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8/17/2019 CT Filters & Freq Resp
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EECE 301
Signals & Systems
Prof. Mark Fowler
Note Set #15
• C-T Systems: CT Filters & Frequency Response
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Ideal Filters
Often we have a scenario where part of the input signal’s spectrum comprises“what we want” and part comprises something we “do not want”. We can use a
filter to remove (or filter out) the “bad part”.
H ( )
( ) x t ( ) y t
Called “a Filter” in this case
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Case #1: )( X
Spectrum of the
Input Signal
In this case, we want
a filter like this:
)( H
otherwise H ,0
,1
)(
Mathematically:
“Passband”
“Stopband”
A filter that
“passes”
low
frequencies
is called a“low-pass
filter”
Undesired
Part
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Case #2:
otherwise H ,1
,0
)(
“Passband”
“Stopband”
)( X
Input SignalSpectrum
Undesired
Part
We then want:)( H
1
A filter that
“passes”
high
frequencies
is called a
“high-pass
filter”
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( ) X
)( H
1
Case # 3:
A filter that “stops” middle
frequencies is called a
“band-stop filter”
Undesired Part
Case #4:
A filter that “passes” middle
frequencies is called a“band-pass filter”
( ) X
)(
H
1
Desired Part
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What about the phase of an IDEAL filter’s H ( )?
Well…we could tolerate a small delay in the output so…
From the time-shift property of the FT then we need:
d t jg e X Y
)()(
H ( )
)(t xg )()( d g t t xt y Put in the signal
we want “passed”
Want to get out the
signal we want“passed”… but we can
accept a “small” delay
d t j
t j
t e H
e H
d
d
)(
1)( For in the“pass band”
of the filter
Thus we should treat the exponential term here as H ( ), so we have:
Line of slope – t d
“Linear Phase”5/14
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So… for an ideal low-pass filter (LPF) we have:
otherwise
e H
d t j
,0
,1)(
)( H
Slope = d t
)( H
Phase is undefined in stop band:
?0
00
je
i.e. phase is undefined
for frequencies outside
the ideal passband
Summary of Ideal Filters
1. Magnitude Response:
a. Constant in Passband
b. Zero in Stopband
2. Phase Response
a. Linear in Passband (negative slope = delay)
b. Undefined in Stopband
p2( )
2( ) ( ) d j t H p e
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)32cos()22cos(3)2cos(59)( t t t t x
H ( )
0 Hz 1 Hz 2 Hz 3 Hz
Filter has
Non-Linear Phase
Filter has FLAT
Passband
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)10
32cos()22cos(3)4
2cos(59)(
t t t t y
Point of this Example
A filter with an ideal magnitude response but non-
ideal phase response can still degrade a signal!!!
Example of the effect of a nonlinear phase but an ideal magnitude
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Are Ideal Filters Realizable? (i.e., can we actually MAKE one?)Sadly… No!!
So… a big part of CT filter design focuses on how to get close to the ideal.
Can’t Get an Ideal Filter… Because they are Non-Causal!!!
For the ideal LPF we had2( ) ( )
d j t H p e
2 sinc 2 / 2t From the FT Table: 22 ( ) p
Now consider applying a delta function as its input: x(t ) = (t ) X ( ) = 1
Then the output has FT 2( ) ( ) ( ) ( ) d j t Y X H p e
Linear Phase
Imparts Delay
Ideal
LPFt
x(t ) =(t ) y(t )
t
Starts before input starts…
Thus, system is non-causal!8/14
So the response to a delta (applied at t = 0) is: ( ) ( / ) sinc ( / )( )d y t t t
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a logarithmic unit of measure for a ratio between two powersDecibel:
decibel
bel
2
110log10
P
P
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Plotting Frequency Response of Practical Filters
Although we’ve previously shown the plots of Freq. Resp. using the actualnumerical values of | H ( )| it is VERY common to plot its decibel values.
P1/P2(non-dB)
P1/P2(dB)
1000 = 103 30 dB
100 = 102 20 dB
10 = 101 10 dB
1 = 100 0 dB
0.1 = 10-1 –10 dB
0.01 = 10-2 –20 dB
0.001 = 10-3 –30 dB
P1/P2 = 2 ~ 3 dB
Another “Rule” to Know!!
P1/P2 = 1/2 ~ -3 dB
0 dB is “P ratio of 1”
10 dB is “P ratio of 10”
20 dB is “P ratio of 100”
30 dB is “P ratio of 1000”
-10 dB is “P ratio of 0.1”
-20 dB is “P ratio of 0.01”
-30 dB is “P ratio of 0.001”
Decibel Power Rules
Know
These!
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But… | H ( )| relates Voltages (or current)… not POWER!!!
h(t)
H( )
)cos( 0 t A ))(cos()( 000 H t H A
Output voltage amplitude = A| H (
0)|
AmplitudeVoltageInput
AmplitudeVoltageOutput|)(| 0 H
Input voltage amplitude = A
Convert to PowersInput Power = A2/2
Output Power = A2|H(0)|2/2
22
10 10 2
2
10
10
( ) / 210 log 10 log
/ 2
10 log ( )
20 log ( )
out
in
A H P
P A
H
H
20 log10
(| H ( )|)Decibel value for | H (
0)|
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In addition to using decibels for the | H ( )| it is also common to use a
logarithmic scale for the frequency axis
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24
68 20
4060
80
10 100
200
We may be just as interested in 0 – 1 kHz as we are in 1 – 10 kHz ─ But the linear axis plot has the 0 – 1 kHz region all “scrunched up”
─ However… the log axis allows us to expand out the lower
frequencies to see them better!
Linear Axis:
Log Axis:
0 1000 2000 3000 4000 5000 6000 7000 8000 9000 100000
f (Hz)
100
101
102
103
104
0
f (Hz)
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Simplest Real-World Lowpass Filter: RC Circuit
v(t ) R C y(t ) Output SignalInput Signal
1. Convert capacitor into impedance:
C j
Z c
1
)( Small impedance at high
Large impedance at low
x Ae j 2. Imagine input as phasor:
3. Now analyze the circuit as if it were a DC circuit with a complex voltage
in (the phasor) and complex resistors (the impedances):
R
1/ j C
y
x Ae j
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Now find the output phasor
as a function of the input
phasor… the thing that
multiplies the input phasor
is ALWAYS the Freq Resp !
( ) 1
( ) 1
c
c
Z y x x
R Z j RC
Here…
use
Voltage
Divider.
1( )
1 H
j RC
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Now… we can plot this
0 1000 2000 3000 4000 5000 6000 7000 8000 9000 1 0000
0
0.5
1
f (Hz)
| H ( f ) |
0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000-1.5
-1
-0.5
0
f (Hz)
< H ( f )
RC=1.5915e-4;
f=0:10:10000;
H=1./(1 + j*2*pi*f*RC);
subplot(2,1,1)
plot(f,abs(H))grid
xlabel('f (Hz)')
ylabel('|H(f)|')
subplot(2,1,2)
plot(f,angle(H))
gridxlabel('f (Hz)')
ylabel('
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Instead… we can plot this using dB and log axes…
RC=1.5915e-4;
f=1:10:10000;
H=1./(1 + j*2*pi*f*RC);subplot(2,1,1)
semilogx(f,20*log10(abs(H)))
grid
xlabel('f (Hz)')
ylabel('|H(f)|')
subplot(2,1,2)semilogx(f,angle(H))
grid
xlabel('f (Hz)')
ylabel('
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