30. Bode Plots. Introduction

30. Bode Plots

Introduction

Each of the circuits in this problem set is represented by a magnitude Bode plot. The network function provides a connection between the Bode plot and the circuit. To solve these problems, we first observe the features of the Bode plot to determine the corresponding network function. Next, we calculate the network function of the circuit by representing and analyzing the circuit in the frequency-domain. Finally, we compare the two network functions.

Network functions are described in Section 13.3 of Introduction to Electric Circuits by R.C. Dorf and J.A Svoboda. Bode plots are described in Section 13.4. Also, Table 10.7-1 summarizes the correspondence between the time-domain and the frequency domain.

Worked Examples

Example 1:

Consider the circuit shown in Figure 1. The input to the circuit is the voltage of the voltage source, vi(t). The output is the node voltage at the output terminal of the op amp, vo(t). The

network function that represents this circuit is

( )

ω

_{( )}

( )

ω ω = o i V H V (1)

The corresponding magnitude Bode plot is also shown in Figure 1. Determine the values of the capacitances, C1 and C2.

Solution: The network function provides a connection between the circuit and the Bode plot. We can determine the network function from the Bode plot and we can also analyze the circuit to determine its network function. The values of the capacitances are determined by equating the coefficients of these two network functions.

First, we make some observations regarding the Bode plot shown in Figure 1:

1. There are two corner frequencies, at 80 and 500 rad/s. The corner frequency at 80 rad/s is a pole because the slope of the Bode plot decreases at 80 rad/s. The corner frequency at 500 rad/s is a zero because the slope increases at 500 rad/s.

2. The coroner frequencies are separated by log₁₀ 500 0.786 decades 80

 _{ =}  

  . The slope of the Bode plot is 15.9 15.9 40 dB/decade

0.796

− −

= − between the corner frequencies.

3. At low frequencies, i.e. at frequencies smaller than the smallest corner frequency, the slope is -1×20 dB/decade so the network function includes a factor

(

jω

)

−1.

Consequently, the network function corresponding to the Bode plot is:

( )

( )

1 1 ₅₀₀ 1 ₅₀₀ 1 1 80 80 j j k j k j j j ω ω ω ω _ω ω ω −  ₊  ₊   = _{ }=    ₊  ₊   _{ }   _{ } H (1)

where k is a constant that is yet to be determined.

Next, we analyze the circuit shown in Figure 1 to determine its network function. A network function is the ratio of the output phasor to the input phasor. Phasors exist in the frequency domain. Consequently, our first step is to represent the circuit in the frequency domain, using phasors and impedances. Figure 2 shows the frequency domain representation of the circuit from Figure 1.

To analyze the circuit in Figure 2, we first write a node equation at the node labeled as node a. (The current entering the non-inverting input of the op amp is zero, so there are two currents in this node equation, the currents in the impedances corresponding to 125 kΩ resistor and capacitor C1.)

( )

( )

( )

3 1 a a 1 125 10 j C ω ω ω ω − = × i V V V

whereV_a

( )

ω is the node voltage at node a. Doing a little algebra gives

( )

_{( )}

1 3 3 a 1 125 10 125 10 j C ω ω ω   =_ + _ ×  ×  i V V Then

( )

(

(

3

)

)

( )

( )

₍

( )

₎

1 3 1 a a 1 125 10 1 125 10 j C j C ω ω ω ω ω ω = + × ⇒ = + × i i V V V V

Next, we write a node equation at the node labeled as node b. (The current entering the inverting input of the op amp is zero, so there are two currents in this node equation, the currents in the impedances corresponding to 10 kΩ resistor and capacitor C2.)

( )

( )

( )

3 2 a a o ₀ 1 10 10 j C ω ω ω ω − + = × V V V

Doing some algebra gives

( )

(

3

)

(

( )

( )

)

2 a ω + j Cω 10 10× a ω − ω 0 V V V_o =

(

)

(

3

)

( )

(

3

)

( )

2 a 2 o 1+ j Cω 10 10× V ω = j Cω 10 10× V ω

(

)

(

3

)

( )

₍

₎

(

3

)

( )

2 3 2 1 o 1 10 10 10 10 1 125 10 j C j C j C ω ω ω ω ω + × = × + × i V V Finally,

( )

_{( )}

( )

₍

₎

(

)

( )

(

(

)

)

3 2 3 3 2 1 1 10 10 1 10 10 1 125 10 j C C j j C ω ω ω ω ω ω   _{+ ×}   = =  _×  _{+ ×}   o i V H V (2)

( )

₍

₎

(

)

( )

(

(

)

)

3 2 3 3 2 1 1 _{1 1 10 10} 500 10 10 1 125 10 1 80 j _{j C} k C j j C j j ω ω ω ω _{ω ω} ω + _{  + ×}   = =    ₊  _ × _ + ×     H

Equating coefficients gives:

(

3

)

1 1 125 10 80=C × ,

(

)

3 2 1 10 10 500=C × and

(

3

)

2 1 500 10 10 k C = = × so

(

)

1 3 1 0.1 F 80 125 10 C = = µ × and 2

(

3

)

1 0.2 F 500 10 10 C = = µ × Example 2:

Consider the circuit shown in Figure 3. The input to the circuit is the voltage of the voltage source, vi(t). The output is the node voltage at the output terminal of the op amp, vo(t). The

network function that represents this circuit is

( )

ω

_{( )}

( )

ω ω = o i V H V (3)

The corresponding magnitude Bode plot is also shown in Figure 3. Determine the values of the capacitances, C1 and C2.

Solution: The network function provides a connection between the circuit and the Bode plot. We can determine the network function from the Bode plot and we can also analyze the circuit to determine its network function. The values of the capacitances are determined by equating the coefficients of these two network functions.

First, we make some observations regarding the Bode plot shown in Figure 3:

1. There are two corner frequencies, at 40 and 160 rad/s. Both corner frequencies are poles because the slope of the Bode plot decreases at both the corner frequencies.

2. Between the corner frequencies the gain is H

( )

ω =26 dB 10= 26 20=20 V/V. 3. At low frequencies, i.e. at frequencies smaller than the smallest corner frequency, the

slope is 1×20 dB/decade so the network function includes a factor

( )

jω 1. Consequently, the network function corresponding to the Bode plot is:

( )

( )

1 1 40 160 k j j j ω ω ω ω =  ₊   ₊       H    (3)

Next, we will analyze the circuit shown in Figure 3 to determine its network function. A network function is the ratio of the output phasor to the input phasor. Phasors exist in the frequency domain. Consequently, our first step is to represent the circuit in the frequency domain, using phasors and impedances. Figure 4 shows the frequency domain representation of the circuit from Figure 3.

Figure 4 The circuit from Figure 3, represented in the frequency domain, using impedances and phasors.

To analyze the circuit in Figure 4, we write a node equation at the node labeled as node a. In doing so, we will treat the series impedances, 20 kΩ and

1 1 j Cω , as an single equivalent impedance equal to 3 1 1 20 10 j Cω

× + . (The node voltage at node a is zero volts because the voltages at the input nodes of an ideal op amp are equal. The current entering the inverting input of the op amp is zero, so there are three currents in this node equation.)

( )

( )

( )

i o o 3 3 1 2 0 1 _{400 10} 1 20 10 j C j C ω ω ω ω ω + + = × × + V V V

Doing some algebra gives

(

)

( )

(

)

( )

1 i 2 o 3 3 1 1 0 400 10 1 20 10 j C j C j C ω ω ω ω ω   +_ + _ = ×   + × V V

(

)

(

)

( )

(

)

(

(

)

)

( )

3 1 i ₃ 2 o 3 1 400 10 1 400 10 1 20 10 j C j C j C ω ω ω ω ω × = − + × + × V V Finally,

( )

_{( )}

( )

(

)

(

)

(

)

(

(

)

)

3 1 3 1 2 400 10 1 20 10 1 400 10 j C j C j C ω ω ω ω _{ω ω} − × = = + × + × o i V H V 3 (4)

The network functions given in Equations 3 and 4 must be equal. That is

( )

_{( )}

(

)

(

)

(

)

(

(

)

)

3 1 3 1 2 400 10 1 20 10 1 400 10 1 1 40 160 j C k j j C j C j j ω ω ω ω ω _{ω ω} − × = =  ₊   ₊  _{+ × + ×}         H 3

Equating coefficients gives:

(

3

)

1 1 20 10 40=C × ,

(

)

3 2 1 400 10 160=C × and

(

)

3 1 400 10 k = −C × so

(

)

1 ₃ 1 1.25 F 40 20 10 C = = µ × and 2

(

₃

)

1 15.625 nF 160 400 10 = = × C also

(

_{400 10}3

) (

_{1.25 10} 6

)(

_{400 10}3

)

_0.5 k = −C × = × − × =