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page 33
Analog System Lab Kit PRO
Design a Band PassandaBand Stop lter.For the BPF, assume
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
and
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
. For the BSF, assume
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
and
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
.
Steady State Response - Apply a square wave input (Try
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
and
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
to both BPF and BSF circuits and observe the outputs.
Band Pass output will output the fundamental frequency of the
square wave multiplied by the gain at the centre frequency. The
amplitude at this frequency is given by
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
, where
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
is the
peak amplitude of the input square wave.
The Band Stop lter’s output will carry all the harmonics of the
square wave, other than fundamental. This illustrates the application
of BSF as a distortion analyzer.
Frequency Response - Apply the sine wave input and obtain the magnitude
and the phase response.
Simulate the circuits and obtain the Steady-State response and Frequency
response.
Take the plots of the Steady-State response and Frequency response from the
oscilloscope and compare it with simulation results.
Frequency Response - Apply a sine wave input and vary its input frequency
to obtain the phase and magnitude error. Use Table 4.2 and 4.3 to note your
readings. The nature of graphs should be as shown above.
The magnitude and phase response of LPF, BPF, BSF, and HPF lters are shown in
Figure4.2.Notethatthelow-passlterfrequencyresponsepeaksat
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
and has a value equal to
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
. The phase sensitivity
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
is maximum at
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
and is given by
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
. This information about phase variation can be used
to tune the lter to a desired frequency
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
. This is demonstrated in the next
experiment.
Forthebandpasslter,themagnituderesponsepeaksat
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
and is given by
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
.Thebandstopltershowsanullmagnituderesponseat
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
.
4.2Specication
4.3 Measurements to be taken
4.4 What you should submit
experiment 4
Table 4-2: Frequency Response of a BPF with
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
,
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
Table 4-3: Frequency Response of a BSF with
d
d
2
1
1
1
1
10
10
1
10
210/
10
1000.1sinsin
N
N
N
N
N
RC
H
V
V
Q
ss
H
V
V
Q
ss
H
s
V
V
Q
ss
H
s
V
V
Q
ss
s
H
Q
Q
HQ
Q
HQ
kHz
Q
kHz
Q
f kHz
f kHz
HQ
V
V
rad s
H
ttt
2
1
2
1
1
1
1
1
2
1
1
4
1
2
4
200
0
03
2
2
0
01
0
2
2
0
0
0
4
th
i
i
i
i
p
p
0
0
0
0
0
2
2
0
2
02
0
0
2
2
0
0
04
0
0
2
2
0
2
2
0
0
2
0
0
0
0
0
0
0
$
$
$
$
$$
$
$
2
~
~
~
~
~
~
~
~
~
~
~
~
~
~
z
~~
~
~
~
~
r
~r
yrr
=
=
++
=
++
=
++
+
=
=
=
=
=
=
=
=
=
=+
-
=
++
+
-
-
-
-
b
b
b
b
a
b
b
_
__
l
l
l
l
k
l
l
i
ii
,
Frequency Response of Filters
1
1
2
3
2
Band Pass Band Stop
S.No. Input Frequency Phase Magnitude Phase Magnitude
1
2
3
4
Band Pass Band Stop
S.No. Input Frequency Phase Magnitude Phase Magnitude
1
2
3
4