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EECS 16B Designing Information Devices and Systems II
Spring 2021 Note 6: Bode Plots
Overview
Having analyzed our first order filters and gone through a design example in the previous Note to show why filter-design is important, we will now plot their transfer functions H (ω) (or frequency responses) using Bode Plots. In the previous Note, we generated tables of |H (ω)|, ]H (ω) at certain key values of ωc, and while this gave some intuition, it didn’t really show what happens at intermediate frequencies. There is immense value in visualizing transfer functions across a wide range of frequencies.
Throughout this section, we will use numerical approximations, which will not only prove useful in plotting a filter’s frequency response but also will help us better understand its behavior.
1 Bode Plots
When we make Bode plots, we plot the frequency and magnitude on a logarithmic scale, and the angle in either degrees or radians. We use the logarithmic scale because it allows us to break up complex transfer functions into its constituent components. Let’s start by generating Bode Plots for low-pass and high-pass first order filters, which will build our intuition. We will soon see how the analysis of more complex transfer functions can be broken down into parts.
1.1 Low-pass Filter
Recall our generalized model of a low-pass filter (perhaps RC or LR):
HLP(ω) =
1 + jω/ωc
We plot the magnitude of the frequency response assuming ωc = 106. Note how |HLP(ω)| is very close to
10 −^4
10 −^2
ω
|H
LP
(ω
Linear Approximation Exact |HLP(ω)|
1 for ω < ωc and |HLP(ω)| starts dropping off with slope −1 after ωc. We can look at 3 distinct regions on the plot:
- ω ωc: =⇒ jω/ωc ≈ 0. So, HLP(ω) ≈ 1 and |HLP(ω)| ≈ 1.
- ω = ωc: =⇒ H(ω) = (^1) +^1 j. So, |HLP(ω)| = √^12.
- ω ωc: =⇒ ω/ωc 1. Therefore HLP(ω) ≈ − j ω ωc. So, |HLP(ω)| ≈ ω ωc. On a log scale, this means that log|HLP(ω)| ≈ log ωc − log ω explaining behavior of dropping off with slope −1.^1
Now let’s plot the phase of HLP(ω): ]HLP(ω) is very close to 0 for ω < 0. 1 ωc and ]HLP(ω) is approxi-
ω
]
H
LP
(ω
Linear Approximation Exact ]HLP(ω)
mately − π 2 for ω > 10 ωc.
- ω 0. 1 ωc =⇒ jω/ωc ≈ 0. So, HLP(ω) ≈ 1 and ]HLP(ω) ≈ 0.
- ω = 0. 1 ωc =⇒ HLP( 0. 1 ωc) = (^1) + 1 j 0. 1 and ]HLP(ω) ≈ − 6 ◦.
- ω = ωc =⇒ HLP(ωc) = (^1) +^1 j and ]HLP(ω) = − 45 ◦.
- ω = 10 ωc =⇒ HLP( 10 ωc) = (^1) +^1 j 10 and ]HLP(ω) ≈ − 84 ◦.
- ω 10 ωc =⇒ ω/ωc 10. So, HLP(ω) ≈ − j · 0 2 and ]HLP(ω) ≈ − 90 ◦.
We can now better understand the values of the magnitude and phase at 0. 1 ωc, ωc, 10ωc (as seen in the tables of Note 5).
1.2 High-pass Filter
We can similarly analyze our generalized high-pass filter model (CR, RL):
HHP(ω) =
jω/ωc 1 + jω/ωc
we plot the magnitude of the frequency response, again assuming ωc = 106.
Here, |HHP(ω)| rises with slope 1 for ω < ωc and |HHP(ω)| ≈ 1 after ωc. We analyze the plot:
- ω ωc, then ω/ωc 1. Therefore HHP(ω) ≈ j (^) ωωc which implies |HHP(ω)| ≈ (^) ωωc. On a log scale, this means that log|HHP(ω)| ≈ log ω − log ωc, which explainsthe rising slope of 1.
- ω = ωc, then H(ω) = (^1) +j j meaning |HHP(ω)| = √^12
- ω ωc, then ω/ωc 1. Therefore HHP(ω) ≈ 1 which implies |HHP(ω)| ≈ 1.
Now let’s plot the phase of the transfer function HHP(ω).
(^1) Recall that the line y = mx + b has slope m. In this case y = log|HLP(ω)| and x = log |ω|. (^2) The magnitude will always be greater than 0, meaning its phase will still be very close to − 90 ◦
−
RL
CL
vcenter(t), V˜center vin(t), V˜in
CH
RH
vout(t), V˜out
Figure 1: A Buffered Band-Pass filter, composed of low-pass and high-pass filter components.
Notice the op-amp serving as a unity gain buffer between the two filters. It is introduced to prevent the second circuit from loading the first.
The input voltage phasor is V˜in at some frequency ω. Suppose the two filters have cutoff frequencies ωLP and ωHP, respectively. Thus, we can write that:
V^ ˜center = HLP(ω)V˜in
Since vcenter(t) is the second filter’s input, the output voltage phasor V˜out is:
V^ ˜out = HHP(ω)V˜center = HHP(ω)HLP(ω)V˜in
Thus, the net transfer function HBP(ω) is:
HBP(ω) = HLP(ω)HHP(ω)
More generally, placing filters in series produces a circuit whose transfer function is the product of the indi- vidual transfer functions. From the properties of complex numbers and logarithms, we see that log |z 1 z 2 | = log |z 1 | + log |z 2 |. So, when plotting |HBP(ω)| on a log-log plot, this corresponds to a graphical addition of the individual transfer functions. Note that we are still taking the product of the actual transfer functions, but it resembles a geomtrical sum: |HLP(ω)| + |HHP(ω)|. We see this idea on display in the examples at the end of this note. Similarly, for ]HBP(ω), we can again plot the sum, ]HLP(ω) + ]HHP(ω).^3
We can compute HBP(ω) symbolically as:
HBP(ω) = HLP(ω)HHP(ω) =
1 + jωRLCL
jωRHCH 1 + jωRHCH
To find the cutoff frequencies of this filter, we can look at the points at which HBP(ωc) = √^12. But based
on our approximations from before and from the cuttof frequencies ωLP = (^) RL^1 CL and ωHP = (^) RH^1 CH , we can
approximate |H(ωLP)| ≈ √^12 · 1 and |H(ωHP)| ≈ 1 · √^12. This approximation holds best when the cuttof frequencies are spaced apart.
We’ve shown a convenient result! The cutoffs for the band-pass filter are identical to the individual cutoffs for the low and high-pass filters. We now plot HBP(ω) with ωLP = 106 and ωHP = 104 to demonstrate the band-pass behavior.
(^3) We must be careful, however, to note that in most of our plots, the x-axis does not correspond to 0, so we can’t simply "stack" the two plots.
10 −^4
10 −^3
10 −^2
10 −^1
ω
|H
BP
(ω
|HBP(ω)|
Our band-pass filter uses an op-amp, but what if we cascade our two filters, causing a loading effect?
RL^ CH
CL
Vmid RH
Vin Vout
We leave the derivation as an exercise (feel free to post your work on Piazza!), but computing the transfer function yields:
H(ω) =
jωRHCH ( 1 + jωRLCL)( 1 + jωRHCH ) + jωRLCH
The loading effect introduces a term of jωRLCH in the denominator. The relative size of this term determines the impact on the circuit. We plot some examples of the band-pass filter with identical low and high cutoff frequencies but different RLCH values to show this loading effect.
10 −^3
10 −^2
10 −^1
ω
|H
HP
(ω
10 −^4
10 −^3
10 −^2
Note how the maximum value of H (ω) decreases as RLCH increases. In addition, the cutoff frequencies move further and further apart from the original ωLP = (^) RL^1 CL and ωHP = (^) RH^1 CH.
3 Higher Order Bode Plots
We will now see how to plot a given transfer function by using straight-line approximations, and we will notice that a specific form of transfer function will make the process of plotting more systematic and simple.
3.1 Rational Transfer Functions
When we write the transfer function of an arbitrary circuit, it always takes the following form, called a "rational transfer function." We also like to factor the numerator and denominator, so that they become easier to work with and plot:
H(ω) = K ·
N(ω) D(ω)
= K
( jω)Nz^0
1 + j (^) ωωz 1
1 + j (^) ωωz 2
1 + j (^) ωωzn
( jω)Np^0
1 + j (^) ωωp 1
1 + j (^) ωωp 2
1 + j (^) ωωpm
To summarize the components, each transfer function is the product of constant gain K, one or more "origin-
poles" (( jω) in a denominator) or "origin zeros" (( jω) in a numerator), and one or more poles (
1 + j (^) ωωp i
in the denominator) or zeros (
1 + j (^) ωωz i
in the numerator).
Here, we define the constants ωz as "zeros" and ωp as "poles."" The zeros are the roots of N(ω) while poles are the roots of D(ω). 5
3.2 Composing Bode Plots
For two transfer functions H 1 (ω) and H 2 (ω), if H(ω) = H 1 (ω) · H 2 (ω),
log|H(ω)| = log|H 1 (ω) · H 2 (ω)| = log|H 1 (ω)| + log|H 2 (ω)| (3) ∠H(ω) = ∠(H 1 (ω) · H 2 (ω)) = ∠H 1 (ω) + ∠H 2 (ω) (4)
As a consequence, when plotting |H(ω)| on a log-log plot, we can simply plot |H 1 (ω)| and |H 2 (ω)| and add them up. This implies that we will be able to add the slopes of each pole and zero to provide a complete plot. In the next section we provide a further analysis on the meaning of zeros and poles and the idea of adding slopes.
3.3 Decibels (Optional)
We define the decibel as the following:
20 log 10 (|H(ω)|) = |H(ω)| [dB]
The origin of the decibel comes from looking at the ratio of the output and input power of the system.
|H(ω)| [dB] = 10 log
∣ Pout Pin
∣ (^) = 10 log
∣Vout Vin
∣^2 = 20 log
∣Vout Vin
This means that 20 dB per decade is equivalent to one order of magnitude. We won’t be using dB when plotting, but understanding the conversion to dB will help when reading other resources on Bode plots.
(^5) Technically if s = jω, then the roots of N(s) and D(s) are −ωz and −ωp. However, when plotting Bode plots, we refer to ωz and ωp as the zero and pole frequencies.
3.4 Poles, Zeros, and Constants
Single Pole, Single Zero
The notion of a pole and zero frequency is a generalization of the term cutoff frequency. Let’s first look back at a plot of our RC low-pass filter:
HP(ω) =
1 + 10 jω 6
In what follows, pay special attention to the Linear Approximations! When drawing Bode plots, we claim that the plot drops off with a slope of −1 after a pole ωp. This transfer function has a single pole at ωp = 106. The magnitude plot has a familiar shape (resembling a low-pass filter’s transfer function!).
10 −^3
10 −^2
10 −^1
ω
|H
(P
ω
Exact |HP(ω)| Linear Approximation
We can look at the phase plot as well:
ω
]
H
(P
ω
Exact ]HP(ω) Linear Approximation
Now let’s take a look at a single zero. HZ (ω) = 1 + jω/ωz (6)
We show that this Magnitude Bode plot rises with a slope of 1 after the zero at ωz.
has a zero at the origin, it will start off with a slope of 1. The phase of a zero at the origin is 90◦^ at all frequencies.
10 −^3 10 −^2 10 −^1 100 101 102
10 −^3
10 −^2
10 −^1
ω
|H
Z, O (ω
|HZ,O(ω)|
Constant Terms
Lastly, we show the plot of a constant K = 100. As expected, the plot remains constant. This implies that multiplication by K will shift up the entire bode plot up by K. Note that positive constants have a constant phase of 0◦^ at all frequencies, while negative constants have a constant phase of 180◦^ ≡ − 180 ◦^ at all frequencies.
10 −^3 10 −^2 10 −^1 100 101 102
10 −^2
ω
|H
K^ (ω
|HK (ω)|
3.5 Examples
We have previously seen examples of how to compute the bode plot of a bandpass filter and for an n−th order low-pass filter. At this point, see if you can go back and compose them yourself with the linear approximations presented above. See if you get the same results!
Transfer Function Example
Now let’s take a look at the Bode plot of a new transfer function.
HT (ω) = 100 ( 1 + jω) ( jω)^2 + 1010 ( jω) + 104
Our first step is to factor this into its rational transfer function form:
HT (ω) = 0. 01
( 1 + jω) ( 1 + jω/ 10 )( 1 + jω/ 103 )
With HT (ω) in its rational form, we see that K = 0. 01 , ωz = 1 , ωp 1 = 10 , ωp 2 = 103. Below is a magnitude plot of each consituent component (following the building-block rules presented above), and the multiplica- tion of all of these provides |HT (ω)|. The linear approximations are omitted to keep the plot legible, but the approximate result will very closely match the exact one.
10 −^1 100 101 102 103 104
10 −^4
10 −^3
10 −^2
10 −^1
ω
|H
(T
ω
|HT (ω)| K = 0. 01 ωz = 1 ωp 1 = 10 ωp 2 = 103
To provide an analysis for this Bode plot, we see that the plot starts off at K = 0. 01. Then at ωz = 1 , it starts rising with slope 1. When it hits the pole at ωp 1 = 10 , the slope of 1 is cancelled out by the −1 slope that the pole provides. Then the Bode plot stays constant until ωp 2 = 103 at which it drops off with a slope of 1. We’ve provided Bode plots of the individual terms to give you a sense of how we “add” Bode plots together.
We can also plot the phase in a very similar way using our building blocks:
10 −^1 100 101 102 103 104 105 106 107 108
ω
]
H
(T
ω
]HT (ω) K = 0. 01 ωz = 1 ωp 1 = 10 ωp 2 = 103
Recall from the note on differential equations that we defined the time constant of a first-order circuit to be the point at which the response vc(t) to a constant input was 1 − e−^1 away from its steady state value. With this in mind, let’s try plugging in an exponential input vin(t) = V 0 e jωt^ into an RC circuit and see what happens.^7
vin(t)
R
C
vout(t)
The differential equation for this circuit is
d dt vout(t) = λ (vout(t) −V 0 e jωt^ ) (8)
for λ = − (^1) τ. In Note 3 we showed that the steady state value of this differential equation is
vss(t) =
−λ jω − λ V 0 e jωt^ (9)
Therefore, plugging in for λ = − (^1) τ , it follows that
vss(t) =
1 + jωτ V 0 e jωt^ (10)
Notice that H(ω) = (^1) +^1 jωτ and the cutoff arises naturally as ωc = (^1) τ. We can also realize that at steady state, H(ω) is in fact the eigenvalue for the differential equation with eigenfunction e jωt^. This is a crucial connection between differential equations and the frequency response of a linear system that you will see in later half of the course and in courses like EE120.
Contributors:
- Neelesh Ramachandran.
- Rahul Arya.
- Anant Sahai.
- Jaijeet Roychowdhury.
- Taejin Hwang.
(^7) We should be inputting vin(t) = V 0 cos(ωt) but we choose e jωt (^) since it provides the same result while simplifying the math.
