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EECS 20 N—March 9, 2001 Lecture 21: Time Responses Laurent El Ghaoui 1 announcements: recall • reading assignment: Chapter 6 of Lee and Varaiya • no lab next week (3/12–3/16) • still, go to lab sessions for problem set discussions • this week’s lab due in two weeks only 2 outline • time-invariance • linearity • impulse response • response to sinusoids look at discrete-time systems only 3 delaying signals given a discrete-time signal, we can associate the same signal, delayed by a time T : a discrete−time sequence 5 4 response 3 2 1 0 0 1 2 3 4 5 6 7 8 9 time (seconds) a discrete−time sequence, delayed 5 original sequence 4 delayed sequence response 3 2 1 0 0 1 2 3 4 5 6 7 8 9 time (seconds) 4 delay operator formally, if x is a function of time, then we deﬁne a new signal DN (x) by ∀ n ∈ Ints, DN (x)(n) = x(n − N ) here, N ∈ Ints is the delay DN is called the delay operator 5 time-invariance if y is the zero-state response to input x of the system s(n + 1) = As(n) + bx(n), n = 0, 1, 2, . . . y(n) = cT s(n) + dx(n) then, for any N ∈ Ints, the response to DN x is DN y i.e., if we denote the zero-state response to input x by y = S(x), then S(DN (x)) = DN (S(x)) this property is called time-invariance 6 time-invariance and output response for the system (denoted S) s(n + 1) = As(n) + bx(n), n = 0, 1, 2, . . . y(n) = cT s(n) + dx(n) recall the zero-state output response formula: n−1 (S(x))(n) = y(n) = cT An−1−m bx(m) + dx(n) m=0 7 if we delay x by N , the output is delayed likewise: n−1−N DN (Sx)(n) = y(n − N ) = cT An−1−m bx(m) + dx(n − N ) m=0 n−1 = cT An−1−m bx(m − N ) + dx(n − N ) m=0 = S(DN (x))(n) we’ll see a more transparent proof next 8 example of a non-time-invariant system 1 if n = 0, y(n) = d(n)x(n), where d(n) = 2 if n = 1 0 if n > 1 is not time-invariant, example: n d x D1 (x) S(x) D1 (S(x)) S(D1 (x)) 0 1 1 0 1 0 0 1 2 2 1 4 1 2 2 0 3 2 0 4 0 3 0 4 3 0 0 0 4 0 5 4 0 0 0 9 a general result the system with time-varying state-space matrices s(n + 1) = A(n)s(n) + b(n)x(n), n = 0, 1, 2, . . . y(n) = c(n)T s(n) + d(n)x(n) is in general not time-invariant when A(n), b(n), c(n) and d(n) are independent of n, then the system is time-invariant 10 impulse response impulse response is the zero-state response with impulse input: 0 if n < 0 δ(n) = 1 if n = 0, 0 if n > 0 we get 0 if n < 0, n−1 y(n) = cT An−1−m bδ(m) + dδ(n) = d if n = 0, m=0 cT An−1 b if n > 0 11 convolution sum the zero-state output response n−1 y(n) = cAn−1−m bx(m) + dx(n) m=0 can be written as a convolution sum n−1 m=+∞ y(n) = h(n − m)x(m) = h(n − m)x(n) m=0 m=−∞ where h is the sequence 0 if n < 0, h(n) = d if n = 0, cAn−1 b if n > 0 h is the impulse response! 12 result for a linear, time-invariant system, the zero-state response to an arbitrary input is the convolution of the input with the impulse response n−1 m=+∞ y(n) = h(n − m)x(m) = h(n − m)x(n) m=0 m=−∞ we denote this with the convolution operator: y=h x what does this mean? 13 LTI systems and convolution assume we are given a discrete-time system y = S(x) that satisﬁes the following properties: • it is linear • it is time-invariant we say the system is LTI given a signal x, what is the form of the output? claim: the output signal can be written as y = h x, where h is the impulse response 14 to get the idea start with x(0), what is y(0)? by linearity, there exist a number h(0) (independent of x(0)!) such that y(0) = h(0)x(0) now, what is the response at time n = 1? by linearity, there exist numbers h(1) and h(0, 1) (independent of x(0), x(1)!) such that y(1) = h(0, 1)x(1) + h(1)x(0) by time-invariance, we must have h(0, 1) = h(0), so that y(1) = h(0)x(1) + h(1)x(0) similarly, at time n = 2 y(2) takes the form y(2) = h(0)x(2) + h(1)x(1) + h(2)x(0) and so on . . . 15 signals as sums of impulses let’s take an arbitrary signal x, can decompose it in a series of impulses: +∞ x(n) = x(m)δ(n − m) m=−∞ this is a weighted sum of delayed impulses 16 response to a delayed impulse take the delayed impulse as input: 0 if n < m, (Dm δ)(n) = δ(n − m) = 1 if n = m, 0 if n > m what is the output response? 17 delayed impulse response by time-invariance, the answer is: simply delay the impulse response! thus, the response to the delayed impulse (with delay m) is (Dm h)(n) = h(n − m) by linearity, response to arbitrary input x is weighted sum of delayed impulse responses +∞ y(n) = m=−∞ x(m)(Dm h)(n) +∞ = m=−∞ x(m)h(n − m) the ﬁnal result: y = h x, uses both time-invariance and linearity 18 complex numbers a complex number is one of the form z = a + i · b, √ where a, b are reals, and i = −1 (so that i2 = −1) a is called the real part of z, b the imaginary part the magnitude of z is deﬁned by |z| = a2 + b 2 it is the Euclidean distance of the 2-D vector a b 19 polar representation of complex numbers √ let |z| = a2 + b 2 , a b z = a + i · b = |z| · +i· , |z| |z| since 2 2 a b a2 + b 2 + = = 1, |z| |z| |z|2 any such number can be represented as z = |z| · (cos θ + i · sin θ), where θ is such that a b cos θ = , sin θ = |z| |z| 20 complex exponentials we can represent a complex number as z = ρeiθ where • ρ = |z| is the magnitude • the complex exponential is deﬁned by eiθ = cos θ + i · sin θ where +∞ un eu = n=0 n! 21 sinusoidal inputs a sinusoidal input is one of the form x(n) = sin(nω + φ), where φ ∈ [0 2π[ is the phase, and ω ∈ Reals is the frequency. can also use a cos in the above (change φ) 22 complex exponentials it is more convenient to work with complex exponentials, for example x(n) = einω or, with a non-zero phase: x(n) = ei(nω+φ) = cos(nω + φ) + i · sin(nω + φ) these signals model a lot of real-life signals (more later) 23 response to a complex exponential consider the output response: n−1 m=+∞ y(n) = h(n − m)x(m) = h(n − m)x(n) m=0 m=−∞ what is the response to a complex exponential x(n) = einω ? 24 response to complex exponential m=+∞ imω y(n) = m=−∞ h(n − m)e m=+∞ i(m−n)ω = m=−∞ h(n − m)e einω = H(w)einω , where H(w) is the complex number m=+∞ H(w) = m=−∞ h(n − m)ei(m−n)ω m=+∞ = m=−∞ h(m)eimω H is called the frequency response 25

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