Inverse Laplace w/ a 2-pole
Lowpass filter
ECE 3100
John Stahl
Western Michigan University
The circuit shown below is a two stage low pass filter with different corner frequencies
1. Derive the Transfer function H(s).
2. Generate the Bode Magnitude using LTSpice, Matlab, and by Hand.
3. For a Step Response show the output Vo(t).
a. By Hand using Partial Fraction Expansion and Plot in Matlab
b. LTSpice Simulation
c. Simulink
Transfer Function
𝑉1 − 0 0 − 𝑉𝑜1
=
3.33𝑘
𝑍𝑒𝑞
−𝑍𝑒𝑞 𝑉𝑜1
=
= 𝐻1 𝑠
3.33𝑘
𝑉1
𝐻1 𝑠 = −
𝑍𝑒𝑞
𝐕𝐨𝟏
𝑋𝑐 ∙ 10𝑘
=
𝑋𝑐 + 10𝑘
𝑋𝑐 ∙ 10𝑘
3.33𝑘 𝑋𝑐 + 10𝑘
𝑋𝑐 =
1
∙ 10𝑘
𝑠𝐶
𝑠𝐶
=−
1
3.33𝑘 𝑠𝐶 + 10𝑘 𝑠𝐶
𝐻1 𝑠 = −3
1
𝑠𝐶
1
𝑠
1 + 1000
The Transfer Function for the
total circuit is the product of the
two stages. The derivation of
the 2nd stage is the same as the
1st.
𝐻 𝑠 =6
𝑠
1 + 1000
1+
𝑠
15000
𝑠1 = −1000
2𝜋𝑓1 = 1000
The angular corner frequencies
of the circuit are 1krad/sec and
15krad/sec. In this context we
call the roots of the
denominator poles or
eigenvalues.
We can draw a Bode plot by
hand just as we did for H(w).
The corner frequencies for the
Bode Plot are 160Hz and
2.39kHz. The Passband gain is
20log(6) = 15.6dB
1
𝑓1 = 160𝐻𝑧
𝑠1 = −15000
2𝜋𝑓2 = −15000
𝑓2 = 2.39𝑘𝐻𝑧
15.6dB
-20dB/dec
-40dB/dec
freq
160Hz
2.39kHz
15.6dB
-20dB/dec
-40dB/dec
160Hz
2.39kHz
Using LTSpice we see the Magnitude of the Bode plot matches the predicted one. The
graph below shows Matlab’s Bode Magnitude.
Step Response
Vin 𝑠
Vo 𝑠
H(s)
𝑉𝑖𝑛 𝑠 ∙ 𝐻 𝑠 = 𝑉𝑜 𝑠
𝑉𝑜 𝑠 =
1
6
∙
𝑠 1+ 𝑠
1000 1 +
Poles
0
𝑠 = −1000
−15000
𝑠
15000
The input to the circuit is a unit
step, u(t). Solve for the output
vo(t).
The Laplace form for the step is
1
𝑉𝑖𝑛 𝑠 = 𝑠 .
We are trying to take the Inverse
Laplace of Vo(s). We are going to
manipulate the function to make
Vo(s) easier to transform back to
the time domain.
Partial Fraction Expansion
requires the poles of the
denominator. Including the one
added by the input.
Partial Fraction Expansion
𝑠
𝑠 1 + 1000
6
𝐴
𝐵
𝐶
= +
𝑠 +
𝑠
𝑠
𝑠
1 + 1000 1 +
1+
15000
15000
𝑠
6= 𝐴 1+
1000
𝑠
𝑠
𝑠
1+
+ 𝐵𝑠 1 +
+ 𝐶𝑠 1 +
15000
15000
1000
Evaluate the equation at each pole to solve for A, B, and C.
@𝑠 = 0
6=𝐴 1+0 1+0 +𝐵∙0+𝐶∙0
𝐴=6
@𝑠 = −1000
6 = 𝐴 ∙ 0 + 𝐵 ∙ −1000
𝐵 = −6.429 ∙ 10−3
1+
−1000
+𝐶∙0
15000
@𝑠 = −15000
6 = 𝐴 ∙ 0 + 𝐵 ∙ 0 + 𝐶 ∙ −15000
−15000
1+
1000
𝐶 = 2.857 ∙ 10−5
𝑏
𝑎+𝑠
We want the components to have this
form for the Inverse Laplace
6 −6.429 ∙ 10−3 1000 2.857 ∙ 10−5 15000
𝑉𝑜 𝑠 = +
∙
+
∙
𝑠
𝑠
𝑠
1000
15000
1 + 1000
1+
15000
𝑉𝑜 𝑠 =
6
−6.429
0.4286
+
+
𝑠 1000 + 𝑠 15000 + 𝑠
ℒ −1
𝑏
= 𝑏𝑒 −𝑎𝑡 𝑢 𝑡
𝑎+𝑠
𝑣𝑜 𝑡 = 6𝑢 𝑡 − 6.429 ∙ 𝑒 −1000𝑡 ∙ 𝑢 𝑡 + 0.4286 ∙ 𝑒 −15000𝑡 ∙ 𝑢 𝑡
Plots
The derived function, y(t) was plotted using Matlab and compared to the simulation
performed in LTSpice. The plots are identical. Links go to source code and SPICE model.
Matlab Plot of y(t)
LTSpice
Simulation Output