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Transfer Functions and Bode Plots
Transfer Functions
For sinusoidal time variations, the input voltage to a filter can be written
vI (t) = Re
Viejωt
where Vi is the phasor input voltage, i.e. it has an amplitude and a phase, and ejωt^ = cos ωt + j sin ωt. A sinusoidal signal is the only signal in nature that is preserved by a linear system. Therefore, if the filter is linear, its output voltage can be written
vO (t) = Re
Voejωt
where Vo is the phasor output voltage. The ratio of Vo to Vi is called the voltage-gain transfer function. It is a function of frequency. Let us denote
T (jω) =
Vo Vi
We can write T (jω) as follows: T (jω) = A (ω) ejϕ(ω)
where A (ω) and ϕ (ω) are real functions of ω. A (ω) is called the gain function and ϕ (ω) is called the phase function. As an example, consider the filter input voltage
vI (t) = V 1 cos (ωt + θ) = Re
V 1 ejθejωt
The corresponding phasor input and output voltages are
Vi = V 1 ejθ
Vo = V 1 ejθA (ω) ejϕ(ω)
It follows that the time domain output voltage is
vO (t) = Re
h V 1 ejθA (ω) ejϕ(ω)ejωt
i = A (ω) V 1 cos [ωt + θ + ϕ (ω)]
This equation illustrates why A (ω) is called the gain function and ϕ (ω) is called the phase function. The complex frequency s is usually used in place of jω in writing transfer functions. In general, most transfer functions can be written in the form
T (s) = K
N (s) D (s)
where K is a gain constant and N (s) and D (s) are polynomials in s containing no reciprocal powers of s. The roots of D (s) are called the poles of the transfer function. The roots of N (s) are called the zeros. As an example, consider the function
T (s) = 4 s/4 + 1 s^2 /6 + 5s/6 + 1
s/4 + 1 (s/2 + 1) (s/3 + 1)
The function has a zero at s = − 4 and poles at s = − 2 and s = − 3. Note that T (∞) = 0. Because of this, some texts would say that T (s) has a zero at s = ∞. However, this is not correct because N (∞) 6 = 0. Note that the constant terms in the numerator and denominator of T (s) are both unity. This is one of two standard ways for writing transfer functions. Another way is to make the coefficient of the highest powers of s unity. In this case, the above transfer function would be written
T (s) = 6 s + 4 s^2 + 5s + 6
s + 4 (s + 2) (s + 3)
Because it is usually easier to construct Bode plots with the first form, that form is used here. Because the complex frequency s is the operator which represents d/dt in the differential equation for a system, the transfer function contains the differential equation. Let the transfer function above represent the voltage gain of a circuit, i.e. T (s) = Vo/Vi, where Vo and Vi, respectively, are the phasor output and input voltages. It follows that (^) μ s^2 6
5 s 6
Vo = 4
³ (^) s 4
Vi
When the operator s is replaced with d/dt, the following differential equation is obtained:
1 6
d^2 vO dt^2
dvO dt
dvI dt
where vO and vI , respectively, are the time domain output and input voltages. Note that the poles are related to the derivatives of the output and the zeros are related to the derivatives of the input.
How to Construct Bode Plots
A Bode plot is a plot of either the magnitude or the phase of a transfer function T (jω) as a function of ω. The magnitude plot is the more common plot because it represents the gain of the system. Therefore, the term “Bode plot” usually refers to the magnitude plot. The rules for making Bode plots can be derived from the following transfer function:
T (s) = K
μ s ω 0
¶±n
where n is a positive integer. For +n as the exponent, the function has n zeros at s = 0. For -n, it has n poles at s = 0. With s = jω, it follows that T (jω) = Kj±n^ (ω/ω 0 )±n, |T (jω)| = K (ω/ω 0 )±n^ and (^6) T (jω) = ±n × 90 ◦. If ω is increased by a factor of 10 , |T (jω)| changes by a factor of 10 ±n. Thus a plot of |T (jω)| versus ω on log—log scales has a slope of log (10±n) = ±n decades/decade. There are 20 dBs in a decade, so the slope can also be expressed as ± 20 n dB/decade. As a first example, consider the low-pass transfer function
T (s) =
K
1 + s/ω 1
This function has a pole at s = −ω 1 and no zeros. For s = jω and ω/ω 1 ¿ 1 ,we have T (jω) ' K, |T (jω)| ' K, and 6 T (jω) ' 0 × 90 ◦^ = 0 ◦. For ω/ω 1 À 1 , T (jω) ' K (jω/ω 1 )−^1 , |T (jω)| ' K (ω/ω 1 )−^1 , and (^6) T (jω) ' − 1 × 90 ◦^ = − 90 ◦. On log − log scales, the magnitude plot for the low-frequency approximation has
a slope of 0 while that for the high-frequency approximation has a slope of − 1. The low and high-frequency
- In any frequency band where a transfer function can be approximated by K (jω/ω 0 )±n, the slope of the Bode magnitude plot is ±n dec/dec. The phase is ±n × 90 ◦.
- Poles cause the asymptotic slope of the magnitude plot to rotate clockwise by one unit at the pole frequency.
- Zeros cause the asymptotic slope of the magnitude plot to rotate counter-clockwise by one unit at the zero frequency.
As a third example, consider the transfer function
T (s) = K
s/ω 1 s/ω 1 + 1
This function has a pole at s = −ω 1 and a zero at s = 0. For s = jω and ω/ω 1 ¿ 1 ,we have |T (jω)| ' K (ω/ω 1 ) and 6 T (jω) ' 90 ◦. For ω/ω 1 À 1 , |T (jω)| ' K and 6 T (jω) ' 0 ◦. On log − log scales, the magnitude plot for the low-frequency approximation has a slope of +1 while that for the high-frequency approximation has a slope of 0. The low and high-frequency approximations intersect when K (ω/ω 1 ) = K, or when ω = ω 1. For ω = ω 1 , |T (jω)| = K/
2 and 6 T (jω) = 90o^ − arctan (1) = 45 ◦. The Bode magnitude and phase plots are shown in Fig. 3. Note that the slope of the asymptotic magnitude plot rotates by − 1 at the pole. The transfer function is called a high-pass function because its gain approaches zero at low frequencies.
Figure 3: Bode plots. (a) Magnitude. (b) Phase.
A shelving transfer function has the form
T (s) = K
1 + s/ω 2 1 + s/ω 1
The function has a pole at s = −ω 1 and a zero at s = −ω 2. We will consider the low-pass shelving function for which ω 1 < ω 2. For s = jω and ω/ω 1 ¿ 1 , we have |T (jω)| ' K and 6 T (jω) ' 0 ◦. As ω is increased, the pole causes the asymptotic slope to rotate from 0 to − 1 at ω 1. The zero causes the asymptotic slope to rotate from − 1 back to 0 at ω 2. For ω/ω 2 À 1 , |T (jω)| ' K (ω 1 /ω 2 ). The Bode magnitude plot is shown in Fig. 4(a). If the transfer function did not have the zero, the actual gain at ω 1 would be K/
The zero causes the gain to be between K/
2 and K. Similarly, the pole causes the actual gain at ω 2 to be between K (ω 1 /ω 2 ) and
2 K (ω 1 /ω 2 ). The actual plot intersects the asymptotic plot at the geometric mean frequency
ω 1 ω 2.
The phase plot has a slope that approaches 0 ◦^ at very low frequencies and at very high frequencies. At the geometric mean frequency
ω 1 ω 2 , the phase is approaching − 90 ◦. If the function only had a pole, the phase at ω 1 would be − 45 ◦, approaching − 90 ◦^ at higher frequencies. However, the zero causes the high-frequency phase to approach 0 ◦. Thus the phase at ω 1 is more positive than − 45 ◦. At the geometric mean frequency
ω 1 ω 2 , the slope of the phase function is zero. The Bode phase plot is shown in Fig. 4(b).
Figure 4: Bode plots. (a) Magnitude. (b) Phase.
Impedance Transfer Functions
RC Network
The impedance transfer function for a two-terminal RC network which contains only one capacitor and is not an open circuit at dc can be written
Z = Rdc
1 + τ (^) zs 1 + τ (^) ps
where Rdc is the dc resistance of the network, τ (^) p is the pole time constant, and τ (^) z is the zero time constant. The pole time constant is the time constant of the network with the terminals open circuited. The zero time constant is the time constant of the network with the terminals short circuited. Figure 5(a) shows the circuit diagram of an example two-terminal RC network. The impedance transfer function can be written by inspection to obtain
Z = R 1
1 + R 2 Cs 1 + (R 1 + R 2 ) Cs
Figure 5: Example RC and RL impedance networks.
Figure 6: Example RC voltage divider networks.
RL Network
The voltage-gain transfer function of a RL voltage-divider network containing only one inductor and having a non-zero gain at dc can be written Vo Vi
= Kdc
1 + τ (^) zs 1 + τ (^) ps
where K∞ is the zero frequency gain (L a short circuit), τ (^) p is the pole time constant, and τ (^) z is the zero time constant. The pole time constant is the time constant of the network with Vi = 0 and Vo open circuited. The zero time constant is the time constant of the network with Vo = 0 and Vi open circuited. Figure 7(a) shows the circuit diagram of an example RL network. The voltage-gain transfer function can be written by inspection to obtain Vo Vi
R 2
R 1 + R 2
×
1 + [L/ (R 2 kR 3 )] s 1 + (L/ [(R 1 + R 2 ) kR 3 ]) s
Figure 7(b) shows the circuit diagram of a second example RL network. The voltage-gain transfer function can be written by inspection to obtain
Vo Vi
R 3
R 1 + R 3
×
1 + (L/R 1 ) s 1 + (L/ [R 1 k (R 2 + R 3 )]) s
Figure 7: Example RL voltage divider circuits.
High-Pass RL Network
The voltage-gain transfer function of a high-pass RL voltage-divider network containing only one inductor can be written Vo Vi
= K∞
τ (^) ps 1 + τ (^) ps
where K∞ is the infinite frequency gain (L an open circuit) and τ (^) p is the pole time constant. The pole time constant is calculated with Vi = 0 and Vo open circuited. Figure 7(c) shows the circuit diagram of a third example RL network. The voltage-gain transfer function can be written by inspection to obtain
Vo Vi
R 2
R 1 + R 2
×
[L/ (R 1 kR 2 )] s 1 + [L/ (R 1 kR 2 )] s
