Time, Frequency And Complex Frequency Domains Of A Sinusoidal Excitation

Time, Frequency And Complex Frequency Domains Of A Sinusoidal Excitation
TECTechnics Classroom TECTechnics Overview

Time, Frequency And Complex Frequency Domains

Analyses of sinusoidal excitations of orders greater than two, are not simple. There are various computer programs that aid in the analyses. Nonetheless, it is very important that one grasps the mathematical tools used to simplify the analyses of sinusiods in the time domain where they exist.
(a) Given that the time, frequency and complex frequency domains are associated with the containerships of sinusoidal excitations; indicate the general identity of a sinusoidal excitation in each of the domains.
(b) What mathematical tools are used to arrive at the general identities of sinusoidal excitations in the frequency and complex frequency domains?
(c) Given an arbitrary sinusoidal excitation in the time domain, derive its representation in the frequency and complex frequency domains. Thus establish the PJProblemStrings Sequence of the domain switch of a sinusoidal excitation.
(d) Determine the complex frequencies that are associated with:
(i) 5e^-4t (ii) cos2ωt (iii) sin (ωt + 2θ)
(iv) 4e^-2tsin(3t - 50^o) (v) e^-3t(2 + cos4t)
(e) Determine the complex frequency s and V(s) associated with:
(i) 5e^-2t (ii) 5e^-2tcos(4t + 10^o) (iii) 4cos(2t - 20^o)
(f) Determine v(t) if:
(i) s = -2, V = 2<0^o (ii) s = j2, V = 12<-30^o (iii) s = -4 + j3, V = 6<10^o
(g) Determine the Laplace transform of the following functions:
(i) e^-atsin(ωt)u(t) (ii) e^-atcos(ωt)u(t)

Time, Frequency And Complex Frequency Domains

The strings: all PjProblems are at play. However, S₇P₁A₁₇ (containership - location), S₇P₂A₂₁ (identity - physical), S₇P₃A₃₂ (force - push) and S₇P₅A₅₁ (change - physical) are in focus.

(a) General representation of a sinusoidal excitation in the time domain: v(t) = Acos(ωt + θ) -------(1)
Where A is the amplitude of the sinusoid; ω is the radian frequency (angular velocity) and θ is the phase angle.

General representation of a sinusoidal excitation in frequency domain: V(jω) = Ae^jθ = A<θ (phasor form) -------(2)
The frequency domain is also called the phasor domain. It is a mathematical space formulated for analytical simplicity. In other words there are no real phasor sinusoids. It should be remembered that the radian frequency ω is implicit in (2) even though it is not explicitly represented.

General representation of a sinusoidal excitation in complex frequency domain (also called the s domain): V(s) = A<θ
Where s = σ + jω -------(3)
In essence the complex frequency domain is an extension of the frequency domain which takes into account the damping of sinusoids. A sinusoid can be negatively damped (σ is negative, figure 7.1(b), exponential decay); or positively damped (σ is positive, figure 7.1(c), exponential growth); or undamped (σ = 0, figure 7.1 (a), steady state sinusoid). There is overdamped, critically damped and under damped within the damped category. All of which has their respective formulas. It is sufficient to identify s and replace jw with s when switching from the frequency domain to the complex frequency domain.

A damped sinusoid in the time domain is represented as: v(t) = Ae^σtcos(ωt + θ) -------(4)

(b) Mathematical tools used to arrive at the general identities of sinusoidal excitations in the frequency and complex frequency domains are:

Euler Identity And Laplace Transform

(c) Consider the complex exponential Ae^{j(ωt + θ)}.
By Euler's Identity, Ae^{(jωt + θ)} = Acos(ωt + θ) + jsin(ωt + θ)
So, v(t) = Acos(ωt + θ) = Re(Ae^{j(ωt + θ)}) = Re(Ae^jθe^jωt) -------(5)
Equation(5) indicates that the general representation of a sinusoid in the time domain can be represented by the real component of a complex vector having argument (angle) = (ωt + θ) and magnitude = peak amplitude of the sinusoid.
If the real notation (Re) and the complex exponential e^jωt are removed from equation(5), we have equation (2), the complex phasor or frequency domain of v(t) = Acos(ωt + θ):
V(jω) = Ae^jθ = A<θ (phasor form). Phasors as previously noted, are mathematical formulations used to simplify analyses. Also the frequency component of equation(5) that was removed for simplification is not discarded. It is implicit in the phasor formulation

Now, define the complex frequency s = σ + jω, and the complex function F(s) = Ae^ste^jθ = Ae^σte^{j(ωt + θ)} ------(6)
Equation (6) is the result of substituting s for jω in equation (5)
So, Re(F(s)) = e^σtRe(Ae^{j(ωt + θ)}) = Ae^σtcos(ωt + θ) = equation (4).
So, essentially the complex frequency domain consists of a damping component and a phasor component.
When σ = 0 (steady-state sinusoid), F(s) = complex frequency domain, V(s). Obtained by substituting s for jω in V(jω) = A<θ. This phasor form is similar to the Laplace transform with respect to circuit analysis.

PjProblemstrings Sequence
S₇P₁A₁₇: establishes the time domain
S₇P₂A₂₁: establishes the identity of v(t)
S₇P₃A₃₂:mathematical action pushes v(t) into frequency domain
S₇P₁A₁₇: frequency domain
S₇P₅A₅₁: change in identity
S₇P₂A₂₁: new mathematical identity in the frequency domain
Thus, PjProblemstrings Sequence is:
S₇P₁A₁₇S₇P₂A₂₁S₇P₃A₃₂S₇P₁A₁₇S₇P₅A₅₁S₇P₂A₂₁

(di) complex frequency s = σ + jω
Examining 5e^-4t in the context of Ae^ste^jθ, indicates σ = -4 and ω = 0
So, s = -4.

(ii) Examining cos2ωt in the context of Ae^ste^jθ, indicates s = j2ω

(iii) sin(ωt + 2θ) = cos(ωt + (2θ - 90^o))
Examining cos(ωt + (2θ - 90^o)) in the context of Ae^ste^jθ, indicates s = jω

(iv) Examining 4e^-2tsin(3t - 50^o), indicates s = -2 + j3

(v) Examining 4e^-3t(2 + cos4t), indicates s = -3 + j4

(ei) v(t) = 5e^-2t, implies s = -2, A (peak magnitude) = 5 phase angle = 0.
So, s = -2 and V(s) = 5<0^o

(ii) v(t) = 5e^-2tcos(4t + 10^o), implies s = -2 + j4, A (peak magnitude) = 5 phase angle = 10^o.
So, s = -2 + j4 and V(s) = 5<10^o

(iii) v(t) = 4cos(2t - 20^o), implies s = j2, A (peak magnitude) = 4 phase angle = -20^o.
So, s = j2 and V(s) = 4<-20^o

(fi) s= -2, V = 2<0^o; implies v(t) = 2e^-2t
(ii) s= j2, V = 12<-30^o; implies v(t) = 12cos(2t - 30^o)
(iii) s= -4 + j3, V = 6<10^o; implies v(t) = 6e^-4tcos(3t + 10^o)

(gi)