Lecture 8: Signal Reconstruction, DT vs CT Processing. 8.1 Reconstruction of a Band-limited Signal from its Samples

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1 EE518 Digital Signal Processing University of Washington Autumn 2001 Dept. of Electrical Engineering Lecture 8: Signal Reconstruction, D vs C Processing Oct 24, 2001 Prof: J. Bilmes <bilmes@ee.washington.edu> A: Mingzhou Song <msong@u.washington.edu> 8.1 Reconstruction of a Band-limited Signal from its Samples he ideal reconstruction is show in the block diagram Fig 8.1(a). It can also be simplified as a discrete-to-continuoustime (D/C) converter as shown in Fig 8.1(b). Conversion from impulse train to discrete time sequence s Ideal Reconstruction Filter r D/C r (a) (b) Figure 8.1: Reconstruction In the D/C converter x s (t) = n= δ[t n ] (8.1) x r (t) = x s (t) h r (t) = [ n= δ[t n ] ] h r (t) = n= h r [t n ] (8.2) he ideal reconstruction filter is an ideal lowpass filter we have mentioned in previous lecture, i.e., Ω < Ωc H r ( jω) = 0 Ω Ω c Ω c is taken to be Ω s 2 = π, which is appropriate to avoid aliasing. hen, h r (t) = sin(πt/ ) πt/ he sinc function can be considered as a form of smoothing or interpolation. So, x r (t) = n= sin[π(t n )/ ] π(t n )/ (8.3) (Interpolation) (8.4) Q: what happens to the points of samples, i.e., x c (n )? By L Hopital s rule, h r (0) = 1. Also h r (n ) = 0 for n = ±1,±2,. So x r (m ) = x[m] = x c (m ) 8-1

2 8-2 which says at t = n where samples are acquired, the reconstruction value is always the same as the original value. But at other locations along the time axis, x r (t) is the same with x c (t) only when there is no aliasing. Frequency Representation of x r (t) Let X r ( jω) be the F of x r (t). By Eq. (8.2), X r ( jω) = n= =H r ( jω) H r ( jω)e jnω e jnω n= (8.5) =H r ( jω)x(e jω ) ω=ω =H r ( jω)x(e jω ) Interpretation: X(e jω ) is scaled on frequency axis and then the base period is selected by the lowpass filter H r ( jω). he band of X r ( jω) is always limited to Ω c of H r ( jω) regardless of whether there is aliasing or not. If Ω c > Ω, there is no damage done. 8.2 Discrete Processing of Continuous-ime Signals We can first sample a continuous signal, process it in discrete-time, and then reconstruct another continuous time signal, as shown in Fig 8.2. Even though the process is done in discrete-time domain, the entire system is equivalent to a continuous time system. he following is a summary of the mathematical representations of converter and c Discrete ime System D/C y (t) r Figure 8.2: Discrete-time processing of continuous time signals. D/C converter: For converter: =x c (n ) X(e jω ) = 1 k= X c ( j(ω kω s )) Ω= ω For D/C converter: y r (t) = n= sin[π(t n )/ ] π(t n )/ Y r ( jω) = H r ( jω)y (e jω ) = Y (e jω ) Ω < π 0 otherwise When the discrete system is LI, the frequency domain analysis can have simple and useful results compared with pure time analysis. Let H(e jω ) be the frequency response of an LI discrete time system, i.e. Y (e jω ) = H(e jω )X(e jω )

3 8-3 then Y r ( jω) = H r ( jω)h(e jω )X(e jω ) = H r ( jω)h(e jω ) 1 X c ( j(ω 2πk k= )) When X c ( jω) = 0 for Ω π, no aliasing occurs, only the k = 0 term in the above summation is selected by H r( jω) of the lowpass filter. hen Let herefore Y r ( jω) = H(e jω )X c ( jω) Ω < π 0 Ω π H e f f ( jω) = H(e jω ) Ω < π 0 Ω π Y r ( jω) = H e f f ( jω)x c ( jω) which means the effective continuous time system is also LI, whose frequency response is H e f f ( jω). In the derivation, we assumed the discrete time system is LI and also we assumed no aliasing occurred. Question: What are sufficient conditions for the effective continuous system to be LI? 1. he discrete time system has to be LI. 2. Input must be bandlimited. 3. Sampling rate must be high enough above Nyquist rate so that there is no aliasing occur. Ex: Ideal continuous time lowpass filter using discrete time lowpass filter. Let For band-limited inputs sampled above Nyquist rate, H(e jω 1 ω < ωc ) = 0 ω c ω π H e f f ( jω) = 1 Ω < ω c 0 Ω ω c he condition for the effective continuous lowpass to be valid is: which guarantees no aliasing occurs when ω < ω c. 2π Ω N > ω c Note: when ω c < Ω N, the condition allows aliasing to occur in ω c < ω < π. Hence it is less strict than what is required, i.e., 2π Ω N > Ω N, for fully reconstruction of the input signal. It makes sense because we can think of lowpass as a partial reconstruction of the input signal, and consequently only a relaxed condition is needed. Ex: consider identity =. If the input is x c (t) = δ(t), then = δ[n]. However, if the input is shifted by half a period, i.e., x c (t) = δ(t 2 ), then = 0, n. herefore the overall system is not time-invariant. he reason is that x c (t) = δ(t) is not bandlimited. Ex: consider aliasing if here exists an aliased component. LI systems 1. do not generate newer frequency components, X c ( jω) 0, 2. only attenuate magnitude and perform phase shift, Ω > π

4 8-4 based on the eigen property of continuous-time systems: y(t) = H(s)e st he normal conditions for no aliasing is 2π Ω N > Ω N Ω N < π/ Now, because of LPF H(e jω ), we can allow some aliases which will eventually be filtered out. hen 2π Ω N > ω c Ω N < (2π ω c )/ i.e., we can have aliases in ω c < ω < π Impulse Invariance We have shown that a D system plus and D/C converters can be an effective C system, i.e. H e f f ( jω) = H(e jω ) Ω < π 0 Ω π Now, if given a bandlimited continuous system with frequency response H c ( jω), how do we choose the D system frequency response H(e jω )? From the above equation, we can let H(e jω ) = H c ( j ω ) ω < π and choose such that H c ( jω) = 0 for Ω π. hen, what is the relationship between h[n] and h c (t)? Since H(e jω ) is a scaled version of H c ( jω), and recall that converter produces a scaled version of the Fourier transform of the input when sampled above Nyquist rate, so we guess that h[n] is close to the sampled version h g [n] of h c (t), i.e., hen we get H g (e jω ) = 1 h g [n] = h c (n ) H c ( j( ω k= 2πk )) if h c (t) is sampled at above Nyquist rate, there will be no aliasing, then H g (e jω ) = 1 H c( j ω ) ω < π Since H(e jω ) = H c ( j ω ), H(e jω ) = H g (e jω ) that is, h[n] = h g [n] = h c (n ) h[n] is called an impulse-invariance version of the continuous system. So the impulse response is invariant between discrete and continuous time systems. Under right conditions: 1. H is bandlimited, 2. is such that no aliasing occurs,

5 8-5 δ(t) H(jΩ) h(t) h[n] Figure 8.3: Reverberation simulation. Big room (European cathedral) Source near field microphone far field microphone x(t) y(t) F X(e^jw) divide Y(e^jw) F H(e^jw) Figure 8.4: Reverb in a room. just sample continuous the system to get a discrete time system. Ex: reverb in a room. See Fig. 8.3 and Fig Ex: Discrete-time lowpass obtained by impulse invariance. 1 Ω < Ωc H c ( jω) = 0 Ω Ω c which corresponds to impulse response which we can sample using such that Ω c < π h c (t) = sin(ω ct) πt and get where ω c = Ω c. So h[n] = h c (n ) = sin(nω c ) πn = sin(nω c) πn H(e jω ) = = n= h[n]e jnω 1 ω < ωc 0 ω c ω π Ex: Apply impulse invariance, we get h c (t) = Ae s0t L u(t) H c (s) = A s s 0 h[n] = h c (n ) = Ae s0n Z A u[n] H(z) = 1 e s0 z 1

6 8-6 hen H(e jω ) = A 1 e s0 e jω H c(s) s= j ω = A jω s 0 Why? Because H c ( jω) is not strictly band-limited signal. Hence H(e jω ) is an aliased version of H c ( jω). But when in fact H c ( jω) falls off at high frequency and the sampling rate is high enough, the aliasing is minimal. So impulseinvariance approach is used in D simulation of C system or design of digital filters. 8.3 Continuous Processing of Discrete-ime Signals his is the complementary of discrete processing of continuous-time signals. We can first convert a discrete time signal to a continuous signal, process it in continuous-time, and then sample the continuous output and get another discrete time signal, as shown in Fig 8.3. Even though the process is done in continuous-time domain, the entire system is equivalent to a discrete time system. D/C c Continuous ime System y (t) c Figure 8.5: Continuous processing of discrete-time signals. In the block diagram, x c (t) = y c (t) = n= n= sin[π(t n )/ ] π(t n )/ sin[π(t n )/ ] π(t n )/ (8.6) (8.7) (8.8) and X c ( jω) = X(e jω ) Ω < π (8.9) Y c ( jω) = H c ( jω)x c ( jω) Ω < π (8.10) Y (e jω ) = 1 Y c( j ω ) = 1 H c( jω/ ) X(e jω ) = H c ( jω/ )X(e jω ) ω < π (8.11) (8.12) i.e., or H(e jω ) = H c ( jω/ ) ω < π (8.13) H c ( jω) = H(e jω ) Ω < π (8.14) Ex: Non-integer delay. Given H(e jω ) = e jω ω < π

7 8-7 When Z, = x[n ], which is an ideal integer delay. When Z, = x[n ] is undefined. If we design a continuous time system and let H c ( jω) = H(e jω ) = e jω we get y c (t) = x c (t ) Now we can get = y c (n ) which is not undefined any more, i.e., = y c (n ) = x c (n ) sin[π(t k )/ ] = x[k] k= π(t k )/ = k= = = h[n] sinπ(n k) x[k] π(n k) sinπ(n ) π(n ) t=n herefore, when = n 0 is an integer, when is not an integer h[n] = δ[n n 0 ] h[n] = sinπ(n ) π(n )