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IV. IIR filters  Examples 
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Note: Do not save the schematics after usage because they may be used in later examples.
Note: Opening the schematics from the links is best done with a Click+Drag
over an existing opened instance of LTspice.
In
.TRAN
there are several
options, displayed in a rather ascending order (simulationwise):
 Use continuoustime signal at the input. This has two possibilities of its own:
 Use
SH=0
⇒ the fastest response, continuoustime output, but unreal. If the input has smooth timesteps (e.g. a sine), the output will be the same (this may be useful with the Hilbert transfomer, for example).
 Use
SH=1
⇒ slower, sampled output, but makes I/O comparison difficult.
 Use sampled input, closer to reality, more choices:
 Don't use a clock, internal (
SH=0
) or external ⇒ there may be spurious spikes, particularly at low levels, though sampled output; more precisely, the input must be properly sampled.
 Don't use internal clock (
f0<0
) while the external clock (SH=1
) is a sine source with SINE(0.5 0.5 {f0} 0 0 {delay})
⇒ smoother timestep but also very prone to spikes and sampling errors;
 Use external sync (i.e. the input's, implies
SH=1
) ⇒ there's always the risk of sampling on the rising edge^{1}, but which can be avoided by setting the signal's S&H ref<0.5
(and/or ref>0.5
for Filter), or by imposing a value for td
(in my experience, td=0.25/f0
seems to work best) ⇒ this is preferred because of the speed gain, i.e. only one clock source instead of more;
 Use internal clock (implies
SH=1
) ⇒ same hidden parameters tinkering as in the above c) point need to be set to avoid sync problems. This has the most stable output, but is also the slowest.
^{1}due to LTspice being a dedicated analog simulator, the transitions are treated as regular transients and tend to get distorted in time.
To help speeding things up a bit for the latest option, a small trick can be employed: since for the sampling process only the rising edge is needed, the pulse source can be set with only Trise
and Tperiod
, resulting in a default 10%
falling edge and one less point (Ton
) to be calculated ⇒ same result, faster time.
As a
"golden rule", in
.AC
→
SH=0
and in
.TRAN
→
SH=1
, but there can be exceptions to the rule.
The following table lists some of the key parameters which, when set to zero, allow various optimizations possible. There can only be one null parameter at a time.

N=0 
Ap=0 
As=0 
Asc=0 

Butterworth_IIR 
minimum order to meet the requirements + stopband optimization 
 
 
passband optimization 
N, Asc > 0 ⇒ stopband optimization 
Chebyshev_IIR 
minimum order to meet the requirements + stopband optimization 
ripple optimization 
 
corner frequency @Ap ^{1} 
N, Ap, Asc > 0 ⇒ stopband optimization 
InvChebyshev_IIR 
minimum order to meet the requirements + stopband optimization 
 
stopband optimization 
passband optimization 
N, As, Asc > 0 ⇒ stopband optimization 
Cauer_IIR 
minimum order to meet the requirements + stopband optimization 
ripple optimization 
stopband optimization 
corner frequency @Ap ^{1} 
N, Asc, Ap, As > 0 ⇒ transitionband optimization 
^{1}only if nT=1
, otherwise it's fp@AscAp
for even orders.
Any of the following examples can be used to verify the table.
A normalized unity gain Chebyshev_IIR LP is chosen, with 1024Hz
sampling frequency, 3dB@128Hz
, 1dB
ripple and 60dB@160Hz
for an .AC
analysis.
 Order automatically determined:
nT=0
sim=x test=x SH=0 Fact=1
f0=1024 fc=0 BWp=128 BWs=160
G=1 Asc=3 Ap=1 As=60 N=0
SH=0
is to disable LTspice's S&H Adevice, it doesn't allow .AC
passing through and Fact=1
is for normal filter behaviour, interpolation/decimation is not needed now.
 Keeping the previous settings but with specified order:
nT=0
sim=x test=x SH=0 Fact=1
f0=1024 fc=0 BWp=128 BWs=160
G=1 Asc=3 Ap=1 As=60 N=12
For a time domain impulse response, the two cards need to be switched. Since the input is already a wellshaped impulse, SH
may be left null.
A comparison to the
Analog active example. The sampling frequency is chosen ten times greater than
ω_{c}
.
N=0
:
sim=x test=x SH=0 Fact=1
f0=10k fc=0 BWp=1k BWs=0.5k
G=1 Asc=1 As=60 N=0
N>0
:
sim=x test=x SH=0 Fact=1
f0=10k fc=0 BWp=1k BWs=0.5k
G=1 Asc=1 As=60 N=11
Optional, the Analog active Butterworth with the same settings can be placed besides for comparison.
An InvChebyshev_IIR with 6dB
gain, 48kHz
sampling frequency, 2.5kHz
center frequency, 1dB@3.75kHz
passband and 100dB@5.25kHz
stopband.
 Minimum order to meet the requirements:
sim=x test=x SH=0 Fact=1
f0=48k fc=2.5 BWp=3.75k BWs=5.25k
G=2 Asc=7 As=106 N=0
 The previously found out order, specified:
sim=x test=x SH=0 Fact=1
f0=48k fc=2.5 BWp=3.75k BWs=5.25k
G=2 Asc=7 As=106 N=15
The
.TRAN
simulation shows its response for a one octave sweep of a sampled sine input. To avoid being the
slowest (2d), option 2b is used.
There may be some sampling errors if the filter is a Cauer_IIR, but they will only be visible for values so small that they shouldn't matter.
For simplicity, the same example from the previous point is used and transformed into a Cauer_IIR BS by renaming InvChebyshev_IIR → Cauer_IIR and by switching the values of BWp
↔ BWs
between themselves. The result can be used to also compare the .TRAN
sweeps.
As
warned, the simulation may take a second or two to start.
A rate of 4
and a familiar sampling frequency of 44.1kHz
is used for a 3dB@20kHz
(the FIR variant will use similar settings) and 120dB
attenuation.
sim=x test=x SH=0 Fact=1
f0=44.1k fc=0 BWp=5k BWs=6025
G=4*1.4142 Asc=3 As=120 N=0
The simulation card has as Stop Frequency
{22049*4}
, 4
representing the rate and 22049
 the π
frequency for unity rate (normal filter), minus 1Hz
; the reason: LTspice, given enough data points, will show the attenuation @π
as a very low value, possibly a few kdB, which is as it should be, but it almost obscures the response of interest, most of it residing within <10dB..110dB>
. This example is not one of the cases, though, as the order is even.
Decimation can also be tried with negative numbers (i.e. Fact=4
).
Visualizing the effect of interpolation can be done by changing the simulation cards and setting SH=1
, where the limit of the audio domain, a 20kHz
signal, is sampled at 44.1kHz
and then interpolated at the above rate of 4. The gain is set to 4⋅1.4142
to allow a better matching for I/O amplitudes.
Setting test=1
will enable voltages at nodes BilTran
and freqs
. To avoid waiting the whole 100ms
, the .TRAN
can be changed to a lower value, say 9ms
⇒ sim=9m
.
Pin
0.2
outputs a voltage in four steps. Following the
explanation,
V(freqs) has four values: the first quarter of the simulation is the value of the lower passband frequency,
1.25kV
(
V →
Hz), then the 2nd time step has the value of the lower stopband frequency,
1kV
, the 3rd is the upper passband frequency,
5kV
and the next is the upper stopband,
6.25kV
.
V(BilTran) goes in the same manner. The time span is first divided in six (9ms/6 = 1.5ms
), one for each term of the bilinear transform, in this order: A_{2}[n], A_{1}[n], A_{0}[n], B_{2}[n], B_{1}[n]
and B_{0}[n]
. Each of these parts is again divided into N
parts: 1.5ms/15 = 0.1ms
(N=15
, this is why .TRAN 9m
has been chosen). Each 0.1ms
represents one value of the coefficients. For example B_{0}[1:15]
(the last time segment, <7.5ms..9ms>
) starts at 7.5ms
until 7.6ms
with B_{0}[0]
, then until 7.7ms
it's B_{0}[1]
and so on, until the last coefficient, <8.9ms..9ms>
, B_{0}[15]
.
If fc=0
, the filter is now a LP and, with test=1
, the second half of the time scale for V(freqs) is zero; for V(BilTran), each 6th part is divided in two and only the first half outputs the values.
The simulation lasts longer now and it does so for Butterworth_IIR and Chebyshev_IIR, too, but the "heavywait" is with Cauer_IIR: it will look as if the window is frozen, patience is really the answer here. Unwrapping all those functions for a behavioural voltage as a function of time does not come cheap.
©Vlad, 2008  2015