using bode plots
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EECE 301Signals & SystemsProf. Mark Fowler
Note Set #37• C-T Systems: Using Bode Plots
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Bode Plot Ideas Can Help Visualize What Circuits Do…
1 1( )
1 1
RC H s
RCs s RC
1( )
1
H
jRC
RC Lowpass Filter
Break Point = 1/ RC
Break Point = 1/ RC
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( )1 1
RCs s H s
RCs s RC
RC Highpass Filter
c= 1/RC
RLC Bandpass Filter
2 2
2( )
2n
n n
s H s
s s
101
102
103
104
105
-60
-50
-40-30
-20
-10
0
10
CT Frequency (rad/sec)
| H ( ) | ( d B )
–172
–5828
0.1
1
3
–1000
–1000
–100+ j995
–100- j995
2
2( )
( )ns
H ss a
2( )
( )( )ns
H ss a s b
All slopes are 20 dB/decade!
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A Better Bandpass Filter? Suppose you want a BPF…. with faster “rolloff”!
1 2
Need an s2
term in the numerator!At 1 we need to change slope by –40 dB/dec… So need double pole @ 1!
At 2 we need to change slope by –40 dB/dec… So need double pole @ 2!
2
2 21 2
( )( ) ( )
Ks H s
s s
There are (at least) three ways to get this!
1 2
1 2 1 2
( )( )( ) ( )( )
K s K s H s
s s s s
1 22 2
1 2
( )( ) ( )
K s K s H s
s s
BPF RLC w/
distinct poles
21 2
2 21 2
( )( ) ( )
K s K H s
s s
HPF RLC w/
Rep. polesBPF RLC w/
repeated poles
LPF RLC w/
Rep. poles
Same exact circuits… just different choices of RLC!!!
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Looks like we could just cascade two of our RLC circuits… Here we cascade BPFs.
Although TF Theory says this will work… the problem is that the second
circuit “Loads” the first one!!!So… one approach would be to re-analyze this cascade and see if it will stillwork but with some “tweeks” on the component choices.
Our “cascade theory” onlyholds when attaching the
2nd system does not change
how the first one behaves!
Another approach is to use an op amp as a “buffer” between the stages!
It may be desirable
to add another
buffer here…
VERY large input resistance
of op amp prevents “loading”
of first stage!
VERY small output resistance
of op amp minimizes impact
on 2nd stage!
1 1 1 2 2 2
Bandpass RLC Bandpass RLC
Remember… there are two ways to choose the components here:1. Each stage has repeated poles 2. Each stage has distinct poles
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2nd Ord Highpass RLC 2nd Ord Lowpass RLC
Another way to make a better BPF:Here we must choose the components so that
each stage has repeated poles.
Although these ideas lead to workable circuits they are not necessarily the best…For one thing… they need inductors (which are big and can’t be made in an IC!)There are other forms… See this link for the form used below.
+
–
+
–
2nd Ord Highpass RC 2nd Ord Lowpass RC
http://pdfserv.maxim-ic.com/en/an/AN1795.pdf
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Design Example using Bode Plot Insight
Suppose you want to build a “treble booster” for an electric guitar.You decide that something like this might work:
100 1000
0dB
20dB)( f H
(Hz) f
100Hz 628 rad/s =
1000Hz 6283 rad/s =
Notice that we are doing our rough“design thinking” in terms of Bode
Plot approximations!!!
The A string on a guitar has a fundamental frequency of 110 Hz
The A note on 17th fret of the high-E string has a fundamental frequency
of 880 Hz
)/1(
)/1()(
)/1(
)/1()(
2
1
2
1
j
j H
s
ss H
From our Bode Plot Insight… we know we can get this from a single real pole,single real zero system… with the “zero first, then the pole”:
with: 1 = 628 rad/s
2 = 6280 rad/s
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A series combination
…has impedance Z (s) = R + 1/Cs
Note: we could get R + sL withan inductor but inductors are
generally avoided when possible
So what do we get if we could some how form a ratio of such impedances?
1
1
/1
/1
)(
)(
22
11
1
2
22
11
2
1
C sR
C sR
C
C
sC R
sC R
s Z
s Z
222
111
/1
/1 : Let
C R
C R
Aha!!! What we want!
Now, how do we get a circuit to do this? Let’s explore!
2
1
1
2
2
1
/1
/1
)(
)(
j
j
C
C
Z
Z
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Okay…how do we build a circuit that has a transfer function
that is a ratio of impedances?!
)(t vin)(t vo
i
f
R
RGain
Ratio of
resistances
Extending the analysis to include impedances we can show that:
)(t vin )(t vo
)(s Z i
)(s Z f
)(
)()(
s Z
s Z s H
i
f
Won’t affect ourmagnitude: | H ( )|
Recall the op-amp inverting amplifier!
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)(t vin)(t vo
But wait!! You then remember that op amps must always have negativefeedback at DC so putting C f here is not a good idea…
So we have to continue…
We also might not like this circuit because it might not give us a verylarge input impedance… and that might excessively “load” the circuit
that you plug into this (e.g., the guitar)
Back to the drawing board!!!
Now, you can choose the R’s & C’sto give the desired frequency points
p. 98, The Art of Electronics,Horowitz & Hill, Cambridge
Press, 1980
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Okay, then you remember there is also Non-Inverting Op-Amp circuit…
)(t vin)(t vo
Infinite input impedance
i
f
R RGain 1
We avoid the “no DC feedback”issue… but we’re not yet sure we’ll
get what we want!
( 1 / ) ( ) 1 /( ) 1
1 / 1 / 1 /
f i i f i f i
i i i i i i
R R C s R R R C s H s
R C s R C s R C s
Applying this gain formula we get:
Oh Cool!! We Get
What We want!
1
1)()(
sC R
sC R Rs H
ii
ii f
rad/s6283
1
rad/s6281
2
1
ii
i f i
C R
C R R
Set: Choose:
F1.0
5.115
i
i f
C
k Rk R
Using standard values
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Computed Frequency Response using Matlab
f (Hz)
| H ( f ) | ( d B )
Discrepancydue to use of
standardvalues
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1. Through insight gained from knowing how to do Bode plots byhand… we recognized the kind of transfer function we needed
2. Through insight gained in circuits class about impedances werecognized a key building block needed: Series R-C
3. Through insight gained in electronics class about op-amps we founda possible solution… the inverting op-amp approach
4. We then scrutinized our design for any overlooked issuesa. We discovered two problems that we needed to fix
5. We used further insight into op-amps to realize that we could fix theinput impedance issue using a non-inverting form of the op-ampcircuit
6. We didn’t give up at first sign that the inverting form might not giveus the form we want…a. Through mathematical analysis we showed that we did in fact
get what we wanted!!!!!!
Summary of Bode-Plot-Driven Design Example