Orthogonal Vector Space and Signal Representation

Orthogonal Vector Space

• Represents a comprehensive set of orthogonal vectors.
• Considers a three-dimensional vector space.
• Considers vector A at point (X₁, Y₁, Z₁).
• Considers three unit vectors (Vx, Vy, Vz) oriented along the X, Y, and Z axes, respectively.
• The unit vectors are mutually orthogonal, which satisfies Vx.Vx = Vy.Vy = Vz.Vz = 1
• Unit vectors also satisfy: Vx.Vy = Vy.Vz = Vz.Vx = 0

Orthonormal Basis Functions

• Mutually orthogonal unit vectors are called orthonormal basis functions.
• Any vector A can be represented by its components and unit vectors as A = X₁Vx + Y₁Vy + Z₁Vz.
• Vectors in a three-dimensional space can be represented by three unit vectors.
• In N-dimensional space, any vector A can be represented as A = X₁Vx + Y₁Vy + Z₁Vz +...+ N₁VN.

Orthogonal Representation of Signal

• Signals can be represented as a vector.
• Signals can also be represented as a weighted sum of orthonormal basis functions; a signal is represented as s(t) = sum(si * pi(t)).
• Two functions are orthonormal if their inner product satisfies (pi(t), pj(t)) = 1 if i = j, or 0 if i ≠ j.
• A circuit produces the signal s(t) from its elements {si} in the signal space.

Elements of a Signal

• Signals are defined on the time interval [a; b].
• Elements si are found by si = integral from a to b of s(t) * pi(t) dt.
• The signal space elements {si} are produced from the signal s(t).

Energy of Signal

• It is represented as (s(t), s(t)) = integral from a to b of s²(t) dt = ||s||² = sum(si²) = Es.
• The length of a vector in the signal space equals the square root of the signal energy.
• The distance between two signal vectors represents the square root of the energy difference between the two signals involved and expressed as dij² = ||si - sj||² = integral from a to b of [si(t) - sj(t)]² dt = Ei + Ej - 2Eij.

One-Dimensional Signal

• Consider two rectangular signals: s₁(t) = −s₂(t) = sqrt(E/T), for 0 ≤ t ≤ T.
• Their bandpass equivalents are s₁(t) = −s₂(t) = sqrt(E/T) * cos(2πfct).
• A two-signal representation shows s₁ = [sqrt(E)] and s₂ = [-sqrt(E)].
• The orthonormal basis function is defined as φ(t) = sqrt(2/T) * cos(2πfct).
• Two signals are s₁(t) = sqrt(E) * φ(t) and s₂(t) = −sqrt(E) * φ(t).

Two-Dimensional Signal

• A two-dimensional signal with s₁(t) = sqrt(2E/T) * cos(ωot) has signals 0 ≤ t ≤ T.
• A two-dimensional signal has s₂(t) = sqrt(2E/T) * sin(ωot) with signals 0 ≤ t ≤ T.
• Basis functions include φ₁(t) = sqrt(2/T) * cos(2πfct) and φ₂(t) = sqrt(2/T) * sin(2πfct).
• Signal points s₁ = [sqrt(E), 0] and s₂ = [0, sqrt(E)], with a distance d₁₂ = sqrt(2E).

Multi-Phase Signal

• Multi-phase signals are represented as si(t) = Re{sqrt(2E/T) * exp(jωot + j(i - 1)(2π/M))}, for i = 1, 2, ..., M; 0 ≤ t ≤ T.
• Vector representation: si = [sqrt(E) * cos((i-1)2π/M), sqrt(E) * sin((i-1)2π/M)].

Channel

• It is the physical path connecting the transmitter and receiver.
• Signal propagates via electrical, light, or electromagnetic energy.
• Coaxial/parallel wire cables use electrical energy.
• Optical Fiber Cables use light energy.
• Wireless channels use electromagnetic energy.

Channel Attenuation

• It is the reduction of signal strength during transmission.
• Attenuation increases with separation between the transmitter (Tx) and receiver (Rx), which directly relates to the channel length.
• Attenuation levels are also dependent on the type of channel.

Channel Noise

• It is an unwanted and unpredictable signal that modifies the original signal.
• Noise is additive, superimposing on the original signal.
• Power spectral density is flat, meaning it has all frequency components.
• Noise voltage is random and can be modeled using a Gaussian probability distribution function.
• Additive White Gaussian Noise (AWGN) describes Channel Noise.
• Signal-to-Noise Ratio (SNR) = Psignal / Pnoise

Channel as a Filter

• Filters gain/attenuation changes with frequency.
• Some frequencies experience heavy attenuation, while others pass through with very low attenuation.
• The frequency band with heavy attenuation is the stop-band of the filter.
• The frequency band that passes with low attenuation is the pass-band of the filter.
• Filter types include Low-pass, High-pass, Band-pass, and Band-stop.

Channel as a Filter (Cont.)

• Communication channel operates as a filter, allowing some bands and rejecting others.
• Coaxial cable acts as a Low Pass Filter.
• OFC (Optical Fiber Cable) and Waveguides act as a High Pass Filter.
• Wireless Channels act as a Band Pass Filter.
• In the time-domain, the signal is convolved with the impulse response of the channel.

Golden Rule for Transmission

• For distortion-free transmission, the transmitted signal band must match the pass band of the channel.
• If signal bandwidth is greater than the channel bandwidth, some frequency components of the signal go missing.

Modulation

• The process through with the wireless channel functions as a passband filter and allows signals within a limited bandwidth.
• To transmit signals effectively through a channel, the signal band must align with the channel's passband.
• Accomplished by multiplying the signal with a sinusoidal carrier that has a frequency at the center of the passband.
• The method using digital signals is digital modulation, including: Amplitude-Shift Keying (ASK), Frequency-Shift Keying (FSK) and Phase-Shift Keying (PSK).

Base Band Signal (Digital PAM)

• Binary information has two possible values.
• Logic 1 is represented as an amplitude A voltage for 0<t<Tb.
• Logic 0 is represented as an amplitude -A voltage for 0<t<Tb.
• A transmission rate Rb = 1/Tb, where Tb represents the bit duration.

M-ary Concept

• Instead of sending one bit at a time, binary information can be sent in groups of k bits.
• Results in an increase of transmission rate by k times.
• The bit sequence is sub-divided into blocks of k-bits called a symbol.
• In each transmitting interval, one symbol (block of k-bits) is transmitted instead of a single bit.
• Grouping bits yields M = 2^k possible symbols.

M-ary PAM

• Each symbol in M-ary PAM is represented with one of the M = 2^k possible amplitude values.
• This signal is represented by s_m(t) = A_m * g_T(t).
• Symbol index: m = 1, 2, ..., M, where 0 ≤ t ≤ Ts.
• g_T(t) is the signal pulse of any arbitrary shape.
• Signal amplitude is symmetric across zero i.e. A_m =(2m - 1 - M).
• For k = 2 with M = 4, the amplitudes are A₁ = -3, A₂ = -1, A₃ = 1, A₄ = 3.

Amplitude Shift Keying (ASK)

• To send digital information through a bandpass channel, the baseband signal sm(t) is multiplied with a carrier signal cos(2πfct).
• The carrier frequency fc corresponds to the center frequency of the passband.
• This method is called Amplitude Shift Keying (ASK).

ASK Signal

• ASK signal is given by u_m(t) = s_m(t) * cos(2πfct) = A_m * g_T(t) * cos(2πfct).
• For Binary ASK, Signal 1: g_T(t) = 1, u₁(t) = A_m * cos(2πfct).
• For Signal 2: g_T(t) = 0, u₂(t) = 0.

Spectrum of ASK signal

• If g_T(t) has a frequency spectrum G_T(f), then the spectrum of the ASK signal u_m(t) is given by U_m(f) = A_m/2 * [G_T(f - fc) + G_T(f + fc)].

Geometric Representation of ASK

• Here, um(t) = Sm(t)ψ(t).
• ψ(t) is an orthonormal basis function with |ψ(t)| = 1.
• ψ(t) = sqrt(2/T) cos 2πfct
• Sm(t) = sqrt(T/2) Am
• Am = 2m - 1 - M

Phase Shift Keying (PSK)

• The method where symbols are differentiated by phase
• All symbols have the same energy level.
• The M-ary PSK signal waveform is given by um(t) = sqrt(2E/T) * cos(2πfct + (2π(m-1))/M), for m = 1, 2, ..., M.
• Uses a change in the value of 'm' to trigger a sudden phase shift in the carrier signal, this occurs at each signaling interval.

Binary PSK (BPSK)

• Has two possible symbols (0 and 1)
• Each symbol represents one bit
• Symbols include: u₁(t) = sqrt(2E/T) * cos(2πfct)
• One of the two possible phase values is assigned to each symbol

Quadrature Phase Shift Keying (QPSK)

• A form of PSK with M=4, meaning that it has four possible symbols.
• In the equation 2^k = M, k = 2.
• Each symbol represents 2 successive bits
• QPSK symbols are: m = 1 (00) -> u1(t) = sqrt(2E/T)cos(2πfct + 0); m = 2 (01) -> u2(t) = sqrt(2E/T)cos(2πfct + π/2); m = 3 (11) -> u3(t) = sqrt(2E/T)cos(2πfct + π)

Geometric representation of PSK

• Vector representation of PSK is given by um(t) = sqrt(2E/T) cos(2πfct + (2π(m-1))/M), m = 1, 2, ..., M.
• Signal can be represented in a two-dimensional signal space using orthonormal basis functions.
• Ortho normal functions: φ1(t) = sqrt(2/T) cos 2πfct and φ2(t) = sqrt(2/T) sin 2πfct
• um(t) = sqrtE [cos ((2π(m-1))/M) φ1(t) − sin ((2π(m-1))/M) φ2(t)]

Constellation Diagrams of PSK

• Constellation diagrams include: BPSK, QPSK and 8PSK.

Euclidean Distance of PSK

• The Euclidean distance between two signal points on a constellation is dmn = sqrt(||sm - sn||²) = sqrt(2E(1 - cos((2π(m-n))/M))).
• Minimum Euclidean distance: dmin = sqrt(2E(1 - cos(2π/M))).
• The value of M controls the performance of M-ary PSK.
• An increase in M causes dimin to decrease, resulting in more bit errors, but the bits per symbol increases which overall increases data transmission rate.

Spectral Analysis of M-ary PSK

• Spectrum of the modulated signal, considering only positive frequency components results in Sm(f) = Esinc²(fc – f)T assuming a rectangular baseband pulse shape.
• T is defined as symbol duration = Tb log2 M with Tb representing bit duration.
• Null to Null bandwidth of the main lobe gives B = 2/T.
• Spectral efficiency is measured in Bits/Hz.
• 2/T Hz spectrum accommodates 1/T symbols.
• 1 Hz spectrum accommodates 0.5 symbols.

Quadrature Amplitude Modulation (QAM)

• In M-ary PSK, the minimum Euclidean distance (dmin) decreases with an increase in value of M.
• It leads to increase in symbol-error-rate at receiver.
• High data rate transmission using M-ary PSK is unsuitable.
• QAM is a modulation scheme using combination of ASK and PSK
• This method aims to keep the dmin value constant with increase in M.

M-ary QAM Formula

• Formula used is u_mn(t) = A_m * g_T(t) * cos(2πfct + θn).
• Signal amplitude level present is M₁ with formula M₁ = Number of amplitude-level present = 2^(k1)
• Signal phase level present is M₂ with formula M₂ = Number of phase level present = 2^(k2)
• Overall number of bits per symbol = k₁ + k₂ = k, and the total number of signals possible is given 2^(k1)* 2^(k1) = 2^k = M₁ * M₂ = M

QAM Formulas Explained

• By using the the trigonometric identitiy, this equation can also be expressed as: u_mn(t) = A_m * g_T(t) * cos(θn) * cos(2πfct) + A_m * g_T(t) * sin(θn) * sin(2πfct) = A_mnI * cos(2πfct) + A_mnQ * sin(2πfct).
• A_mnI = A_m * g_T(t) * cos(θn) represents the In-phase component.
• A_mnQ = A_m * g_T(t) * sin(θn) represents the Quadrature-phase component.
• Using this, we get u_p(t) = u_mn(t) = A_pI * cos(2πfct) + A_pQ * sin(2πfct).
• QAM signal can be viewed as a pair of AM carrier signals

Geometric Representation of QAM signal

• Making use of orthonormal basis functions, φ1(t) = sqrt(2/T) * cos 2πfct ; φ2(t) = sqrt(2/T) * sin 2πfct
• The QAM signal can then be expressed as u_p = (sqrt(Es) Apl, sqrt(Es) ApQ).

QAM Key Points

• Unlike M-ary PSK, in M-ary QAM, the minimum Euclidian distance (distance between two nearest signal points) remains constant with an increase in the value of M.
• Hence, QAM is suitable for high rate data transmission.
• In QAM, the signal energy is not constant for every signals.
• The average signal energy of QAM signaling technique is E_avg = 1/M * sum(||si||²), where i ranges from 1 to M

Frequency Shift Keying (FSK)

• The method where symbols are represented by different frequencies, which produces an M-ary FSK signal.
• Formula used: um(t) = sqrt(2E/T) * cos(2πfct + 2π(2m – 1 – M)∆ft)
• For binary FSK, when M=2: u₁(t) = sqrt(2E/T)*cos(2πf₁t), and u₂(t) = sqrt(2E/T)*cos(2πf₂t)
• In binary, f₁ = fc – ∆f with f₂ = fc + ∆f

BFSK Signal Representation

• Basis functions: φ₁(t) = sqrt(2/T) * cos2πf₁t, and φ₂(t) = sqrt(2/T) * cos2πf₂t
• Signals: u₁(t) = sqrt(E)φ₁(t), u2(t) = sqrt(E)φ2(t)
• Orthogonality condition: ∆f = n/T, where n = 1, 2, ....