PSK Signal Analyzer Bit Rate (bps) 50 Hz Carrier Frequency (Hz) 500 Hz Space Frequency (Hz) 1 Carrier Signal Amplitude (Ac) 1 Upload CSV, .wav, .mp3, or .mp4 Use Test Signal CSV Sample Rate (Hz): No Operation FFT (Spectrum) Amplitude Modulation (AM) Double Sideband Supressed Carrier (DSBSC) Pulse Amplitude Modulation (PAM) Flat To...

FSK Signal Analyzer Bit Rate (bps) 50 Hz Mark Frequency (Hz) 500 Hz Space Frequency (Hz) 1 Carrier Signal Amplitude (Ac) 1 Upload CSV, .wav, .mp3, or .mp4 Use Test Signal CSV Sample Rate (Hz): No Operation FFT (Spectrum) Amplitude Modulation (AM) Double Sideband Supressed Carrier (DSBSC) Pulse Amplitude Modulation (PAM) Flat Top P...

ASK Signal Analyzer Bit Rate (bps) 50 Hz Carrier Frequency (Hz) 500 Hz Message Signal Amplitude (Am) 1 Carrier Signal Amplitude (Ac) 1 Upload CSV, .wav, .mp3, or .mp4 Use Test Signal CSV Sample Rate (Hz): No Operation FFT (Spectrum) Amplitude Modulation (AM) Double Sideband Supressed Carrier (DSBSC) Pulse Amplitude Modulation (PAM) ...

What is a BJT? A BJT (Bipolar Junction Transistor) is a three-terminal semiconductor device used for amplification and switching . The three terminals are: Base (B) Collector (C) Emitter (E) The two main types of BJTs are NPN and PNP . NPN BJT Collector (C) Base (B) Emitter (E) NPN In an NPN transistor, a small base current controls a much larger collector current. Basic BJT Current Relationship I E = I C + I B The collector current is approximately related to the base current by: I C = β I B Here, β (beta) is the DC current gain of the transistor. Example: If β = 100 and I B = 20 μA: I C = 100...

  BJT = transistor &  Common-emitter = one way of connecting that transistor BJT as Common Emitter Here's the important diagram: +VCC R C VOUT VIN C 1 B C E GND Input Input voltage is applied between: Base ↔ Emitter Output Output is taken between: Collector ↔ Emitter  why it's called Common Emitter? The emitter is connected to the common reference (ground) . The emitter is common to both the input and output circuits. What actually happens? Suppose we increase the input voltage increases the base current and it increases the Ic Remember: I C = β I B I_C = \beta I_B Now look at the resistor R C R_C . The voltage across R C R_C is: V R C = I C R C V_{RC}=I_C R_C And: V O U T = V C C − I C R C V_{OUT}=V_{CC}-I_C R_C So if I C I_C increases: I C ↑ I_C↑ then: I C R C ↑ I_CR_C↑ therefo...

We have 3 coins: 🪙 A = HT 🪙 B = HH 🪙 C = TT First pick You pick a coin and get H . So the first coin cannot be TT . There are two possibilities: Possibility 1: You picked HT Remove it → remaining coins are: H H ,   T T HH,\ TT Second pick: HH → H TT → T So chance of H = 1/2 . Possibility 2: You picked HH Remove it → remaining coins are: H T ,   T T HT,\ TT Second pick: HT → H or T, each with probability 1/2 TT → T So chance of H = 1/4 . Now, because you already got H on the first pick, HH was twice as likely as HT : HT chance = 1/3 HH chance = 2/3 Therefore: 1 3 × 1 2 + 2 3 × 1 4 \frac13\times\frac12+\frac23\times\frac14 = 1 6 + 1 6 = 1 3 =\frac16+\frac16 =\boxed{\frac13} Summary: First H → remove that coin → depending on whether it was HT or HH, the remaining coins are different → combining those two cases gives 1 / 3 1/3 . So the answer is 1 / 3 \boxed{1/3} .

PPM Signal Analyzer Message Frequency (Hz) 50 Hz Carrier Frequency (Hz) 500 Hz Message Signal Amplitude (Am) 1 Carrier Signal Amplitude (Ac) 1 Upload CSV, .wav, .mp3, or .mp4 Use Test Signal CSV Sample Rate (Hz): No Operation FFT (Spectrum) Amplitude Modulation (AM) Double Sideband Supressed Carrier (DSBSC) Pulse Amplitude Modulation (PAM) ...

PWM Signal Analyzer Message Frequency (Hz) 50 Hz Carrier Frequency (Hz) 500 Hz Message Signal Amplitude (Am) 1 Carrier Signal Amplitude (Ac) 1 Upload CSV, .wav, .mp3, or .mp4 Use Test Signal CSV Sample Rate (Hz): No Operation FFT (Spectrum) Amplitude Modulation (AM) Double Sideband Supressed Carrier (DSBSC) Pulse Amplitude Modulation (PAM) ...