22-Elec-A3 Signals and Communications · May 2016

Question 4 of 6: AM — Time-Domain Expression, Spectrum and Bandwidth

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 4: AM — Time-Domain Expression, Spectrum and Bandwidth (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) a plot of $m(t)$, (b) $s(t)$ in the time domain, (c) $S(f)$, and (d) the transmission bandwidth.

Approach. Collapse the two-term message into a single phase-shifted cosine to get its peak value, set the carrier amplitude from the peak-envelope constraint, then expand the product into carrier and sideband lines.

1. Collapse the message to one sinusoid. Using $a\cos\theta+b\sin\theta=\sqrt{a^{2}+b^{2}}\cos(\theta-\phi)$ with $\tan\phi=b/a$, and here $a=b=\sqrt2$ so $\phi=\pi/4$, $$m(t)=\sqrt2\cos(2\pi f_mt)+\sqrt2\sin(2\pi f_mt)=2\cos\!\left(2\pi f_mt-\frac{\pi}{4}\right).$$ The message is therefore a single 1 kHz tone of amplitude $$\boxed{\;|m|_{\max}=2\ \text{V},\qquad T_m=1/f_m=1\ \text{ms}\;}$$ lagging a pure cosine by $\pi/4$ rad, i.e. peaking at $t=T_m/8=0.125$ ms.

2. Fix the carrier amplitude from the peak-envelope constraint. Standard AM with modulation index $\mu$ is $$s(t)=A_c\left[1+\mu\,\frac{m(t)}{|m|_{\max}}\right]\cos(2\pi f_ct),$$ whose envelope swings between $A_c(1-\mu)$ and $A_c(1+\mu)$. Setting the maximum to the specified 8 V, $$A_c(1+\mu)=8\;\Rightarrow\;A_c=\frac{8}{1.5}=\boxed{\;\frac{16}{3}=5.333\ \text{V}\;}$$ The envelope minimum is then $A_c(1-\mu)=8/3=2.667$ V, which is positive — confirming there is no over-modulation and that envelope detection would work.
3. Write the AM signal. Substituting $m(t)/|m|_{\max}=\cos(2\pi f_mt-\pi/4)$, $$\boxed{\;s(t)=\frac{16}{3}\Bigl[1+0.5\cos\!\left(2\pi(1000)t-\tfrac{\pi}{4}\right)\Bigr] \cos\!\left(2\pi(10^{6})t\right)\ \text{V}\;}$$ Equivalently, in terms of the message as given, $s(t)=\tfrac{16}{3}\left[1+0.25\,m(t)\right]\cos(2\pi f_ct)$, since $\mu/|m|_{\max}=0.5/2=0.25$.
4. Expand into carrier plus sidebands. Applying $\cos X\cos Y=\tfrac12\left[\cos(X+Y)+\cos(X-Y)\right]$ to the product term, $$s(t)=\underbrace{\frac{16}{3}\cos 2\pi f_ct}_{\text{carrier}} +\underbrace{\frac{4}{3}\cos\!\left(2\pi(f_c+f_m)t-\frac{\pi}{4}\right)}_{\text{USB}} +\underbrace{\frac{4}{3}\cos\!\left(2\pi(f_c-f_m)t+\frac{\pi}{4}\right)}_{\text{LSB}},$$ where the sideband amplitude is $A_c\mu/2=(16/3)(0.5)/2=4/3$ V.
5. State the spectrum. Each cosine of amplitude $C$ and phase $\varphi$ contributes a conjugate pair of impulses of weight $\tfrac{C}{2}e^{\pm j\varphi}$, so $$\boxed{\;\begin{aligned} S(f)=&\ \frac{8}{3}\left[\delta(f-f_c)+\delta(f+f_c)\right]\\ &+\frac{2}{3}\left[e^{-j\pi/4}\delta(f-f_c-f_m)+e^{+j\pi/4}\delta(f+f_c+f_m)\right]\\ &+\frac{2}{3}\left[e^{+j\pi/4}\delta(f-f_c+f_m)+e^{-j\pi/4}\delta(f+f_c-f_m)\right] \end{aligned}\;}$$ The magnitude spectrum is therefore a carrier line of height $8/3$ at $\pm 1$ MHz flanked by two lines of height $2/3$ at $\pm 999$ kHz and $\pm 1001$ kHz.

6. Bandwidth. The occupied band runs from the lower to the upper sideband, $$B_T=(f_c+f_m)-(f_c-f_m)=2f_m=\boxed{\;2\ \text{kHz}\;}$$ independent of the carrier frequency and of the modulation index — a single tone always produces exactly one sideband pair. For reference, the transmitted power is $P=\tfrac{A_c^{2}}{2}\left(1+\tfrac{\mu^{2}}{2}\right)=16.0$ W into 1 Ω, of which only 11 % rides in the sidebands; that poor power efficiency is the price of the transmitted carrier that makes cheap envelope detection possible.