Multistage Rate Change 1/12

Transcription

1 Multistage Rate Change 1/12

2 Motivation for Multi-Stage Schemes Consider Decimation: When M is large (typically > 10 or so) it is usually inefficient to implement decimation in a single step (i.e., in a single stage). x[n] The Culprit: Large M requires the LPF to have a stopband edge of θ s = π/m, which is small for large M Need a LPF with a very narrow passband Requires a long FIR filter Inefficient since long filters require a large # of multiplies Solution: If M can be factored into a product of integers (M = M 1 M 2 M 3 M p ). Then decimation by M can be done by: H 1 (z) M 1 y[n] H 2 (z) M 2 H p (z) M p 1 st Stage 2 nd Stage p th Stage 2/12

3 Trick to Get Efficiency from Multi-Stage The design of H 1 (z) (& other front-end stages) can be relaxed from what you would use for a single-stage design. Certainly, you need H 1 (z) to have θ s = π/m 1 > π/m so no aliasing occurs after M 1. But it is even better than that. Can let θ s > π/m 1 which lets some aliasing occur But only so much aliasing such that the aliasing that occurs gets suppressed by the next filter, H 2 (z) Higher Stopband Edge Shorter Filter More Efficient 3/12

4 Let s See Why for a 2-Stage Case Say that the signal x[n] has spectral content of worth only up to frequency θ = θ p < π/m with M = M 1 M 2. Single-Stage Method Suppose we decimate using a single-stage scheme: x[n] H(z) M y[n] Then we need H(z) Sharp Transition Long Filter θ p π/m π θ 4/12

5 Let s See Why for a 2-Stage Case (cont.) 2-Stage Method x[n] u[n] y[n] H 1 (z) M 1 H 2 (z) M 2 After H 1 (z) but before M 1 we need: H 1 (z) Slow Transition Short Filter θ p θ s,1 = ( 2 2M 1) π M π θ Same Passband as Single-Stage 5/12

6 Let s See Why for a 2-Stage Case (cont.) Let s see the impact of this slower transition on aliasing: After 1 st Filter H 1 (z) θ p ( 2M 2 1) π / M π θ After 1 st Dec H 1 (z) M 1 θ p π 2π θ 2π M 1 (2M = π / M 2 2 1) π / M M 1 (2M 2 1) π / M = 2π ( π / M 2 ) After 2 nd Filter H 2 (z) M 1 θ p π Stopband Suppresses Aliased Replica θ 6/12

7 Design Requirements So say you want to design a 2-stage multirate scheme instead of a 1-stage multirate scheme: If single stage, say the specs need to be: Passband Cutoff = θ p Stopband Cutoff = θ s Passband Ripple = δ p Stopband Level = δ s For a 2-stage scheme, our above results say we need: 1 st Stage θ p,1 = θ p δ p,1 = δ p /2 θ s,1 = (2M 2-1)π/M > θ s δ s,1 = δ s 2 nd Stage θ p,2 = M 1 θ p δ p,2 = δ p /2 θ s,2 = π/m 2 δ s,2 = δ s Passband Ripple is split between 2 filters 7/12

8 Example: How 2-Stage Reduces Computation Goal: Decimation by M = 12 Filter Requirements δ p = 0.01 δ s = to give some desired fidelity (application specific) to limit aliasing to desired level (application specific) θ p = π/16 θ s = π/12 = π/m to pass desired band (application specific) to prevent aliasing for desired decimation rate Single-Stage Method Length of filter determines the # of computations Use ( ) in P&M to estimate filter order needed: N = 20log 10 δ p δ s 2.32θs θ p transition width 13 8/12

9 Example: Single-Stage Method (cont.) Using this order estimate for the given filter requirements gives: N = 244 Length: L = N+1 = 245 Note: # s given below differ slightly from book because it (wrongly) uses N instead of L in its computation estimates Our chosen complexity measure: # Multiplies/Input Sample Each output sample (after decimation): L Multiplies/Output Sample There are M Input Samples/Output Sample (due to decimation) L mult/output M input/output (# Multiplies)/(Input Sample) = = = L M Single-Stage Complexity = 20.4 multiplies/input 9/12

10 Ex. Cont.: Double-Stage Method (M = M 1 M 2 : 12 = 3 4) 1 st Stage θ p,1 = θ p = π/16 δ p,1 = δ p /2 = θ s,1 = (2M 2-1)π/M = 7π/12 δ s,1 = δ s = Estimated filter order gives: N 1 = 11 L 1 = 12 So Mult/Input = L 1 /M 1 = 12/3 = 4 2 nd Stage θ p,2 = M 1 θ p = 3π/16 δ p,2 = δ p /2 = θ s,2 = π/m 2 = π/4 δ s,2 = δ s = Estimated filter order gives: N 2 = 88 L 2 = 89 So Mult/Input = L 2 /(M 1 M 2 ) = 89/12 = 7.4 referenced back to input of whole system Double-Stage Complexity = = 11.4 multiplies/input 2-Stage has ½ Complexity of 1-Stage 10/12

11 Comments on Multistage Method Q: What happens in this example when order of stages is switched? i.e., M 1 = 4 and M 2 = 3 (Left as Exercise!!!) These 2-Stage design ideas can be extended to p-stage designs: M = M 1 M 2 M 3 M p The order of these multiple stages matters Similar ideas can be used for multistage interpolation 11/12

12 Application: Multistage Rate Change Convert Digital Audio Tape (DAT) format to Compact Disk (CD) 3 DAT uses F s = 48 khz = = = 3 CD uses F s = 44.1 khz Single-Stage Approach: L Rate Change Ratio = = M H(z) 160 F s = 48 khz F s = MHz!!!! Multiple-Stage Approach: L = 147 = M = 160 = H 1 (z) 5 7 H 2 (z) 8 3 H 3 (z) 4 12/12