EE518 Digital Signal Processing University of Washington Autumn 2001 Dept. of Electrical Engineering Lecture 2: Properties of LTI Systems, LCCDE and Frequency Response Wed., Oct. 3, 2001
Prof: J. Bilmes
(Continuous From Lecture 1)
Ex: Is x[n] n ≥ 0 y[n] = −x[n] n < 0 time invariant? Let y[n] = T{x[n]} and x1[n] = x[n − nd] Then x1[n] n ≥ 0 x[n − nd] n ≥ 0 T{x[n − nd]} = T{x1[n]} = = −x1[n] n < 0 −x[n − nd] n < 0 But x[n − nd] n ≥ nd y[n − nd] = −x[n − nd] n < nd So T{x[n − nd]} 6= y[n − nd] i.e., T is not time-invariant. Ex: Is y[n] = logx[n] time-invariant? T{x[n − nd]} = logx[n − nd] and y[n − nd] = logx[n − nd] so T is time-invariant. Key Point: Linear and time-invariant are different properties of systems. We will study systems that are both.
More about Linear Systems
Consider response to δ[n − k] T{δ[n − k]} = h[n,k] k is for the fact that the system might respond differently at different times. Due to linear property,
T{δ[n − k1] + δ[n − k2]} = h[n,k1] + h[n,k2]
2-1 2-2
Hence, in general, ( ) T ∑δ[n − k] = ∑h[n,k] k k More over by linear property,
T {x[k1]δ[n − k1] + x[k2]δ[n − k2]} = x[k1]h[n,k1] + x[k2]h[n,k2] Hence, in general, ( ) T ∑x[k]δ[n − k] = ∑x[k]h[n,k] (2.1) k k If T is also time-invariant, delayed input produces same delayed output h[n − k]. It then follows ( ) T ∑x[k]δ[n − k] = ∑x[k]h[n − k] = y[n] = T{x[n]} (2.2) k k The above equation is convolution sum, written as x[n] ∗ h[n] 1 So a linear time-invariant system is equivalent to its impulse response.
Brief Note on Convolution
There is a nice figure (Figure 2.8) in O&S. Make sure you understand. Two ways to view convolution y[n] = ∑x[k]h[n − k] k
1. as a combination by sum of sequences, each a weighted shifted version of impulse response, shown in Figure 2.1.
h[n] h[n]
x[0]
0 0
h[n−1]
x[1] sequence combine {y[n]}
1
h[n−2] x[2]
2
Figure 2.1: Convolution: first way.
2. as a set of computations to compute each value of y[n], shown in Figure 2.2
1This is notational abuse. (x ∗ h)[n] might be better notation to represent the sequence that is the convolution of x[n] and h[n]. 2-3
x[n] x[k] corresponding to pointwise multiply 0 0 k −> and accumulate
y[n]
h[n] h[n−k]
0 0 n
Figure 2.2: Convolution: second way.
2.1 Properties of Linear Time-Invariant Systems
Causality: An LTI system is causal if h[n] = 0 for n < 0. See Figure 2.3. The system is not anticipating if h[n] = 0 for n < 0.
0 n
Figure 2.3: A causal LTI system.
Commutative: An LTI system is commutative because convolution is commutative, i.e., ∞ y[n] = ∑ x[k]h[n − k] k=−∞ ∞ = ∑ x[n − m]h[m](m = n − k) k=−∞ ∞ = ∑ h[m]x[n − m] k=−∞ See Figure 2.4.
x[n] h1 h2 y[n]
x[n] h2 h1 y[n]
Figure 2.4: LTI systems are commutative.
Distributive: See Figure 2.5. BIBO: Whether an LTI system is stable depends on h[n].
∞ ∞ ∞ ∞
|y[n]| = ∑ x[k]h[n − k] ≤ ∑ |x[k]||h[n − k]| ≤ ∑ Bx|h[n − k]| ≤ Bx ∑ |h[k]| k=−∞ k=−∞ k=−∞ k=−∞ 2-4
h1 x[n] + y[n]
h2
y[n] x[n] h1 + h2
Figure 2.5: LTI systems are distributative.
so, if Bx ≤ ∞ then the system is stable if h[n] is absolutely summable, ∞ ∑ |h[k]| < ∞ (2.3) k=−∞ which is actually a both sufficient and necessary condition. Ex: n y[n] = ∑ x[n] = x[n] ∗ h[n] k=−∞ Question: What is h[n]? n 1 n ≥ 0 h[n] = ∑ δ[n] = = u[n] 0 n < 0 k=−∞ ∞ ∞ Is this system stable? No, because ∑n=−∞ |h[n]| = Definition 2.1 (Finite Impulse Response, or FIR). h[n] has a finite number of non-zero samples.
Ex: h[n] = δ[n − nd] (ideal delay) h[n] = δ[n] − δ[n − 1] (Bach difference) 1 h[n] = (u[n + M1] − u[n − M2 − 1]) (Moving average) M1 + M2 + 1 All FIR systems are stable if the coefficients are finite.
∞ n2 ∑ |h[n]| = ∑ |h[n]| < ∞ n=−∞ n=n1 Definition 2.2 (Infinite Impulse Response, or IIR). h[n] has an infinite number of non-zero samples.
Ex: h[n] = u[n] h[n] = anu[n] ω h[n] = e j 0n h[n] = anu[n] is causal. It is stable when |a| < 1 such that ∞ 1 ∑ |a|n = < ∞ n=0 1 − |a| Notice h[n] never becomes exact zero. 2-5
2.2 Linear Constant Coefficient Difference Equations
They are useful for representing many LTI systems.
N M ∑ aky[n − k] = ∑ bmx[n − m] (2.4) k=0 m=0 which is discrete analog to: N dky(t) M dmx(t) ∑ ak k = ∑ bm m (2.5) k=0 dt m=0 dt so, for a particular input xp[n], the solution is
y[n] = yp[n] + yh[n] (2.6) where yp[n] is particular solution and yh[n] is homogeneous solution. The homogeneous equation is:
N ∑ aky[n − k] = 0 (2.7) k=0 It can be shown that the homogeneous solution takes a form of
N n yh[n] = ∑ Amzm (2.8) m=1 where Am (1 ≤ m ≤ N) are arbitrary values, zm (1 ≤ m ≤ N) are the roots of
N −k ∑ akz = 0 k=0
However, this system is underdetermined and needs additional N constraints, e.g., y[−1] = c1,y[−2] = c2,··· ,y[−N] = cN, which are auxiliary values of y but could be values for any time. y[n] can also be computed recursively: ( ) 1 N M y[n] = − ∑ aky[n − k] + ∑ bkx[n − k] (2.9) a0 k=1 k=0
The first term on the righthand side is past outputs, the second term on the righthand side is past and present inputs. So we can compute solution recursively (can do backward as well). Note: For causality, we must have h[n] = 0 for n < 0. So non-zero auxiliary values must be greater than zero for causality. For linearity, zero input produces zero output. So must have zero auxiliary conditions for all n. Time-invariance also depends on initial conditions.
Initial Rest Condition
If x[n] = 0 for n < n0 and y[n] = 0 for n < n0, then the system is linear, time-invariant and causal. 2-6
2.3 Frequency Domain Representation of Discrete-Time Signals and Sys- tems
Consider x[n] = e jωn as input,
∞ ∞ ∞ ∞ ! y[n] = ∑ x[k]h[n − k] = ∑ x[n − k]h[k] = ∑ h[k]e jω(n−k) = e jωn ∑ h[k]e− jωk k=−∞ k=−∞ k=−∞ k=−∞
Define ∞ jω − jωk H(e ) , ∑ h[k]e (2.10) k=−∞ Note: ω is an independent variable representing frequency. This notation will make more sense when we do Z- transform. Then y[n] = H(e jω)e jωn So e jωn is an eigenfunction 2 for the system. Similarly, applying system T{} to some inputs might produce outputs that are identical to multiplication of the inputs by a (complex) constant.
T{e jωn} = y[n] = H(e jω)e jωn where H(e jω) is eigenvalue and e jωn is eigenfunction. H(e jω) is the frequency response of the system by telling how the system response to a particular input frequency. Why is this important? Many signals can be represented as
∞ − jωkn x[n] = ∑ αke k=−∞ which is an infinite sum or as an integral 1 Z ∞ x[n] = X(e jωn)e jωndω 2π −∞ Since the system is linear, ∞ jωkn − jωkn T{x[n]} = ∑ αkH(e )e k=−∞
jωn Ex: ideal delay y[n] = x[n − nd]. When x[n] = e , ω ω ω y[n] = e j n−nd = e j n · e− j nd
In fact, ω ω H(e j ) = e− j nd when h[n] = δ[n − nd]. Note:
jω jω ω |H(e )| = 1 ]H(e ) = − nd jω jω where |H(e )| is the magnitude response and ]H(e ) is the phase response or phase shift. 2Eigenvectors and Eigenvalues in Linear Algebra: let A be a square matrix, x be a vector, λ be a scalar. If Ax = λx for some x and λ, then x is eigenvector and λ is eigenvalue for A. Applying the “system” of matrix multiplication to x is equivalent to multiplying x by a constant. 2-7
Periodicity of Frequency Response
H is periodic with period 2π. Since ∞ H(e jω) = ∑ h[k]e− jωk k=−∞ ∞ H(e j(ω+r2π)) = ∑ h[k]e− j(ω+r2π)k k=−∞ ∞ = ∑ h[k]e− jωk · e− jr2πk k=−∞ jω − jr2πk = H(e )(∵ e = 1)