Signal Transmission & Reception Techniques – Test 2 April 1st, 2020
Last name: First name: ID:
1. Find the nul-to-null bandwidth
of a wide-sense stationary, zero-mean process X(t) whose autocorrelation
function is RX (τ ) = A tri Tτ1 cos(2πf0 τ ), where A > 0 and f0 T1
1. How would your answer change if
RX (τ ) = A rect Tτ1 cos(2πf0 τ ) instead? (4)
Answer. We need to find the PSD, which is the Fourier transform of the autocorrelation:
τ 1
A tri cos(2πf0 τ ) ↔ AT1 sinc2 (T1 f ) ? [δ(f − f0 ) + δ(f + f0 )]
T1 2
AT1
sinc2 (T1 (f − f0 )) + sinc2 (T1 (f + f0 )) ,
=
2
which is a couple of squared sincs centered at f = ±f0 . In the f > 0 range, the squared sinc centered at
f = f0 has zero crossings at f = f0 ± Tk1 , k = 1, 2,. . . Therefore the null-to-null bandwidth is T21 .
If RX (τ ) = A rect Tτ1 cos(2πf0 τ ), then the PSD would be given by
τ 1
A rect cos(2πf0 τ ) ↔ AT1 sinc(T1 f ) ? [δ(f − f0 ) + δ(f + f0 )]
T1 2
AT1
= [sinc(T1 (f − f0 )) + sinc(T1 (f + f0 ))] ,
2
which is not a valid PSD function, since it is negative for some frequencies. Thus RX (τ ) cannot be the
autocorrelation function of any WSS process.
2. A given amplifier has a 15-dB gain and 70-MHz noise equivalent bandwidth. The noise power at its output
is −73 dBm when at its input the PSD of thermal noise is −175 dBm/Hz. Obtain the noise figure of the
amplifier, in dB. [ Boltzmann’s constant: 1.38 · 10−23 J/K; Avogadro’s number: 6.023 · 1023 mol−1 .] (3)
Answer. The noise power at the output is Pno = GPni + Pna , where Pni = N20 · 2Beq is the noise power at the
input, and Pna = GκTe Beq is the noise power introduced by the non-ideal amplifier. Since G = 101.5 = 31.63,
N0 −17.5 = 3.16 · 10−18 mW/Hz, and B = 70 · 106 Hz, we have
2 = 10 eq
GPni = 31.63 × 3.16 · 10−18 × 140 · 106 ≈ 1.4 · 10−8 mW.
On the other hand,
Pna = GκTe Beq = 101.5 × 1.38 · 10−23 × Te × 70 · 106 W = 3.05 · 10−14 Te W = 3.05 · 10−11 Te mW.
Therefore,
1.4 · 10−8 + 3.05 · 10−11 Te = 10−7.3 ≈ 5 · 10−8 mW,
from which the effective noise temperature is Te ≈ 1178.4 K. Finally, the noise figure is
Te 1178.4
F = 10 log10 1 + = 10 log10 1 + ≈ 7 dB.
T0 290
, 3. The following Matlab code is part of a simulation at 2 Msamples/s of an envelope detector. Vector x contains
samples of a DSB-LC signal (modulation index 0.7, carrier frequency 400 kHz, 140 kHz bandwidth) corrupted
by white noise. Fill in the blanks of the first line in order to make sure that the envelope detector works
properly, and show your reasoning. What is the delay, in ms, introduced by the filtering stage? (3)
frecs = ; amps = ;
resp = firpm(80, frecs, amps); y = filter(resp, 1, abs(x));
Answer. Since the bandwidth of the DSB-LC signal is 140 kHz, it is clear that the filter should be lowpass
with a cutoff frequency which should be at least 12 · 140 kHz = 70 kHz. Normalizing this by half the sampling
rate f2s = 1 MHz = 1000 kHz, we obtain the normalized value 1000 70
= 0.07. Therefore, a possible choice is
frecs = [0 0.08 0.2 1]; amps = [1 1 0 0];
in which the transition band has a width of (0.2 − 0.08) f2s = 120 kHz. Of course, other choices are also valid
as long as the cutoff frequency is not smaller than 70 kHz.
80 40
The filter designed by firpm has order 80, and the delay is therefore 2 = 40 samples, or 40 × Ts = fs =
40
2000 kHz = 0.02 ms.
2
Last name: First name: ID:
1. Find the nul-to-null bandwidth
of a wide-sense stationary, zero-mean process X(t) whose autocorrelation
function is RX (τ ) = A tri Tτ1 cos(2πf0 τ ), where A > 0 and f0 T1
1. How would your answer change if
RX (τ ) = A rect Tτ1 cos(2πf0 τ ) instead? (4)
Answer. We need to find the PSD, which is the Fourier transform of the autocorrelation:
τ 1
A tri cos(2πf0 τ ) ↔ AT1 sinc2 (T1 f ) ? [δ(f − f0 ) + δ(f + f0 )]
T1 2
AT1
sinc2 (T1 (f − f0 )) + sinc2 (T1 (f + f0 )) ,
=
2
which is a couple of squared sincs centered at f = ±f0 . In the f > 0 range, the squared sinc centered at
f = f0 has zero crossings at f = f0 ± Tk1 , k = 1, 2,. . . Therefore the null-to-null bandwidth is T21 .
If RX (τ ) = A rect Tτ1 cos(2πf0 τ ), then the PSD would be given by
τ 1
A rect cos(2πf0 τ ) ↔ AT1 sinc(T1 f ) ? [δ(f − f0 ) + δ(f + f0 )]
T1 2
AT1
= [sinc(T1 (f − f0 )) + sinc(T1 (f + f0 ))] ,
2
which is not a valid PSD function, since it is negative for some frequencies. Thus RX (τ ) cannot be the
autocorrelation function of any WSS process.
2. A given amplifier has a 15-dB gain and 70-MHz noise equivalent bandwidth. The noise power at its output
is −73 dBm when at its input the PSD of thermal noise is −175 dBm/Hz. Obtain the noise figure of the
amplifier, in dB. [ Boltzmann’s constant: 1.38 · 10−23 J/K; Avogadro’s number: 6.023 · 1023 mol−1 .] (3)
Answer. The noise power at the output is Pno = GPni + Pna , where Pni = N20 · 2Beq is the noise power at the
input, and Pna = GκTe Beq is the noise power introduced by the non-ideal amplifier. Since G = 101.5 = 31.63,
N0 −17.5 = 3.16 · 10−18 mW/Hz, and B = 70 · 106 Hz, we have
2 = 10 eq
GPni = 31.63 × 3.16 · 10−18 × 140 · 106 ≈ 1.4 · 10−8 mW.
On the other hand,
Pna = GκTe Beq = 101.5 × 1.38 · 10−23 × Te × 70 · 106 W = 3.05 · 10−14 Te W = 3.05 · 10−11 Te mW.
Therefore,
1.4 · 10−8 + 3.05 · 10−11 Te = 10−7.3 ≈ 5 · 10−8 mW,
from which the effective noise temperature is Te ≈ 1178.4 K. Finally, the noise figure is
Te 1178.4
F = 10 log10 1 + = 10 log10 1 + ≈ 7 dB.
T0 290
, 3. The following Matlab code is part of a simulation at 2 Msamples/s of an envelope detector. Vector x contains
samples of a DSB-LC signal (modulation index 0.7, carrier frequency 400 kHz, 140 kHz bandwidth) corrupted
by white noise. Fill in the blanks of the first line in order to make sure that the envelope detector works
properly, and show your reasoning. What is the delay, in ms, introduced by the filtering stage? (3)
frecs = ; amps = ;
resp = firpm(80, frecs, amps); y = filter(resp, 1, abs(x));
Answer. Since the bandwidth of the DSB-LC signal is 140 kHz, it is clear that the filter should be lowpass
with a cutoff frequency which should be at least 12 · 140 kHz = 70 kHz. Normalizing this by half the sampling
rate f2s = 1 MHz = 1000 kHz, we obtain the normalized value 1000 70
= 0.07. Therefore, a possible choice is
frecs = [0 0.08 0.2 1]; amps = [1 1 0 0];
in which the transition band has a width of (0.2 − 0.08) f2s = 120 kHz. Of course, other choices are also valid
as long as the cutoff frequency is not smaller than 70 kHz.
80 40
The filter designed by firpm has order 80, and the delay is therefore 2 = 40 samples, or 40 × Ts = fs =
40
2000 kHz = 0.02 ms.
2
