fc = 25Hz as a cut off frequency for the filter.
Uppgift 4
Construct the filter as a cascade of two filters, one LP filter and one notch filter. Together they should
have at most order 5, which means that we can construct an order 2 notch filter and an order 3 LP filter.
Start with the LP filter. The frequencies between 3400Hz and 4000Hz should be attenuated. In the
sampled frequency this corresponds to ω3400 = 3400 17 π = 4000 20π and ω4000 = π. One (easy) way to
17π
17π
±j ±j construct such filter is to consider zeros at e 20 and -1, and the corresponding poles at Ae 20 and -B,
where both A and B are positive constants less than 1. The transfer function becomes
HLP (z) =
=
(1 + z−1 17π
j )(1 + e 20 z−1 17π
−j )(1 + e 20 z−1 )
(1 + Bz−1 17π
j )(1 + Ae 20 z−1 17π
−j )(1 + Ae 20 z−1 )
1 + (1 − 2 cos 17
20 )z−1 + (1 − 2 cos 17
20 )z−2 + z−3 1 + (B − 2A cos 17
20 )z−1 + (A2 − 2AB cos 17
20 )z−2 + A2Bz −3
With A = 0.8 and B = 0.7 the zero/pole plot and the amplitude response becomes
Imaginary Part
1
0.5
0
−0.5
−1
−1 −0.5 0
Real Part
0.5 1
1.5
1
0.5
0
0 pi
The 50Hz disturber can be cancelled with a notch filter with notch frequency ω50 = 50
transfer function is
HN(z) =
π
(1 + ej 80 z−1 π
−j )(1 + e 80 z−1 )
π
j (1 + Ce 80 z−1 π
−j )(1 + Ce 80 z−1 )
where C is a positive constant less than 1. With C = 0.9 we get
= 1 − 2 cos 1
80 z−1 + z −2
1 − 2C cos 1
80 z−1 + C 2 z −2
4000
1 π = π. The
80

Lösningar till Tentamen i Signaler och kommunikation, ETT080, 2 juni 2006, kl 14–19 4
Imaginary Part
1
0.5
0
−0.5
−1
−1 −0.5 0
Real Part
0.5 1
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0 pi
Using those two filters in cascade we get the following zero/pole plot and amplitude function
Imaginary Part
1
0.5
0
−0.5
−1
−1 −0.5 0
Real Part
0.5 1
The figures of the filters realisations are omitted.
Uppgift 5
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0 pi
(a) There are 48 channels. If each uses M signal alternatives the total number of signals becomes
⎧
M 48 ⎪⎨
=
⎪⎩
2 48 ≈ 10 14 , 2-PAM
4 48 ≈ 10 30 , 4-QAM
16 48 ≈ 10 58 , 16-QAM
64 48 ≈ 10 87 , 64-QAM
(b) Let k ∈ {1, 2, 4, 6} be the number of bits per transmitted signal in one channel. Then, the actual
transmitted bit rate for channel n is
Rb,n = k
·
Ts
3
⎧
0.1875Mbps,
⎪⎨
0.375Mbps,
4
⎪⎩
0.75Mbps,
1.125Mbps,
2-PAM
4-QAM
16-QAM
64-QAM
The maximum and minimum bit rate becomes
Rb,min = 0.1875 · 48 = 9Mbps
Rb,max = 1.125 · 48 = 54Mbps
(c) The signal in channel n is
si,n(t) = ainA cos(2πfnt) − binA sin(2πfnt)

Lösningar till Tentamen i Signaler och kommunikation, ETT080, 2 juni 2006, kl 14–19 5
This is the formula for QAM with a rectangular base pulse. Hence, the spectral density for channel
n is a sinc, see figure below. The sinc has zero for all values of fn + m/Ts, where m is a nonzero
integer. These zeros are used to position other carriers, see the following figure with 4 carriers.
Therefore, at the carrier frequencies, the signals in different channels do not interfear, which is the
origin of orthogonal in OFDM.
1
0
(d) For a rectangular pulse W99 = 20.6/Ts. That means that the contribution on each side of the
main lobe is ∆ = 9.3/Ts. An estimate of the 99% bandwidth of a 48 channel OFDM can be
given by the width of the main lobes plus a ∆ on each side,
W99,OFDM 48
+ 2∆ = 14.3MHz
Ts
The maximum bandwidth efficiency is
ρmax = Rb,max
W99,OFDM
≈ 3.8bps/Hz
The bit rate for a comparable 64-QAM system is
Rb = k
T
· 3
4
= 6
T
· 3
4
= 54Mbps
The symbol time then becomes T = 8.3 · 10 8 s. The 99% bandwidth is (with a rectangular pulse)
W99 = 20.6/T = 247MHz, and the bandwidth efficiency becomes ρ = 0.22bps/Hz. We see in
this comparison that OFDM is an efficient way to use the band width.
1
0