Online Signal Processing Simulations

The cutoff frequency (in GHz) for the dominant TE10 mode of an air-filled rectangular waveguide of ...

2026-11-01T19:01:00.206+05:30

Q. The cutoff frequency (in GHz) for the dominant TE10 mode of an air-filled rectangular waveguide of inner dimension 0.28 inch×0.14 inch is __. (rounded off to two decimal places).

Given Information

Waveguide dimensions = 0.28 inch × 0.14 inch
Wider dimension (a) = 0.28 inch
Smaller dimension (b) = 0.14 inch
Medium = Air
Dominant mode = TE₁₀
Speed of light (c) = 3 × 10⁸ m/s

Step 1: Cutoff Frequency Formula

The cutoff frequency of a rectangular waveguide is:

f_c = (c / 2) √[(m/a)² + (n/b)²]

For the dominant TE₁₀ mode:

m = 1, n = 0

f_c = c / 2a

Step 2: Convert Inch into Meter

We know that 1 inch = 0.0254 meter.

a = 0.28 × 0.0254

a = 0.007112 m

Step 3: Calculate Cutoff Frequency

Substitute the values into the formula:

f_c = c / 2a

= (3 × 10⁸) / (2 × 0.007112)

= 21.0911 × 10⁹ Hz

= 21.0911 GHz

Final Answer

21.09 GHz

Cutoff frequency of the dominant TE10 mode

Important Note

For the dominant TE10 mode, the cutoff frequency depends only on the wider dimension (a), not on the smaller dimension (b).

A wireless digital transmission scheme is using 16-QAM over an additive white Gaussian noise channel ...

2026-11-01T18:48:58.730+05:30

Given Information

Modulation scheme: 16-QAM
Bit rate: R_b = 4 × 10⁶ bits/sec
Channel: Additive White Gaussian Noise (AWGN)
Receiver: Maximum-Likelihood Receiver

Step 1: Calculate Bits per Symbol

For M-ary QAM, the number of bits transmitted per symbol is:

Bits per symbol = log₂(M)

= log₂(16) = 4 bits/symbol

Step 2: Calculate Symbol Rate

Symbol rate is calculated by dividing the bit rate by the number of bits per symbol.

R_s = R_b / log₂(M)

= (4 × 10⁶) / 4

= 10⁶ symbols/sec

Step 3: Calculate Minimum Bandwidth

For ideal Nyquist transmission with zero roll-off factor:

B_min = R_s

= 10⁶ Hz

= 1 MHz

Final Answer

1.00 MHz

Minimum transmission bandwidth required

Note: AWGN and maximum-likelihood detection do not change the minimum Nyquist bandwidth. The calculation assumes ideal zero-roll-off pulse shaping.

What is a Supernode?

2026-11-01T18:31:21.354+05:30

Circuit Analysis

What Is a Supernode?

A supernode is formed when an ideal voltage source connects two non-reference nodes. It allows us to apply Kirchhoff's Current Law (KCL) to the combined circuit boundary.

1. Understanding a Supernode

In nodal analysis, a voltage source between two unknown-voltage nodes makes the current through that source difficult to express directly using Ohm's law. We combine the two nodes and the voltage source into one extended region called a supernode.

Key idea

Apply KCL around the outer boundary of the supernode, then use the voltage source to write an additional voltage equation.

2. Supernode Circuit Diagram

The voltage source connects nodes V_L and V_R. The dashed blue boundary surrounds the supernode. Currents through the external resistors are included in the KCL equation.

Supernode boundary

Voltage source

Circuit connections

Conceptual diagram. The resistor values and exact circuit connections depend on the specific problem being solved.

3. Why Do We Need a Supernode?

An ideal voltage source connects two unknown nodes

The voltage difference is known, but the individual node voltages may be unknown.

Current through the source is unknown

We cannot calculate that current directly from the voltage and resistance of an ideal source.

Combine the nodes and apply KCL

Treat the source and its two terminals as one supernode, and sum the currents crossing the outer boundary.

4. Example From the Given Circuit

In the original circuit, the 6 V source connects the left midpoint V_L and right midpoint V_R. These nodes form a supernode.

Equation 1: KCL at the supernode

(V_L − V_T)/1kΩ + V_L/1kΩ

(V_R − V_T)/1kΩ + V_R/1kΩ = 0

V_T = V_L + V_R

Equation 2: Voltage-source constraint

V_L − V_R = 6 V

These equations are combined with KCL at the top node to find the unknown node voltages.

5. Steps to Solve a Supernode Problem

Identify an ideal voltage source connecting two non-reference nodes.
Combine the two nodes and the source into a supernode.
Apply KCL to the supernode's outer boundary.
Write a voltage constraint from the source polarity and value.
Solve the resulting equations for the unknown node voltages.

Quick Summary

Definition: A combined region formed by two nodes connected by an ideal voltage source.
Main law: Apply KCL around the outer boundary.
Extra equation: Use the voltage source's polarity and voltage value.
Purpose: Simplifies nodal analysis when a voltage source lies between unknown nodes.

Remember

KCL for the supernode plus the voltage-source constraint gives the equations needed to solve the circuit.

Consider the circuit shown in the Figure, where the input Vi(t) is in Volt. The average power (in mW) dissipated in the load resistance of 1 kΩ ...

2026-10-26T19:11:28.383+05:30

Given Values

Input voltage: v(t) = 12 sin(ωt) V
Load resistance: R = 1 kΩ = 1000 Ω
Capacitance: C = 1 μF
Inductance: L = 2.2 mH

Step 1: Calculate RMS Voltage

The peak voltage is V_m = 12 V. Hence,

V_rms = V_m / √2

V_rms = 12 / √2 V

Step 2: Calculate Average Power

The average power dissipated in the resistor is

P = V_rms² / R

P = (12 / √2)² / 1000

P = 144 / 2000 = 0.072 W

Step 3: Convert to Milliwatts

P = 0.072 × 1000 mW

P = 72 mW

Final Answer

72.00 mW

Average power dissipated in the 1 kΩ load resistance at resonance.

The output voltage V0 (in Volt) for the network given in the Figure is ___

2026-10-26T18:08:33.684+05:30

Find the Output Voltage V_o

Solve the given circuit using Kirchhoff's Current Law (KCL), Ohm's law, and the supernode method.

1. Given Circuit Parameters

Current source

1 mA

Resistors

1 kΩ

Each resistor

Voltage source

6 V

Required

V_o

2. Understand the Circuit

The circuit contains a 1 mA current source, four 1 kΩ resistors, and a 6 V source connected between the two midpoint nodes. The output voltage is measured between the top and bottom rails.

Reference node

Take the bottom rail as ground: 0 V. Let the top rail voltage be V_T, and the left and right midpoint voltages be V_L and V_R.

Voltage source polarity

The longer battery plate is on the left. Therefore, V_L − V_R = 6 V.

Apply KCL at the Top Node

The 1 mA current source injects current into the top node. The current flows through the two upper resistors.

(V_T − V_L)/1kΩ

(V_T − V_R)/1kΩ = 1 mA

Multiplying through by 1 kΩ gives:

2V_T − V_L − V_R = 1

Apply KCL Using the Supernode

Since the 6 V source connects the two midpoint nodes, treat both nodes as a single supernode.

Supernode KCL equation:

(V_L − V_T)/1kΩ

V_L/1kΩ

(V_R − V_T)/1kΩ

V_R/1kΩ = 0

Multiplying by 1 kΩ and simplifying:

2V_L + 2V_R − 2V_T = 0

V_T = V_L + V_R

Calculate the Top Rail Voltage

From Step 1:

2V_T − V_L − V_R = 1

Using V_T = V_L + V_R:

2V_T − V_T = 1

Therefore

V_T = 1 V

Find the Midpoint Voltages

The voltage source gives the second equation:

V_L − V_R = 6

From Step 2:

V_L + V_R = 1

Adding the two equations:

2V_L = 7

V_L = 3.5 V

Substituting into V_L + V_R = 1:

V_R = −2.5 V

✓

Final Answer

The output voltage is measured from the top rail to the bottom rail. Since the bottom rail is ground:

V_o = V_T − 0

V_o = 1.00 V

Rounded to two decimal places

SRAM, DRAM, EPROM, MRAM, and DDRAM Explained

2026-10-25T20:25:04.840+05:30

Computer Fundamentals

Types of Memory and Their Technologies

Understand the fundamental relationship between different types of computer memory and the technologies used to implement them.

Fundamental Matching

Memory type, technology, and key characteristics

Memory Type	Technology	Key Characteristic
Cache Memory	SRAM	Very fast; stores frequently accessed data
Main Memory	DRAM	Holds programs and data currently in use
Auxiliary Memory	HDD / SSD	Non-volatile, long-term data storage
Portable Storage	Flash Memory	Used in USB drives and memory cards

Important Memory Technologies

SRAM

Static Random Access Memory. Fast and does not require periodic refreshing while powered. Commonly used for CPU cache.

DRAM

Dynamic Random Access Memory. Requires periodic refreshing and is commonly used as main system memory.

EPROM

Erasable Programmable Read-Only Memory. Non-volatile memory that can be erased using ultraviolet light and reprogrammed.

MRAM

Magnetoresistive Random Access Memory. Non-volatile memory that stores data using magnetic states.

DDRAM means Display Data RAM.

In a character LCD, DDRAM is the LCD controller's small memory area that stores which characters should appear on the screen.

Think of it like this:

DDRAM

Address     Stored character
00H    →       H
01H    →       E
02H    →       L
03H    →       L
04H    →       O

The LCD reads those stored characters and displays:

HELLO

Why do we need DDRAM?

Suppose you tell the LCD:

Put A at address 05H.

The LCD stores:

05H → A

Then the LCD knows that the character at that display position should be A.

So remember:

$\boxed{\text{DDRAM = memory used to store display characters}}$

And:

$\boxed{\text{Address = where the character is stored}}$

$\boxed{\text{Data = which character is stored}}$

For a typical 40×2 LCD, the visible character positions are mapped to DDRAM addresses such as 00H–27H for the first row and 40H–67H for the second row.

DDRAM is not permanent memory.

It is temporary memory (RAM) inside the LCD controller.

Think of it like this:

DDRAM
   ↓
Power ON
   ↓
LCD stores "HELLO"
   ↓
Power OFF
   ↓
Data is lost

So:

DDRAM → temporary → stores characters currently being displayed.
ROM → permanent → stores fixed things, such as the LCD's built-in character patterns.

Example

When you send:

'A'

to the LCD, the LCD puts A into DDRAM.

When power is removed, that stored A is lost.

So:

$\boxed{\text{DDRAM = temporary character storage}}$

$\boxed{\text{ROM = permanent/fixed storage}}$

Also, DDRAM is different from the RAM in your computer—it's a small memory area specifically used by the LCD controller.

Parametric Down-Conversion of Photons

2026-10-25T20:08:56.399+05:30

Nonlinear Optics

One incoming photon is converted into two lower-energy photons, typically in a nonlinear optical medium. Energy is conserved:

E_{in} = E_{1} + E_{2}

Parametric Down-Conversion

A nonlinear optical process in which one pump photon is converted into two lower-energy photons, known as the signal and idler photons.

Energy Conservation

The energy of the pump photon equals the sum of the energies of the signal and idler photons.

\[ \boxed{\hbar\omega_p=\hbar\omega_s+\hbar\omega_i} \] \[ \boxed{\omega_p=\omega_s+\omega_i} \]

ωₚ

Pump angular frequency

ωₛ

Signal angular frequency

ωᵢ

Idler angular frequency

ℏ

Reduced Planck constant

Wavelength Relationship

Using $E=hc/\lambda$, energy conservation can also be expressed in terms of vacuum wavelengths.

\[ \boxed{ \frac{1}{\lambda_p} = \frac{1}{\lambda_s} + \frac{1}{\lambda_i} } \]

Momentum Conservation

Efficient down-conversion requires phase matching. In a common wave-vector description, the pump wave vector equals the vector sum of the signal and idler wave vectors.

\[ \boxed{\mathbf{k}_p=\mathbf{k}_s+\mathbf{k}_i} \]

Here, kₚ, kₛ and kᵢ are the wave vectors of the pump, signal and idler fields, respectively. The phase-matching condition depends on the optical properties of the nonlinear medium.

Degenerate Down-Conversion

In degenerate parametric down-conversion, both emitted photons have the same frequency and, consequently, the same vacuum wavelength.

\[ \omega_s=\omega_i=\frac{\omega_p}{2} \] \[ \boxed{\lambda_s=\lambda_i=2\lambda_p} \]

Example

Suppose the pump photon has a vacuum wavelength of 400 nm. In the degenerate case, each emitted photon has a wavelength of 800 nm.

Pump photon

400 nm

→

Signal

800 nm

Idler

800 nm

Summary

In parametric down-conversion, the pump photon is converted into two photons through a nonlinear optical interaction. It is not ordinary atomic absorption followed by emission. Energy is conserved, and efficient conversion requires appropriate phase-matching conditions.

Active vs Passive Sensors

2026-10-25T19:40:32.651+05:30

Passive sensor: Generates its own electrical signal from the measured physical quantity.

Example: A piezoelectric sensor is a passive sensor.

Reason: A piezoelectric sensor generates an electrical charge or voltage directly when mechanical pressure, force, or vibration is applied to it. It does not require an external power supply to generate this signal.

A piezoelectric sensor used to detect vibration in a machine.

Therefore, a piezoelectric sensor is classified as a passive transducer because it converts mechanical energy into electrical energy directly.

Active sensor: Requires external excitation or power to operate, such as an RTD (which requires a measuring current).

A binary ripple counter is designed to count (0)₁₀ to (64)₁₀. Which of the following is/are the number of flip-flops required ...

2026-10-21T18:44:09.170+05:30

Q. A binary ripple counter is designed to count (0)₁₀ to (64)₁₀. Which of the following is/are the number of flip-flops required to design the counter?

(A) 6 (B) 7 (C) 4 (D) 5

Solution

A binary ripple counter must count from (0)₁₀ to (64)₁₀, including both endpoints.

Step 1: Calculate the total number of states

Since both 0 and 64 are included, add 1 to the difference:

N = Last value − First value + 1
N = 64 − 0 + 1 = 65 states

The counter visits every integer from 0 through 64:

0, 1, 2, 3, ..., 63, 64

There are 65 distinct states in total.

Step 2: Calculate the number of flip-flops

Each flip-flop stores one bit. With n flip-flops, the counter can represent 2ⁿ distinct states.

2⁶ = 64 states (not enough)

2⁷ = 128 states (enough)

n = 7 flip-flops

Final Answer: Option B - 7 Flip-Flops

An S-band radar transmitting at 3 GHz radiates 200 kW of power. Determine the signal power density at a range of 100 ...

2026-10-21T18:20:00.276+05:30

Q. An S-band radar transmitting at 3 GHz radiates 200 kW of power. Determine the signal power density at a range of 100 nautical miles if the effective area of the radar antenna is 9 m².

Solution

Given,

Transmitted power: P_t = 200 kW = 200,000 W
Frequency: f = 3 GHz = 3 × 10⁹ Hz
Range: R = 100 nautical miles
Effective antenna area: A_e = 9 m²
Speed of light: c = 3 × 10⁸ m/s

Step 1: Convert Nautical Miles to Meters

One nautical mile equals 1852 meters.

R = 100 × 1852 = 185200 m

Step 2: Calculate the Wavelength

The wavelength is calculated using the relationship between the speed of light and frequency:

λ = c / f

λ = (3 × 10⁸) / (3 × 10⁹)

λ = 0.1 m

Step 3: Calculate the Antenna Gain

The relationship between effective aperture and antenna gain is:

G = (4πAₑ) / λ²

G = (4π × 9) / (0.1)²

G ≈ 11309.73

Antenna gain is dimensionless when expressed as a linear ratio.

Step 4: Calculate the Signal Power Density

In the far field, the power density at a distance R from a transmitting antenna is:

S = (PₜG) / (4πR²)

Substituting the antenna gain formula gives a simpler expression:

S = (PₜAₑ) / (λ²R²)

S = (200000 × 9) / [(0.1)² × (185200)²]

S ≈ 0.005248 W/m²

Step 5: Convert to Milliwatts per Square Meter

Since 1 W = 1000 mW:

S = 0.005248 × 1000

S = 5.248 mW/m²

Final Answer: Option 4

5.248 mW/m²

The electric field of a monochromatic plane wave travelling in a lossless isotropic and homogenous medium is given by E(z, t) = ...

2026-10-20T17:38:33.917+05:30

Q. The electric field of a monochromatic plane wave travelling in a lossless isotropic and homogenous medium is given by E(z, t) = E₀ [x̂ cos(ωt − kz) + ŷ sin(ωt − kz)] in a right-handed orthogonal co-ordinate system. Which of the following is the correct polarization of the electromagnetic wave?

(A) Right-handed circularly polarized

Correct answer under the IEEE convention.

(B) Left-handed circularly polarized

(C) Linearly polarized

(D) Linearly polarized at −45° to x̂

Solution

1. Identify the electric-field components

Let ψ = ωt − kz. The components along the x and y directions are:

Eₓ = E₀ cos ψ

Eᵧ = E₀ sin ψ

Both components have equal amplitudes, the same frequency, and a phase difference of 90°. These are the conditions for circular polarization.

2. Verify the magnitude of the electric field

|E⃗| = √(Eₓ² + Eᵧ²)

Substituting the components:

|E⃗| = E₀ √(cos² ψ + sin² ψ)

|E⃗| = E₀

Since cos² ψ + sin² ψ = 1, the magnitude remains constant. The electric-field vector rotates in the xy-plane, confirming circular polarization.

3. Determine the handedness

The phase is ψ = ωt − kz, so the wave propagates in the +z direction. At a fixed position, z = 0:

At t = 0: E⃗ = E₀ x̂

For a small positive time: Eₓ decreases while Eᵧ becomes positive.

Therefore, the field rotates from +x̂ toward +ŷ as time increases.

Under the IEEE convention, this rotation for propagation in the +z direction corresponds to right-handed circular polarization.

Final answer

Option (A) — Right-handed circularly polarized

Summary

Equal amplitudes and a 90° phase difference indicate circular polarization.
Unequal amplitudes with a 90° phase difference generally indicate elliptical polarization.
Components in phase or 180° out of phase indicate linear polarization.
The handedness depends on the propagation direction, phase convention, and the polarization convention being used.

GATE EC Previous Year Papers with Solutions →

Electronics and Communication Study Material →

The relation between the input current (I) and the output voltage (V) of a circuit is governed by the equation: 𝐶𝑑𝑉 𝑑𝑡 =𝐼(𝑡)−𝑚(𝑡)...

2026-10-18T13:34:13.982+05:30

Q. The relation between the input current (I) and the output voltage (V) of a circuit is governed by the equation: 𝐶𝑑𝑉 𝑑𝑡 =𝐼(𝑡)−𝑚(𝑡). The circuit is excited by I(t) = q(t), where q is a real valued constant. V at 𝑡=0− is V0. Which of the following is an equivalent representation of the above case?

Solution

The relationship between input current I(t) and output voltage V(t) is governed by the equation:

C dV/dt = I(t) − m(t)

The circuit is excited by an impulse current:

I(t) = qδ(t)

Given: V(0⁻) = V₀

Step 1: Substitute the input current

Since I(t) = qδ(t), substitute it into the circuit equation:

C dV/dt = qδ(t) − m(t)

Step 2: Integrate across the impulse

Integrate both sides over the interval from 0⁻ to 0⁺:

C ∫ dV/dt dt = q ∫ δ(t) dt − ∫ m(t) dt

Assuming m(t) is finite and has no impulse at t = 0, its integral over this infinitesimal interval vanishes. Also, the integral of the Dirac delta impulse is 1.

C [V(0⁺) − V(0⁻)] = q

Step 3: Calculate the voltage jump

Using the initial condition V(0⁻) = V₀:

C [V(0⁺) − V₀] = q

V(0⁺) − V₀ = q/C

V(0⁺) = V₀ + q/C

Step 4: Find the equation after the impulse

For t > 0, the impulse term δ(t) is zero. Therefore:

C dV/dt = −m(t)

Correct option

Option (A)

C dV/dt = −m(t)

V(0⁺) = V₀ + q/C

The impulse current produces an instantaneous voltage jump of q/C across the capacitor.

The capacitor voltage jumps by $q / C$ , while the ordinary signal $m (t)$ does not contribute a finite amount of charge over that infinitesimal interval.

GATE EC Previous Year Papers with Solutions →

Electronics and Communication Study Material →

Consider a discrete memoryless source with an alphabet of four source symbols. s(t) is a multi-level (-1, 0, +1, +2) ...

2026-10-18T13:05:52.138+05:30

Q. Consider a discrete memoryless source with an alphabet of four source symbols. The signal s(t) has four levels (-1, 0, +1, +2) and generates 10⁴ symbols per second. Find the equivalent Nyquist bandwidth.

(A) 10 kHz (B) 64 kHz (C) 5 kHz (D) 20 kHz

Given

Number of signal levels: M = 4
Symbol rate: R_s = 10⁴ symbols/s
Signal levels: -1, 0, +1, +2

Calculate Nyquist bandwidth

For ideal Nyquist signaling, the minimum bandwidth can be expressed in terms of the symbol rate:

B = R_s / 2

B = 10,000 / 2 = 5,000 Hz

Correct answer

Option (C) — 5 kHz

Using the symbol-rate definition of Nyquist bandwidth, B = R_s/2 gives 5 kHz.

Don't confuse symbol rate with bit rate

The bit rate is Rb = Rs*log2(M) = 20 kbps, while the symbol rate is 10 kbaud. These are different quantities because each symbol represents 2 bits.

Symbol-rate form: B = R_s/2 = 5 kHz

Bit-rate form: B = R_b/2 = 10 kHz

The formulas use different rate conventions. For this question, use the symbol-rate convention if that is what your course intends by equivalent Nyquist bandwidth.

GATE EC Previous Year Papers with Solutions →

Electronics and Communication Study Material →

ADC Conversion Time Comparison: Flash, SAR, Ramp & Dual-Slope ADC

2026-10-17T20:02:59.208+05:30

ADC Conversion Time Comparison: Flash, SAR, Ramp & Dual-Slope ADC

ADC Conversion Time: Flash, Digital Ramp, SAR and Dual-Slope ADC

Analog-to-digital converters (ADCs) convert continuous analog signals into digital values. Different ADC architectures require different amounts of time to complete this conversion. This guide explains the working principles, conversion-time formulas, and differences between Flash, Digital Ramp, Successive Approximation, and Dual-Slope ADCs.

What Is ADC Conversion Time?

ADC conversion time is the time required by an analog-to-digital converter to produce a digital output corresponding to its analog input. It depends on the converter architecture, resolution, clock frequency, circuit delays, and conversion method.

An important distinction is whether conversion time depends on the magnitude of the input voltage. Flash and conventional SAR ADCs use a fixed conversion process for a given configuration, whereas Digital Ramp and Dual-Slope ADCs have conversion times that vary with the input magnitude.

Flash ADC and Successive Approximation ADC have conversion times that are independent of input magnitude. Digital Ramp ADC and Dual-Slope ADC have input-dependent conversion times.

1. Flash ADC Conversion Time

A Flash ADC, also called a parallel ADC, is one of the fastest types of analog-to-digital converters. It uses multiple comparators to compare the input voltage against different reference levels simultaneously.

Flash ADC block diagram

Figure 1: Simplified Flash ADC block diagram using parallel comparators.

How does a Flash ADC work?

The analog input is applied to the comparator bank.
Each comparator checks the input against a reference level.
All comparators operate in parallel.
An encoder converts the comparator pattern into a digital code.

An n-bit Flash ADC typically requires 2ⁿ − 1 comparators. For example, a 4-bit Flash ADC requires 15 comparators.

Flash ADC conversion-time formula

T_conv ≈ T_comparator + T_encoder

The conversion time is approximately constant for a given configuration because the comparisons take place simultaneously. The input magnitude does not determine how many comparison steps are required.

Key point: Flash ADC is extremely fast, but it requires many comparators as resolution increases.

2. Digital Ramp ADC Conversion Time

A Digital Ramp ADC, also called a counter-type ADC, uses a counter, a digital-to-analog converter (DAC), and a comparator. The counter increments until the DAC output reaches the input voltage.

Digital Ramp ADC block diagram

Figure 2: Digital Ramp ADC using a counter, DAC, and comparator.

How does a Digital Ramp ADC work?

The counter starts at zero.
The DAC converts the counter value into an analog voltage.
The comparator compares the DAC voltage with the input.
The counter stops when the comparison condition is satisfied.

Digital Ramp ADC conversion-time formula

T_conv ≈ N × T_clk

Here, N is the number of clock counts required and T_clk is the clock period.

A small input voltage generally requires fewer counts, whereas a larger input voltage requires more counts. Therefore, the conversion time depends on the input magnitude.

For an n-bit counter, the maximum count is normally 2ⁿ − 1. The exact conversion time depends on the counting convention, circuit delays, and whether the initial count is included.

Key point: Digital Ramp ADC is relatively simple, but its conversion time can be long and variable.

3. Successive Approximation ADC (SAR ADC) Conversion Time

A Successive Approximation ADC uses a successive approximation register (SAR), a DAC, and a comparator. It determines the digital output through a sequence of binary decisions, similar to a binary search.

SAR ADC block diagram

Figure 3: Simplified SAR ADC architecture with a comparator, DAC, and register.

How does a SAR ADC work?

The SAR initially tests the most significant bit (MSB).
The DAC generates the voltage corresponding to the trial code.
The comparator determines whether the trial voltage is too high or too low.
The SAR retains or clears the tested bit and moves to the next bit.
After the required bit decisions, the digital output is available.

SAR ADC conversion-time formula

T_conv ≈ n × T_bit

Here, n is the ADC resolution in bits and T_bit is the time required for one bit decision. Actual timing depends on the implementation and may include acquisition and settling time.

A 12-bit SAR ADC generally needs 12 bit decisions. The number of decisions remains the same whether the input is close to zero, midway through the range, or close to full scale.

Key point: SAR ADC provides a good balance of speed, resolution, and hardware complexity.

4. Dual-Slope Integrating ADC Conversion Time

A Dual-Slope ADC is an integrating converter commonly used in precision measurement instruments, including digital multimeters. It integrates the unknown input voltage for a fixed period and then integrates a known reference voltage of opposite polarity until the integrator returns to its starting level.

Dual-Slope ADC block diagram

Figure 4: Simplified Dual-Slope ADC showing the integrator, comparator, control logic, clock, and counter.

How does a Dual-Slope ADC work?

Run-up phase: The unknown input voltage is integrated for a fixed time T₁.
Run-down phase: A known reference voltage with opposite polarity is applied.
The counter measures how long the integrator takes to return to zero.
The measured run-down time is proportional to the magnitude of the input voltage, assuming a fixed reference and integration period.

Dual-Slope ADC conversion-time formula

T_conv = T₁ + T₂

Ideally, the run-down time is:

T₂ = (|V_in| / |V_ref|) × T₁

Therefore, the total conversion time varies with the input magnitude because the run-down time T₂ varies. The run-up interval T₁ is fixed.

Dual-Slope ADCs are relatively slow, but they offer excellent accuracy and rejection of certain types of noise, particularly when the integration period is chosen to match the noise interference frequency.

Key point: Dual-Slope ADC is ideal for precision measurements where accuracy is more important than speed.

ADC Conversion Time Comparison Table

The following table summarizes how conversion time behaves in the four commonly studied ADC architectures.

ADC Type	Operating Principle	Conversion-Time Behavior	Depends on Input Magnitude?	Typical Advantage
Flash ADC	Parallel voltage comparisons	Very fast and approximately constant	No	Highest conversion speed
Digital Ramp ADC	Counter and DAC staircase	Varies with the number of counts	Yes	Simple architecture
Successive Approximation ADC	Binary-search bit decisions	Approximately fixed for a given resolution	No	Good speed and resolution balance
Dual-Slope ADC	Fixed integration followed by de-integration	Fixed run-up plus variable run-down	Yes	High accuracy and noise rejection

Correct Answer for the ADC Conversion-Time MCQ

Question: Which analog-to-digital converters have conversion times independent of the magnitude of the analog input?

A. Flash ADC
B. Digital Ramp ADC
C. Successive Approximation ADC
D. Dual-Slope Integrating ADC

Correct option: A and C only (Option 4).

Flash ADC performs comparisons in parallel, while SAR ADC uses a fixed number of bit decisions for a given resolution. Digital Ramp ADC requires a variable number of counter steps, and Dual-Slope ADC has a variable run-down interval.

Frequently Asked Questions About ADC Conversion Time

1. Which ADC has the fastest conversion time?

Flash ADC is generally the fastest conventional ADC architecture because it performs multiple comparisons simultaneously. Its hardware complexity and power consumption increase significantly with resolution.

2. Which ADC has conversion time independent of input magnitude?

Flash ADC and conventional Successive Approximation (SAR) ADC have conversion times independent of input magnitude for a fixed configuration and resolution.

3. Why does Digital Ramp ADC conversion time vary?

The counter must generate enough counts for the DAC output to reach the input voltage. A larger input generally requires more counts and therefore more conversion time.

4. Why is Dual-Slope ADC relatively slow?

It needs a fixed integration interval followed by a de-integration interval. The process takes time, but provides good measurement accuracy and noise rejection.

5. What is the conversion time of an n-bit SAR ADC?

A conventional n-bit SAR ADC requires approximately n bit-decision cycles. Its total conversion time also depends on the clock rate, DAC settling time, comparator delay, and implementation.

6. Is conversion time the same as sampling time?

No. Sampling or acquisition time is the time required to acquire and settle the analog input. Conversion time is the time required to determine the digital result. Total latency may include both.

Conclusion

ADC conversion time is an important parameter when selecting an analog-to-digital converter for a particular application. Flash ADCs are suitable for very high-speed conversion, SAR ADCs offer a practical balance between speed and resolution, Digital Ramp ADCs are simple but have input-dependent conversion time, and Dual-Slope ADCs are useful for accurate measurements.

For exam preparation, remember this distinction: Flash and SAR ADCs have conversion times independent of input magnitude, whereas Digital Ramp and Dual-Slope ADCs have input-dependent conversion times.

CMOS Logic Gates: Working, Types, Truth Tables and Applications

2026-10-15T19:00:43.602+05:30

CMOS logic gates are digital electronic circuits built using Complementary Metal-Oxide-Semiconductor (CMOS) technology. They use two types of MOSFET transistors, NMOS and PMOS, to perform Boolean logic operations such as NOT, NAND and NOR. CMOS technology is widely used in modern digital integrated circuits because it offers low static power consumption, high input impedance and good noise immunity.

1. How CMOS Logic Gates Work

CMOS Logic Gates: Block Diagram

CMOS logic gates use complementary PMOS and NMOS transistors to implement digital logic functions.

CMOS circuits use complementary NMOS and PMOS transistors to control the output voltage. In a typical CMOS logic gate, the PMOS network provides a path from the output to the positive supply voltage (VDD), while the NMOS network provides a path from the output to ground (GND).

PMOS transistor: Generally turns ON when its gate voltage is sufficiently low relative to its source.
NMOS transistor: Generally turns ON when its gate voltage is sufficiently high relative to its source.

The transistor networks are arranged so that the output represents the required Boolean function. In a standard static CMOS gate, one network pulls the output HIGH while the complementary network pulls it LOW.

When the circuit is in a stable logic state, ideally there is negligible direct current from VDD to GND. During switching, however, CMOS circuits consume dynamic power because of capacitance charging and discharging, along with short-circuit and leakage currents.

2. Common CMOS Logic Gates

2.1 CMOS NOT Gate (Inverter)

CMOS NOT Gate Circuit Diagram

A CMOS inverter consists of one PMOS transistor and one NMOS transistor. The PMOS transistor is connected to VDD, and the NMOS transistor is connected to GND. Their drains are connected together to form the output.

When the input is LOW (0), the PMOS transistor turns ON and the NMOS transistor turns OFF, producing a HIGH output (1). When the input is HIGH (1), the PMOS transistor turns OFF and the NMOS transistor turns ON, producing a LOW output (0).

Boolean expression: Y = NOT A = Ā

Input A	Output Y
0	1
1	0

2.2 CMOS NAND Gate

CMOS NAND Gate Circuit Diagram

A NAND gate produces a LOW output only when all its inputs are HIGH. For a two-input static CMOS NAND gate, the PMOS transistors are connected in parallel, while the NMOS transistors are connected in series.

If either input is LOW, at least one PMOS transistor provides a pull-up path, and the series NMOS path is interrupted. When both inputs are HIGH, the NMOS series path conducts and the output becomes LOW.

Boolean expression: Y = NOT (A AND B) = (A · B)̅

Input A	Input B	Output Y
0	0	1
0	1	1
1	0	1
1	1	0

Important: NAND is called a universal logic gate because any Boolean logic function can be implemented using NAND gates alone.

2.3 CMOS NOR Gate

CMOS NOR Gate Circuit Diagram

A NOR gate produces a HIGH output only when all its inputs are LOW. In a two-input static CMOS NOR gate, the PMOS transistors are connected in series, while the NMOS transistors are connected in parallel.

If either input is HIGH, the corresponding NMOS transistor provides a path to ground, making the output LOW. When both inputs are LOW, the PMOS series path conducts and the output becomes HIGH.

Boolean expression: Y = NOT (A OR B) = (A + B)̅

Input A	Input B	Output Y
0	0	1
0	1	0
1	0	0
1	1	0

NOR is also a universal logic gate, meaning that any Boolean logic function can be implemented using NOR gates alone.

2.4 CMOS AND and OR Gates

AND and OR functions can be implemented by combining CMOS gates. For example, an AND gate can be built by connecting a NAND gate to a NOT gate. Similarly, an OR gate can be built by connecting a NOR gate to a NOT gate.

AND: Y = A · B
OR: Y = A + B

3. CMOS Logic Gate Truth Table Summary

The following table summarizes the output of the common two-input CMOS logic gates.

A	B	AND	OR	NAND	NOR
0	0	0	0	1	1
0	1	0	1	1	0
1	0	0	1	1	0
1	1	1	1	0	0

4. Advantages of CMOS Technology

Low static power consumption: Ideally, a stable CMOS logic state draws negligible direct current between the supply rails.
High input impedance: MOSFET gate terminals draw very little steady-state input current.
Good noise immunity: Properly designed CMOS circuits can tolerate a useful range of input-voltage disturbances.
High integration density: CMOS technology enables large numbers of transistors to be integrated into a small chip.
Scalability: CMOS processes have supported the development of increasingly compact and complex digital circuits.

5. Applications of CMOS Logic Gates

CMOS logic gates form the building blocks of many digital electronic systems, including:

Microprocessors and microcontrollers
Memory circuits, including SRAM and CMOS-based logic in memory systems
Digital signal processors (DSPs)
Arithmetic logic units (ALUs)
Digital communication systems
Control circuits and embedded systems
Computers, smartphones and other digital devices

6. Why CMOS Is Important in Digital Electronics

CMOS technology combines complementary transistor networks to implement logic functions with low static power consumption and high integration density. These properties make CMOS a fundamental technology for modern digital integrated circuits. Understanding CMOS inverters, NAND gates and NOR gates helps students understand how complex digital systems are constructed from basic transistor-level circuits.

Frequently Asked Questions (FAQs)

What does CMOS stand for?

CMOS stands for Complementary Metal-Oxide-Semiconductor. It refers to a technology that uses complementary NMOS and PMOS transistors.

Why are both NMOS and PMOS transistors used in CMOS?

NMOS and PMOS transistors provide complementary switching behavior. Together, they form pull-down and pull-up networks that implement logic functions efficiently.

Why is NAND called a universal gate?

NAND is a universal gate because NOT, AND, OR and other Boolean functions can be constructed using NAND gates alone.

Do CMOS circuits consume power?

Yes. Although ideal CMOS logic draws negligible static current in stable states, practical circuits consume power due to switching, leakage and other non-ideal effects.

Conclusion

CMOS logic gates use NMOS and PMOS transistors to perform digital logic operations. The inverter, NAND gate and NOR gate are fundamental CMOS circuits, while AND and OR functions can be constructed from these basic gates. Their low static power consumption, noise immunity and high integration density make CMOS technology essential to modern digital electronics.

Consider the Friis’ transmission equation 𝑃𝑅=(𝑃𝑇𝐺𝑇𝐺𝑅𝜆^2) (4𝜋𝐷)^2 ⁄ , where P𝑅 and 𝑃𝑇 are the received and the transmitted powers ...

2026-10-13T07:34:01.640+05:30

Q. Consider the Friis’ transmission equation 𝑃𝑅=(𝑃𝑇𝐺𝑇𝐺𝑅𝜆2) (4𝜋𝐷)2 ⁄ , where P𝑅 and 𝑃𝑇 are the received and the transmitted powers, respectively. G𝑇 and 𝐺𝑅 are the gain of transmitting and receiving antennas, respectively, 𝐷 is the distance between the transmitting and receiving antennas, and 𝜆 is the wavelength in free space. Given: 𝐺𝑇=𝐺𝑅=1.0,𝜆=0.30 m and PT = +10 dBm. Choose the distance (D), in km, from the following options at which the received power, PR = −90 dBm?

Solution

Given,

$P_{T} = + 10 dBm$
$P_{R} = - 90 dBm$
$G_{T} = G_{R} = 1$ (linear gain)
$λ = 0.30 m$

$P_{R} (in dBm) - P_{T}$ (in dBm): Power difference in dB

dB = 10^(dB/10) (in ratio)

Pr = Pt.Gt.Gr.λ^2 / (4.π.D)^2
Or, Pr/Pt = Gt.Gr.λ^2 / (4.π.D)^2
Or, 10^(-90-10)/10 = Gt.Gr.λ^2 / (4.π.D)^2
Or, 10^-10 = 1.1.(0.3)^2 / (4.π.D)^2 
Or, D = 15/2π
Answer: Option B

Further Reading
Friis' Law

GATE EC Previous Year Papers with Solutions →

Electronics and Communication Study Material →

In the circuit shown in the Figure, A and B are logic inputs and Y is the logic output. Which of the following logic operations is realized by the circuit?

2026-10-13T07:02:28.549+05:30

Q. In the circuit shown in the Figure, A and B are logic inputs and Y is the logic output. Which of the following logic operations is realized by the circuit?

(A) NOR (B) OR (C) XOR (D) AND

Step 1: Identify the transistors

Your diagram has 4 transistors.

2 PMOS

The two transistors at the top, connected to $V_{DD}$ .

2 NMOS

The two transistors at the bottom, connected to ground.

Step 2: Remember the ON/OFF rule

This is the most important part!

Input at gate	PMOS	NMOS
0 (LOW)	ON	OFF
1 (HIGH)	OFF	ON

So, PMOS works opposite to NMOS.

Step 3: Understand the connections

The two PMOS transistors are in series. Current can flow from $V_{DD}$ to $Y$ only when both PMOS are ON.
The two NMOS transistors are in parallel. The output can connect to ground if at least one NMOS is ON.

Step 4: Check all input combinations

Case 1: A = 0, B = 0

Both PMOS are ON.
Both NMOS are OFF.
The output is connected to $V_{DD}$ , not ground.

Therefore, $Y=1$ .

Case 2: A = 0, B = 1

PMOS A is ON, but PMOS B is OFF.
NMOS A is OFF, but NMOS B is ON.
The series PMOS path is broken, and the NMOS path connects the output to ground.

Therefore, $Y=0$ .

Case 3: A = 1, B = 0

PMOS A is OFF, and PMOS B is ON.
NMOS A is ON, and NMOS B is OFF.
The series PMOS path is broken, and the output connects to ground.

Therefore, $Y=0$ .

Case 4: A = 1, B = 1

Both PMOS are OFF.
Both NMOS are ON.
The output connects to ground.

Therefore, $Y=0$ .

Step 5: Write the truth table

A	B	Y
0	0	1
0	1	0
1	0	0
1	1	0

The output is 1 only when both inputs are 0.

That is exactly the behavior of a NOR gate.

\boxed{Y=\overline{A+B}}

Answer: Option A

A control system is shown in the Figure. Which option represents the correct transfer function of the system?

2026-10-11T10:18:35.321+05:30

Q. A control system is shown in the Figure. Which option represents the correct transfer function of the system?

Output of the upper block, C(s) = X(s)*1/(s+4)

Output of the lower summing junction = 0. So, feedback signal is zero.

So, final transfer function is 1/(s+4)

Answer: Option B

Consider the circuit shown in the Figure with 𝑉𝑖=3 V and 𝑉𝐶𝐶=12 V. Assume 𝑉𝐵𝐸=0.7 V and 𝛽𝑑𝑐=99 for the BJT ...

2026-10-11T09:41:33.927+05:30

Q. Consider the circuit shown in the Figure with 𝑉𝑖=3 V and 𝑉𝐶𝐶=12 V. Assume 𝑉𝐵𝐸=0.7 V and 𝛽𝑑𝑐=99 for the BJT. Which of the following options is the correct value of the current 𝑰𝒐?

Here, feedback is negative

so, V+ = V-

Or, Vi = VE = 3V

Then, IE = 3 / 1 mA = 3 mA

On the other hand, Ic = β.IB and IE = Ic + IB
Or, IB = IE - Ic
Or, IB = 3 - β.IB
Or, IB(1+β) = 3
Or, IB = 3/100 mA = 30 micro-Ampere

Answer: Option D

Baseband vs Passband M-QAM Signals

2026-10-10T15:44:13.788+05:30

For M-QAM (M-ary Quadrature Amplitude Modulation), the baseband symbols are complex numbers with different amplitudes and phases, unlike M-PSK, where the amplitude is constant.

The general baseband symbol is

\boxed{s_k=I_k+jQ_k}

where $I_k$ is the in-phase component and $Q_k$ is the quadrature component.

1. 4-QAM

4-QAM has 4 symbols, with 2 possible values for each component:

I,Q\in\{-1,+1\}

The baseband symbols are

\boxed{S=\{-1-j,-1+j,1-j,1+j\}}

Symbol	$I$	$Q$
$-1-j$	-1	-1
$-1+j$	-1	+1
$1-j$	+1	-1
$1+j$	+1	+1

Each symbol represents $\log_2 4=2$ bits.

2. 16-QAM

16-QAM has 16 symbols. The possible values of each component are

I,Q\in\{-3,-1,+1,+3\}

Therefore,

\boxed{s_k=I+jQ}

for all 16 combinations of $I$ and $Q$ .

16-QAM constellation

Each point represents one symbol $I+jQ$ . There are 4 × 4 = 16 points.

Examples of symbols:

\begin{aligned} s_1&=-3-j3\\ s_2&=-3-j1\\ s_3&=-1+j3\\ s_4&=1+j1\\ s_5&=3+j3 \end{aligned}

There are 16 possible combinations in total. Each symbol represents $\log_2 16=4$ bits.

3. 32-QAM

32-QAM has 32 symbols. Unlike 16-QAM, it does not have one universally fixed constellation layout.

One common rectangular arrangement uses 4 levels on the $I$ -axis and 8 levels on the $Q$ -axis:

I\in\{-3,-1,+1,+3\}

Q\in\{-7,-5,-3,-1,+1,+3,+5,+7\}

The baseband symbols are

\boxed{s_k=I+jQ}

for all combinations of these values, giving $4\times8=32$ symbols.

Rectangular 32-QAM constellation

This is one possible 32-QAM layout: 4 × 8 = 32 constellation points.

Examples of symbols:

\begin{aligned} s_1&=-3-j7\\ s_2&=-1-j5\\ s_3&=1+j3\\ s_4&=3+j7 \end{aligned}

Each symbol represents $\log_2 32=5$ bits.

Final comparison

Modulation	Possible $I$ levels	Possible $Q$ levels	Bits/symbol
4-QAM	$\pm1$	$\pm1$	2
16-QAM	$\pm1,\pm3$	$\pm1,\pm3$	4
32-QAM (rectangular)	$\pm1,\pm3$	$\pm1,\pm3,\pm5,\pm7$	5

Passband M-QAM

The passband signal at the transmitter is given by

s(t)=I(t)\cos(2\pi f_ct)-Q(t)\sin(2\pi f_ct)

where $I(t)$ and $Q(t)$ denote the in-phase and quadrature components of the baseband signal, respectively, and $f_c$ represents the carrier frequency. The cosine and negative sine terms form two orthogonal carrier signals.

In a typical transmission system, digital data symbols are mapped into I/Q sample sequences and passed through pulse-shaping filters. The resulting signals modulate the corresponding carrier components, which are combined to produce the final passband signal for transmission.

Baseband vs Passband M-PSK Signals

2026-10-10T15:43:46.586+05:30

In M-PSK (M-ary Phase Shift Keying), the baseband symbols are complex numbers representing different phases of the carrier. The general formula is

\boxed{s_k= A e^{j\frac{2\pi k}{M}}}

where:

$M$ = number of symbols
$A$ = symbol amplitude
$k=0,1,\ldots,M-1$
$j=\sqrt{-1}$

Using Euler's formula:

s_k=A\left[\cos\left(\frac{2\pi k}{M}\right) +j\sin\left(\frac{2\pi k}{M}\right)\right]

1. 4-PSK (QPSK)

Here, $M=4$ , so there are 4 symbols with a phase difference of $360^\circ/4=90^\circ$ .

s_k=Ae^{j\frac{2\pi k}{4}}

$k$	Phase	Baseband symbol
0	$0^\circ$	$A+j0$
1	$90^\circ$	$jA$
2	$180^\circ$	$-A+j0$
3	$270^\circ$	$-jA$

For $A=1$ , the symbols are

\boxed{\{1,\ j,\ -1,\ -j\}}

2. 8-PSK

Here, $M=8$ , so the phase difference is $360^\circ/8=45^\circ$ .

s_k=Ae^{j\frac{2\pi k}{8}}

$k$	Phase	Baseband symbol ( $A=1$ )
0	$0^\circ$	$1$
1	$45^\circ$	$0.707+j0.707$
2	$90^\circ$	$j$
3	$135^\circ$	$-0.707+j0.707$
4	$180^\circ$	$-1$
5	$225^\circ$	$-0.707-j0.707$
6	$270^\circ$	$-j$
7	$315^\circ$	$0.707-j0.707$

Thus,

\boxed{ \begin{aligned} S=\{&1,\ 0.707+j0.707,\ j,\\ &-0.707+j0.707,\ -1,\\ &-0.707-j0.707,\ -j,\\ &0.707-j0.707\} \end{aligned} }

3. 16-PSK

Here, $M=16$ , so the phase difference is $360^\circ/16=22.5^\circ$ .

s_k=Ae^{j\frac{2\pi k}{16}}

There are 16 symbols, one at each phase angle:

$k$	Phase	Baseband symbol ( $A=1$ )
0	$0^\circ$	$1+j0$
1	$22.5^\circ$	$0.924+j0.383$
2	$45^\circ$	$0.707+j0.707$
3	$67.5^\circ$	$0.383+j0.924$
4	$90^\circ$	$j$
5	$112.5^\circ$	$-0.383+j0.924$
6	$135^\circ$	$-0.707+j0.707$
7	$157.5^\circ$	$-0.924+j0.383$
8	$180^\circ$	$-1+j0$
9	$202.5^\circ$	$-0.924-j0.383$
10	$225^\circ$	$-0.707-j0.707$
11	$247.5^\circ$	$-0.383-j0.924$
12	$270^\circ$	$-j$
13	$292.5^\circ$	$0.383-j0.924$
14	$315^\circ$	$0.707-j0.707$
15	$337.5^\circ$	$0.924-j0.383$

Comparison at a glance

Modulation	Number of symbols	Bits per symbol	Phase separation
4-PSK	4	2	$90^\circ$
8-PSK	8	3	$45^\circ$
16-PSK	16	4	$22.5^\circ$

Passband M-PSK Signal

The transmitted passband signal can be expressed as

s (t) = I (t) \cos (2 π f_{c} t) - Q (t) \sin (2 π f_{c} t)

where:

$I (t)$ represents the in-phase baseband signal.
$Q (t)$ represents the quadrature baseband signal.
$f_{c}$ denotes the carrier frequency.
$\cos (2 π f_{c} t)$ and $- \sin (2 π f_{c} t)$ are orthogonal carrier components.

In a practical transmitter, digital symbols are first mapped into in-phase and quadrature (I/Q) sample values. These samples are then processed by pulse-shaping filters before modulating the respective orthogonal carriers to generate the transmitted passband signal.

Baseband to Passband Conversion Simulator

2026-10-10T11:31:51.331+05:30

M-PSK Digital Modulation Simulator

Enter a binary bit stream, select the PSK order, and observe the mapped baseband I/Q signals and the modulated passband waveform.

Input Bit Stream

Modulation Order (M)

Symbol Rate (Baud)

Carrier Frequency (Hz)

Symbol Amplitude

Samples per Carrier Cycle

1. Input Bit Stream

2. Baseband I/Q Signals

3. M-PSK Passband Signal

M-ary QAM Modulation Simulator

Enter a bit stream, select the modulation order and baud rate, and observe the baseband I/Q signals and passband waveform.

Input Bit Stream

Modulation Order (M)

Symbol Rate (baud)

Carrier Frequency (Hz)

Amplitude (A)

Samples per Symbol

Modulation Parameters

1. Input Bit Stream

2. QAM Constellation

3. Baseband I/Q Signals

4. QAM Passband Signal

Symbol Mapping

Symbol	Input Bits	I Level	Q Level	Symbol Amplitude

In practical digital communication systems, we usually convert the $I$ and $Q$ symbols into two baseband signals, then modulate them using two orthogonal carriers. This applies to both M-PSK and M-QAM.

The key idea is that changing the phase of a carrier and using the I/Q modulation equation are two equivalent ways of representing PSK, when the I/Q values are chosen appropriately.

1. The practical I/Q modulation equation

The transmitted passband signal is

\boxed{s(t)=I(t)\cos(2\pi f_ct)-Q(t)\sin(2\pi f_ct)}

where:

$I(t)$ : in-phase baseband signal.
$Q(t)$ : quadrature baseband signal.
$f_c$ : carrier frequency.
$\cos(2\pi f_ct)$ and $-\sin(2\pi f_ct)$ : orthogonal carriers.

In a practical transmitter, the digital symbols are converted into I/Q sample values, passed through pulse-shaping filters, and then used to modulate the two carriers.

Input bits

Symbol mapper

Bits → I and Q values

I branch

Pulse shaping

× cos(2πfct)

Q branch

Pulse shaping

× −sin(2πfct)

Adder → DAC/RF chain → Antenna

2. For M-PSK, do we vary the carrier phase or vary I and Q?

Both descriptions are correct.

For M-PSK, the symbol amplitude remains constant, and the phase changes according to the input symbol.

The complex baseband symbol is

\boxed{a_k=I_k+jQ_k=Ae^{j\phi_k}}

Therefore,

I_k=A\cos\phi_k

Q_k=A\sin\phi_k

Substituting these values into the I/Q modulator gives

s_k(t)=A\cos\phi_k\cos(2\pi f_ct) -A\sin\phi_k\sin(2\pi f_ct)

Using the identity

\cos(x+y)=\cos x\cos y-\sin x\sin y

we obtain

\boxed{s_k(t)=A\cos(2\pi f_ct+\phi_k)}

So the I/Q modulation equation produces the same result as directly changing the carrier phase.

3. Example: 4-PSK (QPSK)

Assume unit amplitude, $A=1$ , and the four phases are $0^\circ,90^\circ,180^\circ,270^\circ$ .

Phase	$I_k=\cos\phi_k$	$Q_k=\sin\phi_k$	Passband signal
$0^\circ$	1	0	$\cos(2\pi f_ct)$
$90^\circ$	0	1	$-\sin(2\pi f_ct)$
$180^\circ$	-1	0	$-\cos(2\pi f_ct)$
$270^\circ$	0	-1	$+\sin(2\pi f_ct)$

For example, if the input bits map to the symbol $I=0,Q=1$ , then

s(t)=0\cos(2\pi f_ct)-1\sin(2\pi f_ct)

Hence,

\boxed{s(t)=-\sin(2\pi f_ct)}

This is simply a carrier with a $90^\circ$ phase shift under this convention.

Visualizing QPSK symbols

Illustration of four possible I/Q combinations. For conventional QPSK with phases at 0°, 90°, 180°, and 270°, the symbols lie on the axes instead; the diagonal arrangement is a common phase-offset QPSK constellation.

4. What changes for 8-PSK, 16-PSK and M-QAM?

The same I/Q hardware can be used for all of them. Only the mapper's output values change.

Modulation	I/Q symbol values	What changes?
4-PSK	$I=A\cos\phi_k,\ Q=A\sin\phi_k$	4 possible phases
8-PSK	$I=A\cos\phi_k,\ Q=A\sin\phi_k$	8 possible phases
16-PSK	$I=A\cos\phi_k,\ Q=A\sin\phi_k$	16 possible phases
16-QAM	$I,Q$ selected from multiple amplitude levels	Both phase and amplitude can vary
32-QAM	$I,Q$ selected from the specified 32-point constellation	Both phase and amplitude can vary

For M-PSK, the phase is typically

\phi_k=\frac{2\pi k}{M}

although a system may use a reference phase offset.

For M-QAM, the mapper directly selects the appropriate $I_k,Q_k$ constellation coordinates.

5. How it works in a real transmitter

The input bitstream is grouped into symbols, such as 3 bits per symbol for 8-PSK.
A mapper converts each group into a complex symbol $I_k+jQ_k$ .
Pulse-shaping filters generate continuous-time or oversampled signals $I(t)$ and $Q(t)$ .
The I/Q modulator combines the two branches using cosine and sine carriers.
The resulting passband signal is converted and amplified for transmission.

The main takeaway: In a mathematical description, M-PSK changes the carrier phase. In a typical practical digital transmitter, that same phase change is implemented through the I/Q values and the equation

\boxed{s(t)=I(t)\cos(2\pi f_ct)-Q(t)\sin(2\pi f_ct)}

OFDM Baseband to Passband Conversion: A Complete Guide

2026-10-09T18:52:17.260+05:30

The main idea is that an OFDM transmitter first creates many modulated subcarriers digitally at baseband, combines them into one complex signal using an IFFT, and then converts that complex baseband signal into a real RF waveform using I/Q upconversion.

1. The overall OFDM transmitter

Input bits

0s and 1s

QAM / QPSK mapping

Bits → complex symbols $X[k]$

IFFT

Frequency-domain symbols → time-domain samples $x[n]$

Cyclic prefix + DAC

Prepare the discrete signal for transmission

I branch

$I(t)\cos(2\pi f_ct)$

Q branch

$-Q(t)\sin(2\pi f_ct)$

RF output

s(t)=I(t)\cos(2\pi f_ct)-Q(t)\sin(2\pi f_ct)

The transmitter has two conceptually different stages:

OFDM generation: construct a complex baseband signal containing many orthogonal subcarriers.
I/Q upconversion: shift that complex baseband signal to the radio carrier frequency $f_c$ .

Let's examine the equations behind each stage.

2. Step 1: Modulate the data onto subcarriers

Suppose an OFDM symbol uses $N$ subcarriers. Each subcarrier carries a complex modulation symbol:

X[0],X[1],\ldots,X[N-1]

For example, with QPSK, each $X[k]$ might take one of the values

X[k]\in \left\{ \frac{1+j}{\sqrt2}, \frac{1-j}{\sqrt2}, \frac{-1+j}{\sqrt2}, \frac{-1-j}{\sqrt2} \right\}.

Each symbol has two components:

X[k]=A_k+jB_k

where $A_k$ is the in-phase component and $B_k$ is the quadrature component.

Importantly, these complex symbols are not yet the RF waveform. They represent the data assigned to the OFDM subcarriers.

For example, if $N=4$ , you might have four symbols $X[0]$ , $X[1]$ , $X[2]$ , and $X[3]$ , each carrying its own data.

3. Step 2: The IFFT combines all subcarriers

This is the heart of OFDM.

The $N$ frequency-domain symbols are converted into $N$ time-domain samples using an inverse discrete Fourier transform:

\boxed{ x[n]=\frac{1}{\sqrt N} \sum_{k=0}^{N-1}X[k]e^{j2\pi kn/N} }

for $n=0,1,\ldots,N-1$ .

Here:

$X[k]$ : data symbol assigned to subcarrier $k$ .
$x[n]$ : complex baseband sample at time index $n$ .
$N$ : number of IFFT points.
$e^{j2\pi kn/N}$ : the complex sinusoid corresponding to subcarrier $k$ .

Using Euler's identity,

e^{j\theta}=\cos\theta+j\sin\theta,

we can expand the IFFT:

x[n]=\frac{1}{\sqrt N}\sum_{k=0}^{N-1} X[k]\left[ \cos\left(\frac{2\pi kn}{N}\right) +j\sin\left(\frac{2\pi kn}{N}\right) \right].

Since $X[k]=A_k+jB_k$ , each subcarrier contributes to both the real and imaginary components of $x[n]$ .

We can write the final result as

\boxed{x[n]=I[n]+jQ[n]}

where $I[n]$ and $Q[n]$ are the real and imaginary parts of the IFFT output.

Important: The IFFT has already combined all the subcarriers into a single complex time-domain signal. We do not need a separate RF oscillator for every subcarrier.

The continuous-time interpretation

Let the useful OFDM symbol duration be $T_u$ . The subcarrier spacing is

\boxed{\Delta f=\frac{1}{T_u}}

Ignoring the cyclic prefix, the continuous-time complex baseband waveform can be written as

\boxed{ x(t)=\frac{1}{\sqrt N} \sum_{k=0}^{N-1}X[k]e^{j2\pi k\Delta f t} }

for $0\le t<T_u$ .

Each term represents a different subcarrier. Because their frequencies are integer multiples of $\Delta f$ , the subcarriers are orthogonal over the useful symbol interval:

\int_0^{T_u} e^{j2\pi(k-m)\Delta f t}\,dt = \begin{cases} T_u,&k=m\\ 0,&k\ne m. \end{cases}

This orthogonality allows the receiver to separate the subcarriers using an FFT, even though their spectra overlap.

4. Step 3: Separate the complex baseband into I and Q

Suppose the IFFT produces

x(t)=I(t)+jQ(t).

For example, at a particular instant, suppose

x(t)=0.6+j0.8.

Then

I(t)=0.6,\qquad Q(t)=0.8.

These are not two separate OFDM transmissions. They are the two real-valued components of the same complex waveform.

The transmitter sends both components through separate signal paths:

The I path carries $I(t)$ .
The Q path carries $Q(t)$ .

The key point is that $I(t)$ and $Q(t)$ are generated by the same IFFT. They are generally independent components of the complex data waveform, not a signal and its Hilbert transform.

In a practical transmitter, the digital samples are converted to analog waveforms using two DAC paths, followed by filtering and RF upconversion.

5. Step 4: Upconvert the complex baseband to the RF carrier

Now we reach the equation you originally asked about.

Let the RF carrier frequency be $f_c$ . The transmitter uses two quadrature carriers:

c_I(t)=\cos(2\pi f_ct)

c_Q(t)=-\sin(2\pi f_ct).

The I branch is multiplied by the cosine:

s_I(t)=I(t)\cos(2\pi f_ct).

The Q branch is multiplied by the negative sine:

s_Q(t)=-Q(t)\sin(2\pi f_ct).

Adding both branches gives the transmitted real signal:

\boxed{ s(t)=I(t)\cos(2\pi f_ct) -Q(t)\sin(2\pi f_ct) }

Now substitute $I(t)+jQ(t)=x(t)$ :

\begin{aligned} s(t) &=\Re\left\{ [I(t)+jQ(t)]e^{j2\pi f_ct} \right\}\\ &=\Re\left\{ x(t)e^{j2\pi f_ct} \right\}. \end{aligned}

Therefore,

\boxed{s(t)=\Re\{x(t)e^{j2\pi f_ct}\}}

This is the standard complex-baseband-to-real-passband conversion equation.

Notice that the transmitter does not need to calculate the Hilbert transform of $I(t)$ to produce $Q(t)$ . Both components already exist from the complex IFFT output.

6. What does the actual transmitted OFDM waveform look like?

Let's substitute the IFFT expression directly into the upconversion equation.

We have

x(t)=\frac{1}{\sqrt N} \sum_{k=0}^{N-1}X[k]e^{j2\pi k\Delta f t}.

The transmitted waveform is

s(t)=\Re\{x(t)e^{j2\pi f_ct}\}.

Substituting,

s(t)=\Re\left\{ \frac{1}{\sqrt N} \sum_{k=0}^{N-1}X[k] e^{j2\pi(f_c+k\Delta f)t} \right\}.

Hence,

\boxed{ s(t)=\frac{1}{\sqrt N} \sum_{k=0}^{N-1} \Re\left\{ X[k]e^{j2\pi(f_c+k\Delta f)t} \right\} }

This equation gives us a very useful interpretation:

Each complex OFDM symbol is translated to its corresponding RF subcarrier frequency, and all the resulting real waveforms are added together.

For example, suppose the carrier is $f_c=2.4\text{ GHz}$ , and the subcarrier spacing is $\Delta f=15\text{ kHz}$ . The subcarrier frequencies are

f_k=f_c+k(15\text{ kHz}).

For subcarrier indices $k=0,1,2,3$ , the frequencies are 2.400000 GHz, 2.400015 GHz, 2.400030 GHz, and 2.400045 GHz.

In a practical OFDM system, subcarrier indices may include negative frequencies relative to the baseband center, and some bins may be reserved for guard bands or a DC null. The same principle still applies.

7. Where does the cyclic prefix fit in?

Before upconversion, OFDM transmitters usually add a cyclic prefix (CP) to each time-domain OFDM symbol.

If the useful IFFT output is

x[0],x[1],\ldots,x[N-1],

a cyclic prefix of length $N_{\rm CP}$ copies the last $N_{\rm CP}$ samples to the beginning:

x[N-N_{\rm CP}],\ldots,x[N-1], x[0],\ldots,x[N-1].

The cyclic prefix helps handle multipath propagation and enables simple frequency-domain equalization when the channel delay spread is within the CP duration.

After adding the CP, the transmitter performs the digital-to-analog conversion and filtering, then applies I/Q upconversion to generate the real RF waveform. Some implementations perform equivalent operations in a different order, but the underlying signal model is the same.

Summary

The OFDM transmitter performs two conceptually separate operations:

IFFT: combines independently modulated subcarriers into one complex baseband signal, $x(t)=I(t)+jQ(t)$ .
I/Q upconversion: translates that signal to RF using \[ s(t)=I(t)\cos(2\pi f_ct)-Q(t)\sin(2\pi f_ct). \]

What Is an Analytical Signal? Definition, Applications & Examples

2026-10-09T18:41:36.576+05:30

Analytical signal is sometime referred as real passband because here you generate Q(t) part from real baseband m(t). The analytic signal is useful when you want more control over the spectrum, especially when generating SSB or working with complex I/Q signals.

Multiplying a real baseband signal by a cosine generally creates both upper and lower sidebands around the carrier.

For example, if the baseband contains a frequency $f_m$, then

\[\cos(2\pi f_mt)\cos(2\pi f_ct)\]

using the product-to-sum identity gives

\[\frac12\cos(2\pi(f_c+f_m)t) +\frac12\cos(2\pi(f_c-f_m)t)\]

So you get frequencies at $f_c+f_m$ and $f_c-f_m$.

That's double-sideband modulation.

2. What does the analytic signal add?

Suppose you start with a real message $m(t)$. The Hilbert transform generates a quadrature version:

\[m_a(t)=m(t)+j\hat m(t)\]

Here, $\hat m(t)$ is the Hilbert transform of $m(t)$.

You can then modulate it using

\[s(t)=\Re\{m_a(t)e^{j2\pi f_ct}\}\]

Expanding,

\[s(t)=m(t)\cos(2\pi f_ct) -\hat m(t)\sin(2\pi f_ct)\]

The extra sine branch is not just decoration. It allows the two sidebands to cancel each other when the appropriate analytic-signal relationship is used.

Ordinary AM / DSB

Both sidebands are present. The required bandwidth is generally about twice the highest baseband frequency for a real low-pass message.

Hilbert-based SSB

One sideband can be suppressed, reducing the required bandwidth approximately by half.

Finding Object Velocity and Range Using FMCW Radar

2026-10-08T06:42:48.076+05:30

FMCW Radar: The Mathematical Foundation

How one radar waveform can reveal an object's range, velocity, and direction using frequency, phase, Fourier transforms, and antenna geometry.

An FMCW radar does not directly measure range, velocity, and angle. Instead, it measures changes in frequency and phase. Those measurements are transformed into physical quantities.

Dimension	What changes?	Information
Samples within a chirp	Beat frequency	Range
Chirp to chirp	Doppler phase	Velocity
Receiver to receiver	Spatial phase	Angle

1. Start With an FMCW Chirp

FMCW stands for Frequency-Modulated Continuous Wave. Instead of transmitting a single frequency, the radar continuously sweeps its frequency.

A simple up-chirp can be written as:

[ f(t) = f_c + St ]

where:

(f_c) = carrier frequency
(S) = chirp slope in Hz/s
(t) = time

The slope is:

[ S = \frac{B}{T_c} ]

where (B) is bandwidth and (T_c) is chirp duration.

Numerical Example

Suppose:

Carrier frequency = 77 GHz
Bandwidth = 1 GHz
Chirp duration = 50 μs

Then:

[ S = \frac{1\times10^9}{50\times10^{-6}} = 2\times10^{13}\text{ Hz/s} ]

So the radar's frequency increases by approximately 20 THz every second.

2. Range: Why Does the Beat Frequency Give Distance?

Electromagnetic waves travel approximately at the speed of light:

[ c \approx 3\times10^8\text{ m/s} ]

If an object is at distance (R), the signal must travel to the object and back.

The round-trip delay is therefore:

[ \tau = \frac{2R}{c} ]

Because the transmitted chirp is changing frequency with time, the delayed signal has a slightly different instantaneous frequency from the signal currently being transmitted.

Mixing the transmitted signal with the received signal produces the beat frequency:

[ f_b = S\tau ]

Substituting the delay:

[ f_b = S\frac{2R}{c} ]

Therefore:

[ \boxed{R=\frac{c f_b}{2S}} ]

Numerical Range Example

Assume:

(S=2\times10^{13}) Hz/s
Object range (R=30) m
(c=3\times10^8) m/s

The round-trip delay is:

[ \tau = \frac{2(30)}{3\times10^8} = 2\times10^{-7}\text{ s} = 200\text{ ns} ]

The beat frequency is:

[ f_b = (2\times10^{13})(2\times10^{-7}) = 4\times10^6 = \boxed{4\text{ MHz}} ]

Therefore, a target 30 m away produces approximately a 4 MHz beat frequency.

3. Why Do We Take an FFT for Range?

The ADC does not tell us directly:

'The target is 30 metres away.'

Instead, it gives us a sequence of voltage samples representing a signal that oscillates at approximately 4 MHz.

For example:

[ x[n] = A\cos(2\pi fb nTs) ]

The question becomes:

What frequencies are present in these samples?

The Fourier Transform answers exactly that question.

In practice, we use the Fast Fourier Transform (FFT), which is an efficient algorithm for calculating the Discrete Fourier Transform.

The DFT is:

[ X[k] = \sum_{n=0}^{N-1}x[n]e^{-j2\pi kn/N} ]

The FFT efficiently calculates these frequency components. A strong peak at frequency (f_b) means that the received signal contains a strong component at that beat frequency.

Important: The FFT is not magically calculating distance. It is finding frequency. We then use the relationship (R=cf_b/(2S)) to convert frequency into distance.

4. Range Resolution

An important property of FMCW radar is that range resolution is primarily determined by bandwidth.

[ \boxed{\Delta R = \frac{c}{2B}} ]

For (B=1) GHz:

[ \Delta R = \frac{3\times10^8}{2\times10^9} = \boxed{0.15\text{ m}} ]

So approximately 15 cm range resolution is possible under ideal assumptions.

More bandwidth → better range resolution.

5. Velocity: The Doppler Effect

Now imagine the target is moving. Motion changes the frequency of the returned wave. This is the Doppler effect.

The Doppler frequency for a monostatic radar is:

[ \boxed{f_D = \frac{2v}{\lambda}} ]

where:

(v) = radial velocity
(\lambda) = wavelength

The wavelength is:

[ \lambda = \frac{c}{f_c} ]

Numerical Velocity Example

For a 77 GHz radar:

[ \lambda = \frac{3\times10^8}{77\times10^9} \approx 3.896\text{ mm} ]

Suppose a car is approaching at (v=20) m/s.

[ f_D = \frac{2(20)}{0.003896} \approx \boxed{10.27\text{ kHz}} ]

6. Why Do We Take Another FFT for Velocity?

This is where FMCW radar becomes especially interesting.

We already used the samples within one chirp to determine range.

Now we look at the same range bin across many chirps.

Suppose the radar transmits chirps every (T_c) seconds.

The Doppler signal across chirps can be approximated as:

[ x[m] = Ae^{j2\pi fDmTc} ]

Notice what happened:

[ \text{Fast time} \rightarrow f_b \rightarrow \text{Range} ] [ \text{Slow time} \rightarrow f_D \rightarrow \text{Velocity} ]

Because the Doppler signal is also a frequency component, we again use an FFT.

First FFT: frequency across ADC samples → range.
Second FFT: frequency across chirps → velocity.

Numerical Slow-Time Example

Suppose chirps are separated by:

[ T_c=50\ \mu s ]

For (v=20) m/s we calculated:

[ f_D\approx10.27\text{ kHz} ]

The phase change between consecutive chirps is:

[ \Delta\phi = 2\pi fD Tc ]

Therefore:

[ \Delta\phi = 2\pi(10270)(50\times10^{-6}) \approx 3.23\text{ rad} ]

So the complex radar signal rotates by roughly 3.23 radians per chirp.

That phase rotation is the mathematical origin of Doppler estimation in the slow-time dimension.

7. Range-Doppler Map

After the first FFT, every chirp contains a set of range bins. Stack those chirps together:

[ \begin{bmatrix} \text{chirp 1}\ \text{chirp 2}\ \text{chirp 3}\ \vdots\ \text{chirp M} \end{bmatrix} ] [ \Downarrow ] [ X[\text{range},\text{chirp}] ]

Then perform another FFT along the chirp dimension.

[ X[n,m] \xrightarrow{\text{FFT in }n} X[kR,m] \xrightarrow{\text{FFT in }m} X[kR,k_D] ]

The result is a range-Doppler map.

Axis	Physical meaning
X-axis	Range
Y-axis	Velocity / Doppler
Brightness	Signal strength

8. Angle: Why Can Two Antennas Tell Direction?

Now consider two receiving antennas separated by distance (d).

A wave arriving at an angle does not reach both antennas at exactly the same time.

The path difference is approximately:

[ \Delta r = d\sin\theta ]

A path difference creates a phase difference:

[ \Delta\phi = \frac{2\pi}{\lambda}\Delta r ]

Substituting:

[ \boxed{ \Delta\phi = \frac{2\pi d\sin\theta}{\lambda} } ]

Solving for angle:

[ \boxed{ \theta = \sin^{-1} \left( \frac{\lambda\Delta\phi}{2\pi d} \right) } ]

Numerical Angle Example

Suppose:

Radar frequency = 77 GHz
(\lambda\approx3.896) mm
Antenna spacing (d=\lambda/2)
Target angle (\theta=30^\circ)

Then:

[ \Delta\phi = \frac{2\pi(\lambda/2)\sin(30^\circ)} {\lambda} ]

[ \Delta\phi = \pi(0.5) = \boxed{\frac{\pi}{2}\text{ rad}} ]

So a target at 30° produces approximately a 90° phase difference between the two receivers.

9. Why Only Two Receivers Are Limited

With two antennas, we essentially get one spatial phase measurement:

[ \Delta\phi = \phi2-\phi1 ]

That gives us an estimate of one angle.

But with more antennas, we obtain many spatial measurements:

[ \phi1,\phi2,\phi3,\ldots,\phiN ]

This allows us to perform spatial beamforming or a spatial FFT to obtain a more precise angular spectrum.

More antennas → more spatial samples → better angular discrimination.

10. Why MIMO Helps

MIMO stands for Multiple Input Multiple Output.

Multiple transmitters and receivers can create a larger virtual antenna array.

For example, with:

[ N{Tx}\times N{Rx} ]

appropriately arranged channels, the radar can obtain many virtual spatial samples.

This improves angular resolution and allows the radar to distinguish targets that are close together in angle.

11. Putting Everything Together

We can now see why the same raw radar data contains all three measurements.

Range

[ \boxed{f_b=S\frac{2R}{c}} ] [ \boxed{R=\frac{cf_b}{2S}} ]

Find the beat frequency using an FFT across samples within a chirp.

Velocity

[ \boxed{f_D=\frac{2v}{\lambda}} ] [ \boxed{v=\frac{\lambda f_D}{2}} ]

Find the Doppler frequency using an FFT across chirps.

Angle

[ \boxed{ \Delta\phi = \frac{2\pi d\sin\theta}{\lambda} } ] [ \boxed{ \theta = \sin^{-1} \left( \frac{\lambda\Delta\phi}{2\pi d} \right) } ]

Compare the phase received by spatially separated antennas.

12. The Mathematical Picture

The most useful way to think about FMCW radar is as a multidimensional signal-processing problem.

[ \boxed{ \text{Fast Time} \xrightarrow{\text{FFT}} \text{Range} } ] [ \boxed{ \text{Slow Time} \xrightarrow{\text{FFT}} \text{Velocity} } ] [ \boxed{ \text{Spatial Dimension} \xrightarrow{\text{Phase / FFT}} \text{Angle} } ]

Therefore, the radar data cube can conceptually be represented as:

[ \boxed{ X[\text{range},\text{velocity},\text{angle}] } ]

In real radar systems there are many additional complications: noise, clutter, multiple targets, acceleration, phase noise, leakage, range-Doppler coupling, antenna calibration, windowing, CFAR detection, and more.

13. Numerical Example

Consider a 77 GHz FMCW radar with:

Bandwidth (B=1) GHz
Chirp duration (T_c=50) μs
Target range (R=30) m
Target velocity (v=20) m/s
Two receivers separated by (d=\lambda/2)
Target angle (\theta=30^\circ)

Step 1 — Chirp slope

[ S= \frac{B}{T_c} = \frac{10^9}{50\times10^{-6}} = 2\times10^{13}\text{ Hz/s} ]

Step 2 — Range

[ f_b = S\frac{2R}{c} = (2\times10^{13}) \frac{60}{3\times10^8} = \boxed{4\text{ MHz}} ]

Step 3 — Wavelength

[ \lambda = \frac{3\times10^8}{77\times10^9} \approx \boxed{3.896\text{ mm}} ]

Step 4 — Doppler

[ f_D = \frac{2(20)}{0.003896} \approx \boxed{10.27\text{ kHz}} ]

Step 5 — Angle

[ d=\frac{\lambda}{2} ] [ \Delta\phi = \frac{2\pi d\sin30^\circ}{\lambda} = \boxed{\frac{\pi}{2}} ]

The radar therefore observes approximately:

Measurement	Observed quantity	Physical result
Fast-time FFT	4 MHz beat frequency	30 m range
Slow-time FFT	10.27 kHz Doppler	20 m/s velocity
Rx phase difference	π/2 rad	30° angle

14. Summary

FMCW radar turns distance into frequency, motion into Doppler frequency, and direction into spatial phase difference; FFTs are used because they efficiently reveal the frequency components hidden inside the sampled radar signal.


  %% Simple FMCW Radar: Fast-Time FFT + Slow-Time FFT
clear;
clc;
close all;

%% Radar parameters
c  = 3e8;              % Speed of light (m/s)
fc = 77e9;             % Carrier frequency (Hz)

B  = 1e9;              % Bandwidth = 1 GHz
Tc = 50e-6;            % Chirp duration = 50 us

S = B/Tc;              % Chirp slope (Hz/s)

%% Target
R = 30;                % Target range = 30 m
v = 20;                % Target velocity = 20 m/s

lambda = c/fc;

%% Sampling parameters
Fs = 20e6;             % ADC sampling frequency = 20 MHz
Ns = 1000;              % Samples per chirp
Nc = 128;               % Number of chirps

Ts = 1/Fs;

%% Create data matrix
% Rows    = fast time (samples within chirp)
% Columns = slow time (different chirps)

radarData = zeros(Ns, Nc);

%% Generate received beat signal
for m = 1:Nc

    % Time within this chirp
    t = (0:Ns-1).' * Ts;

    % Target range changes because target is moving
    Rm = R - v*(m-1)*Tc;

    % Round-trip delay
    tau = 2*Rm/c;

    % Beat frequency caused by range
    fb = S*tau;

    % Doppler frequency
    fd = 2*v/lambda;

    % Beat signal
    radarData(:,m) = exp(1j*2*pi*(fb + fd)*t);

end


%% ============================================================
%  FAST-TIME FFT
%  Samples within ONE chirp -> Range
% =============================================================

% Take FFT along rows
rangeFFT = fft(radarData, [], 1);

% Keep only positive frequencies
rangeFFT = rangeFFT(1:Ns/2, :);

% Frequency axis
fRange = (0:Ns/2-1).' * Fs/Ns;

% Convert beat frequency to range
rangeAxis = c*fRange/(2*S);


%% Plot Fast-Time FFT
figure;

plot(rangeAxis, abs(rangeFFT(:,1)));

xlabel('Range (m)');
ylabel('Magnitude');
title('Fast-Time FFT → Range');
grid on;

xlim([0 50]);


%% ============================================================
% SLOW-TIME FFT
% Same range bin across MANY chirps -> Velocity
% =============================================================

% Pick the range bin containing the target
[~, rangeBin] = max(abs(rangeFFT(:,1)));

% Extract this range bin across all chirps
slowTimeSignal = rangeFFT(rangeBin, :);

% FFT across chirps
dopplerFFT = fftshift(fft(slowTimeSignal, Nc));

% Doppler frequency axis
PRF = 1/Tc;

fDoppler = (-Nc/2:Nc/2-1) * PRF/Nc;

% Convert Doppler frequency to velocity
velocityAxis = lambda*fDoppler/2;


%% Plot Slow-Time FFT
figure;

plot(velocityAxis, abs(dopplerFFT));

xlabel('Velocity (m/s)');
ylabel('Magnitude');
title('Slow-Time FFT → Velocity');
grid on;

xlim([-30 30]);


%% ============================================================
% RANGE-DOPPLER MAP
% =============================================================

% FFT across slow-time/chirp dimension
RD = fftshift(fft(rangeFFT, Nc, 2), 2);

figure;

imagesc(velocityAxis, rangeAxis, abs(RD));

xlabel('Velocity (m/s)');
ylabel('Range (m)');
title('Range-Doppler Map');

colorbar;
axis xy;

ylim([0 50]);
xlim([-30 30]);

Online Signal Processing Simulations

The cutoff frequency (in GHz) for the dominant TE10 mode of an air-filled rectangular waveguide of ...

Q. The cutoff frequency (in GHz) for the dominant TE10 mode of an air-filled rectangular waveguide of inner dimension 0.28 inch×0.14 inch is __. (rounded off to two decimal places).

Given Information

Step 1: Cutoff Frequency Formula

Step 2: Convert Inch into Meter

Step 3: Calculate Cutoff Frequency

Final Answer

Important Note

A wireless digital transmission scheme is using 16-QAM over an additive white Gaussian noise channel ...

Given Information

Step 1: Calculate Bits per Symbol

Step 2: Calculate Symbol Rate

Step 3: Calculate Minimum Bandwidth

Final Answer

What is a Supernode?

1. Understanding a Supernode

2. Supernode Circuit Diagram

3. Why Do We Need a Supernode?

An ideal voltage source connects two unknown nodes

Current through the source is unknown

Combine the nodes and apply KCL

4. Example From the Given Circuit

Equation 1: KCL at the supernode

Equation 2: Voltage-source constraint

5. Steps to Solve a Supernode Problem

Quick Summary

Consider the circuit shown in the Figure, where the input Vi(t) is in Volt. The average power (in mW) dissipated in the load resistance of 1 kΩ ...

Given Values

Step 1: Calculate RMS Voltage

Step 2: Calculate Average Power

Step 3: Convert to Milliwatts

The output voltage V0 (in Volt) for the network given in the Figure is ___

1. Given Circuit Parameters

2. Understand the Circuit

Apply KCL at the Top Node

Apply KCL Using the Supernode

Calculate the Top Rail Voltage

Find the Midpoint Voltages

Final Answer

Further Reading

SRAM, DRAM, EPROM, MRAM, and DDRAM Explained

Types of Memory and Their Technologies

Fundamental Matching

Important Memory Technologies

SRAM

DRAM

EPROM

MRAM

Why do we need DDRAM?

Example

Parametric Down-Conversion of Photons

Parametric Down-Conversion

Energy Conservation

Wavelength Relationship

Momentum Conservation

Degenerate Down-Conversion

Summary

Active vs Passive Sensors

Passive sensor: Generates its own electrical signal from the measured physical quantity.

Active sensor: Requires external excitation or power to operate, such as an RTD (which requires a measuring current).

A binary ripple counter is designed to count (0)₁₀ to (64)₁₀. Which of the following is/are the number of flip-flops required ...

Solution

Step 1: Calculate the total number of states

Step 2: Calculate the number of flip-flops

Final Answer: Option B - 7 Flip-Flops

Further Reading

An S-band radar transmitting at 3 GHz radiates 200 kW of power. Determine the signal power density at a range of 100 ...

Solution

Step 1: Convert Nautical Miles to Meters

Step 2: Calculate the Wavelength

Step 3: Calculate the Antenna Gain

Step 4: Calculate the Signal Power Density

Step 5: Convert to Milliwatts per Square Meter

Final Answer: Option 4

The electric field of a monochromatic plane wave travelling in a lossless isotropic and homogenous medium is given by E(z, t) = ...

Solution

1. Identify the electric-field components

2. Verify the magnitude of the electric field

3. Determine the handedness