MOSFET Transconductance (g m ) Transconductance g m is the rate at which the drain current changes with respect to the gate-source voltage: g m = ∂I D / ∂V GS 1. Linear (Triode) Region Condition: V DS < V GS − V T Drain Current: I D = μ n C ox (W/L) [(V GS − V T )V DS − V DS 2 /2] Taking derivative with respect to V GS : g m = μ n C ox (W/L)V DS Let: k n = μ n C ox (W/L) Then: g m = k n V DS In the linear region, transconductance is directly proportional to V DS . 2. Saturation Region Condition: V DS ≥ V GS − V T Drain Current: I D = (1/2)k n (V GS − V T ) 2 Taking derivative with respect to V GS : g m = k n (V GS − V T ) Alternative forms: g m = 2I D / (V GS − V T ) g m = √(2k n I D ) Summary Region Drain Current (I D ) Transconductance (g m ) Linear (Triode) k n [(V GS −V T )V DS − V DS 2 /2] ...

OFDM 16-QAM Visual Lab OFDM Visual Lab: 16-QAM High Spectral Efficiency (4 bits/symbol) using Gray-Coded Mapping Subcarriers (N) (Choose 2,4,8,16,etc) Cyclic Prefix (CP) Channel SNR (dB) Update Simulator Step 1: Parallel Bit Groups In 16-QAM, each subcarrier $k$ carries 4 bits . Two bits determine the In-phase (Real) amplitude, and two determine the Quadrature (Imaginary) amplitude. $$\text{Bits per Subcarrier: } \{b_0, b_1, b_2, b_3\}$$ Step 2: Gray-Coded Amplitude Mapping ...

OFDM 4-QAM: All Steps Visualizer OFDM 4-QAM: Step-by-Step Multiplexing Visualizing Frequency-to-Time transformations Subcarriers (N) (Choose 2, 4, 8, 16, etc) CP Length SNR (dB) Update Step 1: Serial-to-Parallel Bits Generating two parallel streams (In-phase and Quadrature). Step 2: 4-QAM Mapping $$X_k = \frac{1}{\sqrt{2}}(A_{I} + jA_{Q})$$ Step 3: Multiplexing (IFFT) Converting frequency subcarriers into a time-domain wave. $$x[n] = \text{IFFT}\{X_k\}$$ Step 4: Guard Interval (Cyclic Prefix) Prepending the last $N_{cp}$ samples to the front to prevent ISI. ...

OFDM QPSK Lab OFDM Visual Lab: QPSK Quadrature Phase Shift Keying (2 bits/symbol) Subcarriers (N) (Choose 2, 4, 8, 16, etc) SNR (dB) Run QPSK Simulation Step 1: Mapping 2 Bits per Subcarrier QPSK uses four points. Each subcarrier $k$ carries two bits $(b_{2k}, b_{2k+1})$. $$X_k = \frac{1}{\sqrt{2}} \left[ (2b_{2k}-1) + j(2b_{2k+1}-1) \right]$$ Step 2: Frequency to Time (IFFT) Step 3: Received Constellation Note how the points cluster around $(\pm 0.7, \pm 0.7)$ due to the $1/\sqrt{2}$ normalization. Return to Online Simulations Main Page → OFDM Main Page →

OFDM BPSK Step-by-Step OFDM Visual Lab: BPSK Modulation A complete visual walkthrough of the Multiplexing and Demultiplexing process. Subcarriers (N) (Choose 2, 4, 8, 16, etc) Cyclic Prefix (Choose value SNR (dB) Update Simulation Step 1: Information Bit Generation We start with raw binary data. For BPSK, each bit represents one symbol. $$\text{Data Vector: } \mathbf{b} = [b_0, b_1, \dots, b_{N-1}], \quad b_i \in \{0, 1\}$$ Step 2: BPSK Constellation Mapping Bits are mapped to complex voltages. 0 becomes -1 and 1 becomes +1. $$X_k = (2b_k - 1) + 0j$$ Step 3: Inverse Fast Fourier Transform (IFFT) This is where the magic happens. We transfo...

6. The feedback network used in Colpitts oscillator is given below: 20 mH C₂ C₁ 4 pF The value of \(C_2\) and feedback factor '\(\beta\)' for oscillating frequency of 1 MHz is: 1. 0.8 pF, 0.2 2. 4 pF, 0.9 3. 1.8 pF, 0.45 4. 2.2 pF, 0.9 Answer: 3 Solution: The oscillating frequency \(f_0\) of a Colpitts oscillator is given by: \(f_0 = \frac{1}{2\pi\sqrt{L_{eq}C_{eq}}}\) where \(L_{e...

Colpitts Oscillator A Colpitts oscillator is an LC sinusoidal oscillator that uses: One inductor (L) Two capacitors (C₁ and C₂) connected in series The two capacitors form a capacitive voltage divider that provides the positive feedback required for oscillation. Circuit Diagram Simplified transistor Colpitts oscillator: L +----^^^^----+ | | C1 C2 | | +-----+------+ | GND Transistor amplifier provides gain and feedback Working Principle The transistor amplifies a small noise signal. The LC tank circuit resonates at its natural frequency. C₁ and C₂ divide the voltage and feed part of the output back to the input. If the loop gain satisfies the Barkhausen criterion, continuous oscillations are produced. Frequency of Oscillation...

  Which of the following is an intersegment indirect mode of control transfer instruction in 8086? 1. JMP SHORT LABEL 2. JMP 5000H:2000H 3. JMP [2000H] 4. JMP [BX] Answer; 3 Solution 1. JMP SHORT LABEL Intrasegment (within the same code segment). Direct jump. ❌ Not intersegment indirect. 2. JMP 5000H:2000H Intersegment (far jump because both CS and IP are specified). Direct jump (address is explicitly given). ❌ Not indirect. 3. JMP [2000H] The destination address is taken from memory location 2000H. This is indirect. In 8086, a far indirect jump can use a memory operand containing both IP and CS (depending on operand size), making it an intersegment indirect jump. ✅ Correct answer. 4. JMP [BX] Indirect jump through memory addressed by BX. Usually intrasegment (near indirect jump). ❌ Not intersegment. Browse All Solved Papers (2012 - 2025) → UGC-NET : Electronics Science Study Material (Subject: 088) → ...