12V L1 10mH R1 47KQ 65 % Key=A C3 Q1 4.7µF Vout C1 R4 4.7kQ 1µF 2N2222A V1 10 Vpk 100kHz R2 222 kΩ R3 -1.0kΩ C2 =4.7pF v2 5 Vpk 5kHz 0° Figure 5. AM transistor modulator

How can you produce one sideband only? Suggest a method to obtain

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12V<br>L1<br>10mH<br>R1<br>47KQ<br>65 %<br>Key=A<br>C3<br>Q1<br>4.7µF<br>Vout<br>C1<br>R4<br>4.7kQ<br>1µF<br>2N2222A<br>V1<br>10 Vpk<br>100kHz<br>R2<br>222 kΩ<br>R3<br>-1.0kΩ<br>C2<br>=4.7pF<br>v2<br>5 Vpk<br>5kHz<br>0°<br>Figure 5. AM transistor modulator<br>

Extracted text: 12V L1 10mH R1 47KQ 65 % Key=A C3 Q1 4.7µF Vout C1 R4 4.7kQ 1µF 2N2222A V1 10 Vpk 100kHz R2 222 kΩ R3 -1.0kΩ C2 =4.7pF v2 5 Vpk 5kHz 0° Figure 5. AM transistor modulator
R1 D1 D1N4002 500 2 V2 VOFF = 1 L1 VAMPL = 0.8 FREQ = R2 2.5mH 1k C1 100n 300HZ 1 V1 VOFF = 0 VAMPL = 2 FREQ = 20KHZ 0. Figure 3. AM diode modulator.

R1<br>D1<br>D1N4002<br>500<br>2<br>V2<br>VOFF = 1<br>L1<br>VAMPL = 0.8<br>FREQ =<br>R2<br>2.5mH<br>1k<br>C1<br>100n<br>300HZ<br>1<br>V1<br>VOFF = 0<br>VAMPL = 2<br>FREQ =<br>20KHZ<br>0.<br>Figure 3. AM diode modulator.<br>

Extracted text: R1 D1 D1N4002 500 2 V2 VOFF = 1 L1 VAMPL = 0.8 FREQ = R2 2.5mH 1k C1 100n 300HZ 1 V1 VOFF = 0 VAMPL = 2 FREQ = 20KHZ 0. Figure 3. AM diode modulator.

Jun 11, 2022

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12V L1 10mH R1 47KQ 65 % Key=A C3 Q1 4.7µF Vout C1 R4 4.7kQ 1µF 2N2222A V1 10 Vpk 100kHz R2 222 kΩ R3 -1.0kΩ C2 =4.7pF v2 5 Vpk 5kHz 0° Figure 5. AM transistor modulator

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