Lecture 9

# Lecture 9 - ECE 3101 Signals and Systems Reading Assignment...

This preview shows pages 1–7. Sign up to view the full content.

Lecture 9, 2/22/11 ECE 3101, Copyright P. B. Luh 1 Reading Assignment: Sections 3.7-3.8 Problem Set 3: Due today Problem Set 4: Due next Tuesday The next makeup class: Wed., 3/2/11, 6:30 pm, MSB 411 Also, I will be at NSF on Tue. 3/1/11. That class and the discuss session on Mon. 3/14/11 will be swapped Last Time: Classical Solution of Differential Equations y(t) = y ZI (t) + y ZS (t) = Σ c k e λ k t + h(t) u(t) = y n (t) + y φ (t) Q(D)y φ (t) = P(D)u(t) u(t) = (t r + α r-1 t r-1 +.. + α 0 )e ζ t y φ (t) = ( β r t r + β r-1 t r-1 +.. + β 0 )e ζ t The Method of Undetermined Coefficients ECE 3101 Signals and Systems

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Lecture 9, 2/22/11 ECE 3101, Copyright P. B. Luh 2 Time-Domain Analysis of LTID Systems y[k+N] + a 1 y[k+N-1] + … + a N-1 y[k+1] + a N y[k] = b 0 u[k+N] + b 1 u[k+N-1] + … + b N-1 u[k+1] + b N u[k] Q[E]y[k] = P[E]u[k] x[k+1] = Ax[k] + Bu[k], y[k] = Cx[k] + Du[k] Zero Input Response Q( γ ) = ( γ - γ 1 ) ( γ - γ 2 ) . . ( γ - γ N ) All roots are distinct: y 0 [k] = c 1 γ 1 k + c 2 γ 2 k + . . + c N γ N k With r-repeated root ( λ - λ 1 ) r : c 11 γ 1 k + c 12 k γ 1 k + . .+ c 1r k r-1 γ 1 k With complex roots | γ |e ± j β : c| γ | k cos( β k+ θ ) Examples
Lecture 9, 2/22/11 ECE 3101, Copyright P. B. Luh 3 Today 3.7 Unit Impulse Response 3.8 Zero State Response Convolution of Discrete-Time Signals Properties of Convolution 3.9 Classical Solution of Linear Difference Equations (Read yourself) Next Time: 4.1 The Laplace Transform 4.2 Properties of the Laplace Transform

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
4 Unit Impulse Response Unit impulse response: Response of the system to δ [k] when the system is at rest ( E N + a 1 E N-1 + . . + a N-1 E + a N )y[k] = ( b 0 E N + b 1 E N-1 + . . + b N-1 E + b N )u[k] (Q[E]y[k] = P[E]u[k]) with u[k] = δ [k], and h[-1] = h[-2] = . . = h[-N] = 0 Example: Find h[0], h[1], and h[2] δ [k] Input System Output = = δ otherwise 0 0 k 1 ] k [ k δ [k] 1 h[k] y[k] = 0.5y[k-1] + u[k] h[0] = 1, h[1] = 1/2, h[2] = 1/4 ~ Bootstrapping In fact, h[k] = (0.5) k U[k] For a general problem, how to find h[k]? u[k] + z -1 y[k] + 0.5 y[k-1]
5 Recall that for the continuous-time case : h(t) ~ b 0 δ (t) + zero input response due to IC at 0 + h(t) = b 0 δ (t) + [P(D)y n (t)]U(t), where y n (t): Linear combination of characteristic modes with y n (N-1) (0) = 1 and all other ICs = 0 How about for discrete-time systems? (E N + a 1 E N-1 + . . + a N-1 E + a N )y[k] = (b 0 E N + b 1 E N-1 + . . + b N-1 E + b N )u[k] Impulse may come through. The response afterward is then caused by a linear combination of characteristic modes IWBS that if a N 0 , then h[k] = (b N /a N ) δ [k] + y c [k]U[k] y c [k]: linear combination of characteristic modes with coefficients determined by h[0], h[1], . ., h[N-1]

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
6 Proof Suppose that h[k] = A 0 δ [k] + y c [k]U[k], want to find A 0 Q[E]y[k] = P[E]u[k] Q[E](A
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 24

Lecture 9 - ECE 3101 Signals and Systems Reading Assignment...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online