Fast Frequency And Response Measurements Using Ffts

  • Uploaded by: snowjazz
  • 0
  • 0
  • July 2020
  • PDF

This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA


Overview

Download & View Fast Frequency And Response Measurements Using Ffts as PDF for free.

More details

  • Words: 1,181
  • Pages: 33
Fast Frequency and Response Measurements using FFTs Alain Moriat, Senior Architect Fri. 12:45p Pecan (9B) www.natinst.com

Accurately Detect a Tone  What is the exact frequency and amplitude of a tone embedded in a complex signal?  How fast can I perform these measurements?  How accurate are the results?

www.natinst.com

Presentation Overview       

Why use the frequency domain? FFT – a short introduction Frequency interpolation Improvements using windowing Error evaluation Amplitude/phase response measurements Demos

www.natinst.com

Clean Single Tone Measurement Time signal

2 Volt 1

FFT Spectrum

20 dBV 0

0

-20

-1

-40

-2 0.0

-60

0.2

0.4

0.6

 Clean sine tone  Easy to measure

www.natinst.com

0.8ms 1.0

kHz 0

5 10 15 20 25 30 35 40 45 50

 Clean tone spectrum

Noisy Tone Measurement Time signal

2 Volt 1

Our signal

FFT Spectrum

20 dBV 0

0

-20

-1

-40

-2 0.0

-60

0.2

0.4

0.6

0.8ms 1.0

 Noisy signal  Difficult to measure in the time domain www.natinst.com

kHz 0

5 10 15 20 25 30 35 40 45 50

 Noisy signal spectrum  Easier to measure

Fast Fourier Transform (FFT) Fundamentals (Ideal Case) Time signal

2 Volt 1

0

0

-20

-1

-40

-2 0.0

FFT Spectrum

20 dBV

0.1

0.2

0.3

Fsampling =100 kHz Time res =10 us

0.4ms 0.5

-60 0

5 10 15 20 25 30 35 40 45 50 kHz

Record size =50 samples Freq. res =2 kHz

 The tone frequency is an exact multiple of the frequency resolution (“hits a bin”) www.natinst.com

FFT Fundamentals (Realistic Case) Time signal

2 Volt 1

0

0

-20

-1

-40

-2 0.0

FFT Spectrum

20 dBV

0.1

0.2

0.3

Fsampling =100 kHz Time res =10 us

0.4ms 0.5

-60 0

5 10 15 20 25 30 35 40 45 50 kHz

Record size =50 samples Freq. res =2 kHz

 The tone frequency is not a multiple of the frequency resolution www.natinst.com

Input Frequency Hits Exactly a Bin 0 dB

 Only one bin is activated

-10 -20 -30 -40 -50 -60 -6

-5

-4

www.natinst.com

-3

-2

-1

0

1

2

3

4

5 Bin 6

Input Frequency is +0.01 Bin “off” 0 dB

 More bins are activated

-10 -20 -30 -40 -50 -60 -6

-5

-4

www.natinst.com

-3

-2

-1

0

1

2

3

4

5 Bin 6

Input Frequency is +0.25 Bin “off” 0 dB

Real top

-10

Highest Bin

-20

Next Highest Bin

-30 -40 -50 -60 -6

-5

-4

www.natinst.com

-3

-2

-1

0

1

2

3

4

5 Bin 6

Input Frequency is +0.50 Bin “off” 0 dB

 Highest side-lobes

-10 -20 -30 -40 -50 -60 -6

-5

-4

www.natinst.com

-3

-2

-1

0

1

2

3

4

5 Bin 6

Input Frequency is +0.75 Bin “off” 0 dB

 The Side lobe levels decrease

-10 -20 -30 -40 -50 -60 -6

-5

-4

www.natinst.com

-3

-2

-1

0

1

2

3

4

5 Bin 6

Input Frequency is +1.00 Bin “off” 0 dB

 Only one bin is activated

-10 -20 -30 -40 -50 -60 -6

-5

-4

www.natinst.com

-3

-2

-1

0

1

2

3

4

5 Bin 6

The Envelope Function 1.1 1.0

Real top

0.8

Highest Bin = a

0.6

Next highest Bin = b

0.4 0.2 0.0 -0.2 -0.3 -4

-3

www.natinst.com

-2

-1

0

1

2

3

Bin 4

The Mathematics  Envelope function:

Sin( π ⋅ bin) Env = (π ⋅ bin)

 Bin offset:

b Δbin = ± (a + b)

 Real amplitude:

(π ⋅ Δbin) Amp = a ⋅ Sin( π ⋅ Δbin)

www.natinst.com

Demo  Amplitude and frequency detection by Sin(x) / x interpolation

www.natinst.com

Aliasing of the Side-Lobes Highest Bin = Bin 4

0 dB -10 -20

Aliased Bin = “Negative Bin 4”

-30 -40 -50 -60

0

1

2

www.natinst.com

3

4

5

6

7

8

9 Bin 10

Weighted Measurement  Apply a Window to the signal 2

2

Volt 1

Volt 1

0

0

-1

-1

-2 0.0

-2 0.0

0.1

0.2

0.3

0.4ms 0.5

Hanning window – one period of ( 1 - COS ) www.natinst.com

0.1

0.2

0.3

0.4ms 0.5

Weighted Spectrum Measurement  Apply a Window to the Signal 20 dBV 0

Without Window

20With dBV 0

-20

-20

-40

-40

-60 0

-60 0

5

www.natinst.com

10

15

20kHz25

Hanning Window

5

10

15

20kHz25

Rectangular and Hanning Windows 0 dB -10 -20 -30 -40 -50 -60 -6

-5

-4

www.natinst.com

-3

-2

-1

0

1

2

3

4

5 Bin 6

 Side lobes for Hanning Window are significantly lower than for Rectangular window

Input Frequency Exactly Hits a Bin 0 dB

 Three bins are activated

-10 -20 -30 -40 -50 -60 -6

-5

-4

www.natinst.com

-3

-2

-1

0

1

2

3

4

5 Bin 6

Input Frequency is +0.25 Bin “off” 0 dB

 More bins are activated

-10 -20 -30 -40 -50 -60 -6

-5

-4

www.natinst.com

-3

-2

-1

0

1

2

3

4

5 Bin 6

Input Frequency is +0.50 Bin “off” 0 dB

 Highest side-lobes

-10 -20 -30 -40 -50 -60 -6

-5

-4

www.natinst.com

-3

-2

-1

0

1

2

3

4

5 Bin 6

Input Frequency is +0.75 Bin “off” 0 dB

 The Side lobe levels decrease

-10 -20 -30 -40 -50 -60 -6

-5

-4

www.natinst.com

-3

-2

-1

0

1

2

3

4

5 Bin 6

Input Frequency is +1.00 Bin “off” 0 dB

 Only three bins activated

-10 -20 -30 -40 -50 -60 -6

-5

-4

www.natinst.com

-3

-2

-1

0

1

2

3

4

5 Bin 6

The Mathematics for Hanning ... Sin( π ⋅ bin)  Envelope: Env = (π ⋅ bin) ⋅ (1 − bin 2 ) (a - 2b)  Bin Offset: Δbin = ± (a + b) (π ⋅ Δbin)  Amplitude: Amp = a ⋅ ⋅ (1 −Δbin 2 ) Sin( π ⋅ Δbin) www.natinst.com

A LabVIEW Tool

 Tone detector LabVIEW virtual instrument (VI) www.natinst.com

Demo  Amplitude and frequency detection using a Hanning Window (named after Von Hann)  Real world demo using:  The NI-5411 ARBitrary Waveform Generator  The NI-5911 FLEXible Resolution Oscilloscope

www.natinst.com

Frequency Detection Resolution 1000.00

Freq error (ppm)

ppm

100.00 10.00 1.00 0.10 0.01 1

www.natinst.com

10

Signal periods

100

Amplitude Detection Resolution 1000.00

Amplitude error (ppm)

ppm

100.00 10.00 1.00 0.10 0.01 1

www.natinst.com

10

Signal periods

100

Phase Detection Resolution 1000.00

Phase error (mdeg)

mdeg.

100.00 10.00 1.00 0.10 0.01 1

www.natinst.com

10

Signal periods

100

Conclusions  Traditional counters resolve 10 digits in one second  FFT techniques can do this in much less than 100 ms  Another example of 10X for test  Similar improvements apply to amplitude and phase

www.natinst.com

Conclusions (Notes Page Only)  Traditional Counters Resolve 10 digits in one second  FFT Techniques can do this in much less than 100 ms  Another example of 10X for test  Similar improvements apply to Amplitude and Phase

www.natinst.com

Related Documents


More Documents from ""