2260
F 09
HOMEWORK #3 prob 1 solution
EX:
R=100 kΩ
+
vi
+
vo
C=1 µF
–
a)
b)
c)
–
Determine the transfer function Vo/Vi.
Plot |Vo/Vi| versus ω.
Find the cutoff frequency, ωc.
SOL'N: a) Filters are voltage-dividers. The output is measured across the C, so the
transfer function is the impedance of the C divided by the total impedance
of R and C.
V
1/ jωC
H( jω) = o =
Vi R + 1/ jωC
or
€
H( jω) =
1
1+ jωRC
b) For the plot, we compute the numerical value of RC:
€
RC = 100k *1µ s = 0.1 s
The following Matlab code makes the plot:
€
% ECE2260F09_HW3p1Matlab.m
%
% Plot of filter's frequency response curve
figure(1)
omega = 0:1:100;
s = j * omega;
FilterResp = 1./(1 + j * (1/10)*omega);
plot(omega,abs(FilterResp))
axis([0, max(omega), 0, 1])
xlabel('omega')
ylabel('|H|')
title('HW 3 prob 2 Frequency Response')
c) The cutoff frequency, ωc, occurs where the magnitude of H(jω) is 1/ 2
times the maximum value attained by H(jω) for any ω > 0.
From the plot in (b), the maximum magnitude of H(jω) occurs when ω = 0
€
and has a value equal to one. Thus, the cutoff frequency will occur where
H(jω) = 1/ 2 .
H( jω c ) =
€
1
1
max H( jω) =
2 ω
2
€
In part (a) we derived the following convenient form for the transfer
function:
H( jω) =
1
1+ jωRC
So we want to solve the following equation:
€
H( jω c ) =
1
1
=
1+ jω c RC
2
This form is convenient because the magnitude of H( jω) is 1/ 2 when
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the imaginary part of the denominator is equal to one.
€
€
H( jω) =
1
1
1
=
when H( jω) =
1+ jωRC
1+ j
2
This holds because 1+ j = 2 .
€ We conclude that the cutoff€frequency is found by solving the following
equation:
€
ω c RC = 1
or
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ωc =
NOTE:
1
1
=
= 10 r/s
RC 0.1 s
Because
€
1
1
=
, it is always convenient to express
1+ j
2
transfer functions in the following form when finding cutoff
frequencies:
1
€
H( jω) =
1+ jX
where X ≡ real number or quantity
€