This document summarizes the results of 7 problems completed as part of an electronics engineering assignment. 1) The output of a half-wave rectifier circuit with and without a capacitor was examined and the capacitor value was found to affect the output waveform. 2) The frequency response of a circuit was measured and found to behave as a low-pass filter with a phase shift to -120 degrees at higher frequencies. 3) Key parameters of a low-pass filter circuit including cutoff frequency and quality factor were calculated. 4) The conditions for oscillation in a circuit were derived and used to solve for a component value. 5) The frequency responses of first and second-order low-pass filters using

1. ASSIGNMENT
EE204
NAME:PRADEEP GODARA
ROLL NO:EE1200223
Problem 1)
Green graph is output and blue graph for input
1) When capacitor not connected
It’s behave like half wave rectifier
Output across R1 is green graph

2. Fig 1
2)
When capacitor connected to parallel to R1

3. a)When c1 take 1uf
Output very between 10 to 8 volt
Now I reduces capacitor value
b)c1 =100nf
Output is between 10 to 4 volt

4. c)c1 = 10n f
c) c1=1nf
As we decrease capacitor value output behave like half wave rectifier
Because time constant T=RC
When C decrease ,T decrease ,then discharging time decrease and when C take 1nf
then then capacitor take very less time to discharge so output same as input in
positive half cycle

5. Problem 2:
Fig 2:

6. Magnitude and phase response
From graph of problem 2 got that circuit is low pass filter
And at frequency 1 Hz phase is zero and as increase frequency phase go to negative to
-120

7. Problem 3:
Circuit :Low-pass filter
Phase and magnitude reaponse
Form graph get cutoff frequency Fo =3.4kHz ,
So Wo =2*pi*Fo =21.352k rad/s
From circuit Wo =1/( squareroot of L*C) =22.360k rad/s
Quality factor Q= Wo RC =21352*500*10^(-6) = 10.67

8. Problem 4:
A=150 v/v
Oscillation condition =>
(1+R2/R1)/1+(1/A)(1+R2/R1) =1+R3/R4+C2/C1
A=150 v/v,take R3=R4=1k ohm,c1=c2=.1uf,
Then got R2/R1=2.08,

9. Time period is 0.6293ms .
Freq.= 1/T =1.589 kHz.
Problem 5:
In problem 5 I take R=500 ohm ,C=1uf for both circuit
In 2nd
order circuit L=1mH
Circuit for 1st
order low pass filter

10. Magnitude and phase response for 1st
order
Circuit for 2nd order low pass filter

11. Magnitude and phase response for 2nd order
Here for same value of R &C :
In 1st
order phase shift very slowly ,
In 2nd
order phase shift suddenly after max gain get
in 2nd
order we get more gain ~=40 dB.(max)
But in 1st
order get 7dB
2nd
order curve is more stiffer then 1st
order .
So we can find good low pass in 2nd
order,
That’s why 2nd
order

12. Problem 6:
Low pass Butterworth filter of order three
V1 for N =1,V2 for N=2,V3 for N=3,…
As value of N increase we get more stiffer so we get good low pass filter.

13. Problem 7:
here I take V1=V2=10 volts then output get
Vo =squareroot ofV1*V2 =10 volts
But here Vo is 9.993 volts almost equal to 10 volts
Circuit for geometric mean
Output of geometric mean