Multirate Signal Processing* Tutorial using MATLAB** By, Deborah Goshorn dgoshorn@cs.ucsd.edu I. Signal processing background II. Downsample Example III. Upsample Example * Multrate signal processing is used for the practical applications in signal processing to save costs, processing time, and many other practical reasons. ** MATLAB is an industry standard software which performed all computations and corresponding figures in this tutorial
I. Signal processing background
Receive an analog signal • Receive an analog signal at 5 Hz (as pictured below left, there are 5 wave cycles in one second.) • The highest frequency component (5 Hz) of the signal is called the signal’s bandwidth, BW, since in the examples in this presentation, the minimum frequency component is 0Hz . • This signal can be represented in two ways: 600 1 500 0.5 Signal Strength Signal Value 400 Peak signal 0 strength at 5 Hz 300 200 -0.5 100 -1 BW 0 0 0.2 0.4 0.6 0.8 1 0 5 10 15 20 25 Time (sec) Frequency (Hz) time representation (sec) frequency representation (Hz)
Add high frequency components 600 1 500 Signal Strength 0.5 400 1. Original Signal Value 300 5 Hz signal 0 200 -0.5 Adding high BW = 5 Hz 100 -1 0 frequency 0 5 10 15 20 25 0 0.2 0.4 0.6 0.8 1 Frequency (Hz) Time (sec) components 600 creates 2. Add a 2 500 jagged 1.5 Signal Strength 400 10 Hz edges in Signal Value 1 300 component the original 0.5 200 0 BW = 10 Hz 5 Hz signal. 100 -0.5 0 -1 0 5 10 15 20 25 Frequency (Hz) 0 0.2 0.4 0.6 0.8 1 Time (sec) 600 3 3. Then add 500 2 a 15 Hz Signal Strength 400 Signal Value 1 component! 300 200 0 BW = 15 Hz 100 -1 0 0 5 10 15 20 25 0 0.2 0.4 0.6 0.8 1 Frequency (Hz) Time (sec)
Sampling the signal: N yquist Rate • In order to sample the signal without losing information, use a sampling rate (S R ) of at least the Nyquist Rate ( N R ), which is 2 x BW of the received analog signal. 600 500 400 Signal Strength 300 200 Signal Nyquist Rate N R bandwidth = 2 x 15Hz BW = 15 Hz = 30 Hz 100 0 0 5 10 15 20 25 30 35 Frequency (Hz) RULE: Sampling Rate S R ≥ Nyquist Rate N R
Sampling the signal: N yquist Rate Since Bandwidth BW = 15 Hz, the Nyquist Rate N R = 2 x 15Hz = 30Hz. RULE #1: Sampling Rate S R ≥ Nyquist Rate N R 200 3 2 150 Signal Strength Signal Value 1 100 Signal 0 bandwidth 50 Sample Rate BW = 15 Hz S R = 40 Hz -1 0 0 0.2 0.4 0.6 0.8 1 0 5 10 15 20 25 30 35 40 Time (sec) Frequency (Hz) Let Sample Rate S R = 40 Hz, Nyquist Rate N R = 30 Hz so sample signal every 0.025 sec (25 milliseconds).
Sampling the signal: Nyquist Freq • The Nyquist Frequency ( N F ) is equal to half of the sampling rate ( S R ). The N F must be equal to or greater than the bandwidth BW of the desired signal to reconstruct. Rule #2: Nyquist Frequency N F ≥ Bandwidth BW 200 150 Signal Strength 100 Nyquist Freq N F Signal = 40/2 50 bandwidth = 20 Hz BW = 15 Hz Sample Rate 0 S R = 40 Hz 0 5 10 15 20 25 30 35 40 Frequency (Hz)
II. Downsample Example
Recall, our original signal at 5Hz… 600 1 1. Original 500 0.5 Signal Strength 400 Signal Value 5 Hz signal 0 300 BW = 5 Hz 200 -0.5 100 -1 0 0 0.2 0.4 0.6 0.8 1 0 5 10 15 20 25 Time (sec) Frequency (Hz) 2. We added 3 600 10 & 15 Hz 500 2 Signal Value Signal Strength 400 components! 1 300 0 BW = 15 Hz 200 -1 100 0 0.2 0.4 0.6 0.8 1 0 Time (sec) 0 5 10 15 20 25 Frequency (Hz) 200 3 3. Then we 2 150 sampled at Signal Strength Signal Value 1 S R1 = 40Hz 100 0 50 BW = 15 Hz -1 0 0 0.2 0.4 0.6 0.8 1 0 5 10 15 20 25 Time (sec) Frequency (Hz)
Resample the sampled signal: downsampling Downsample by 4 means to retain only every 4 th sample 3 3 2 2 Signal Value Signal Value 4 1 1 0 0 -1 -1 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 Time (sec) Time (sec) Sample Rate 1 S R1 = 40Hz Sample Rate 2 S R2 = 10 Hz N F1 = 20Hz > N F2 = 5Hz < 15Hz = BW 15Hz = BW GOOD! BAD!
Nyquist Freq < Bandwidth Cannot recover original signal bandwidth, since new Nyquist Frequency (5Hz) is less than the desired signal bandwdidth BW (15Hz). N F2 < BW means we cannot recover 15Hz BW signal 200 150 Signal Strength 100 Signal Nyquist Freq 1 Nyquist Freq 2 50 bandwidth N F1 = 40/2 N F2 = 10/2 BW = 15 Hz = 20 Hz = 5 Hz 0 0 5 10 15 20 25 30 35 40 Frequency (Hz) Is the original 5Hz signal recoverable? It should be, since N F2 ≥ BW 5 Hz
Why 5Hz signal not recoverable: High Frequency band causes aliasing when downsampled Will wrap down to 0Hz 200 150 Signal Strength 100 Nyquist Freq 1 N F1 = 40/2 Signal = 20 Hz Nyquist Freq 2 50 bandwidth N F2 = 10/2 BW = 15 Hz = 5 Hz 0 0 5 10 15 20 25 30 35 40 Frequency (Hz) High frequency band N F2 N F1 High frequency band will wrap down to 0Hz when downsampled
Why 5Hz signal not recoverable: Aliasing Effects Due to the high frequency components at 10Hz and 15Hz that show up at 0Hz when the signal is downsampled , the 5Hz component is not recoverable . 180 3 160 Recovered 5Hz 140 2 component 120 Signal Value Signal Strength 1 100 80 0 Aliasing effects: high 60 frequency components 40 -1 wrapped around to 0Hz! 20 0 0.2 0.4 0.6 0.8 1 Time (sec) 0 0 5 10 15 20 25 30 35 40 Frequency (Hz) S R2 = 10 Hz … unless we remove the high frequency components before downsampling .
How to Remove the High Frequency components before downsampling using a low-pass filter • A low-pass filter (LPF) removes high frequency components by only letting low frequency components pass through. 200 100 150 80 Signal Strength Signal Strength LPF 60 100 40 50 20 0 0 5 10 15 20 25 30 35 40 0 Frequency (Hz) 0 5 10 15 20 25 30 35 40 It removes the jagged edges that were due to Frequency (Hz) high frequencies. 3 1 2 0.5 Signal Value Signal Value LPF 1 0 0 -0.5 -1 -1 0 0.2 0.4 0.6 0.8 1 4 4.2 4.4 4.6 4.8 5 Time (sec) Time (sec)
Proof in the pudding : No more aliasing effects when using low pass filter! 200 5 150 4 Signal Strength 4 Signal Strength 3 LPF 100 2 50 1 0 0 5 10 15 20 25 30 35 40 Frequency (Hz) 0 0 5 10 15 20 25 30 35 40 Frequency (Hz) S R1 = 40 Hz S R2 = 10 Hz The original 5Hz signal is successfully recovered!
Proof in the pudding : LPF+downsampling <==> multirate polyphase filter resampling Sample Rate 2 S R2 = 10 Hz 5 4 Signal Strength 4 3 LPF 200 2 150 Signal Strength 1 100 0 0 5 10 15 20 25 30 35 40 Frequency (Hz) 50 Sample Rate 2 MATLAB’S * 0 0 5 10 15 20 25 30 35 40 S R2 = 10 Hz Frequency (Hz) Polyphase-filter Implemented Resample (by 1/4) Sample Rate 1 S R1 = 40 Hz Function * MATLAB is an industry standard software which performed all computations and corresponding figures in this presentation
III. Upsampling example
Assume our original signal at 5Hz… 600 1 1. Original 500 0.5 5 Hz signal Signal Strength 400 Signal Value 0 300 200 BW = 5 Hz -0.5 100 -1 0 0 0.2 0.4 0.6 0.8 1 0 5 10 15 20 25 Time (sec) Frequency (Hz) Nyquist Rate N R = 2 x BW 5Hz = 10Hz, so sample at sampling rate S R = 15Hz 250 1 200 2. We sample Signal Strength 0.5 150 at S R1 = 15Hz Signal Value 0 100 -0.5 50 BW = 5Hz 0 -1 0 5 10 15 20 25 0 0.2 0.4 0.6 0.8 1 Frequency (Hz) Time (sec)
Resample the sampled signal: upsampling If you need to increase the number of samples in a given time by a factor of 5, you upsample by 5 ( insert 5-1=4 zeros between each sample ). 0.2 sec 0.2 sec 1 1 0.5 0.5 Signal Value Signal Value 5 0 0 -0.5 -0.5 -1 -1 0 0.2 0.4 0.6 0.8 1 0 0.05 0.1 0.15 0.2 Time (sec) Time (sec) Sample Rate 1 Sample Rate 2 S R1 = 15 Hz S R2 = 75 Hz
Upsampled signal in frequency representation 0.2 sec 0.2 sec 1 1 0.5 0.5 Signal Value Signal Value 0 0 -0.5 -0.5 -1 -1 0 0.2 0.4 0.6 0.8 1 0 0.05 0.1 0.15 0.2 5 Time (sec) Time (sec) Sample Rate 2 Sample Rate 1 S R2 = 75 Hz S R1 = 15 Hz Oops! 120 250 100 200 80 Signal Strength Signal Strength 150 60 100 40 50 20 0 0 0 5 10 15 20 25 30 35 40 0 5 10 15 20 25 30 35 40 Frequency (Hz) Frequency (Hz)
Upsampling causes aliasing in higher frequencies Upsampling causes copies of the original 5Hz component at multiples of original sampling rate, 15Hz, plus/minus 5Hz Mirror Images at: 15 – 5 = 10 Hz 15 + 5 = 20 Hz 120 2*15 - 5 = 25 Hz 250 100 2*15 + 15 = 35 Hz 5 200 … 80 Signal Strength Signal Strength 150 60 100 40 Signal bandwidth 50 BW = 5 Hz 20 0 0 0 5 10 15 20 25 30 35 40 0 5 10 15 20 25 30 35 40 Signal Frequency (Hz) Frequency (Hz) Sample Rate 1 bandwidth S R1 = 15 Hz BW = 5 Hz How do we remove these extra high frequency components?
How to remove the extra high frequency components caused by upsampling using a low-pass filter 8 120 100 6 Signal Strength 80 Signal Strength 4 60 LPF 40 2 20 0 0 5 10 15 20 25 30 35 40 0 0 5 10 15 20 25 30 35 40 Frequency (Hz) Frequency (Hz) Low pass filter removes these extra high frequency components
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