DSBSC Waveform Generator Standard: Double Sideband Suppresed Carrier (DSBSC) Message Frequency (Hz) 50 Hz Carrier Frequency (Hz) 500 Hz Message Signal Gain 1 Step 1: messagePlot $m(t) = A_m\cos(2\pi f_m t)$ Step 2: carrierPlot $c(t) = A_c\cos(2\pi f_c t)$ Step 3: superimposed $x(t) = [c(t) + m(t)]$ Step 4: modulatedPlot $s(t) = A_c\,m(t)\cos(2\pi f_c t)$ Perform DSBSC Demodulation ...

  Q. The area bounded by the curves y = |x – 1| + |x – 2| and y = 3 is equal to   (1) 3   (2) 4   (3) 5   (4) 6 Area Between the Curves We are given: \[ y = |x-1| + |x-2| \] and \[ y = 3 \] 1. Find the points of intersection At the points of intersection: \[ |x-1| + |x-2| = 3 \] Case 1: \(x \leq 1\) Here both expressions inside the modulus are negative: \[ |x-1| = 1-x \] \[ |x-2| = 2-x \] Therefore: \[ (1-x)+(2-x)=3 \] \[ 3-2x=3 \] \[ x=0 \] First intersection: \[ \boxed{(0,3)} \] Case 2: \(1 \leq x \leq 2\) In this interval: \[ |x-1|=x-1 \] ...

  Carson’s Rule For an FM signal, the approximate bandwidth is B W ≈ 2 ( Δ f + f m ) \boxed{BW \approx 2(\Delta f+f_m)} where: Δ f \Delta f = maximum frequency deviation f m f_m = highest frequency in the modulating/message signal It gives a practical estimate of the bandwidth containing roughly 98% of the FM signal power .  Where does the derivative come in? An FM signal can be written as s ( t ) = A c cos ⁡ ( 2 π f c t + ϕ ( t ) ) s(t)=A_c\cos\left(2\pi f_ct+\phi(t)\right) The instantaneous phase is θ ( t ) = 2 π f c t + ϕ ( t ) \theta(t)=2\pi f_ct+\phi(t) and instantaneous frequency is obtained by differentiating phase: f i ( t ) = 1 2 π d θ ( t ) d t \boxed{f_i(t)=\frac{1}{2\pi}\frac{d\theta(t)}{dt}} Simple example Suppose an FM transmitter has: Δ f = 75   k H z \Delta f=75\,kHz and the highest modulating frequency is f m = 15   k H z . f_m=15\,kHz. Then B W ≈ 2 ( 75 + 15 ) BW\approx2(75+15) B W ≈ 180   k H z \boxed{BW\approx180\,kHz} So you'd allocate roughly 180 kHz ...

  Difference Between DC Motor and Induction Motor Feature DC Motor Induction Motor Power Supply Direct Current (DC) Alternating Current (AC) Working Principle Magnetic force on a current-carrying conductor Electromagnetic induction Rotor Current Supplied directly Induced automatically Brushes Required (except BLDC motors) Not required Commutator Present Not present Maintenance Higher Lower Efficiency High Very High Speed Control Easy and precise Requires Variable Frequency Drive (VFD) Starting Torque High Moderate to High Cost Usually higher Generally lower Reliability Good Excellent Common Applications Electric vehicles, cranes, robotics, conveyors Pumps, fans, compressors, industrial machinery Advantages of DC Motors Excellent speed control. High starting torque. Quick acceleration and deceleration. Suitable for battery-powered applications. Ideal for precision control systems. Advantages of Induction Motors Simple construction. Low maintenance. No brushes or commutator. Long se...

Electric motors are essential components in industrial machines, household appliances, electric vehicles, robotics, and automation systems. Among the most widely used motors are the DC motor and the induction motor . Although both convert electrical energy into mechanical energy, they differ significantly in their operating principles, construction, and applications. What Is a DC Motor? A Direct Current (DC) motor is an electrical machine that converts direct current (DC) electrical energy into mechanical rotation. It operates based on the interaction between a magnetic field and a current-carrying conductor. How Does a DC Motor Work? The working principle of a DC motor is based on Lorentz Force , which states that when a current-carrying conductor is placed in a magnetic field, it experiences a force. This force creates torque that rotates the motor shaft. Working Process DC voltage is supplied to the armature winding. The armature current produces ...

  A traditional ceiling fan is an example of an induction motor , specifically a single-phase induction motor . Traditional ceiling fan: usually a single-phase induction motor , commonly a capacitor-run motor . The capacitor creates a phase difference between the main and auxiliary windings, producing a rotating magnetic field.  A single AC wire is enough to produce a magnetic field, but not a naturally rotating magnetic field.  Suppose we connect one coil to AC:  So the field simply gets stronger, weaker, and reverses direction. It does not continuously rotate around the motor.  Because of this, a single-phase induction motor has zero starting torque when stationary (in the ideal case). Role of a Capacitor in a Ceiling Fan  A typical AC fan has two windings:  - Main winding → connected to AC - Auxiliary winding → AC through a capacitor  The capacitor shifts the current in the auxiliary winding, creating a phase difference between the two magne...

  The key point is that the magnetic field is continuously changing inside the stator. In an AC induction motor: AC current changes direction every half cycle. Therefore, the magnetic field also changes direction . In a 3-phase induction motor , the three phase currents are 120° apart, so their combined magnetic field produces a continuously rotating magnetic field . The rotor experiences this continuously changing magnetic field, so current is induced in the rotor. That induced rotor current produces torque , keeping the rotor rotating. So even though the AC current becomes zero momentarily , the motor doesn't stop. The rotating magnetic field has inertia/momentum in the rotor, and the field immediately changes direction and continues rotating. But in a 3-phase motor,  the magnetic field is changing continuously . That's why a 3-phase induction motor runs continuously rather than starting and stopping 50 times per second. A single-phase induction motor is differe...

  If V r e f V_{ref} is the output reference value and V f s V_{fs} is the ideal full-scale output voltage when all digital inputs are 1. Which of the following equations is correct? A. V r e f = V f s ( 1 − 1 2 n ) V_{ref} = V_{fs}\left(1-\frac{1}{2^n}\right) B. V f s = V r e f ( 1 2 n − 1 ) V_{fs} = V_{ref}\left(\frac{1}{2^n}-1\right) C. V f s = V r e f ( 1 − 1 2 n ) V_{fs} = V_{ref}\left(1-\frac{1}{2^n}\right) D. V r e f = V f s ( 1 2 n − 1 ) V_{ref} = V_{fs}\left(\frac{1}{2^n}-1\right) The correct answer is C : V f s = V r e f ( 1 − 1 2 n ) \boxed{V_{fs}=V_{ref}\left(1-\frac{1}{2^n}\right)} Why? For an n-bit DAC , the largest digital input is: 111 … 111 = 2 n − 1 111\ldots111 = 2^n-1 The DAC output is: V o = V r e f Digital input 2 n V_o=V_{ref}\frac{\text{Digital input}}{2^n} At the maximum input: V f s = V r e f 2 n − 1 2 n V_{fs}=V_{ref}\frac{2^n-1}{2^n} Separate the fraction: V f s = V r e f ( 2 n 2 n − 1 2 n ) V_{fs}=V_{ref}\left(\frac{2^n}{2^n}-\frac{1}{2^n}\r...