DSP Lecture 20-21

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Lectures 20-21 EE-802 ADSP SEECS-NUST

EE 802-Advanced Digital SignalProcessing

Dr. Amir A. Khan

Office : A-218, SEECS

9085-2162; [email protected]

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Lecture Outline

Wrap-up of Last Lecture Linear Filtering Approach to Computation

of DFT

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Radix-2 DIT (Complexity Reduction)

Total computations

N/2.log2N

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Radix-2 DIT Bit-reversed Input Ordering

i/p Indx b2 b1 b0 b2' b1' b0' Bit-reversed orderInput placement

x[0] 0 0 0 0 0 0 0 0

x[1] 1 0 0 1 1 0 0 4

x[2] 2 0 1 0 0 1 0 2

x[3] 3 0 1 1 1 1 0 6

x[4] 4 1 0 0 0 0 1 1

x[5] 5 1 0 1 1 0 1 5x[6] 6 1 1 0 0 1 1 3

x[7] 7 1 1 1 1 1 1 7

x[0] goes in location 0; x[1] goes in location 4; x[6] goes in location 3,

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Radix-2 DIT In-Place Computations

A = a + WNr b

B = a - WNr b

Once (a,b) are used to calculate (A,B), there is no need for (a,b)

Hence (A,B) results can be saved in the same memory location as (a,b)

resulting in efficient memory utilization and the name in-place computation

2N memory registers required

Condition : inputs (sequence) /outputs (DFT coefficients) be stored in bit-

reversed order

a

b

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Direct DFT vs FFT Round-up

Understanding DSP (Lyons)

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Further Recommended Reading

Understanding DSP (Lyons) Chapter 4 (Section 4.2: Hints on using FFTs in Practice)*

Other FFT Algorithms

Radix-4

Split Radix FFT

Other DFT algorithms (Discussion in Next lecture)

Goertzel Algorithm (Tone-detector)

Chirp Transform Algorithm

* Highly recommended

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Linear Filtering/Convolution Approach to

Computation of Fourier Transform

The Geortzel Algorithm (Oppenheim)

The Chirp-Transform Algorithm (Oppenheim)

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The Goertzel Algorithm

Suitable if a particular frequency or a small number offrequencies need to be detected

Algorithm exploits periodicity of phase factors

Express computation of DFT as a linear filtering operation

Define

Using x[n]=0 for nN-1, DFT can be calculated as

1

2/2 kjNkNjkN

N eeW

1

0

1

0

1

0

][][][N

r

rNk

N

N

r

kr

N

kN

N

N

r

kr

N WrxWrxWWrxkX

r

rnk

NkrnuWrxny ][

Nn

k nykX

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The Goertzel Algorithm-Complexity Reduction

Complexity

Real multiplications for poles 2 per input sample

Real additions for poles 4 per input sample

Where do we gain then?

Implementation of zeros 1 complex mult. at Nth iter. 4 real multiplications

4 real additions

Total Complexity Poles+Zeros

2N+4 = 2(N+2) real multiplications

4N+4 = 4(N+1) real additions

Multiplications reduced by half as compared to Direct Computation

vk[n]

nxnvnvN

knv kkk

21

2cos2

1][

NvWNvnykX kk

NkNn

k

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Single Tone Detection Example (Lyons)

Tone Detector

To detect 30 KHz tone.Assume Sampling Frequency fs = 128KHz

N = 64 point FFT

Solution

k = ftone/N fs = 15

2Cos (2k/N) = 0.196

-exp(-j2k/N)=-0.098 + j 0.995

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DTMF Tone Detection-Goertzel Algorithm

Dual-Tone Multi Frequency (DTMF) signaling system uses a

combination of two tones to represent a specific digit, character orsymbol

1 2 3

654

7 8 9

#0*

A

B

C

D

1209Hz 1336Hz 1477Hz 1633Hz

697Hz

770Hz

852Hz

941Hz

1 2 3

654

7 8 9

#0*

A

B

C

D

1336770770770770 133613361336

Output

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DTMF Tone Detection-Goertzel Algorithm

For tone detection, use either a filter bank (band-pass filters) or Goertzel

algorithm (more efficient) Determine the value of k for using Goertzel Algorithm

Minimum DTMF tone duration required is 40 ms, with a sampling rate of

8kHz results in 0.04 x 8000 = 320 samples

Standard DTMF number of samples in input tone, N = 205

N

kff sanalysis N

f

fk

s

tone

Frequency k

1633 421477 38

1336 34

1209 31

941 24852 22

770 20697 18

Feed-back (pole) section would be repeated N=205 times

Evaluation of Goertzel algorithm zero equation at k bins

For detecting presence/absence of tone, we detect only

the magnitude of the filter output (ignore the phase)

xnvnvNknv kkk

212cos2

1][

NvWNvnykX kk

NkNnk

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Frequency resolution of FFT?

Same as that of DFT = 2/N, with N the length of the input sequence.

What should we do if resolution is not sufficient enough for agiven application?

zero padding is an option

Potential problem with zero padding? may increase the complexity unnecessarily

Practically increased resolution is required only for a certain bandof frequencies

Assume we need to determine the frequency of a sinusoid 0with

good accuracy from a given N point sequence Compute DFT only around the frequency bins susceptible to

capture 0 Chirp transform does exactly this job to compute DFT in a

zoomed region

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Algorithm does not reduce the computational complexity over FFT

But, allows computing DTFT over any equally spaced finite set of

frequencies

Consider x[n], an N point sequence with DTFT X(ej)

Evaluate M samples of X(ejw

) equally spaced in frequency on theunit circle at frequencies

Results in DFT when

w 0 and M = N

k = 0 + k where k=0,1,.M-1

= 2k/M

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DTFT is sampled atw

=w

k

Defining Was

Chirp transform represents X(ejwk) as a convolution

By substituting

Defining

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Re-formulating for easier interpretation as convolution

Convolution of g[n] with W-n/2

Complex exponential with frequencyn

Linearly increasing frequency signal called chirpused extensively in radars/sonars

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Chirp Transform Algorithm-FIR form

Assuming that g[n] is of finite duration

W is an infinite duration exponential

For above equation

Finite portion of W-n is used for conv

Over interval n=0 M-1

Portion of W-n required is

-(N-1) to M-1

Defining an FIR filter

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Chirp Transform Algorithm-Advantages

Advantages N = M condition is not required as in FFT

Neither N nor M needs to be a composite number (they can

be prime numbers)

Parameter 0 is arbitrary, thus providing flexibility

The convolution sum involved in chirp transform can be

implemented efficiently using FFT with a size

M+N-1

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Chirp Transform with Causal FIR filter

Non-causal

Causal

Delay h[n] by N-1 samples

The Fourier Transform is thus advancing the output by N-1

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Chirp Transform Algorithm - Example

Example: Suppose we have a finite length sequence x[n] that is

non-zero over the interval n=0,,25 and we wish to compute16 samples of DFT (Xej) at frequencies k = 2/27 +

2k/1024 for k = 0,,15

Solution

Let M = 16 and N = 26, 0

= 2/27 and = 2/1024

For the Causal Impulse Response

h[n] = [e-j2/1024](n-25)/2 for n=0,.40 (n= 0 to N+M-2)

Output y[n] will be samples beginning at index n+25

y[n+25] = X(ejwn)|wn = 2/27 + 2n/1024

CTA Hi h R l ti N b d F

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CTA- High Resolution Narrowband Frequency

Analysis

DSP Lecture 20-21

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