SUBJECT:
EEE222 ECA II
INSTRUCTOR
:
SYED BILAL JAVED
SUBMITTED BY:
UMAIR AHMAD
REGISTRATION:
FA18
-
BEE
-
155
L
AB REPORT#11
COMSATS UNIVERSITY ISLAMABAD (CUI)
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ECA 2 LAB 11 REPORT (Filter Circuits), Lab Reports of Electrical Circuit Analysis
RC/RL Low Pass and High Pass Filters, first order and higher order
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2020/2021
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SUBJECT: EEE222 ECA II
INSTRUCTOR: SYED BILAL JAVED
SUBMITTED BY: UMAIR AHMAD
REGISTRATION: FA18-BEE- 155
LAB REPORT# 11
COMSATS UNIVERSITY ISLAMABAD (CUI)
COMSATS University Islamabad
Department of Electrical and Computer Engineering
Lab 11
ECA II
Name: ______UMAIR AHMAD____ Reg. No: ______FA18-BEE- 155 ____
- To design RC low pass & high pass filter circuit with cutoff frequency of 155
KHZ, also draw a frequency response theoretically on semi-log plots: (a)
magnitude response (b) phase response.
- To design RL low pass & high pass filter circuit with cutoff frequency of 155
KHZ,, also draw a frequency response theoretically on semi-log plots: (a)
magnitude response (b) phase response.
RC Low/High Pass Filter
RC Low / High Pass Filter:
The cutoff frequency of a RC low / high pass filter is:
𝑐
1
𝑅𝐶
Taking the value of capacitor C = 10uF and f c
= 155 KHz, we get:
1
𝐶
1
2
( 155 × 10
3
)( 10 × 10
− 6
)
RC Low Pass Filter:
So our required circuit is:
RC High Pass Filter:
Our required circuit is:
The magnitude transfer function and phase of RC High-Pass filter circuit is:
| H(ω)| =
𝑅
√
𝑅
2 +(
1
𝜔𝐶
)
2
= 90 − tan
− 1
𝑅𝐶
For R=102.681m, C=10uF we have:
ω |H(ω)| (dB)
ω = 0 20log( 0 ) 90
ω = ∞ 20log( 1 ) 0
ω = ω c
= 2 * 155 Khz 20log(1/ √
Bode Plot:
RL Low/High Pass Filter
RL Low / High Pass Filter:
The cutoff frequency of a RL low / high pass filter is:
𝑐
𝑅
𝐿
Taking the value of inductor as L = 1mH and f c
= 155 KHz, we get:
𝐶
𝑅 = 2 ( 155 × 10
3
1 × 10
− 3
RL Low Pass Filter:
So our required circuit is:
RL High Pass Filter:
Our required circuit is:
The magnitude transfer function and phase of RL High-Pass filter circuit is:
| H(ω)| =
2
𝑅
𝐿
2
= 90 − tan
− 1
𝐿
𝑅
For R=973.9, L=1mH we have:
ω |H(ω)| (dB)
ω = 0 20log( 0 ) 90
ω = ∞ 20log( 1 ) 0
ω = ω c
= 2 * 155 Khz 20log(1/ √
Bode Plot:
1KHz 10KHz 100KHz
V(vout)/V(vin) V(vout)/V(vin) --- C:\Users\Useless Noob\Documents\Umair\Sem 5th\ECA 2\LAB\Lab 11\Pre Lab.raw --- 5.4068849KHz,999.39184m 154.95959KHz,707.24435m Fc = Stop Band Pass Band Here we observe that at the cutoff frequency Fc, magnitude is 70.7% of the maximumm gain (1) Also for frequecy lower than cutoff, gain aproches max value and for values greater than cutoff, Gain aproaches zero, which matches with our theoretical results. 154.95959KHz,-44.985054° We observe that at cutoff phase is equal to -44.98 deg which is apporximatley equal to our theoretical results
AC 5
V
R
102.681m
C
10μF
Vin Vout
.ac dec 50 1KHz 500KHz
RC HighPass Filter
--- C:\Users\Useless Noob\Documents\Umair\Sem 5th\ECA 2\LAB\Lab 11\Pre Lab.asc ---
AC 5
V
R
L
1mH
Vin Vout
.ac dec 50 1KHz 500KHz
RL Low Pass Filter
--- C:\Users\Useless Noob\Documents\Umair\Sem 5th\ECA 2\LAB\Lab 11\Pre Lab.asc ---
1KHz 10KHz 100KHz
V(vout)/V(vin) V(vout)/V(vin) --- C:\Users\Useless Noob\Documents\Umair\Sem 5th\ECA 2\LAB\Lab 11\Pre Lab --- Stop Band Pass Band 154.95959KHz,707.24761m Here we observe that at the cutoff frequency Fc, magnitude is 70.7% of the maximumm gain (1) Also for frequecy lower than cutoff, gain aproches max value 1 and for values greater than cutoff, Gain aproaches zero, which matches with our theoretical results. Fc = We observe that at cutoff phase is equal to -44.98 deg which is apporximatley equal to our theoretical results 154.95959KHz,-44.98473°
1KHz 10KHz 100KHz 0m 90m 180m 270m 360m 450m 540m 630m 720m 810m 900m 990m
V(vout)/V(vin) V(vout)/V(vin) --- C:\Users\Useless Noob\Documents\Umair\Sem 5th\ECA 2\LAB\Lab 11\Pre Lab --- Pass Band Stop Band Here we observe that at the cutoff frequency Fc, magnitude is 70.7% of the maximumm gain (1) Also for frequecy lower than cutoff, gain aproches zero and for values greater than cutoff, Gain aproaches max value 1, which matches with our theoretical results. 155.19492KHz,707.40876m Fc = 155.19492KHz,44.971691° We observe that at cutoff phase is equal to 44.97 deg which is apporximatley equal to our theoretical results
Post Lab
1) Why are capacitors preferred over inductors in filter design?
The basic reason is physical and economical. In electronics, all the parts should be as small as
possible. An average capacitor is much smaller and cheaper as compared to that of an inductor.
Moreover manufacturing of an inductor having tolerance less than 10% is very difficult and its
inductance may vary due to stray-magnetic field in the surroundings. whereas a capacitor can be
manufactured of specific value having very much less tolerance.
Inductors are very hard to realize in IC circuits and occupy a very large area. A capacitor on the
other hand can be realized using a varactor diode, MOS based capacitors etc.
2) Create Bode plots of the magnitude transfer functions of your low-pass
and high-pass filters. The theoretical plots (using measured values of
resistors and capacitors) should be drawn as lines. Include your data
(taken at 1/10, ½, 1, 2, and 10 times the cutoff frequency) as points.
RC Low/High Pass Filter
The magnitude transfer function and phase of RC Low-Pass filter circuit is:
| H(ω)| =
1
√ 1 +(𝜔𝑅𝐶)
2
= − tan
− 1
( 𝑅𝐶)
ω |H(ω)| (dB)
ω = 0 20log( 1 )=0 0
ω = ∞
20log( 0 ) - 90
ω = ω c
= 2 *155Khz 20 log(1/ √
The magnitude transfer function and phase of RC High-Pass filter circuit is:
| H(ω)| =
𝑅
√
𝑅
2 +(
1
𝜔𝐶
)
2
= 90 − tan
− 1
𝑅𝐶
ω |H(ω)| (dB)
ω = 0 20log( 0 ) 90
ω = ∞ 20log( 1 ) 0
ω = ω c
= 2 *155Khz 20log(1/√ 2 )
RC High Pass Filter:
Our required circuit is:
Fc = 155/10 = 15 .5KHz:
The cutoff frequency of a RC low / high pass filter is:
𝑐
1
𝑅𝐶
Taking the value of capacitor C = 10uF and f c
= (155)/10 = 15 .5 KHz, we get:
1
𝐶
1
2 ( 15. 5 × 10
3
)( 10 × 10
− 6
)
RC Low Pass Filter:
So our required circuit is: