A symbol stream contains alternate QPSK and 16-QAM symbols. If symbols from this stream are transmitted at the rate of 1 mega-symbols per second, the raw (uncoded) data rate is _______ mega-bits per second (rounded off to one decimal place).   Answer:  Since alternative QPSK and 16-QAM used, data rate of uncoded data is (2+4)/2 = 3 Mbps Number of bits in one QPSK symbol is 2 and number of bits in one 16-QAM symbol is 4 Read more about Modulation Techniques → GATE EC Previous Year Papers with Solutions → Electronics and Communication Study Material →

Question: An ideal MOS capacitor (p-type semiconductor) is shown in the figure. The MOS capacitor is under strong inversion with V G1 = 2 V. The corresponding inversion charge density (Q inv1 ) is 2.2 µC/cm 2 . Assume oxide capacitance per unit area (C ox ) as 1.7 µF/cm 2 . For V G2 = 4 V, the value of Q inv2 is ______ µC/cm 2 (rounded off to one decimal place). The Catch: Once a MOS capacitor enters Strong Inversion , the surface potential (φ s ) becomes pinned. Any further increase in gate voltage (V G ) results in a linear increase in inversion charge (Q inv ), while the depletion charge remains constant. Step 1: Identify Given Parameters Initial Gate Voltage (V G1 ): 2 V Initial Inversion Charge (Q inv1 ): 2.2 µC/cm 2 Oxide Capacitance (C ox ): 1.7 µF/cm 2 Target Gate Voltage (V G2 ): 4 V Step 2: Use the Strong Inversion Relat...

  Let 𝑥1(𝑡)=𝑒^(-t)𝑢(𝑡) and 𝑥2(𝑡)=𝑢(𝑡)−𝑢(𝑡−2), where 𝑢(⋅) denotes the unit step function. If 𝑦(𝑡) denotes the convolution of 𝑥1(𝑡) and 𝑥2(𝑡), then lim(x→0) 𝑦(𝑡) = _________ (rounded off to one decimal place). Solution Let \[ x_1(t)=e^{-t}u(t) \] and \[ x_2(t)=u(t)-u(t-2) \] where \(u(t)\) is the unit step function. \[ y(t)=x_1(t)*x_2(t) \] Find the final value of \(y(t)\). Laplace Transforms \[ \mathcal{L}\{e^{-t}u(t)\} =\frac{1}{s+1} \] \[ \mathcal{L}\{u(t)-u(t-2)\} =\frac{1}{s}-\frac{e^{-2s}}{s} =\frac{1-e^{-2s}}{s} \] Using Convolution Property \[ Y(s)=X_1(s)X_2(s) \] \[ Y(s) = \frac{1}{s+1} \cdot \frac{1-e^{-2s}}{s} \] Applying Final Value Theorem \[ \lim_{t\to\infty}y(t) = \lim_{s\to0}sY(s) \] \[ = \lim_{s\to0} s \left( \frac{1}{s+1} \cdot \frac{1-e^{-2s}}{s} \right) \] \[ = \lim_{s\to0} \frac{1-e^{-2s}}{s+1} \] Using the limit we obtain \[ \boxed{\lim_{t\to\inf...

  A simple closed path 𝐶 in the complex plane is shown in the figure if \[ \oint_C \frac{2^z}{z^2-1}\,dz=-i\pi A \] where 𝑖=√−1, then the value of 𝐴 is ______ (rounded off to two decimal places). \[ \oint_C \frac{2^z}{z^2-1}\,dz=-i\pi A \] \[ \oint_C \frac{2^z}{z^2-1}\,dz = \oint_C \frac{2^z}{(z+1)(z-1)}\,dz \] Residue at \(z=1\): \[ \operatorname{Res}_{z=1} \frac{2^z}{(z+1)(z-1)} = \lim_{z\to1} (z-1) \frac{2^z}{(z+1)(z-1)} = \frac{2}{2} =1 \] \[ \oint_C \frac{2^z}{z^2-1}\,dz = 2\pi i \] Residue at \(z=-1\): \[ \operatorname{Res}_{z=-1} \frac{2^z}{(z+1)(z-1)} = \lim_{z\to-1} (z+1) \frac{2^z}{(z+1)(z-1)} = \frac{2^{-1}}{-2} = -\frac14 \] \[ \oint_C \frac{2^z}{z^2-1}\,dz = -\frac14(2\pi i) = -\frac12\pi i \] Given: \[ \oint_C \frac{2^z}{z^2-1}\,dz = -i\pi A \] Therefore, \[ -i\pi A = -\frac12\pi i \] \[ A=\frac12=0.50 \] Note: Here only \(z=-1\) lies inside the contour. GATE EC Previous Year...

The bar graph shows the frequency of the number of wickets taken in a match by a bowler in her career. For example, in 17 of her matches, the bowler has taken 5 wickets each. The median number of wickets taken by the bowler in a match is __________ (rounded off to one decimal place).   Solving for the Median Fundamental Concepts Median: The "middle" value in a sorted data set. Frequency: How many times a specific value occurs. Cumulative Frequency: The running total of frequencies. It helps us locate the middle position in a large data set without listing every single number. Step 1: Create a Frequency Table We extract the values from the bar graph and calculate the cumulative frequency (CF). Number of Wickets (x) Frequency (f) Cumulative Frequency (CF) ...

  Differential Pulse Position Modulation (DPPM) Differential Pulse Position Modulation (DPPM) is an advanced, high-efficiency variant of PPM where the position of each pulse is determined relative to the previous pulse rather than a fixed clock reference. DPPM Step-by-Step Example Imagine we want to send the sequence of numbers: 2, 1, 0 . In DPPM, each number represents the number of empty slots to wait after the last pulse. Step 1: Start Initial Reference: t = 0 ms Step 2: Send Data "2" Wait 2 units from the previous pulse. Pulse sent at: 0 + 2 = 2 ms Step 3: Send Data "1" Wait 1 unit from the previous pulse (at 2ms). Pulse sent at: 2 + 1 = 3 ms Step 4: Send Data "0" Wait 0 units from the previous pulse (at 3ms). Pulse sent at: 3 + 0 = 3.5 ms (Immediate pul...

  Pulse Position Modulation Example (Digital Method) Bits per symbol: m = 3 Number of slots per symbol duration: M = 2^m = 2^3 = 8 Example symbol: 101 → decimal 5 → pulse sent in slot 5 (out of 0…7) Symbol duration: T_s = 8 ms → each time slot: T_slot = T_s / M = 8 / 8 = 1 ms Pulse for symbol 101 is transmitted between 5 ms and 6 ms . Timeline representation of 3-bit PPM Slot: 0 1 2 3 4 5 6 7 Time: 0-1 1-2 2-3 3-4 4-5 5-6 6-7 7-8 ms Pulse → █ Return to Pulse Modulation Techniques Main Page →

GATE Differential Equations - One Day Revision Formula Sheet 1. First Order ODE Standard linear equation: \[ \frac{dy}{dx}+P(x)y=Q(x) \] Integrating factor: \[ \boxed{ IF=e^{\int P(x)dx} } \] Solution: \[ \boxed{ y(IF)=\int Q(IF)dx+C } \] 2. Exact Differential Equation Equation: \[ Mdx+Ndy=0 \] Condition: \[ \boxed{ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} } \] Solution: \[ \boxed{ \int Mdx+\phi(y)=C } \] 3. Second Order Linear ODE General form: \[ ay''+by'+cy=0 \] Auxiliary equation: \[ \boxed{ am^2+bm+c=0 } \] Roots: Roots Solution \(m_1,m_2\) \[ C_1e^{m_1x}+C_2e^{m_2x} \] Repeated root m \[ (C_1+C_2x)e^{mx} \] \(\alpha\pm i\beta\) \[ e^{\alpha x} (C_1\cos\beta x+C_2\sin\beta x) \] 4. Particular Integral (PI) For: \[ F(D)y=X \] PI: \[ \boxed{ \frac1{F(D)}X } \] Important rule: If RHS is: \[ e^{ax} \] replace: ...