STMicroelectronics AN2182 Filters using the st10 dsp library Datasheet

AN2182
Application note
Filters using the ST10 DSP library
Introduction
The ST10F2xx family provides a 16-bit multiply and accumulate unit (MAC) allowing control-oriented
signal processing and filtering widely used in digital applications.
An ST10 DSP software library, developed by STMicroelectronics, contains a set of basic arithmetic
operations such as multiplication as well as two main filter functions, FIR (finite impulse response) and IIR
(infinite impulse response), mainly used in digital signal processing.
The first chapter of this application note describes a theoretical digital implementation of four different
filters:
■
Low-pass filter
■
High-pass filter
■
Passband filter
■
Cut-band filter
The method adopted for each filter is the approximation of the ideal filter model by a FIR filter. This theory
aims to compute the FIR’s coefficients by truncating the real signal with a known window.
The second chapter illustrates a practical implementation of a low-pass filter using the ST10 DSP library,
its results and its limitations.
July 2007
Rev 1
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18
AN2182
Contents
Contents
1
ST10 DSP library . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
Digital filtering principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.1
Fourier transform of a sampled signal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.2
Linear filtering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.3
Finite impulse response filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3
Low-pass filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
4
High-pass filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
5
Passband filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
6
Cutoff band filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
7
Implementation example using the ST10 DSP library . . . . . . . . . . . . . . . 12
7.1
Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
7.2
Sampling frequency and FIR coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
7.3
Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
7.3.1
Frequency response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
7.3.2
Phase response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
8
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
9
Revision history . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
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1
ST10 DSP library
ST10 DSP library
The ST10 DSP free library is a set of arithmetic and signal processing functions based on the
ST10 MAC unit. These functions are callable from C and fully compatible with the Tasking
compiler.
This library manipulates signed integers coded on 16 or 32 bits. These integers represent
numbers belonging to the interval [–1, 1[. We name these formats: Q1.31 and Q1.15.
Table 1.
Examples of integer representations
–1
-0.5
-0.25
0
0.25
0.5
1 - 1/215
1 - 1/231
Q1.15
0xFFFF
0xC000
0xA000
0x0000
0x2000
0x4000
0x7FFF
0x7FFF
Q1.31
0xFFFF
FFFF
0xC000
0000
0xA0000 0x00000 0x20000 0x40000
000
000
000
000
0x7FFF
0000
0x7FFFFFFF
For a detailed description of the ST10 DSP library, please refer to the application note AN1442
“Signal processing with ST10-DSP”.
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AN2182
Digital filtering principles
2
Digital filtering principles
Assume a continuous signal x(t) (the complex form corresponds to the signal’s phase and
magnitude at the instant t) with a pass band B. Assume that this signal will be filtered using a
filter with a continuous impulse response h(t).
When digital processing has to be used, it is necessary to sample the input signal with a
frequency of Fs = 1/Ts (Ts being the sampling period). The output signal is then reconstituted
from the samples obtained at the filter’s output.
Figure 1.
Example of input and output signals
Filter h
Input signal
Output signal
The Shannon theorem states that when sampling a signal at discrete intervals, the sampling
frequency Fs should be greater than twice the highest frequency of the input signal.
2.1
Fourier transform of a sampled signal
Signals are converted from the time domain to the frequency domain usually through the
Fourier transform. With the Fourier transform, the signal is converted to a magnitude and phase
at each frequency.
The Fourier transform of the sampled signal x(k) has the following expression:
∞
X(f) =
∑x ( k )e
i2kπf
–∞
The time representation can be computed from the X(f) as follows:
∞
x(k ) =
∑X ( f )e
– i 2kπf
–∞
2.2
Linear filtering
Using the notations defined in the previous section, the output signal y(n) is the convolution of
the input signal x(k) and the filter impulse response h(k) .
∞
y(n) =
∑h ( n – k )x ( k )
= (h(k) ⊗ x(k))
n
–∞
where x(n) = x(nTs), h(n – k) = h((n – k)Ts) and y(n) = y(nTs).
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Digital filtering principles
The output signal frequency response is given by the following expression:
Y(f) = H(f) ⋅ X(f)
where H(f), X(f) and Y(f) are the respective Fourier transforms of h(k), x(k) and y(k).
2.3
Finite impulse response filters
The FIR (Finite Impulse Response) are non-recursive filters, meaning that the output signal y(i)
is a linear combination of N input samples x(k) in the case of a N–1 order filter. Its equation is
N–1
⎛
⎞
⎜
y(n) =
a k x ( n – k )⎟
∑
⎜
⎟
⎝k = 0
⎠
where ak are the FIR’s coefficients.
A FIR filter is characterized by its order and its coefficients and can be used to implement any
kinds of filters (low-pass, high-pass, pass-band or cutoff band).
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Low-pass filter
3
Low-pass filter
The aim of this section is to create a low-pass filter with a cutoff frequency Fc and a gain G = 1,
by determining a FIR filter using the digital approach. The FIR coefficients correspond to h(n)
where h is its continuous time response and h(n) = h(nTs).
In the frequency domain, the ideal filter corresponding to these criteria has the following
response:
Ideal low-pass filter frequency response
Magnitude
Figure 2.
1
-Fc
0
Fc
Frequency
This filter’s impulse response in the time domain is
∞
(1)
h(t ) =
∫
H ( f )e
2πift
df = 2F c sin c ( 2tF c )
–∞
This response is sampled with a rate Fs= 1/Ts (sampling frequency), so the discrete response
has the following expression
∞
( 2)
∑h ( nTs )δ( t – nTs )
hs ( t ) =
–∞
The impact of sampling h(t) with a rate Fs is a periodization of the analog signal spectrum
around Fs and a gain of Fs. In fact, the frequency response corresponding to the sampled
impulse response is
∞
H s ( f ) = F s ∑H ( f – nF s )
–∞
Therefore, to obtain a gain of 1, the hs response filter should be divided by Fs. Using equations
(1) and (2), the low-pass filter time response becomes:
∞
(3)
hs ( t ) =
Fc
sin c ( 2nF c T s )δ( t – ns )
∑2 -----F
–∞
s
The impulse response is a sinus cardinal (sinc) function centered at the origin.
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Low-pass filter
Figure 3.
Ideal low-pass filter impulse response
h(kTs)
h(k/2Fc) = 0
h(0)= 2Fc
k>=1
Two issues can be observed at this stage:
First, the filter length is infinite. This means that in order to have the ideal low-pass filter, an
infinite number of filter coefficients is required. In practice, this cannot be done because of the
calculation complexity. To solve this feasibility problem, the filter’s impulse response should be
truncated by a known window. In this application note, we will truncate this response using a
rectangular window.
Truncating the filter’s impulse response, however, changes the ideal rectangular response;
more fluctuations are observed in the pass band. In order to reduce this effect, the FIR’s
coefficients’ numbers can be increased.
Effect of truncating the ideal impulse with a rectangular window
Magnitude
Figure 4.
W(n)
1
Time
-Fc
0
Fc
Frequency domain
Frequency
Time domain
In the following sections, we will note:
●
W is the window’s time response
●
N the number of coefficients included in the window W. This means that the window’s
length is (N–1)Ts. The window should be large enough to include at least the first lobe of
the sinus cardinal.
Second, the truncated response is finite but not causal because there are non-null coefficients
on the negative time axis. This means that the filter reacts before being excited by the input
signal. Practically, this filter is not feasible. To solve this issue, we will shift the filter’s impulse
response by (N–1)Ts/2. By doing so, all coefficients on the negative time axis are 0 and the filter
becomes feasible.
f
The impact of shifting the filter’s time response is a phase shift of ( N – 1 ) ⋅ π ⋅ -----F s in the
frequency domain.
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Low-pass filter
By fixing a number of points N and a sampling period Ts, the FIR coefficients h(n)
corresponding to a low-pass filter with a cutoff frequency Fc are
(4)
Fc
–1
N–1
h ( n ) = 2 ------ sin c ⎛ 2F c ⎛ n – N
-------------⎞⎠ T s⎞⎠ ⋅ W ⎛⎝ n – -------------⎞⎠
⎝
⎝
Fs
2
2
if n = 0 .. N – 1 and 0 if n > N – 1.
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4
High-pass filter
High-pass filter
The aim of this section is to create a high-pass filter with a cutoff frequency Fc and a gain G = 1,
by determining a FIR h filter using the digital approach. The FIR coefficients correspond to h(n)
where h is its continuous time response and h(n) = h(nTs).
In the frequency domain, the ideal high-pass filter corresponding to these criteria has the
following response:
Figure 5.
Ideal high-pass filter frequency response
1
1
-Fs
-Fc
0
Fc
Fs
-Fs
-Fc
0
Fc
Fs
High-pass filter Fc
Low-pass filter Fc
If we consider HH(f) the frequency response of the high-pass filter with a cutoff frequency Fc
and HL(f) the frequency response of the low-pass filter with a cutoff frequency Fc, we can easily
notice that
(5)
HH ( f ) = 1 – HL ( f )
In the time domain, this gives
(6)
Fc
⎛
⎞
–1
N–1
N–1
h H ( n ) = ⎜ δ⎛ n – N
-------------⎞⎠ – 2 ------ sin c ⎛⎝ 2F c ⎛⎝ n – -------------⎞⎠ T s⎞⎠ ⎟ ⋅ W ⎛⎝ n – -------------⎞⎠
⎝
Fs
2
2
2
⎝
⎠
where N is the number of the filter’s coefficients, N–1 is the filter’s order and W is the window’s
time function.
In the case of a high-pass filter, N should be odd because of the dirac.
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Passband filter
5
Passband filter
The aim of this section is to create a passband filter with a pass frequency F1, a cutoff
frequency F2> F1, and a gain G=1, by determining a FIR h filter using the digital approach.
In the frequency domain, the ideal passband filter corresponding to these criteria has the
following response:
Ideal passband filter frequency response
Magnitude
Figure 6.
1
-F2
0
-F1
F2
F1
Frequency
This filter’s impulse response in the time domain is
∞
h(t) =
∫
– F1
H ( f )e
–∞
2πift
df =
∫
F2
e
2πift
df +
– F2
∫
e
2πift
df = 2F 2 sin c ( 2F 2 t ) – 2F 1 sin c ( 2F1 t )
F1
Here, the same method described in the section entitled Low-pass filter is handled. Please refer
to this section for more details.
If we fix N the number of the filter’s coefficients and Ts the sampling period then we obtain
(7)
F1
⎛ F2
⎞
N–1
– 1⎞ T ⎞ – 2 -----–1
h ( n ) = ⎜ 2 ------- sin c ⎛ 2F 2 ⎛ n – N
- sin c ⎛ 2F 1 ⎛ n – N
------------------------⎞⎠ T s⎞⎠ ⎟ ⋅ W ⎛⎝ n – -------------⎞⎠
s
⎝
⎠
⎝
⎝
⎠
⎝
F
F
2
2
2
⎝ s
⎠
s
with n = 0.. N – 1 and where
10/18
●
N is the number of the FIR coefficients
●
N – 1 is the FIR’s order
●
W is the window time response
AN2182
Cutoff band filter
The aim of this section is to create a cutoff band filter with a cutoff frequency F1, a pass
frequency F2> F1, and a gain of G=1, by determining a FIR filter using the digital approach.
The FIR coefficients correspond to h(n) where h is the continuous time response of the filter
and h(n) = h(nTs).
In the frequency domain, the ideal cutoff band filter corresponding to these criteria has the
following response:
Ideal cut-band filter frequency response
Magnitude
Figure 7.
Magnitude
6
Cutoff band filter
1
-F1
-F2
0
1
F1
F2
Frequency
-F2
-F1
0
F1
F2
Frequency
Cut-band filter
Passband filter
If we consider
●
HP(f) the frequency response of the passband filter with a pass frequency F1 and cutoff
frequency F2
●
HC(f) the frequency response of the cutoff band filter with a cutoff frequency F1 and pass
frequency F2
●
Ts is the sampling period
then, we can easily notice that
(8)
HC ( f ) = 1 – HP ( f )
In the time domain, this gives
F
F1
⎞
–1 ⎛ 2
N–1
N–1
h C ( n ) = δ⎛⎝ n – N
-------------⎞⎠ – ⎜ 2 ------- sin c ⎛⎝ 2F 2 ⎛⎝ n – -------------⎞⎠ T s⎞⎠ – 2 ------- sin c ⎛⎝ 2F 1 ⎛⎝ n – -------------⎞⎠ T s⎞⎠ ⎟
Fs
2
2
2
⎝ Fs
⎠
with n = 0.. N – 1 and where
●
N is the number of FIR coefficients
●
N – 1 is the FIR’s order
●
W is the window time response
In the case of a cutoff band filter, N should be odd because of the dirac, that is the FIR’s order
should be even.
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AN2182
Implementation example using the ST10 DSP library
7
Implementation example using the ST10 DSP library
7.1
Overview
In this example, an input signal is fed to an analog input of the ST10F27x, processed by a
digital low-pass filter and then output on a PWM output setup as a digital to analog converter.
The digital filter is a low-pass filter with a cutoff frequency equal to 2 kHz and an order of 14
implemented on the ST10F27x using the DSP library provided by STMicroelectronics.
P5.0
Signal
ADC
module sampled
Lowpass
filter
Signal
filtered
PWM
module
DSP
P7.0
27 kΩ
ST10F276
Output signal
Input signal
P5.0
Low-pass filter modules
220pc
Figure 8.
The whole implementation includes the following elements:
12/18
●
ADC (analog-to-digital converter): the 10-bit ADC of the ST10F27x is used. The input
signal is entered on channel 0 (P5.0) of the ST10F27x. The ADC conversion time is
4.85 µs (ADCST = ADCTC and ST10 frequency = 40 MHz). When the conversion is
complete, the result stored in the ADDAT register is provided to the FIR filter module.
●
Digital filter module: it provides a digital filtered sample. The function used in this
application is the fir_q15_q15_q15() provided in the LibST10.h. This means that the
inputs, outputs and the FIR coefficients are fractional between –1 and 1 and are coded in
the format Q1.15.
●
DAC (digital-to-analog converter): this module converts the digital filtered signal into an
analog one. An 8-bit DAC is implemented using the ST10F27x PWM and an analog lowpass filter RC. The PWM frequency is 158.8 kHz. The RC filter is used to retain the
continuous components and to remove the frequency of the PWM module. In our case, the
RC cutoff frequency is 26.8 kHz. The digital filtered sample sets the pulse width of the
PWM signal.
AN2182
Implementation example using the ST10 DSP library
The following flowchart summarizes the software implementation of the low-pass filter:
Figure 9.
Low-pass filter flowchart
PWM initialization (PP0 = 0xFF)
FIR structure initialization(1)
ADC initialization
Circular buffer initialization(2)
A
No
Yes
PWMIR = 1
Clear PWMIR
Trigger ADC conversion
No
Yes
ADCIR = 1
ADCIR = 0
Update the circular buffer with (ADDAT << 5)(3)
Apply the filter->Output
PW0 = PP0 +1 - (Output >> 7)(4)
B
1. Please refer to the technical note ST10 DSP library for a detailed description of the FIR structure used and its
fields.
2. If the filter’s order is 14, 15 input samples are needed to compute an output. A circular buffer is used to store
the 15 input samples of the input signal. The size of the circular buffer is 2 x (filter’s order +1), which is 30 in this
example. An input sample is stored twice in the circular buffer: at buffer[i] and buffer[i + filter’s order +1] where
0<= i<= filter’s order.
3. The ST10F27x ADC converts analog voltages between 0 and 5V into 10-bit digital values. The inputs for the
fir_q15_q15_q15 should be fractional. To normalize these samples, they are shifted to the left.
4. The FIR function delivers a fractional output coded on 16 bits(Q1.15). The DAC realized with the ST10F27x is
an 8-bit one.
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Implementation example using the ST10 DSP library
7.2
Sampling frequency and FIR coefficients
Using the flowchart in the previous section, the sampling period is the time needed between
point A and point B.
To calculate the FIR’s coefficients, the sampling period should be known. We measure the time
between points A and B and the result is 17.5 kHz.
Note that to reconstitute the filtered signal, the input signal should not contain frequencies
greater than 8.75 kHz.
The FIR’s coefficients are then computed according to this sampling frequency
···
h( 0)
– 1402
0.23 sin c ( – 1.61 )
– 0.0428
h( 1)
– 1605
0.23 sin c ( – 1.37 )
– 0.049
h( 2)
– 895
0.23 sin c ( – 1.14 )
– 0.0273
h( 3)
701
0.23 sin c ( – 0.914 )
0.0214
h( 4)
2956
0.23 sin c ( – 0.68 )
0.0902
h( 5)
5174
0.23 sin c ( – 0.46 )
0.1579
h( 6)
6898
0.23 sin c ( – 0.23 )
0.2105
=
= 7537
h( 7) =
0.23 sin c ( 0 )
0.23
h( 8)
6898
0.23 sin c ( 0.23 )
0.2105
h( 9)
5174
0.23 sin c ( 0.46 )
0.1579
h ( 10 )
2956
0.23 sin c ( 0.68 )
0.0902
h ( 11 )
701
0.23 sin c ( 0.914 )
0.0214
h ( 12 )
– 895
0.23 sin c ( 1.14 )
– 0.0273
h ( 13 )
– 1605
0.23 sin c ( 1.37 )
– 0.049
h ( 14 )
– 1402
0.23 sin c ( 1.61 )
– 0.0428
7.3
⋅ 2
– 15
Results
To analyze the filter’s frequency and phase responses, a sinusoidal signal of a 2V amplitude is
used. The input voltage varies between 0 and 2 V.
7.3.1
Frequency response
To draw the filter’s frequency response, the amplitude of the output signal is measured.
The following figure gives the filter’s response.
Figure 10. Filter’s frequency response
Gain in dB
10
0
-10
1
10
100
-20
-30
Frequency
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AN2182
Phase response
The following figure gives the phase response.
Figure 11. Phase response
120
Phase in °
7.3.2
Implementation example using the ST10 DSP library
20
-80
1
10
100
1000
10000
-180
Frequency
The code implemented to generate this digital low-pass filter can be used to implement a highpass, a passband or a cutoff band filter. One simply needs to change the filter’s coefficients and
the number of coefficients. However, with this implementation, care should be taken with the
sampling frequency, which changes with the filter’s coefficients number and the ST10F27x
frequency.
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AN2182
References
8
16/18
References
1.
ST10F276 user manual
2.
ST10F276 datasheet
3.
AN1442 Signal processing with ST10-DSP, application note
AN2182
9
Revision history
Revision history
Table 2.
Document revision history
Date
Revision
19-Jul-2007
1
Changes
Initial release.
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AN2182
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