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Engineering, 15.02.2021 19:50 nakeytrag
"In a mobile radio system (e. g., cell phones), there is one type of degradation that can be modeled easily with sinusoids. This is the case of multipath fading caused by reflections of the radio wave interfering destructively at some locations. Suppose that a transmitting tower sends a sinusoid signal, and a mobile user receives not one but two copies of the transmitted signal: a direct-path transmission and a reflected path signal (e. g. from a large building).
The received signal is the sum of the two copies, and since they travel different distances they have different time delays. If the transmitted signal is s(t), then the received signal is
r(t)=s(tβt1)+s(tβt2)
In a mobile phone scenario, the distance between the mobile user and the transmitting tower is always changing. Suppose that the direct-path distance is
d1=x2+106ββββββββ
where x is the position of a mobile user who is moving along the x axis. Assume that the reflected-path distance is
d2=(xβ55)2+106βββββββββββββ+55
The amount of the delay (in seconds) can be computed for both propagation paths, using the fact that the time delay is the distance divided by the speed of light (3 x 10^8 m/s).
Assume that the transmitted signal is
s(t)=cos(300Γ106Οt)
The amplitude of the received signal is a measure of its strength. Show that as the mobile user moves, it is possible to find position where the signal strength is zero. Find one such location."
So, here's what I did:
The only thing needed from the received signal function is the amplitude, so I turn each direct and reflected signal into their phasor representation:
S1=eβjt1
S2=eβjt2
Then I sum the phasors:
S=S1+S2=eβjt1+eβjt2=cos(βt1)+cos(βt 2)+j(sin(βt1)+sin(βt2))
Then the amplitude can be achieved this way:
A=(cos(t1)+cos(t2))2+(sin(t1)+sin(t 2))2βββββββββββββββββββββββββββββββ ββ
And make the amplitude equal to zero:
0=(cos(t1)+cos(t2))2+(sin(t1)+sin(t 2))2βββββββββββββββββββββββββββββββ ββ
But when I try to solve for x (in which is obtained from t = d(x)/c) using Wolfram Alpha, I got no solution. What am I doing wrong here?
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Answers: 2
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