Major Exam No. 1 — Guided Practice: AM and FM

1. Given The amplitude factor \(6+1.5\cos(2\pi\cdot1200\,t)~\text{V}\) and carrier \(\cos(2\pi\cdot240000\,t)\)
2. Find \(f_c\), \(f_m\), \(A_c\), \(\mu\) and the envelope
3. Solution Factor out the average level:
   \[ 6+1.5\cos(2\pi\cdot1200\,t)=6\left[1+\frac{1.5}{6}\cos(2\pi\cdot1200\,t)\right] \]
   Compare with \(A_c\left[1+\mu\cos(2\pi f_mt)\right]\cos(2\pi f_ct)\): \[ A_c=6~\text{V} \] \[ \begin{aligned} \mu &=\frac{1.5~\text{V}}{6~\text{V}} \\ &=0.25 \end{aligned} \] \[ \begin{aligned} f_m &=1200~\text{Hz} \\ &=1.2~\text{kHz} \end{aligned} \] \[ \begin{aligned} f_c &=240000~\text{Hz} \\ &=240~\text{kHz} \end{aligned} \]
4. Result The envelope is \[ E(t)=\left[6+1.5\cos(2\pi\cdot1200\,t)\right]~\text{V}. \] It varies between \[ 6-1.5=4.5~\text{V} \] \[ 6+1.5=7.5~\text{V} \]
5. What it means The amplitude changes. The carrier stays a \(240~\text{kHz}\) cosine. The envelope stays positive, so it follows the message. The student's \(1.5\) is a swing in volts, not \(\mu\). The \(1200~\text{Hz}\) is the message frequency, not the carrier frequency.

1. Given spectral lines at \(448~\text{kHz}\), \(450~\text{kHz}\) and \(452~\text{kHz}\)
2. Find the line names, \(f_c\) and \(f_m\)
3. Solution The carrier is halfway between the sidebands, and the message frequency is half of their separation: \[ \begin{aligned} f_c &= \frac{f_{\mathrm{USB}}+f_{\mathrm{LSB}}}{2} \\ &= \frac{452+448}{2}~\text{kHz} \\ &= 450~\text{kHz} \end{aligned} \] \[ \begin{aligned} f_m &= \frac{f_{\mathrm{USB}}-f_{\mathrm{LSB}}}{2} \\ &= \frac{452-448}{2}~\text{kHz} \\ &= 2~\text{kHz}. \end{aligned} \] Check: \(452-450=2~\text{kHz}\).
4. Result \[ \text{LSB}=f_c-f_m=448~\text{kHz} \] \[ \text{carrier}=f_c=450~\text{kHz} \] \[ \text{USB}=f_c+f_m=452~\text{kHz} \] Put these labels under the matching stems. The draft had the two sideband labels reversed.
5. What it means The \(2~\text{kHz}\) message multiplies the \(450~\text{kHz}\) carrier and creates the difference and sum frequencies. The carrier line stays in the middle. Both sidebands carry the message. The carrier carries no new information. The span from the lower sideband to the upper sideband is \(452-448=4~\text{kHz}\).

1. Given channel \(732\text{–}740~\text{kHz}\); \(f_c=736~\text{kHz}\); \(B_m=3.4~\text{kHz}\), later \(4.6~\text{kHz}\)
2. Find whether the signal fits before and after the new tone, the effect of changing \(\mu\), and the largest \(B_m\)
3. Solution The channel is \(740-732=8~\text{kHz}\) wide. The AM bandwidth is \(B_{\mathrm{AM}}=2B_m\), and the outer edges are at \(f_c\pm B_m\). For \(B_m=3.4~\text{kHz}\): \[ \begin{aligned} B_{\mathrm{AM}} &=2B_m \\ &=2(3.4) \\ &=6.8~\text{kHz} \end{aligned} \] \[ \begin{aligned} 736-3.4 &=732.6~\text{kHz} \\ 736+3.4 &=739.4~\text{kHz} \end{aligned} \] For \(B_m=4.6~\text{kHz}\): \[ \begin{aligned} B_{\mathrm{AM}} &=2B_m \\ &=2(4.6) \\ &=9.2~\text{kHz} \end{aligned} \] \[ \begin{aligned} 736-4.6 &=731.4~\text{kHz} \\ 736+4.6 &=740.6~\text{kHz} \end{aligned} \] \[ \begin{aligned} B_{m,\max} &=\min(736-732,\ 740-736) \\ &=4~\text{kHz} \end{aligned} \]
4. Result The original message fits (\(6.8<8~\text{kHz}\)). With the added tone it does not fit (\(9.2>8~\text{kHz}\)). The highest message frequency that fits is \(4~\text{kHz}\).
5. What it means Cutting \(\mu\) in half does not help. The \(4.6~\text{kHz}\) offsets from the carrier stay the same. To fit, you must remove audio above \(4~\text{kHz}\) or use a wider channel. Removing audio loses part of the message, so reducing amplitude is not a fix for a bandwidth problem.

1. Given \[ f_c=90~\text{MHz}, \] \[ k_f=8~\text{kHz/V}, \] \[ A_m=1.5~\text{V} \]
2. Find \(f_i(t)\) at P, Q and R, and \(\Delta f\)
3. Solution \[ f_i(t)-f_c=k_fm(t) \] \[ \begin{aligned} \Delta f &=k_fA_m \\ &=8\times1.5 \\ &=12~\text{kHz} \end{aligned} \] The offsets at P, Q and R are \[ \begin{aligned} \text{P: }(8~\text{kHz/V})(+1.5~\text{V}) &=+12~\text{kHz} \\ \text{Q: }(8~\text{kHz/V})(0~\text{V}) &=0 \\ \text{R: }(8~\text{kHz/V})(-1.5~\text{V}) &=-12~\text{kHz} \end{aligned} \]
4. Result \[ \begin{aligned} \text{P: } f_i&=90.012~\text{MHz}\\ \text{Q: } f_i&=90.000~\text{MHz}\\ \text{R: } f_i&=89.988~\text{MHz} \end{aligned} \] \[ \Delta f=12~\text{kHz}. \]
5. What it means The center is \(90~\text{MHz}\). The instantaneous frequency moves up and down around it, and \(\Delta f\) is the largest distance from the center. A positive message gives a frequency above the center, zero gives the center, and a negative message gives a frequency below it. In ideal FM the amplitude stays constant. The message changes how close together the cycles are, not how tall they are. The full swing from lowest to highest is \(2\Delta f=24~\text{kHz}\).

1. Given \[ f_{\min}=83.992~\text{MHz}, \] \[ f_{\max}=84.008~\text{MHz}, \] \[ A_m'=1.5A_m \]
2. Find \(f_c\), \(\Delta f\), and the new \(f_{\max}'\) and \(f_{\min}'\)
3. Solution \[ \begin{aligned} f_c &= \frac{f_{\max}+f_{\min}}{2} \\ &= \frac{84.008+83.992}{2} \\ &= 84.000~\text{MHz} \end{aligned} \] \[ \begin{aligned} \Delta f &= \frac{f_{\max}-f_{\min}}{2} \\ &= \frac{84.008-83.992}{2} \\ &= 0.008~\text{MHz} \\ &= 8~\text{kHz} \end{aligned} \] \[ \begin{aligned} \Delta f' &= 1.5\,\Delta f \\ &= 1.5(8) \\ &= 12~\text{kHz} \\ &= 0.012~\text{MHz} \end{aligned} \]
4. Result \[ \begin{aligned} f_{\min}' &=84.000-0.012 \\ &=83.988~\text{MHz} \end{aligned} \] \[ \begin{aligned} f_{\max}' &=84.000+0.012 \\ &=84.012~\text{MHz} \end{aligned} \]
5. What it means A larger message amplitude gives a wider frequency swing. The center stays at \(84~\text{MHz}\), and the swing repeats at the same rate. In ideal FM the amplitude of the RF signal stays constant. A bigger message does not make a bigger RF signal.

1. Given \[ \Delta f=0.40~\text{kHz}, \] \[ f_{m,\mathrm{P}}=8~\text{kHz}, \] \[ f_{m,\mathrm{Q}}=0.10~\text{kHz} \]
2. Find \[ \beta_{\mathrm{P}},\qquad \beta_{\mathrm{Q}} \]
3. Solution \[ \begin{aligned} \beta_{\mathrm{P}} &=\frac{\Delta f}{f_{m,\mathrm{P}}} \\ &=\frac{0.40}{8} \\ &=0.05 \end{aligned} \] \[ \begin{aligned} \beta_{\mathrm{Q}} &=\frac{\Delta f}{f_{m,\mathrm{Q}}} \\ &=\frac{0.40}{0.10} \\ &=4 \end{aligned} \]
4. Result Test P: \(\beta=0.05\ll1\), so it is NBFM. Test Q: \(\beta=4>1\), so it is WBFM. Both have the same peak deviation, \(400~\text{Hz}\), and the same full swing, \(800~\text{Hz}\).
5. What it means \(\beta\) is larger in test Q because its message changes more slowly, not because its frequency moves farther. Doubling only \(f_c\) changes neither \(\Delta f\) nor \(f_m\), so both values of \(\beta\) and both answers stay the same. Also, \(\beta>1\) is not “overmodulation” as in AM.

1. Given \[ \Delta f=18~\text{kHz}, \] \[ f_{m,\mathrm{A}}=2~\text{kHz}, \] \[ f_{m,\mathrm{B}}=6~\text{kHz}, \] \[ B_{\text{available}}=46~\text{kHz} \]
2. Find \(B_{\mathrm{A}}\), \(B_{\mathrm{B}}\), and the largest \(\Delta f\) for the \(6~\text{kHz}\) tone
3. Solution \[ \begin{aligned} B_{\mathrm{A}} &\approx2(\Delta f+f_{m,\mathrm{A}}) \\ &=2(18+2) \\ &=40~\text{kHz} \end{aligned} \] \[ \begin{aligned} B_{\mathrm{B}} &\approx2(\Delta f+f_{m,\mathrm{B}}) \\ &=2(18+6) \\ &=48~\text{kHz} \end{aligned} \] \[ \begin{aligned} \Delta f_{\max} &\le\frac{B_{\text{available}}}{2}-f_m \\ &=\frac{46}{2}-6 \\ &=17~\text{kHz} \end{aligned} \]
4. Result Test A fits, with \(6~\text{kHz}\) to spare. Test B is \(2~\text{kHz}\) over the available bandwidth. For the \(6~\text{kHz}\) tone, the largest deviation that fits is \(17~\text{kHz}\), because \(2(17+6)=46~\text{kHz}\).
5. What it means The same deviation does not give the same bandwidth. Carson's rule gives a practical band that holds about 98% of the signal power, as taught in class. It does not say there is zero energy outside the band. Lowering \(\Delta f\) to make the signal fit changes the FM design, and this calculation does not tell us what happens to noise performance.

1. Given AM: \(B_m=3~\text{kHz}\). FM: \(B_m=9~\text{kHz}\) and \(\Delta f=24~\text{kHz}\). Maximum allowed RF bandwidth: morning \(12~\text{kHz}\), evening \(80~\text{kHz}\).
2. Find \(B_{\mathrm{AM}}\), \(B_{\mathrm{FM}}\), and one choice for each session
3. Solution \[ \begin{aligned} B_{\mathrm{AM}} &=2B_m \\ &=2(3) \\ &=6~\text{kHz} \end{aligned} \] \[ \begin{aligned} B_{\mathrm{FM}} &\approx2(\Delta f+B_m) \\ &=2(24+9) \\ &=66~\text{kHz} \end{aligned} \] An option must meet both the audio requirement and the maximum allowed RF bandwidth.
4. Result Morning: choose AM. Its \(3~\text{kHz}\) audio is enough for announcements, and \(6<12~\text{kHz}\). FM needs \(66>12~\text{kHz}\), so it does not fit. Evening: choose FM. Its \(9~\text{kHz}\) audio is enough for music, and \(66<80~\text{kHz}\), with \(14~\text{kHz}\) to spare. The AM filter stops at \(3~\text{kHz}\), so it cannot carry audio up to \(9~\text{kHz}\).
5. What it means Morning AM saves spectrum and needs a simpler receiver, but amplitude disturbances still affect it. Evening FM uses much more RF bandwidth and a more complex receiver. In return, the limiter removes amplitude variations, and the message stays in the instantaneous frequency. FM does not remove all noise, and we do not claim a numerical noise improvement here.

A complete evening claim: “Choose the fixed FM configuration. It passes audio up to \(9~\text{kHz}\), and its \(66~\text{kHz}\) Carson estimate fits within the \(80~\text{kHz}\) maximum allowed RF bandwidth. Its limiter helps against the moderate amplitude disturbances. The tradeoff is a wider RF channel and a more complex receiver.”