Time Domain Analysis of Linear Systems Ch2 University of Central Oklahoma Dr. Mohamed Bingabr
Outline Zero-input Response Impulse Response h(t) Convolution Zero-State Response System Stability
System Response System x(t) = 0 Initial Conditions are not zeros (stored energy) y(t) = zero input response x(t) System Initial Conditions are zeros (No stored energy) y(t) = zero state response Total response = Zero input response + Zero state response
System Response dd 2 yy dddd 2 + RR LL dddd dddd + 1 LLLL yy = 1 dddd LL dddd What is the zero-input response? dd 2 yy dddd 2 + BB MM dddd dddd + kk MM yy = 1 MM xx DD 2 + RR LL DD + 1 LLLL yy = 1 LL DDDD DD 2 + BB MM DD + kk MM yy = 1 MM xx For N order system DD NN + aa 1 DD NN 1 + aa 2 DD NN 2 + + aa NN yy = bb 0 DD MM + bb 1 DD MM 1 + bb 2 DD MM 2 + + bb MM xx
Zero-Input Response DD NN + aa 1 DD NN 1 + aa 2 DD NN 2 + + aa NN yy(tt) = 0 A solution to the above differential equation is yy tt = ccee λλλλ DDDD = λλccee λλλλ DD 2 yy = λλ 2 ccee λλλλ DD NN yy = λλ NN ccee λλλλ λλ NN + aa 1 λλ NN 1 + aa 2 λλ NN 2 + + aa NN ccee λλλλ = 0 Characteristic Polynomial QQ λλ Characteristic Equation QQ λλ = 0 λλ NN + aa 1 λλ NN 1 + aa 2 λλ NN 2 + + aa NN = 0
Zero-Input Response Next step is to solve the characteristic equation λλ NN + aa 1 λλ NN 1 + aa 2 λλ NN 2 + + aa NN = 0 λλ λλ 1 λλ λλ 2 λλ λλ NN = 0 If the characteristic roots λs are distinct real then a possible solutions to the differential equation are cc 1 ee λλ 1tt cc 2 ee λλ 2tt cc NN ee λλ NNtt A general solution of the zero-input response is yy tt = cc 1 ee λλ1tt + cc 2 ee λλ2tt + + cc NN ee λλ NNtt
Zero-Input Response The zero-input response is yy tt = cc 1 ee λλ 1tt + cc 2 ee λλ 2tt + + cc NN ee λλ NNtt Characteristic Roots: λ 1 λ N - Also called the natural frequencies or eigenvalues - For stable system all λ 0 - Can be real distinct, repeated roots, and/or complex roots Characteristic Modes: cc 1 ee λλ 1tt cc NN ee λλ NNtt, determine the system s behavior The constants c 1,, c n are arbitrary constants that will be determined by the initial conditions (initial states of the system)
Example For the LTI system described by the following differential equation (D 2 + 3D + 2) y(t) = D x(t) The initial conditions are y(0) = 0 and yy 0 = 5 Find a) The characteristic equation b) The characteristic roots c) The characteristic modes d) The zero-input response Answer: yy tt = 5ee tt + 5ee 2tt
Repeated Characteristic Roots If the characteristic equation λλ λλ 1 λλ λλ 2 λλ λλ NN = 0 has this form λλ λλ NN 1 = 0, then the N characteristic roots has the same value λ 1. The zero-input solution will be yy tt = cc 1 ee λλ 1tt + cc 2 ttee λλ 1tt + cc 2 tt 2 ee λλ 1tt + + cc NN tt NN 1 ee λλ 1tt
Example For the LTI system described by the following differential equation (D 2 + 6D + 9) y(t) = (3D +5) x(t) The initial conditions are y(0) = 3 and yy 0 = 7 Find a) The characteristic equation b) The characteristic roots c) The characteristic modes d) The zero-input response Answer: yy tt = 3ee 3tt + 2tttt 3tt
Complex Characteristic Roots If the characteristic equation λλ λλ 1 λλ λλ 2 λλ λλ NN = 0 has a complex characteristic root σ + jω then its conjugate σ - jω is also a characteristic root. This is necessary for the system to be physically realizable. The zero-input solution for a pair of conjugate roots is yy tt = cc 1 ee σσ+jjjj tt + cc 2 ee σσ jjjj tt If c 1 and c 2 are complex then c 2 is conjugate of c 1. yy tt = 0.5ccee jjθθ ee σσ+jjjj tt + 0.5ccee jjθθ ee yy tt = ccee σσtt cccccc ωωωω + θθ σσ jjjj tt
Example For the LTI system described by the following differential equation (D 2 + 4D + 40) y(t) = (D +2) x(t) The initial conditions are y(0) = 2 and yy 0 = 16.78 Find a) The characteristic equation b) The characteristic roots c) The characteristic modes d) The zero-input response Answer: yy tt = 4ee 2tt cos(6tt ππ 3 )
Meanings of Initial Conditions If the output of a system y = 5x + 3 then dy/dx = 5. To find y from the differential equation dy/dx we integrate both side but the answer will be y = 5x + C. The constant C is the value of y when x was zero. To find C we need an initial condition to tell us what is the value of y when x was zero. For N th order differential equation we need N initial conditions (auxiliary conditions) to solve the equation.
The Meaning of t = 0 - and t = 0 + t = 0 x(t) System y(t) y(0 - ): At t = 0 - the input is not applied to the system yet so the output is due only to the initial conditions. y(0 + ): At t = 0 + the input is applied to the system so the output is due the input and the initial conditions.
Impulse Response h(t) x(t) = δ(t) System y(t) = h(t)
Unit Impulse Response h(t) Reveal system behavior Depends on the system internal characteristic modes Helps in finding system response to any input x(t)
Derivation of Impulse Response h(t) 1- Find the differential equation of the system DD NN + aa 1 DD NN 1 + + aa NN yy = bb 0 DD MM + bb 1 DD MM 1 + + bb MM x QQ DD yy tt = PP DD xx(tt) 2- Find the natural response y n (t) using the same steps used to find the zero-input response. yy nn tt = cc 1 ee λλ 1tt + cc 2 ee λλ 2tt + + cc NN ee λλ NNtt 3- To find the constants c, set all initial conditions to zeros except N-1 derivative, set it to equal 1. yy nn 0 = yy nn 0 = yy nn 0 = = yy NN 2 nn 0 = 0 yy NN 1 nn 0 = 1 4- h tt = PP DD yy nn tt uu(tt) If M = N then h(t) = b 0 δ(t)+ [P(D) y n (t)] u(t)
Example Determine the impulse response h(t) for the system (D 2 + 3D + 2) y(t) = (D +2) x(t) HW3_Ch2
Zero-State Response x(t) h(t) y(t) = x(t) * h(t) convolution
Zero-State Response Any signal can be represented as a train of pulses of different amplitudes and at different locations xx tt = lim Δττ 0 nn xx nn ττ pp tt nn ττ xx tt = lim Δττ 0 nn xx nn ττ pp tt nn ττ ττ ττ xx tt = lim Δττ 0 nn xx nn ττ δδ tt nn ττ ττ
Zero-State Response
Zero-State Response lim Δττ 0 nn x(t) xx nn ττ δδ tt nn ττ ττ lim Δττ 0 nn y(t) xx nn ττ h tt nn ττ ττ y(t) is the sum of all curves. Each curve represent the output for one value of n yy tt = lim Δττ 0 nn xx nn ττ h tt nn ττ ττ = xx ττ h tt ττ dddd
Zero-State Response The Convolution Integral y(t) = x(t) * h(t) x(t) h(t) yy tt = xx ττ h tt ττ dddd y(t) = h(t) * x(t) yy tt = h ττ xx tt ττ dddd
Example For a LTI system with impulse response h(t) = e -2t u(t), determine the response y(t) for the input x(t) = e -t u(t). Answer: yy tt = (ee tt ee 2tt )uu(tt) =
The Convolution Properties Commutative: x 1 * x 2 = x 2 * x 1 Associative: x 1 * [x 2 * x 3 ] = [x 1 * x 2 ] * x 3 Distributive: x 1 * [x 2 + x 3 ] = [x 1 * x 2 ] + [x 1 * x 3 ] Impulse Convolution: x(t) * δ(t) = xx ττ δδ tt ττ dddd = x(t)
The Convolution Properties Shift Property: if x 1 (t) * x 2 (t) = c(t) then x 1 (t -T) * x 2 (t) = x 1 (t) * x 2 (t -T) = c(t -T) also x 1 (t T 1 ) * x 2 (t T 2 ) = c(t T 1 T 2 ) Width Property: if the width of x 1 (t) is T 1 and the width of x 2 (t) is T 2 then the width of x 1 (t ) * x 2 (t) is T 1 +T 2
Natural Response Forced Response Example Find the total response for the system D 2 y + 3Dy + 2y = Dx for input x(t)=10e -3t u(t) with initial condition y(0) = 0 and y ( 0) = 5 Answer y( t) ( t 2t ) ( t 2t 3t 5e + 5e + 5e + 20e 15e ) = For t 0 Zero-input Response Zero-state Response y( t) = ( t 2t ) ( 3t 10e + 25e + 15e )
External Stability (BIBO) System Stability If the input is bounded then the output is bounded. h ( τ ) dτ < Internal Stability (Asymptotic) If and only if all the characteristic roots are in the LHP Unstable if, and only if, one or both of the following conditions exist: At least one root is in the RHP There are repeated roots on the imaginary axis Marginally stable if, and only if, there are no roots in the RHP, and there are some unrepeated roots on the imaginary axis.
Example Investigate the asymptotic & BIBO stability of the following systems: a) (D+1)(D 2 + 4D + 8) y(t) = (D - 3) x(t) b) (D-1)(D 2 + 4D + 8) y(t) = (D + 2) x(t) c) (D+2)(D 2 + 4) y(t) = (D 2 + D + 1) x(t) HW4_Ch2
Example Find the zero-state response y(t) of the system described by the impulse response h(t) for an input x(t), shown below. 4 h(t) 2 x(t) 2 0 4
Convolution Demonstration h(τ) x(t-τ) 4 t - 4 t 2 τ