Gain of simple lowpass
CMPT 889: Decay Rate Tutorial/Implementation
Tamara Smyth, tamaras@cs.sfu.ca
School of Computing Science,
Simon Fraser University
• What is the decay rate of the original simple low pass
filter y(n) = 0.5x(n) − 0.5x(n − 1)?
November 13, 2009
• The steps are as follows
1. find the gain of the filter;
2. find the attenuation due to the filter in the context
of the model with a known period;
3. find the associated time constants τ and t60;
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CMPT 889: Computational Modelling for Sound Synthesis: Decay Rate Tutorial
Gain
Attenuation of simple lowpass
• Consider the frequency response of the filter:
• Consider the attenuation caused by this filter in the
context of a model with delay N + 1/2 or with
fundamenatal frequency
−jωT
1+e
2
ejωT /2 + e−jωT /2 −jωT /2
=
e
2
= cos(ωT /2)e−jωT /2.
H(ejωT ) =
2
f1 =
1
fs
.
=
(N + 1/2)Ts (N + 1/2)
• Consider the number of trips through the loop after n
samples (t = nTs seconds):
• The gain is given by the magnitude of the frequency
response:
M=
G(ω) = | cos(ωT /2)e−jωT /2|
= cos(ωT /2).
n
nTsfs
tfs
=
=
= tf1.
N + 1/2 N + 1/2 N + 1/2
• The attenuation at time t = nTs is then given by
α(t) , G(f1)tf1 = cos(πf1T )tf1 .
• The gain at frequency f is given by
• In general the attenuation at time t = nTs is given by
G(f ) = cos(πf T ).
tfs
α(t) , G(f1) N+P (f1)
where P (f1) is the phase delay of the lowpass filter at
frequency f1.
CMPT 889: Computational Modelling for Sound Synthesis: Decay Rate Tutorial
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CMPT 889: Computational Modelling for Sound Synthesis: Decay Rate Tutorial
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Time constant τ
Note Duration (t60)
• The time constant τ , i.e. time to decay by 1/e, is
given by
• The t60 (time to decay by 60 dB) is the time after
which the note can no longer be heard, i.e., the note
duration.
α(t) = e−t/τ
t
ln α(t) = −
τ
τ = −
• An attenuation of 60 dB corresponds to an mplitude
of 0.001.
t
1
=−
ln α(t)
f1 ln(cos(πf1T ))
• To convert τ to t60:
0.001 = e−t60/τ
t60
ln(0.001) = −
τ
t60 = − ln(0.001)τ
≈ 6.91τ
• or more generally
τ =−
t
1
= − fs
ln α(t)
ln G(f1)
N +P (f )
CMPT 889: Computational Modelling for Sound Synthesis: Decay Rate Tutorial
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Shortening the Note Duration
CMPT 889: Computational Modelling for Sound Synthesis: Decay Rate Tutorial
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Setting ρ for a desired t60
• The lowpass filter 0.5y(n) + 0.5y(n − N ) has a set
t60 for a note at frequency f1.
• For a desired t60, determine the corresponding time
constant τ
t60
t60
=
.
τ =−
ln(0.001) ln(1000)
• Consider multiplying by a factor 0 < ρ <= 1.
1 + e−jωT
.
H(ejωT ) = ρ
2
• The time constant is given by
1
f1 ln(ρ cos(πf1T ))
1
ln(ρ cos(πf1T )) = −
f1
τ
1
ρ cos(πf1T ) = exp −
f1τ
−1/(f1τ )
e
ρ = −
cos(πf1T )
• This changes the gain at frequency f1 to
τ = −
G(f ) = ρ cos(πf1T ),
and the attenuation at time t to
α(f ) = (ρ cos(πf1T ))tf1 .
• As this is just a scalar gain, it does not effect the
phase delay of 1/2 a sample (nor the frequency f1).
• The time constant is given by
1
τ =−
f1 ln(ρ cos(πf1T ))
CMPT 889: Computational Modelling for Sound Synthesis: Decay Rate Tutorial
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CMPT 889: Computational Modelling for Sound Synthesis: Decay Rate Tutorial
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Decay Stretching
• To find the stretch factor S
τ = −
• The new filter is
H =
=
=
=
=
ln(G(S; f )) = −
(1 − S) + Se−jwT
[(1 − S)ejωT /2 + Se−jωT /2]e−jωT /2
[ejωT /2 − S(ejωT /2 − e−jωT /2]e−jωT /2
[cos(ωT /2) + j sin(ωT /2) − 2jS sin(ωT /2)]e−jωT /2
[cos(ωT /2) + j sin(ωT /2)(1 − 2S)]e−jωT /2
• The gain of the filter is
q
G(S; f ) = cos2(ωT /2) + (1 − 2S)2 sin2(ωT /2)
N +S
fs ln(G(S; f )
N +S
fsτ
G(S; f ) = e−(N +S)/(fsτ )
G(S; f )
= e−N/(fsτ )
e−S/(fsτ )
• As you can see this gets a little harrowing. Since
eS/(f sτ ) ≈ 1 for the values of S we are considering,
we may use an approximation.
• The attenuation at time t is
q
tf
2
2
2
α(t) =
cos (ωT /2) + (1 − 2S) sin (ωT /2)
• The time constant is
N +S
1
1
=−
τ =−
= − fs
f ln(G(f ))
f
s ln(G(f ))
N +S ln(G(S; f ))
CMPT 889: Computational Modelling for Sound Synthesis: Decay Rate Tutorial
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Finding approximate Stretch Factor S
• The Stretch factor S for a give frequency f and time
constant τ (omitting phase delay) is approximated by
1
f ln(G(S; f )
1
ln(G(S; f )) = −
fτ
G(S; f ) = e−1/(f τ )
τ =−
q
cos2(πf T ) + (1 − 2S)2 sin2(πf T ) = e−1/(f τ )
cos2(πf T ) + (1 − 2S)2 sin2(πf T ) = e−2/(f τ )
e−2/(f τ ) − cos2(πf T )
(1 − 2S)2 =
sin2(πf T )
s
e−2/(f1τ ) − cos2(πf T )
(1 − 2S) = ±
sin2(πf T )
s
e−2/(f1τ ) − cos2(πf T )
2S = 1 ±
sin2(πf T )
s
e−2/(f1τ ) − cos2(πf T )
1
S= ±
2
4 sin2(πf T )
CMPT 889: Computational Modelling for Sound Synthesis: Decay Rate Tutorial
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CMPT 889: Computational Modelling for Sound Synthesis: Decay Rate Tutorial
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