Electronics Study Material

Electronics 1
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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 √ 100.1 ×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