What is a Filter?
Input
Signal
Amplitude
Output
Signal
Frequency
Time Sequence
Low Pass Filter
Time Sequence
What is a Filter
Input
Signal
Amplitude
Output
Signal
Frequency
Signal
Noise
Frequency Transform
Signal
Low Pass Filter
Noise
Frequency Transform
FIR and IIR Filters
• A finite impulse response (FIR) filter produces an
output that will go to zero when the input goes to
zero and stays zero
• Its response to an impulse function input is finite
• An infinite impulse response (IIR) filter produces
an output that may continue indefinitely even
after its input goes to zero and stays zero
• Its response to an impulse function input is infinite
FIR and IIR Filters
• An FIR filter’s z-Transform has zeroes and poles
only at 0 + j0, so it is always stable
• An IIR filter’s z-Transform has pole(s) and may
also have zeroes, so it may be unstable if any of
the poles are outside the unit circle
• To get the same sharpness of filter shape, an
FIR filter may require more delay elements and
more computations than an equivalent IIR filter
• But an IIR filter won’t have linear phase
FIR Filters
• A simple example of a moving average FIR filter
Input
C0
In
z-1
C1
In-1
z-1
C2
In-2
Sum
Output
Output = (C0 + C1 * z-1 + C2 * z-2) Input Where C0 = C1 = C2 = 1/3
Output/Input = (z2 + z + 1) / (3 z2 )
Two zeros at -1/2 + j sqrt(3)/2 and -1/2 - j sqrt(3)/2
Two poles at 0 + j0
FIR Filters
• Pole-Zero Plot and Frequency Response
http://en.wikipedia.org/wiki/Finite_impulse_response
FIR Filters
• Program Output
Enter the integer sampling rate (samples per second):
[5000]
Enter number of poles and number of zeros
[2 2]
Amplitude at 0 Hz: 3.000, phase: 0 degrees
Enter the real and imag values for 2 poles:
Amplitude at 200 Hz: 2.937, phase: -14 degrees
Pole 0: [0 0]
Amplitude at 400 Hz: 2.753, phase: -29 degrees
Pole 1: [0 0]
Amplitude at 600 Hz: 2.458, phase: -43 degrees
Enter the real and imag values for 2 zeros:
Amplitude at 800 Hz: 2.072, phase: -58 degrees
Zero 0: [-0.5 0.866]
Amplitude at 1,000 Hz: 1.618, phase: -72 degrees
Zero 1: [-0.5 -0.866]
Amplitude at 1,200 Hz: 1.126, phase: -86 degrees
Amplitude at 1,400 Hz: 0.625, phase: -101 degrees
Amplitude at 1,600 Hz: 0.148, phase: -115 degrees
Amplitude at 1,650 Hz: 0.037, phase: -119 degrees
Amplitude at 1,700 Hz: 0.072, phase: -302 degrees
Amplitude at 1,800 Hz: 0.275, phase: -310 degrees
Amplitude at 2,000 Hz: 0.618, phase: -324 degrees
Amplitude at 2,200 Hz: 0.860, phase: -338 degrees
Amplitude at 2,400 Hz: 0.984, phase: -353 degrees
Amplitude at 2,500 Hz: 1.000, phase: 0 degrees
FIR Filters
• How do we get values for FIR filter coefficients?
• We find the impulse response of the system
either by calculation or experiment and sample it
• The sample values are the coefficients
T
System
C0
…
Cn
IIR Filters
• A difference equation is a digital approximation
for an analog differential equation
• A typical difference equation (with a0=1, m <= n):
a0Yt + a1Yt-1 + . . . + anYt-n = b0Xt + b1Xt-1 + … + bmXt-m
• An IIR filter is implemented as the solution for a
difference equation
• The z-Transform for the IIR filter:
Y/X = (b0zn + b1zn-1 + … + bmzn-m)/(a1zn + a2zn-1 + . . . + an)
IIR Filters
• A simple example of a moving average IIR filter
X
X
+
Y
X
z-1
1Y = X + (1 - ) Y z-1
Where 0 <
Y/X = z / (z – (1 - ))
Zero at z = 0 + j 0
Pole at 1 -
<1
IIR Filters
• Pole-Zero Plot and Frequency Response
Plot the amplitude
0 + j1
-1 + j0
O
Imaginary
X
Sweep vector
points over all
frequencies
Real
1 + j0
f
Plot the phase shift
f
0 – j1
IIR Filters
• Program Output
Inverse z-Transform Calculation
Enter the integer sampling rate (samples per second):
[5000]
Amplitude at 0 Hz: 4.000, phase: 0 degrees
Enter number of poles and number of zeros
Amplitude at 200 Hz: 3.020, phase: -34 degrees
[1 1]
Amplitude at 400 Hz: 2.008, phase: -47 degrees
Enter the real and imag values for 1 poles:
Amplitude at 600 Hz: 1.460, phase: -49 degrees
Pole 0: [.75 0]
Amplitude at 800 Hz: 1.148, phase: -47 degrees
Enter the real and imag values for 1 zeros:
Amplitude at 1,000 Hz: 0.954, phase: -43 degrees
Zero 0: [0 0]
Amplitude at 1,200 Hz: 0.825, phase: -38 degrees
Amplitude at 1,400 Hz: 0.736, phase: -33 degrees
Amplitude at 1,600 Hz: 0.674, phase: -27 degrees
Amplitude at 1,800 Hz: 0.630, phase: -21 degrees
Amplitude at 2,000 Hz: 0.600, phase: -15 degrees
Amplitude at 2,200 Hz: 0.582, phase: -9 degrees
Amplitude at 2,400 Hz: 0.573, phase: -3 degrees
Amplitude at 2,500 Hz: 0.571, phase: 0 degrees
First Order Filters
• A first order IIR filter has only one delay z-1
• It tracks well against a step function input
(constant level after t=0), but not against a
ramp function (constant first derivative) input
Steady State
Error is zero
Time
Steady state lag is
inversely prop to
phase gain
Time
First Order Filters
• Coefficient
is a compromise between:
– Large value for small steady state error, but with
wider bandwidth letting in more noise
– Small value to exclude more noise, but larger
steady state error with a constant derivative input
• A second order filter can be used to avoid this
compromise
Second Order Filters
• A second order IIR filter learns frequency offset
• It tracks well against a ramp function input, but
may ring around a step or ramp function input
unless coefficients are “critically damped”
Steady State
Error is zero,
but may ring
Steady State
Error is zero
One way to implement 2nd Order Filter
• Coefficients and are usually selected so that
phase difference is learned faster than frequency
offset is learned
• Coefficient relationships: 0 < < < 1
X(T n)
Y(T n)
+
z -1
+
F(T n)
z -1
Implement Second Order Filter
z-Plane
Difference Equations
F(T n) = (1 – ) F( T n-1) +
Y(T n) = (1 – ) Y(T n-1) +
X(T n)
X(T n) + F(T n)
Stable if all
Poles inside
unit circle
j
z = +f
Real
-1 + j0
1 + j0
z-Transform
Y(z) =
z [( + ) z – (1 – )]
[z – (1 – )] [ z – (1 – )]
X(z)
Two Poles at:
z = (1 – ) + j 0
z = (1 – ) + j 0
Two Zeros at:
z=0+j0
z = (1 – )/( + ) + j 0
-j
z = -f
Critically damped
since both poles
are on the real axis
so it won’t ring
Implement Second Order Filter
• Program Output for
= 7/8 and
= 1/8
Inverse z-Transform Calculation
Enter the integer sampling rate (samples per second):
[1000]
Enter number of poles and number of zeros
[2 2]
Enter the real and imag values for 2 poles:
Pole 0: [.875 0]
Pole 1: [.125 0]
Enter the real and imag values for 2 zeros:
Zero 0: [0 0]
Amplitude at 0 Hz: 2.139, phase: 0 degrees
Zero 1: [.766 0]
Amplitude at 100 Hz: 1.105, phase: -135 degrees
Amplitude at 200 Hz: 0.984, phase: -197 degrees
Amplitude at 300 Hz: 0.904, phase: -251 degrees
Amplitude at 400 Hz: 0.854, phase: -303 degrees
Amplitude at 500 Hz: 0.837, phase: -354 degrees
Second Order Filters
• Compromise between noise exclusion and steady
state error for constant derivative is not needed
• However, it has a constant steady state error for
a constant second derivative input function
Steady State
Error is zero
Steady state lag
• If that is a problem, use a cascade of second
order filters to get a higher order filter