STEM Problem Detail

Problem #95

Computer Science Hard

Problem Statement

Prompt

Please provide me step by step solution for the given Example - 5.3.

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**Solution**

Assume a mobile traveling at a velocity of 10 m/s receives two components at a carrier frequency of 1000 MHz.

Given:

\[
\lambda = \frac{3 \times 10^8}{1000 \times 10^6} = 0.3\ \text{m}
\]

The signal is received by the receiver from two paths with equal amplitude.

Note: For $$t = 0$$, the phases of both multipath components are $$0^\circ$$.

The narrowband instantaneous power is

\[
|r(t)|^2 = \left|\sqrt{50}\exp(j0) + \sqrt{50}\exp(j0)\right|^2 = 291\ \text{pW}
\]

At $$t = 0.1$$ s, the phase of the first component is

\[
\theta_1 = 2\pi d = 2\pi (10 \times 0.1 / 0.3) = 120^\circ
\]

Thus $$\theta_1 = 120^\circ$$, $$\theta_2 = -120^\circ$$, and the instantaneous power is

\[
|r(t)|^2 = \left|\sqrt{50}\exp(j120^\circ) + \sqrt{50}\exp(-j120^\circ)\right|^2 = 79.3\ \text{pW}
\]

Similarly, at $$t = 0.2$$ s,

\[
|r(t)|^2 = 79.3\ \text{pW}
\]

Similarly, at $$t = 0.3$$ s,

\[
|r(t)|^2 = 291\ \text{pW}
\]

It follows that at $$t = 0.4$$ s, $$|r(t)|^2 = 79.3$$ pW and at $$t = 0.5$$ s, $$|r(t)|^2 = 79.3$$ pW.

The average narrowband received power is equal to

\[
\frac{2(291) + 4(79.3)}{6} = 149\ \text{pW}
\]

Using Equation (5.19), the wideband power is given by

\[
E[a_r(t)] = \sum |a_i|^2 = 50 + 50 = 100\ \text{pW}
\]

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