2. CONVOLUTION. Convolution sum. Response of d.t. LTI systems at a certain input signal
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1 2. CONVOLUTION Convolution sum. Response of d.t. LTI systems at a certain input signal Any signal multiplied by the unit impulse = the unit impulse weighted by the value of the signal in 0: xn [ ] δ [ n] = x[ 0] δ[ n] Delayed unit impulse: x[ n] δ[ n k] = x[ k] δ[ n k] Any signal can be "broken" into a sum of shifted and weighted unit impulses [ ] = [ ] δ[ ] xn xk n k k = 2
2 x n x k n k k = 3 [ ] = [ ] δ[ ] The impulse response: response to δ[n] δ[n] h[n] S d LTI system δ[n-k] h[n-k] S d LTI system 4 2
3 Convolution sum [ ] [ ] [ ] y n = x k h n k = k = [ ] [ ] [ ] [ ] = xn hn= hn xn x[n] y[n] S d LTI system 5 Example ) ( x y)[ n] = x[ k] y[ n k] k = two signals with finite durations N and N 2 Convolution - duration of N + N
4 Example 2) [ ] = n σ[ ] x n a n hn [ ] =σ[ n] σ[ n N 2 ] n+ k a 0 n N2 ( x h)[ n] = a = k = a n N2 k n N ( )[ ] 2 + a n N2 x h n = a = a k= n N2 + a 7 Causal Discrete-Time Signals If the input signal and the system are causal then the output signal is also causal. n x[ n] 0 and h[ n] 0, n< 0 y[ n] = x[ k] h[ n k] k = 0 If the system causal: [ ] [ ] [ ] [ ] hn 0, n< 0 hn= hn σ n, n Z [ ] = [ ] [ ] = [ ] [ ] y n x n k h k x k h n k k= 0 k= n 8 4
5 BIBO stability bounded input bounded output (BIBO) stable system BIBO stability condition A discrete time LTI system is stable if and only if its impulse response is absolutely summable k = [ ] < [ ] hk,hn l 9 The stability of the accumulator h[n] - the unit step σ[n], bounded signal finite difference eq. (x[n] causal) : [ ] [ ] [ ] [ ] [ ] yn= xn δ n yn= xk n k = 0 0 5
6 The corresponding output signal : y n [] n = = n + = k = 0 The output signal is not bounded. The accumulator is unstable. Used in practice, limited interval of summation n, or, from time to time, the output set on zero. 2 6
7 Properties of convolution Neutral element unit impulse δ[n] is neutral element. x[n]* δ[n] = x[n], for any signal x[n] δ[n] h[n] h[n] h[n] - the system s impulse response. 3 Identity system impulse response δ[n] input δ[n] output δ[n]. 4 7
8 Delay system Impulse respone h[n]=δ[n-n 0 ]. Response: a time shifted variant of the input signal [ ] δ [ ] = [ ] x n n n x n n Associativity. Series interconnection of 2 d.t. LTI systems two LTI systems h [n], h 2 [n] connected in series [ ] = [ ] [ ] [ ] = [ ] 2[ ] = ( )[ ] 2[ ] [ ] = [ ] ( )[ ] y n x n h n yn y n h n x h n h n y n x n h h n 2 6 8
9 Associativity The convolution is associative. ( h )[ n] h [ n] = x[ n] ( h [ n] h [ n] ) x 2 2 The response of the equivalent system at the unit impulse signal [ ] = [ ] [ ] hn h n h n 2 7 Stability for series interconnection of two 2 d.t. LTI systems By the cascade connection of two stable systems another stable system is obtained. h [] n l, h [ n] l ( h h )[ n] 2 2 l x[n] h [n] y [n] h2 [n] y[n] 8 9
10 Commutativity The convolution sum is commutative. The equivalent system is [ ] = [ ] [ ] = [ ] [ ] h n h n h n h n h n 2 2 their order is not important in the cascade connection!!! 9 The order in series connection x[n] h [n] y [n] h2 [n] y[n] x[n] h 2 [n] y 2 [n] h [n] y[n] 20 0
11 The Inverse System Two systems connected in series The output of the second system - the original input signal [ ] [ ] =δ[ ] hn h n n i 2 Inverse system - example delay system: h n [ ] = δ[ n ] inverse system of a delay system time shift in the other sense of the time axis. not a causal system n 0 [ ] = δ [ + ] h n n n 0 22
12 Distributivity. Parallel interconnection of dt LTI systems convolution is distributive with respect to addition x h n + x h n = x h + h n ( )[ ] ( 2)[ ] ( ( 2) )[ ] 23 y = = = Proof [] n = y [] n + y [ n] = ( x h )[ n] + ( x h )[ n] k = x k = x [] k h [ n k] + x[] k h [ n k] [] k ( h [ n k] + h [ n k] ) ( x ( h + h ))[] n 2 2 k = 2 2 = 2 = = 24 2
13 Response of a d.t. LTI system at the unit step. The unit step response The unit step response s[n]. [ ] = σ [ n] xn, [ ] [ ] [ ] σ [ ] [ ] yn= sn= hn n= hk [ ] = [ ] [ ] hn sn sn n k = 25 For causal systems: [ ] 0, for 0 [ ] [ ] hn n< sn= hk k = 0 For an accumulator the unit step response a ramp signal (see previous slides). n [ ] = σ [ ] [ ] = + xn n sn n 26 3
14 Finite Impulse (FIR) and Infinite Impulse (IIR) Response d.t. Systems There are two types of impulse responses with finite duration (FIR) and with infinite duration (IIR). 27 Finite Impulse (FIR) and Infinite Impulse (IIR) Response d.t. Systems FIR IIR b n, 0 n M = 0, otherwise [ ] a0 hn 28 4
15 FIR The finite difference equation N M [ ] = [ ] ayn k bxn k k k= 0 k= 0 k a FIR system has the property: a0 a a2 a N 0 and = =... = = 0 29 FIR the output signal using the finite diff.eq.: y M b a k [] n = x[ n k] k = 0 0 the output signal using convolution: y [] n = h[][ k x n k] k = 30 5
16 Identification : h [ k] FIR bk, k = 0,,..., M = a0 0, in rest The impulse response has finite duration M+ hence the name finite impulse response system 3 IIR systems If the last proposition is not verified, for example if a and a then we obtain a IIR system. 32 6
17 IIR systems - example [ ] 0.5 [ ] = [ ] y n y n x n [ ] = δ [ ] [ ] = [ ] x n n y n h n n=0 h[0]= n= h[]=0.5 n=2 h[]=0.5 2 n=3 h[]=0.5 3, and so on [ ] = 0.5 n σ [ n] hn initial condition y[-]=0 -infinite duration 33 Implementation of d.t. LTI systems described by finite differences equations with constant coefficients 34 7
18 Implementation D.t. LTI systems are mathematically described by finite difference equations with constant coefficients. Implemented using standard sub-systems: delay systems, multipliers with constant, adders. 35 Implementation st order LTI system a y n + a y n = b x n + b x n [] [ ] [ ] [ ] 0 0 z [ n] a delay sub-system: input signal x[n] into x[n-] two multipliers with constants b 0 and b :b 0 x[n] and b x[n-] one adder Implemented using the system a) 36 8
19 Direct form I Moving average part y a [] n = ( z[] n a y[ n ] ) 0 Autoregressive part 37 a [] n + a y[ n ] = b x[ n] + b x[ n ] 0 y
20 N th order system N M ak y k k = 0 k = 0 [ n k] = b x[ n k] ; z M [] n = bk x[ n k] k = 0 y N a = a0 k = a k [] n z[] n y[ n k] 0 delay systems, multipliers & adders 39 For a FIR discrete-time system : y a [] n = ( z[] n ) 0 sub-system 2 = multiplier with constant /a
21 FIR system - transversal form substitute the M adders from subsystem with a single adder with M inputs we have supposed that a 0 =. 4 Direct form II For st order LTI system direct form I: series interconnection of two sub-systems, and 2. in the series interconnection, the systems order is not important Reverse order and have an equivalent form of the direct form I : series interconnection of the sub-system 2 with the sub-system. 42 2
22 Direct form II Systems order is not important (series interconnection) Reverse order - series connection of subsystem 2 with subsystem Equivalent form of the direct form I : direct form II Remove redundant delay system 43 two times the same signal v[n-] -system a). So, a delay subsystem is redundant. Remove the delay subsystem implementation b). direct form II of a st order LTI d.t. system. minimum number of subsystems (delay systems, multipliers and adders)
23 Direct form II for N th order system M>N a N+, a N+2,, a M = 0 M<N b M+, b N+2,, b N = 0 45 We have supposed that M=N If M>N then we can suppose that a N+, a N+2,, a M = 0 and we can use the same implementation. If M<N then we can suppose that b M+, b N+2,, b N = 0 and we can use the same implementation
24 The convolution product. The response of c.t. LTI systems at a specified input signal 47 Convolution product. C.t. signals C.t. LTI system, mathematically described by the operator S. find the output signal y(t) when the input signal x(t) is known. x(t) S y(t) LTI system 48 24
25 Convolution product. C.t. signals Reminder (chp.): Filtering property of the Dirac unit impulse, δ(t) ( ) ( ) d = ( 0) ϕτδτ τ ϕ Function test x(τ), and as impulse the shifted version with t (time is τ) ( τ) δ ( τ) τ = ( ) x t d x t 49 () = ( ) ( ) x t x τ δ t τ dτ S y(t) the output signal { } () = S x( t) y t ( ) ( ) LTI system = S x( τ ) δ( t τ ) dτ const. const. { } = x τ S δ t τ dτ 50 25
26 Convolution product. Definition y t = x t h t = x τ h t τ dτ () () () ( ) ( ) convolution product of continuous-time signals x and h. compute the response of a given system, with known impulse response ( h ) to a known input signal ( x ) 5 Neutral element Neutral element for convolution: Dirac distribution, δ(t). () δ () = ( ), for any signal ( ) x t t xt xt 52 26
27 Commutativity With the notation: t u = τ, we have: ( f g)( t) = g( u) f ( t u) du = ( g f )( t) So, the convolution product is commutative almost everywhere. 53 Some remarks If the input signal x(t) L and h(t) L (the system is stable) Then the convolution exists and the output signal y(t)=x(t)*h(t) L 54 27
28 Some remarks If the input signal x(t) L 2 and the impulse response system h(t) L 2 then the convolution product x(t)*h(t) exists, is bounded and continuous 55 Some remarks If the input signal x(t) L 2 and the system is stable h(t) L Then the output signal has a finite energy y(t)=x(t)*h(t) L
29 ( )( ) Example ) ( ) = σ ( ) σ ( ) () = σ () σ ( ) f t t t T g t t t T length of the convolution s support = T +T 2 (sum of lengths of the supports of the two functions) 0, t < 0 t, 0 t < T2 f g t = T2, T2 t T T + T t, T t T + T 0, t > T+ T Denote g(-τ) = z(τ) Time-shifting of z(τ) with t to the right t < 0, f τ g t τ = 0; ( ) ( ) 2 ( )( ) 0 < t T, f g t = dτ = t; ( )( ) T t T, f g t = dτ = T ; 2 2 t T 0 ( )( ) T t T + T, f g t = dτ = T + T t; 2 2 t T 2 2 ( τ) ( τ) ( )( ) t > T + T, f f t = 0, f g t = 0. t t T
30 Example 2) T T f () t = σ t+ σ t ; 2 2 g t ; () = f () t t ( ); T ( f g)( t) = T σ ( t+ T) σ ( t T) 59 Example 3) f(0) ; f(t) and f(t) are even. function f belongs to L f () t =, f () t f () t? t t + = du π f () t dt = 2 = 2arctgu 2 0 = 2 = π < u f * f is convergent a.e.; but not for t=0 f f f f d = τ τ τ = d τ τ ( )( 0) ( ) ( ) ( τ + )
31 Example 4) T T f () t = σ t+ σ t ; 2 2 t () σ () 2 ( f g) L g t = a t, 0 < a< ; ( )( ) T t 2 a t. has infinite support. T t+ 2 T f g t = a σ t + 2 ln a T σ 2 6 t < -T/2, f*g(t) = 0. t > -T/2 partial superposition - for T/2 <t <T/2 : t T = = + T ln 2 a t+ t τ 2 ( f g)( t) a dτ a t > -T/2 complete superposition - for t T/2 : ( )( ) T 2 t τ f g t = a dτ = a a T ln 2 a T T t t
32 Example 5) the integrator Not bounded impulse response = σ(t). () =δ() () = () = δ( τ) τ=σ() step response = ramp signal: x t t y t h t d t ( σ σ)() t = σ ( τ) σ ( t τ) dτ = = t σ () t t t, t 0 0, t < The result is not bounded, so the integrator is not stable. Despite this fact, the integrators can be used in practice, but only for finite duration input signals (case in which the output is bounded)
33 Associativity The convolution is associative ( f () t g() t ) h( t) = f ( t) g( t) h( t) ( ) If. f, g, h L ; 2. f, g, h L loc and two of them have compact support; 3. f, g, h L loc and all three have like support a closed set included in [0,). 65 Unit step response of an LTI system response to the unit step signal. t () () () ( ) s t = h t σ t = h τ dτ Its derivative is the impulse response. s'( t) = h( t) 66 33
34 response of the system h to the ramp signal xt () = tσ () t y ( t) = S{ tσ ( t) } = h ( tσ ( t) ); The 2nd derivative of y(t) is h(t). ( ) () σ ( ) y" t = h t t " = h δ = h. () = σ () y ( t) = h( t) tσ ( t) x t t t S LTI system y ( t) = h( t) 67 find the impulse response h(t) of a system the derivation of its step response or the double derivation of its response to a ramp signal
35 The unit step response of a causal LTI system for a causal system, impulse response = zero for t<0 or h( t) = h( t) σ ( t) () () () ( ) s t = h t σ t = h τ dτ Causal input signal and system causal output y t () t = x( t τ ) h( τ ) 0 t 0 dτ 69 BIBO stability for continuous-time LTI systems bounded input bounded output (BIBO) stable system BIBO stability condition A continuous time LTI system is stable if and only if its impulse response is absolutely integrable ( τ) τ< () h d,h t L 70 35
36 An example the integrator () = () ; () = ( ) ht σ t L yt xτ dτ t () = ( τ) y t x dτ (input signal is causal) 0 () σ () () Bounded input x t = t unbounded output y t = dτ = t. () yt () If xt has finite duration then is bounded. t t 0 the integrator is not stable. It can be used in practice because all the practical signals have finite duration. 7 Practical significance of the convolution product s properties Equivalent system for series interconnection of 2 systems has impulse response h t = h t h t = h t h t () ( ) ( ) ( ) ( )
37 Inverse system. Identity system Two systems connected in series Output original input signal h(t)* h i (t) = δ(t) The system connected in series with h(t) and h i (t) is an identity system y(t)=x(t) 73 Parallel interconnection of LTI systems Equivalent system for parallel interconnection of 2 systems has impulse response ht = h t+ h t ( ) ( ) ( )
38 Implementation of c.t. LTI systems with linear differential equations & constant coefficients 75 Direct form II with differentiators. N k = 0 a k k d y dt k ( t) N d x( t) k = k = 0 b k dt direct form II: delay systems (discrete-time) differentiators (cont. time). difficult to construct differentiators in continuoustime k a N
39 Direct form II with integrators. it is preferable to use integrators. Integrate N times a differential equation an integral equation. With the notations y ( 0), y( ),..., y( N ), x( 0), x( ),..., x( N ) we obtain the integral equation : N ak k = 0 y ( N k )() t bk x( N k )(). t = N k = 0 77 N ( )() = 0 () N k () d x( t) k d y t a = b a 0 k k k k N k= 0 dt k= 0 dt y t y t, ()() = () () = ( ) y t y t σ t y τ dτ y t y t σ t σ t y τ dτ dτ ( )() = () () 2 () = ( )... ( )() = ( )() k k () = ( ) ( )() = () ()() = 0 () σ () 2 k 2 2 y t y t σ t... y τ dτ dτ... dτ dτ x t x t ; x t x t t,... t t k t τ τ τ τ 2 k k 78 39
40 79 Example for 2nd order c.t. system N ay t bx t N ( )() = n k k ( n k)() k k= 0 k= 0 LCy t RCy t y t x t ( )() + ()() + 0 ( 2)() = ( 2)(), 80 40
41 identification : a = ; a = RC ; a = LC b 0 = Using these coefficients we can write the corresponding integral equation we obtain the canonical form II with integrators. 82 4
42 Transversal structure for FIR systems continuous-time FIR systems have the impulse response : h () t h δ ( t kt ). = N k k= 0 implemented using the transversal structure 83 y y h () t = x() t h + x( t T ) h x( t NT ) N () t = x() t h δ( t kt ) = x() t h() t N () t = h δ( t kt ). k = 0 Transversal structure. k 0 k = 0 k, h N, 84 42
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