1. Filter Designs
By Katherine Quinn

2. Obtain the spectrum of a sinusoid with an amplitude of 1 volt peak-to-peak (Vpp), frequency of 10khz, and a phase of 0.1
rad. First, find the exact spectrum using the Fourier Transform, and then compute the spectrum using DFT. Assume that
the signal is sampled at 4 times the Nyquist rate. You can use MATLAB or any other programming language to compute
DFT. Plot both spectra. Use dBv to represent spectra.
Exact Spectrum Double Sided Spectrum
Single Sided Spectrum
Repeat the previous steps when the sinusoid is replaced with a square wave with peak-to peak amplitude of 2 volts ,zero
dc level, and a period of sec10 mT  . In order to determine the Nyquist frequency, assume that the spectrum of a square
wave is negligible for where is the fundamental frequency of the square wave.
Exact Spectrum Double-Sided Spectrum
Single Sided Spectrum
0
0
1
T
f 03 ff 

3. dB
fH
fH
c
20
)(
)(
log10
2
10   dB
f
f
c
20
)(1
1
log10
2
10 


 2
10
20
1
1
log 210 







n
kHz
kHz
 2
21
1
log 210 
 n 
100
1
21
1
2

 n
 n2
21100   n2
299       n2
2ln99ln  
 
 
n
2ln2
99ln
 31467831.3n  4n
))(( af
n
kk
g
C 
f
an
k
kg
L
))((
 ))(( RkR a
))((
1
1
af kk
g
C  
)50)(100002(
7654.0
1

C  9
1 106343869.243 
C  nFC 2431 
)(
))(( 2
2
f
a
k
kg
L  
)100002(
)50)(8478.1(
2

L  3
2 10470432519.1 
L  mHL 47.12 
))((
3
3
af kk
g
C  
)50)(100002(
8478.1
3

C  9
3 101730077.588 
C  nFC 5883 
)(
))(( 4
4
f
a
k
kg
L  
)100002(
)50)(7654.0(
4

L  3
4 106090859672. 
L  mHL 609.4 
Design a maximally flat, low pass filter with the cut-off frequency of 10kHz and an attenuation of 20 dB at 5kHz. Use
amplitude scaling so that the filter is terminated onto a 50 Ω resistance. Use PSpice (free student version is available
online) to simulate/ validate the response of your circuit.
Decibel attenuation to find order:
Maximally Flat Low Pass Filter (4th Order)
Component Calculations:
element values for n=4 order low pass filter:
g1=0.7654 g2=1.8478 g3=1.8478 g4=0.7654 g5=1
Maximally Flat Low pass Filter Design

4. dB
fH
fH c
20
)(
)(
log10
2
10   dB
f
fc
20
)(1
1
log10
2
10 


 2
5
10
1
1
log 210 







n
kHz
kHz
 2
21
1
log 210 
 n 
100
1
21
1
2

 n
 n2
21100   n2
299       n2
2ln99ln  
 
 
n
2ln2
99ln
 31467831.3n  4n
))(( 1
1
f
a
kg
k
L  
)100002)(7654.0(
50
1

L  3
1 1003968476.1 
L  mHL 11 
))()((
1
2
2
fa kkg
C  
)8478.1)(50)(100002(
1
2

C  9
2 102642527.172 
C  nFC 1722 
))(( 3
3
f
a
kg
k
L  
)100002)(8478.1(
50
3

L  3
3 104306606318. 
L  mHL 431.3 
))()((
1
4
4
fa kkg
C  
)7654.0)(50)(100002(
1
4

C  9
4 108739041.415 
C  nFC 1722 
Design a maximally-flat, High pass filter with the cut off frequency of 10 kHz and an attenuation of 20 dB at 5 kHz. Use
amplitude scaling so that the filter is terminated at 50Ω resistance. Use PSpice to simulate/ validate the response of your
circuit.
Decibel attenuation to find order:
Maximally Flat High Pass Filter (4th Order):
Component Calculations:
element values for n=4 order high pass filter
g1=0.7654 g2=1.8478 g3=1.8478 g4=0.7654 g5=1
Maximally Flat High Pass Filter Design:
))((
'
fn
a
kg
k
LC 
))()((
1
'
fan kkg
CL 

5. 











12
022
2
21
1
)0(
)(
ff
ff
Tk
H
fH
n
1
1
))(coscos(
))(coshcosh(
{
1
1
)(





x
x
xn
xnn xT


  
 
 
10
1
2
))((
gf
k
L a





)6703.1100002(
)50(1.0
1



L  6
1 106426220.47 
L  HL 6.471 
)2( 0
2
2


f
kg
L a


)1.0)(100002(
)1926.1(50
2

L  3
2 10490409257.9 
L  mHL 49.92 
))(100002)(( 2
2
akg
C


 
)50)(1926.1)(100002(
)1.0(
2

C  9
2 1069041474.26 
C  nFC 7.262 
))(2( 0
1
1
akf
g
C




)50)(1.0)(100002(
6703.1
1

C  6
1 10316730026.5 
C  FC 3.51 
30
3
2
))((
gf
k
L a





)3661.2100002(
)50(1.0
3



L  6
3 1063233656.33 
L  HL 6.333 
))(2( 0
3
3
akf
g
C




)50)(1.0)(100002(
3661.2
3

C  6
3 10316730026.5 
C  FC 3.53 
)2( 0
4
4


f
kg
L a


)1.0)(100002(
)8419.0(50
4

L  3
4 10699627329.6 
L  mHL 7.64 
))(100002)(( 2
4
akg
C


 
)50)(8419.0)(100002(
)1.0(
4

C  9
2 1080851481.37 
C  nFC 8.372 
205.99)9841.1)(50()( 5  gkR a
 100R
Design an ER-BPF with maximum ripple of 0.5 dB with a center frequency of 10 kHz and 10% bandwidth ( 1.0 ).
Assume that the filter provides an attenuation of 20 dB at 0ff  = .
*Ripple:
*In this case
* For n = 4:
Equal Ripple Band Pass Filter:
Component Calculations:
*element values for n=4 order of 0.5dB Ripple
g1=1.6703 g2=1.1926 g3=2.3661 g4=0.8419 g5=1.9841
 


02
'
f
L
LinductorL

in series with capacitor (C')
Lf02 



Cf
LinductorC
02
)'(




in parallel with capacitor (C')


02 f
C

   020 22 ffff 
))(coshcosh()(2 1
xnxTx n

 dB
n
20
))2(cosh(cosh122.1
1
log10 1210 






 
dBdB 20
898.1148
1
log1020
))2(cosh4(cosh122.1
1
log10 101210 












 
)1(log10 2
10 kRipple  )1(log105.0 2
10 k 205.0
110 k 122018454.02
k
)060281473.3(10  dB6.30
TruedBdB  206.30 4n

6. Equal Ripple Band Pass Filter Design:
Simulation Results:
Low Pass Filter
High Pass Filter
Band Pass Filter

7. Experiment Results:
Sine Wave Output Square Wave Output
Low Pass Filter: High Pass Filter:
Band Pass Filter
4kHz 10kHz 20kHz