OSpecpm fft Opm
(3.2.7.12)0 1 10 10
0 0.01 0.02 0.03
Output signal spectrum max OSpecpm
OSpecpmk Specpmk
fcpm
k fspm
N0gd
0 1 10 10
4
2 0 2 4
Phase spectrum
0 arg OSpecpm
k
arg Specpm
k
fcpm
kfspm
N0gd
Fig.:3.2.7.5 Fig.:3.2.7.6
Wdbω3c20 log Wlp j ω3c
Wdbω3c 5.85 dB ωc is the angular frequency of the carrier100 1 10 3 1 10 4 1 10 5 1 10 6 1 10 7
40
20 0 20 40
Magnitude of W(ω)
20 log A3
30 20 log Wlp j ω
( )
Wlpp
ω31cfm ω3
ω
Fig.:3.2.7.7
3.3
Equivalent Digital Low Pass Filter (I°order)
3.3.1) Z-transfer function of the I° Order Low Pass Digital Filter
Chosen sampling period: T3s 1388.89 ns place: Ts T3s Given the transfer function: Wlp s( ) A3 ω3
sω3
= , I can find its z-transform in this way:
with the change of variable: s 1 z 1
= Ts ,we can place: 1 Ts
ωs 2 π
= , s ωs
2 π
1 z 1
= ,
A3A3 Ts Ts ω3 ω3 s s zz Transfer function z-transform:
H1o z
A3 ω3s ω3
substitute s 1 z 1
= Ts
collect z
A3 Ts ω3z z Ts ω3
1
1
and after some algebraic manipulation and the definition of the following parameters:
α0 A3 ω3 Ts Ts ω3 1
β0
1 ω3 Ts
1 A310α01.157482795 , β0 0.88425172 , you get the following result for the t. f. as a function of z:
H1o z
α0 1 1 β0 z 1
= Ts 1388.89 ns
3.3 Equivalent Digital Low Pass Filter (I°order)
3.3.2) Difference equations (Low Pass filter(I°order)). Canonical form
Given the z transfer function H1o z( ) α0 1 1 β0 z 1
= , I can split it so that:
H1o z
Y z
X z
= Y z
W_ z
W_ z
X z
=
and place: 1) Y z
W_ z
=α0 Y z
=α0 W_ z
2) W_ z
X z
=
1 β0 z 1 1
X z
=
1 β0 z 1
W_ z
=W_ z
β0 z 1W_ z
which, inverting the z transform, gives :
x n
=w_ n
β0 w_ n 1
The corresponding set of difference equations is:
1) w n
=x n
β0 w n 1
2) y n
=α0 w n
α0 α0 β0 β0
z ∞ α0 1
1 β0 z 1
lim
α0
Fig.:3.3.2.1
DELPF1OCF
3.3 Equivalent Digital Low Pass Filter (I°order)
For each test signal there would be shown the following results:
1) Sequence of the periodic response,
2) Digital first order low pass filter difference equations, 3) Schematic,
4) Graphics,
5) Comparison of the Bode plots of the z and s transfer functions
3.3.3)
Sequence of the Voltage step response Digital first order low pass filter difference equations:dimensionless input signal: u1k Vstpsl n3
kVpp
vi1 k
u1k volt
10 0 10 20 30
0 2 4
Unit Pulses Sequence.
0 Vpp u1k
0 8
k Vpp 4500 mV
rows u1
256fig.:3.3.3.1)
w1y1 DELPF1OCF vi1 A3 T3s
sec ω3 rad sec
N0gd
w1w1y1 0 y1 w1y1 1 α1
w1y1 2
0 β1
w1y1 3
00 25 50 75 100 0
10 20 30 40
Sequence of the function w
w1k
k
0 25 50 75 100
50
40
30
20
10 0
Sequence of the response A3 Vi 0 y1k
k
fig.:3.3.3.2 fig.:3.3.3.3
t1 0 τ3 0 τ3 100 τ3 10000
100 τ3
0 4 10 5 8 10 5
0.06
0.04
0.02
Graph of the step response y(t)
time as multiple of τ
Output amplitude
A3 Vi
e11
A3 Vi ysr t1( )
yas t1( ) 0
τ3 5 τ3
t1
fig.:3.3.3.4
3.3 Equivalent Digital Low Pass Filter (I°order)
3.3.4) Sequence of the short Voltage pulse response.
TsTs3sp
t τpw2 τpw2 4 τpw
5000
2 τpw
2105 0 2 10 5 4 10 5 0
0.002 0.004 0.006
Sampled Input
Vi Vw t()
u44k
τ3svp τpw τ3svp
t ns3sp k fig.:3.3.4.1 Digital first order low pass filter difference equations:
dimensionless input signal: vi2 k
u44kw2y2 DELPF1OCF vi2 A3 Ts3sp
s ω3 s
rad
N0gd
w2w2y2 0 y2w2y2 1 α2
w2y2 2
0 β2
w2y2 3
0Fig.:3.3.4.2
0 2 10 5 4 10 5 0
0.05 0.1 0.15
Sequence of w
w2k
τ3svp τpw τ3svp
ns3spk
0 2 10 5 4 10 5
0.04
0.02 0
Sequence of the response
A3 Vi y2k
τ3svp τpw τ3svp
ns3spk
fig.:3.3.4.3 fig.:3.3.4.4
t τpw2 τpw2 4 τpw
5000
2 τpw
0 2 10 5 4 10 5
0.04
0.02 0
Graph of the Short Pulse Response
time as multiple of τ
Output amplitude
Vi
A3 Vi τpw τ3svp
τ3svp
Vi 0.01 V Vpp 4.5 V
fig.:3.3.4.5
3.3 Equivalent Digital Low Pass Filter (I°order)
3.3.5)
Sequence of the Bipolar Pulse train response TsTssqw Ttest 13.33 μs Ts 138.89 ns τ3 10.61 μs fs 1 Ts ωs 2 π fs Chosen test signal period, Ttest 13333.33 ns 1
Ttest 0.08 MHz Laplace transform of the test signal: Vip s
Vis tanh Ttest s
4
u3k Vsqwb nsqw
k
Pulse amplitude ±Vi:
0 100 200
5103 0 5 10 3
Pulses Sequence.
0 Vi
u3k
0
k fig.:3.3.5.1 Digital first order low pass filter difference equations:
dimensionless input signal: vi3 k
u3kw3y3 DELPF1OCF vi3 A3 Tssqw
s ω3 s
rad
N0gd
w3 w3y3 0 y3 w3y3 1 α3
w3y3 2
0 β3
w3y3 3
0α30.12920836 , β3 0.987079164 , you get the following result for the t. f. as a function of z:
H1o z
α3 1 1 β3 z 1
Ts 138.89 ns
N0gd 256
Fig.:3.3.5.2
0 63.75 127.5 191.25 255
0.1
Sequence of the response
0
ωtest 0.47 Mrads
sec
Normalized Magnitude of H(z) and W(jω) 03
3.3 Equivalent Digital Low Pass Filter (I°order)
3.3.6)
Sequence of the Cusp test signal response. Definition:u4k vincspk
TsTocsp A310 t3 Ttest4 Ttest4 8 Ttest
1000
4 Ttest
0 1.333 10 2.667 105 54 10 55.333 10 5 0
2 4
Test signal
Vi fx1 t3( )
pwtcTtest
t3
0 1 10 52 10 53 10 54 10 5 0
2 4
Sampled test signal
Vi V u4k
Ttest pwtc
nincspk
Fig.:3.3.6.1 Fig.:3.3.6.2
Cusps sequence of amplitude Vi:
Digital first order low pass filter recurrence relations:
dimensionless input signal: vi4 k
u4k fsf3c 96 w4y4 DELPF1OCF vi4 A3 Tocsp
s ω3 s
rad
N0gd
w4 w4y4 0 y4w4y4 1 α4
w4y4 2
0 β4
w4y4 3
0α40.12920836 , β4 0.987079164 , you get the following result for the t. f. as a function of z:
H1o z
α4 1 1 β4 z 1
Ts 138.89 ns
Fig.:3.3.6.3
0 1 10 52 10 53 10 54 10 5 0
50 100 150
Sequence of the state function w
Fig.:3.3.6.4
0 1 10 52 10 53 10 54 10 5
20
15
10
5 0
Sequence of the periodic response A3 Vi y4k
Vocsp nincsp
k
nincspk Fig.:3.3.6.5
1.333 10 6 1.433 10 5 2.733 10 5 4.033 10 5 5.333 10 5
Normalized Magnitude of H(z) and W(jω) 03
3.3 Equivalent Digital Low Pass Filter (I°order)
3.3.7)
Sequence of the Sawtooth response Ts Tssw Sawtooth sequence of amplitude Vi:Digital first order low pass filter recurrence relations:
fs f3c 96 dimensionless input signal: vi5 k
u10k V
w5y5 DELPF1OCF vi5 A3 Tssw
s ω3 s you get the following result for the t. f. as a function of z:
H1o z
α5 1Sequence of the response
0 y5k
voswc nsw
knswk
Fig.:3.3.7.2 Fig.:3.3.7.3 Spec5o FFT y5
0 1 10 6 2 10 6 3 10 6 0
1 2
Amplitude Spectrum max Spec5o
Spec5ok ftest
k fs
N0gd
0 1 10 6 2 10 6 3 10 6
4
2 0 2 4
Phase spectrum
arg Spec5o
k
0 ftestk fs
N0gd
Fig.:3.3.7.4 Fig.:3.3.7.5
max Spec5o
max Spec5o
1 Ts 0.14 μs 100 1 10 3 1 10 4 1 10 5 1 10 6 1 10 7 1 10 8
40
20 0
Normalized Magnitude of H(z) and W(jω) 03
20 log H1o e
j ω Ts
A3
20 log Wlp j ω( )
A3
20 log H1o e
j ωtest Ts
A3
ω3 ωtest
ω Fig.:3.3.7.6
3.3 Equivalent Digital Low Pass Filter (I°order)
3.3.8)
Sequence of the (single tone) Frequency Modulated carrier response mfm 8 dimensionless input signal: vi6 k
u8k TsTsfmw6y6 DELPF1OCF vi6 A3 Tsfm
s ω3 s
rad
N0gd
w6w6y6 0 y6 w6y6 1 α6
w6y6 2
0 β6
w6y6 3
0α60.03131754 , β6 0.996868246 ,
Digital first order low pass filter difference relations:
Fig.:3.3.8.1
you get the following result for the t. f. as a function of z:
H1o z
α6 1 1 β6 z 1
Ts 33.33 ns
0 4 10 6 8 10 6
0.2
0.1 0 0.1 0.2
FM signal
Vfm tfm
0 Tcfmtfm 020 10 0
2 4 6 8 10
FM Spectrum (sinus. test signal)
A 4 A Jn ki mfm
ki
Fig.:3.3.8.2 Fig.:3.3.8.3
X8 fft u8
0 100 200
0.2
0.1 0 0.1 0.2
Sampled FM signal
u8k
k 0 5 10 6 1 10 7 1.5 10 7
0 0.2 0.4
FM spectrum max X8( )
X8k
fs f3c
k fsfm
N0gd
Fig.:3.3.8.4 Fig.:3.3.8.5
fsfm f3c 400
0 100 200
1
0.5 0 0.5
Sequence of the function w
0 100 200
0.02
0.01 0 0.01 0.02
Sequence of the response
Fig.:3.3.8.6 Fig.:3.3.8.7
f3c 0.08 MHz fs
f3c 96 Spec6fft y6
mfm 8 ωfmm 942477.81 s
0 2 10 6 4 10 6 6 10 6 0
0.01 0.02 0.03 0.04
FM Signal spectrum max Spec6( )
Spec6k f3c
k fsfm
N0gd
0 5 10 6 1 10 7 1.5 10 7
4
2 0 2 4
Phase spectrum
arg Spec6
k
0 f3ck fsfm
N0gd
Fig.:3.3.8.8 Fig.:3.3.8.9
max Spec6
0.09100 1 10 3 1 10 4 1 10 5 1 10 6 1 10 7 1 10 8
60
40
20 0
Normalized Magnitude of H(z) and W(jω) 03
20 log H1o e
j ω Ts
A3
20 log Wlp j ω( )
A3
20 log H1o e
j ωtest Ts
A3
ω3 ωtest
ω Fig.:3.3.8.10
3.3 Equivalent Digital Low Pass Filter (I°order)
3.3.9)
Sequence of the Phase Modulated carrier response. mpm 6 dimensionless input signal: vi7 k
u9k TsTspmw7y7 DELPF1OCF vi7 A3 Tspm
s ω3 s
you get the following result for the t. f. as a function of z:
H1o z
α7 1 1 β7 z 1
Ts 0.02 ns
107.4999999964.9999999922.4999999871.66799996 10 8
20
PM Spectrum (sinus. test signal)
A 4 A Jn ki mpm
ki
mpm 6 Fig.:3.3.9.2 Fig.:3.3.9.3
0 100 200
Sequence of the state function w
w7k
Sequence of the response
y7k
3.3 Equivalent Digital Low Pass Filter (I°order)
3.3.10)
BODE plot (Low Pass Analog v. s. Digital filter(I°order)) 20 log α7
96.08 Ts 0.02 ns Normalized Magnitude of H(z) and W(jω) 03
The Phase of H(z)and W(jω)
π
For each test signal would be shown the following results:
1) Sequence of the periodic response,
2) Digital first order low pass filter difference equations, 3) Schematic,
4) Graphics,
5) Comparison of the Bode plots of the z and s transfer functions
α0 and β0 are functions of the sampling period which in turn it depends from the kind of signal used.
TsT3s
Numerator degree Nn 1 Denominator degree Md 1
N1Nn Md N0gd 256 h1k0
A generic first order transfer function in the z domain takes this form:
H z
b0b1z1a0a1z1
=
The coefficients of the numerator and denominator can be defined as the elements of two vectors, namely a and b, hence:
Numerator coeffs Denominator coeffs n11 N0gd 1 bn1=0.0 an1=0.0
b0=α0 a0=1
b1=0 a1=β0
and divide the two polynomial by means of the following algorithm:
N12 h10 b0
y8ν 0 ν
k
if ν k
0h1ku1νk0
IACAN
tfs1 IACAN u1
VA3T3sω3N0gd
α0
tfs1 0
0 β0
tfs1 0
1 S1
tfs1 0
2 E11
tfs1 0
3 h11 tfs1 1a
tfs1 2
b
tfs1 3
T. F. Numerator coefficients:
aT 1 -0.88 0 0 0 0 0 0 ...
T. F. Denominator coefficients:
bT 0 1 2 3 4 5 6 7 8
0 -1.16 0 0 0 0 0 0 0 ...
Sequence of the Impulse response:
h1T 0 0 0 0 0 ...
Stability (S1<∞): S1
0 rows h1( ) 1
k
h1k
= S110
Energy of the sequence h1: E11
0 rows h1( ) 1
k
h1k
2
= E116.14
0 5 10 5 1 10 4 1.037
106
7.54105 4.712
105 1.885
105
9.425 10 4 Impulse Response
A3 ω3 e
A3 ω3 w t1( )
A3ω3
t1 ω3 1
0
τ3 5τ3
t1
0 10 20 30 40 50
1.5
1
0.5 0
Impulse Response Sequence
h11k
k
3.4 Transfer Function Sequence obtained by an Algorithm. Convoluting Output.