se dotted lines, draw a sinusoid cos(w)t. That's it. You've got your AM signal. Analog Communications Ac(1 + mx(t)) cos(Wct) Figure 12.4 (a) Input to AM modulator (b) Output of AM modulator (m=1) = Example 12.1 Plot the output of an AM modulator using m = 1 and A =2, = when the input is that in Figure E12.1. Solution: The sent signal, which in general corresponds to Equation (12.2), this time looks like S (t) = 2(1+x(t)) cos Wct (E12.1) The 2(1 + x(t)) looks like the dotted line drawn at the top of Figure E12.2. This creates the envelope of the sent signal s(t), which is shown in the solid "jiggly" line of Figure E12.2. Chapter Twelve Figure E12.1 Input to AM modulator 2(1 +x(t)) Figure E12.2 Output from AM modulator 12.2.2 AM Modulation-in Frequency Let's see what the sent signal s(t) looks like in the frequency domain-that is, let's look at the Fourier transform of s(t), S(f). First, we know (12.10) (12.11) Now, turning to the frequency domain, and using basic properties of Fourier trans- forms, we have (12.12) (12.13) A picture is worth a thousand words, SO let's see what S(f) looks like. Let's say we have an X(f) as shown in Figure 12.5(a). According to Equation (12.13), we then have an S(f) as shown in Figure 12.5(b). The different parts of S(f) in Equation (12.13) are pointed out in the plot of Figure 12.5(b). Analog Communications Ac (S(f++c) X(f+fc) Acm x(f-fo) Figure 12.5 (a) Information signal input to AM modulator in frequency domain, X(f) (b) Signal output by AM modulator in frequency domain, s(f) (shown with m = 0.5) One thing engineers noticed when they looked at what was going on in the frequency domain was that transmission power was being spent in sending the im- pulses in S(f)-the s(f+f) and s(f-f)-across the channel. They decided to come up with a measure to figure out what percent of power was being spent sending these impulses, called modulation efficiency, and defined as follows: modulation efficiency, n = percent of total power that is being used to convey information; or, more mathematical