ENG632d1 Electronics Assignment: Filter Design, PLL Analysis (2018/19)

This assignment solution focuses on electronics, specifically addressing filter design and Phase-Locked Loop (PLL) analysis. It begins by determining the required order of a Butterworth filter based on given specifications, including passband and stopband frequencies and attenuation levels. The solution then proceeds to design a Butterworth passive filter, detailing the prototype circuit, component values, and actual filter component values. Furthermore, it analyzes a PLL with a first-order low-pass filter, calculating the frequency at which the PLL reaches its maximum response. The assignment also includes the design of a Chebyshev passive filter and the conversion of a filter circuit into the S-domain for frequency response analysis. Finally, the solution calculates the loop gain, time constant, and rise time for a given system, along with the phase detector output when the input frequency changes. Desklib offers this and many more solved assignments.

Electronics 1
ELECTRONICS
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Electronics 2
QUESTION ONE
Given that information vf low pass filter has the following specification;
Fp=2.4 MHz
Fs=MHZ
Lp=3dB
Ls= 40dB
There,
Fp= 2.4 MHz = Ωp=2A*106 = 15.07*106 rad/sec
Fe = 6MHz = ΩA=2A*6*106 = 37.7 *106 rad/sec
Lp= 3dB
Therefore the cutoff frequency is
Ωc= Ωp
¿ ¿
But for ∝p= dB , the term 1 00.1 ∝ p-1 = 1 00.1∗3-1 = 1
Therefore Ωc(cut off frequency ) = Ωp ( pass band frequency)
For αp= 3dB
Now, the order of the filter for butter, worth approximation is given by

Electronics 3
N =
log √ 1 00.1∝ p −1
1 00.1∝ s−1
log (Ωp
Ωs )
N =
log √ 1 00.1× 3−1
1 00.1× 4−1
log ( 15.07 ×1 06
57.6 ×1 06 )
= 5.928
N≈ 6
QUESTION 2
Here, N=6, the characteristics equation is given by
Ga (s) = (s2+0.518s+1)(s2+1.414s+1)(s2+1.432s+1)
Therefore the transfer function of analog prototype filter is given by (NAIR, 2014).
Ha(s)= 1
Ga(s)
( Prototype : A low pass filter with Ωc= 1rad/sec
Ha (s) = 1
( S2 +0518+1)(S2 +1.414 S+1)¿ ¿
To design the required low pass filter from prototype filter, replace S with S
Ωc in Ha (s)
Therefore
But Ωc = Ωp= 15.07 ×106 rad/sec

Electronics 4
QUESTION THREE
Given that wn = 4rad/sec
Ɛ=0.5
Wr=? For maximum peak (1 1+(5+w)
Consider the phase locked loop with first order RC filter as follows
Open loop filter used as PLL filter
Closed loop system
The closed loop transfer function given characteristics equation (Paarmann, 2011).
S2 +2 Ɛ WnS +Wn2 =0

Electronics 5
Where Wn = Natural frequency
Ɛp= damping factor
As ∅ out ( s)
∅ ∈(s ) =
KpKv /R
S2+ 1
RC S+ KpKv
RC
Where Wn = √ KpKv
RC
Ɛ= 1
2 √ KpKvRc
Maximum Maginitude (H(Jw)) at Wr = wn √1−2 Ɛ 2
So the maximum amplitude of the response will occur t frequency Wr given by
Wr=Wn √ 1−2 Ɛ2
Wr= 4× √1−0.52 ×2
Wr= 4× √ 1−0.25 ×2
Wr= 4× √0. 5 = 4
√ 2 = 2√2
Wr= 2.82 rad/sec
QUESTION FOUR
A) PART 1
Chebyshev filter design
It provides more efficient approximation than Butterworth filter
Given that Fp=8KHz ,
Fs = 6.8 KHz
Lp=Amax = 0.5 dB , Ls = A min =26dB
R1=RL=KΩ
A max = 10 log (1+ε 2 ) where ε=
√ 1 0
A Max
10 −1
1 0
0.5
10 .−1
= 0.349
By trying various values of N in expression
A(Ws)=10 log [1+ ε2 cosh 2 (N cosh−¿1 (ws/wp)]

Electronics 6

Electronics 7
PART 2
QUESTION FOUR
PART B
Convert the given filter circuit into S domain.
i.
Vout
Vin =
R
RCS+1
(R+ 1
CS )+( R
RCS +1 )
Vout
Vin =
S
RC
S2+ 3 S
RC + 1
R2 C2
Putting the given values, the transfer function will be

Electronics 8
H(s) = Vout
Vin = 1 03 S
S2+3× 103 S+1 06
ii. Now for a maximum magnitude response , we know that
H(jw) = 1
H(jw) = 1 03 jw
−W 2 +3 ×1 03 jw +106
H(jw) = 103 jw
√ ( 1 06 −w2 ) 2+ (3 × 103 ) 2
103 w = √ ¿ ¿
103w2=1012+w4 −∝× 1 06w2+9×1 06
W4−¿3×1 06w2+(9×1 06 +1012) =0
Solving the quadratic equation we obtain
W1=618.04 and w2= 1618.03
W1= 2 πf1 and w2 = 2 πf2
F1= 98.414Hz
F2= 2 57.65Hz
These are two frequencies at which filter have maximum magnitude
iii.
Now for the gain of the filter consider the circuit as below;

Electronics 9
Vout
Vin = WRC
W 2 R2 C2+3 WRC +1
W= ∝ πf
Vout
Vin = Gain =
2 πf
RC
¿ ¿
Now for DC , f= 0
Gain =0
For f = 1
2 πRC
= 1
¿ ¿ = 1
4+ ¿ ¿

Electronics 10
= 0.2 5 ×10−6
Similarly at f = 100
2 πRC
Gain = 0.97×1 0−2
R=2kΩ
C=0. 5μF
PART C
I
Magnitude response of the N th order low pass Butterworth filter .
Butterworth filter provides a maximally flat response
This filter is designed so that at zero frequency , the first 2n−¿ 1 derivatives for the power
function with respect to frequency are zero (S.Chitode, 2010).
ButterWorth filter frequency response
Where f
=frequency at which alculations are made
F0= the cutoff frequency
Vin = input voltage
Vout= Output voltage
N= Number of elements in the filter
Transfer function H (jw)
It is assumed that filter has zero gain

Electronics 11
Chebyshev filter
Chebyshev filter has a steeper roll off and more pass band ripple than Butterworth filter
Transfer function (H(jw)) =
1
√1+ ε2 Tn2 ×( W
W 0 )
Where Ɛ= ripple factor
W0= Cutoff frequency, Tn = Chebyshev polynomial of the nth order
ii.
From the above, we can observe that Chebyshev filter has more rolloff rate in stop band.
Iii,

Electronics 12
Attenuation for 7th order filter at 1jw=w is given by
= 1
26 .8 6
Attenuation =0.03723
For first order
Attenuation factor = 1
√1+¿ ¿ ¿
= 1
√3. 56
= 0.886
7th order attenuation factor = 0.03723
1st order attenuation factor = 0.5302
Therefore as the order increases attenuation decreases (Wanhammar, 2012).
QUESTION FIVE
a).
The loop gain is Kv= KdKo
Kv=( 1. 5)×2 π × 4 5× 1000