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

Fig 1

2)

When capacitor connected to parallel to R1

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

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

Problem 2:

Fig 2:

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

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

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,

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

Magnitude and phase response for 1st order

Circuit for 2nd order low pass filter

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

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.

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