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Sec. 4–19 Study-Aid Examples 301<br />

Note: An alternative method of computing (P s ) norm is to calculate the area under the PDF for s(t).<br />

That is, by using Eq. (4–125),<br />

2<br />

(P s ) norm = (V s ) rms<br />

q<br />

= P s (f) df = 165 kW<br />

L<br />

-q<br />

(4–126b)<br />

Using Eq. (4–126a) or Eq. (4–126b), we obtain the actual average power dissipated in the<br />

50-Ω load: † (4–127)<br />

(P s ) actual = (V s) rms<br />

=<br />

R L<br />

2<br />

1.65 * 105<br />

50<br />

= 3.3 kW<br />

SA4–4 PEP for an AM Signal If the AM voltage signal of SA4–1 appears across a<br />

50-Ω resistive load, compute the actual peak envelope power (PEP).<br />

Solution.<br />

Using Eq. (4–18), we get the normalized PEP:<br />

(P PEP ) norm = 1 2 [ max |g(t)|]2 = 1 2 A c 2 [1 + max m(t)] 2<br />

= 1 2 (500)2 [1 + 0.8] 2 = 405 kW<br />

(4–128)<br />

Then the actual PEP for this AM voltage signal with a 50-Ω load is<br />

(P PEP ) actual = (P PEP) norm<br />

R L<br />

=<br />

4.50 * 105<br />

50<br />

= 8.1 kW<br />

(4–129)<br />

SA4–5 Sampling Methods for Bandpass Signals Suppose that a bandpass signal s(t) is<br />

to be sampled and that the samples are to be stored for processing at a later time. As shown<br />

in Fig. 4–32a, this bandpass signal has a bandwidth of B T centered about f c , where f c B T and<br />

B T 7 0. The signal s(t) is to be sampled by using any one of three methods shown in Fig. 4–32. ‡<br />

For each of these sampling methods, determine the minimum sampling frequency (i.e., minimum<br />

clock frequency) required, and discuss the advantages and disadvantages of each method.<br />

Solution.<br />

Method I Referring to Fig. 4–32a, we see that Method I uses direct sampling as described in<br />

Chapter 2. From Eq. (2–168), the minimum sampling frequency is (f s ) min = 2B, where B is the<br />

highest frequency in the signal. For this bandpass signal, the highest frequency is B = f c + B T 2.<br />

Thus, for Method I, the minimum sampling frequency is<br />

(f s ) min = 2f c + B T Method I<br />

(4–130)<br />

For example, if f c = 100 MHz and B T = 1 MHz, a minimum sampling frequency of (f s ) min =<br />

201 MHz would be required.<br />

† If s(t) is a current signal (instead of a voltage signal), then (P s ) actual = (I s ) 2 rms R L .<br />

‡ Thanks to Professor Christopher S. Anderson, Department of Electrical and Computer Engineering,<br />

University of Florida, for suggesting Method II.

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