Skip to main content

Constellation Diagram of FSK in Detail


 

Binary bits '0' and '1' can be mapped to 'j' and '1' to '1', respectively, for Baseband Binary Frequency Shift Keying (BFSK). Signals are in phase here. These bits can be mapped into baseband representation for a number of uses, including power spectral density (PSD) calculations. For passband BFSK transmission, we can modulate signal 'j' with a lower carrier frequency and signal '1' with a higher carrier frequency while transmitting over a wireless channel.

Let's assume we are transmitting carrier signal fc1 for the transmission of binary bit '1' and carrier signal fc2 for the transmission of binary bit '0'.

Simulator for 2-FSK Constellation Diagram

Simulator for 2-FSK Constellation Diagram

Energy per bit (Eb):

For transmission of binary ‘1’

Starting from the passband signal \( s_1(t)=A_c\cos(2\pi f_1 t) \) over \(0\le t\le T_b\):

\[ E_b \;=\; \int_{0}^{T_b}\!\big(A_c\cos 2\pi f_1 t\big)^2\,dt \;=\; \int_{0}^{T_b}\!\frac{A_c^2}{2}\,dt \;+\; \int_{0}^{T_b}\!\frac{A_c^2}{2}\cos(4\pi f_1 t)\,dt \] \[ \;=\; \int_{0}^{T_b}\!\frac{A_c^2}{2}\,dt \;+\; 0 \quad \text{(second term averages to 0 over a full cycle)} \;=\; \frac{A_c^2}{2}\,T_b . \]

For transmission of binary ‘0’

Similarly for \( s_2(t)=A_c\cos(2\pi f_2 t) \):

\[ E_b \;=\; \int_{0}^{T_b}\!\big(A_c\cos 2\pi f_2 t\big)^2\,dt \;=\; \int_{0}^{T_b}\!\frac{A_c^2}{2}\,dt \;+\; \int_{0}^{T_b}\!\frac{A_c^2}{2}\cos(4\pi f_2 t)\,dt \] \[ \;=\; \int_{0}^{T_b}\!\frac{A_c^2}{2}\,dt \;+\; 0 \;=\; \frac{A_c^2}{2}\,T_b . \]

Amplitude in terms of \(E_b\)

\[ A_c \;=\; \sqrt{\frac{2E_b}{T_b}} \;. \]

Constellation Diagram of FSK

In Binary FSK (BFSK), two orthogonal signals represent binary symbols:
\( s_1(t) = \sqrt{\frac{2E_b}{T_b}} \cos(2\pi f_1 t), \quad 0 \leq t \leq T_b \)
\( s_2(t) = \sqrt{\frac{2E_b}{T_b}} \cos(2\pi f_2 t), \quad 0 \leq t \leq T_b \)

Symbol 0: \( f_1 \)
Symbol 1: \( f_2 \)

The points lie on orthogonal axes because \( s_1(t) \) and \( s_2(t) \) are orthogonal signals.







Fig 1: Constellation Diagram of FSK

 In the above figure values are in terms of the normalized functions. √(2/Tb).cos2Пf1t and √(2/Tb).cos2Пf2t are orthogonal functions in the interval (0, Tb). And the distance between signaling points, d12 = √(2Eb)
By interpreting these functions as vectors, the phase angle between the resulting vectors will be 90 degrees.   

Using more frequency shifts to display multiple symbols or bits of digital data is known as high-order frequency shift keying (FSK). Every frequency in FSK corresponds to a distinct symbol or collection of bits. Higher data rates are possible with high-order FSK schemes, but they may also be more vulnerable to channel impairments and noise. 

 

Also read about

  1.  Constellation Diagram of ASK in detail
  2. Constellation Diagram of PSK in detail
  3. Baseband ASK, FSK, and PSK
  4. Theoretical BER vs SNR for binary ASK, FSK, and PSK

BER vs SNR from Constellation Diagram of FSK

The Bit Error Rate (BER) versus Signal-to-Noise Ratio (SNR) can be derived from the Euclidean distance between constellation points. Once the minimum distance between the symbols is known, the probability of making an incorrect decision in an Additive White Gaussian Noise (AWGN) channel can be computed using the Q-function.

The Q-function gives the probability that Gaussian noise exceeds the decision threshold, causing the received symbol to cross into the neighboring decision region and resulting in an incorrect bit detection.

Example: Coherent Orthogonal FSK

For coherent orthogonal Frequency Shift Keying (FSK), the theoretical BER is

BER = Q(√(Eb / N0))

The Euclidean distance between the two orthogonal signal points is

d = √(2Eb)

Substituting this distance into the general AWGN error probability formula gives

BER = Q(d / √(2N0))

= Q(√(2Eb) / √(2N0))

= Q(√(Eb / N0))

Why is BER approximately 0.15 at 0 dB?

At an SNR of 0 dB,

Eb / N0 = 100/10 = 1

Therefore,

BER = Q(√1)

= Q(1)

Using the standard Gaussian Q-function table,

Q(1) ≈ 0.1587

Thus, at an SNR of 0 dB, the theoretical BER for coherent orthogonal FSK is approximately

BER ≈ 0.1587 ≈ 0.15

This matches the theoretical BER curve for coherent orthogonal FSK in an AWGN channel.

Note: If you instead use the expression Q(√(2Eb/N0)), the BER at 0 dB becomes Q(√2) = Q(1.4142) ≈ 0.0786, which corresponds to coherent BPSK (and QPSK), not coherent orthogonal FSK.

Read More: Learn how the Q-function is used to derive theoretical BER curves for various digital modulation schemes in AWGN channels.



Contact Us

Name

Email *

Message *

Popular Posts

UGC NET Electronic Science Previous Year Question Papers with Solutions

Home / Engineering & Other Exams / UGC NET 2026 PYQ ⬇️ Download Papers and Solutions 📋 Exam Pattern 💡 Preparation Tips ❓ FAQs 📊 Exam Highlights: Electronic Science (88) Feature Details Junior Research Fellowship (JRF) ₹37,000 + HRA per month Eligibility M.Sc/M.Tech in Electronics (55%) Validity of Certificate JRF (3 Years) | Lectureship (Lifetime) 📥 Download UGC NET Electronics PDFs Complete collection of previous year question papers, answer keys and explanations for Subject Code 88. Start Downloading 📂 View All Question Papers June 2025 - Question Paper Download PDF June 2025 - Solved Paper + Explanation ...

Hybrid Beamforming | Page 1

Beamforming Techniques Hybrid Beamforming... Page 1 | Page 2 | Hybrid Beamforming: Hybrid beam formation was developed to address some of the limitations of digital pre-coding approaches. Every antenna element is connected to an RF chain in digital pre-coding (beam forming) method. We also know that each RF chain is in charge of providing a separate data stream between the transmitter and the receiver. We know that a larger number of independent data streams leads to higher data rates. It has a spatial multiplexing feature for MIMO. As a result, we may assume that switching from MIMO to massive MIMO will benefit us more in terms of spatial multiplexing in massive MIMO, where each antenna is coupled to a single RF chain. We'll proceed with a definition of hybrid beam forming. Overview of hybrid beam forming with example: Unlike digital beam forming, more than one antenna element is connected to a single RF chain in hybr...

Design of CMOS XOR/XNOR Gates

Design of CMOS XOR/XNOR Gates The semiconductor industry has experienced rapid integration of multimedia applications into mobile electronics, leading to very high integration density in CMOS VLSI. As operating frequencies increase, power consumption, speed, silicon area, and reliability become critical considerations. The XOR-XNOR circuits are fundamental building blocks in arithmetic circuits (Full Adders, Multipliers), compressors, comparators, parity checkers, code converters, error-detecting/correcting codes, and phase detectors. Their performance directly impacts the complex circuits they are used in. Design goals include full output voltage swing, low power consumption, reduced transistor count, minimal delay, and simultaneous non-skewed outputs. Static Logic (Static CMOS) Stat...

UGC NET Electronic Science June 2025 Question Paper with Answer Key & Detailed Solutions

Home / UGC NET PYQ / June 2025 Solved UGC NET Electronic Science June 2025 Question Paper with Answer Key and Full Explanations 📥 Download Question Paper (PDF) 2025 2024 2023 2022 2021 2020 Explanations 1.  Answer: Option (3) For forming a p-type semiconductor, the dopant must be a trivalent impurity (three valence electrons) so that it creates acceptor levels and holes become the majority carriers. Among the given elements, boron (B) is a group-III element (trivalent). Arsenic (As) and phosphorus (P) are group-V (pentavalent) donors that produce n-type material, and germanium (Ge) is a group-IV element usually used as the semiconductor, not as an acceptor dopant. Hence, doping an intrinsic semiconductor with B produces a p-type semiconductor. 2.  Answer: Option (4) The ohmic resistance of a JFET at zero gate bias is given by the standard relation: R DS(on) = V P / I DSS ...

MIMO Channel Matrix | Rank and Condition Number

MIMO / Massive MIMO MIMO Channel Matrix | Rank and Condition...   The channel matrix in wireless communication is a matrix that describes the impact of the channel on the transmitted signal. The channel matrix can be used to model the effects of the atmospheric or underwater environment on the signal, such as the absorption, reflection or scattering of the signal by surrounding objects. When addressing multi-antenna communication, the term "channel matrix" is used. Let's assume that only one TX and one RX are in communication and there's no surrounding object. Here, in our case, we can apply the proper threshold condition to a received signal and get the original transmitted signal at the RX side. However, in real-world situations, we see signal path blockage, reflections, etc.,  (NLOS paths [↗]) more frequently. The obstruction is typically caused by building walls, etc. Multi-antenna communication was introduced to address this issue. It makes diversity app...

Online Simulator for ASK, FSK, and PSK Signal Generation

Interactive Digital Signal Processing (DSP) Tutorial and Simulator for ASK, FSK, and BPSK modulation techniques. Try our new Digital Signal Processing Simulator!   •   Interactive ASK, FSK, and BPSK tools updated for 2025. Start Now Digital Modulation Visualizer: ASK, FSK, & BPSK Simulator Learn and visualize binary modulation techniques (ASK, FSK, BPSK) in real-time with adjustable carrier and sampling parameters. Perfect for DSP students and engineers. 📡 ASK Simulator 📶 FSK Simulator 🎚️ BPSK Simulator 📚 More Topics ASK Modulator FSK Modulator BPSK Modulator More Topics 1. ASK (Amplitude Shift Keying) Simulat...