Unit Impulse Function Lesson #2 2CT.2,4, 3CT.2 Appendix A BME 333 Biomedical Signals and Systems - J.Schesser 17 Complex Numbers • Constants: s = a + jb Rectangular Form Complex Plane a is called the Real part of s b is called the Imaginary part of s = a 2 + b2 e Imaginary axis b tan-1(b/a) a −1 j tan ( b ) a = a 2 + b 2 ∠ tan −1 ( b ) Polar Form a • Functions: Example : e jω t Imaginary axis Real axis Rotating Unit Vector at rate w = cos ω t + j sin ω t jθ (recall: e = cosθ + j sin θ ) BME 333 Biomedical Signals and Systems - J.Schesser Real axis 18 Complex Exponential Function as a function of time j 2π (1) t j 2πt z ( t ) = 1 e = e = cos 2πt + j sin 2πt • Let’s look at this t=2/8 seconds t=8/8 seconds arg(z(t))=2π x8/8 = 2π ; z(t)= 1+ j0 arg(z(t))=2π x2/8= π /2; z(t)= 0 + j1 t=3/8 seconds t=1/8 seconds Im{z} arg(z(t))=2π x3/8 = 3 π /4; arg(z(t))=2π x1/8=π/4; z(t)=0.707+j 0.707 z(t)= -0.707+ j0.707 t=0 seconds t=4/8 seconds 45o arg(z(t))=2π x0=0; z(t)=1+ j0 arg(z(t))=2π x4/8 = π; z(t)= -1+ j0 Re{z} t=5/8 seconds t=7/8 seconds arg(z(t))=2π x5/8 = 5π /4; arg(z(t))=2π x7/8= 7π /4; z(t)= -0.707 - j0.707 t=6/8 seconds z(t) = 0 .707- j0.707 arg(z(t))=2π x6/8 = 3π /2; z(t) = 0 - j BME 333 Biomedical Signals and Systems - J.Schesser 19 Phasor Representation of a Complex Exponential Signal • Using the multiplication rule, we can rewrite the complex exponential signal as z (t ) = Ae j (ωot +φ ) = Ae jωot e jφ = Ae jφ e jωot = Xe jωot = Xe j 2π Fot where X is a complex number equal to X = Ae jφ • X is complex amplitude of the complex exponential signal and is also called a phasor BME 333 Biomedical Signals and Systems - J.Schesser 20 Phasors • • Note that the real sinusoidal function f(t)=Acos (ωt+ϕ) can be represented by a complex function f(t)=A cos (ωt+ϕ) = Re[Ae j(ωt+ϕ)] Let’s represent this function by a phasor which is its magnitude and phase angle: f (t ) = A cos(ωt + φ ) = Re[ Ae j (ωt +φ ) ] = Re[ Ae jφ e jωt ] ⇒ A∠φ • Therefore, we can use phasors to represent complex functions which makes it easy to solve and calculate system solutions BME 333 Biomedical Signals and Systems - J.Schesser 21 Example Using ODE with Trigonometry • Let’s calculate the current I(t) assuming V(t)= A cos ωt dI (t ) = V (t ) = A cos ωt dt Use Trigonometric functions RI (t ) + L R dI (t ) = − I ω sin(ωt + θ ) dt RI cos(ωt + θ ) − I ω L sin(ωt + θ ) = A cos ωt Let I (t ) = I cos(ωt + θ ); I(t) V(t) L To solve for I and θ , use the identities: cos( A + B) = cos A cos B − sin A sin B; sin( A + B) = sin A cos B + cos A sin B RI [cos ωt cos θ − sin ωt sin θ ] − I ω L[sin ωt cos θ + cos ωt sin θ ] = A cos ωt 0 sin θ −ω L −ω L = tan θ = ⇒ θ = tan −1 ( ) cos θ R R A A RI cos θ − I ω L sin θ = A ⇒ I = = R −ω L R cos θ − ω L sin θ R − ω L( ) 2 2 2 2 R + (ω L) R + (ω L) RI [− sin θ ] − I ω L cos θ = 0 ⇒ R[sin θ ] = −ω L cos θ ⇒ I= A = R + (ω L) 2 2 R + (ω L) 2 I (t ) = A R + (ω L) 2 R θ -ωL R2 + (ω L)2 2 2 A R 2 + (ω L) 2 cos(ωt + tan −1 ( −ω L )) R MESSY !!!! BME 333 Biomedical Signals and Systems - J.Schesser 22 Example Using ODE with Complex Exponentials • Let’s calculate the current I(t) assuming R V(t)= A cos ωt dI (t) = V (t) = Acos ω t dt Use complex exponent functions RI (t) + L V(t) I(t) Let I (t) = I cos(ω t + θ ) = ℜe{Ie jθ e jω t };Let V (t) = Acos(ω t) = ℜe{Ae jω t }; L 0 dI (t) = jω Ie jθ e jω t dt RIe jθ e jω t + jω LIe jθ e jω t = Ae jω t RIe jθ + jω LIe jθ = A A Ie = = R + jω L A jθ I (t) = ℜe{ R + (ω L) 2 A R 2 + (ω L) 2 e 2 − j tan −1 e ωL R − j tan −1 ωL R e jω t } = A R 2 + (ω L) 2 cos(ω t − tan −1 BME 333 Biomedical Signals and Systems - J.Schesser ωL ) R 23 A Special Function – Unit Impulse Function • The unit impulse function, δ(t), also known as the Dirac delta function, is defined as: δ(t) δ(t) = 0 for t ≠ 0; = undefined for t = 0 and has the following special property: 0 -100 ∞ ∫ f (t)δ (t −τ )dt = f (τ ) −∞ ∞ ∴ ∫ δ (t)dt =1 −∞ BME 333 Biomedical Signals and Systems - J.Schesser -50 -25 -1 0 1 25 50 100 24 Unit Impulse Function Continued • A consequence of the delta function is that it can be approximated by a narrow pulse as the width of the pulse approaches zero while the area under the curve = 1 δ(t) lim δ (t) ≈1/ε for -ε / 2 < t < ε / 2; = 0 otherwise. ε →0 10 1 0.5 -1 BME 333 Biomedical Signals and Systems - J.Schesser -.5 -.05 .05 .5 1 25 Unit Impulse Function Continued ∞ ∫ f (t)δ (t − τ ) dt −∞ Let's approximate δ (t − τ ) with a pulse of height ∞ ∫ f (t)δ (t − τ ) dt ≈ ∫ τ −ε 2 −∞ τ τ +ε 2 1 and width ε ε 1/ ε 1 f (t) dt ε If we take the limit of this integral as ε → 0, τ - ε /2 τ τ + ε/2 the approximation integral approaches the original integral ∞ ∫ −∞ τ +ε 2 f (t)δ (t − τ ) dt = lim ∫ ε →0 τ −ε 2 1 1 f (t) dt → lim f (τ ) ε = f (τ ), ε ε ε →0 since as ε → 0, the integral is zero except at t = τ BME 333 Biomedical Signals and Systems - J.Schesser 26 Uses of Delta Function • Modeling of electrical, mechanical, physical phenomenon: – point charge, – impulsive force, – point mass – point light BME 333 Biomedical Signals and Systems - J.Schesser 27 Another Special Function – Unit Step Function • The unit step function, u(t) is defined as: u(t) = 1 for t ≥ 0; 1 = 0 for t < 0. t and is related to the delta function as follows: t u(t ) = ∫−∞ δ (τ )dτ BME 333 Biomedical Signals and Systems - J.Schesser 28 Integration of the Delta Function • δ(t) • u(t) • tu(t) . . . • u(t) tu(t) t 2 u(t) 2! t n u(t) n! 1st order 2nd order nth order BME 333 Biomedical Signals and Systems - J.Schesser 29 Signal Representations using the Unit Step Function • x(t) = e-σt 1.2 cos(ωt)u(t) 1 0.8 0.6 0.4 0.2 -0.5 0 -0.2 0 0.5 1 1.5 2 2.5 3 -0.4 -0.6 -0.8 • x(t) = t u(t) – 2 (t-1)u(t-1) + (t-2) u(t-2) 6 tu(t) 4 (t-2)u(t-2) 2 0 -2 -4 0 1 2 3 4 5 6 x(t) -2(t-1)u(t-1) -6 BME 333 Biomedical Signals and Systems - J.Schesser 30 Homework • Complex numbers – – – – – • Convert 1+j1 to its magnitude/angle representation (phasor) Convert 1/(1+j1) to a phasor Draw ejωt and ej(ωt+α) in the complex plane For the series R-L circuit in class, calculate the voltage across the inductor. Appendix A.4, A.7 Unit Impulse and Unit Step Functions – – – – – Using unit step functions, construct a single pulse of magnitude 10 starting at t=5 and ending at t=10. Repeat problem 1) with 2 pulses where the second is of magnitude 5 starting at t=15 and ending at t=25. Is the unit step function a bounded function? Is the unit impulse function a bounded function? 2CT.2.4a,b BME 333 Biomedical Signals and Systems - J.Schesser 31
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