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FREQUENCY RESPONSE

Module by: Dinh Sy Hien

Summary: We begin by considering the frequency response of simple circuits using their transfer functions. We then consider Bode plots which are the industry-standard way of presenting frequency response. We also consider series and parallel resonant circuits and encounter important concepts such as resonance, quality factor, cutoff frequency and bandwidth. We discuss different kinds of filters and network scaling. In the last section, we consider one practical application of resonant circuits and two applications of filters.

INTRODUCTION

In sinusoidal circuit analysis, we have learned how to find voltages and currents in a circuit with a constant frequency source. If we left the amplitude of the sinusoidal source remain constant and vary the frequency, we obtain the circuit’s frequency response. The frequency response may be regarded as a complete description of the sinusoidal steady-state behavior of a circuit as a function of frequency.
The frequency response of a circuit is the variation in its behavior with change in signal frequency.
The sinusoidal steady-state frequency responses of circuits are of significance in many applications, especially in communications and control systems. A specific application is in electric filters that block out or eliminate signals with unwanted frequencies and pass signals of the desired frequencies. Filters are used in radio, TV, and telephone systems to separate one broadcast frequency from another.
We begin this chapter by considering the frequency response of simple circuits using their transfer functions. We then consider Bode plots which are the industry-standard way of presenting frequency response. We also consider series and parallel resonant circuits and encounter important concepts such as resonance, quality factor, cutoff frequency and bandwidth. We discuss different kinds of filters and network scaling. In the last section, we consider one practical application of resonant circuits and two applications of filters.

TRANSFER FUNCTION

The transfer function is a useful analytical tool for finding the frequency response of a circuit. In fact, the frequency response of a circuit is the plot of the circuit’s transfer function H( ωω size 12{ω} {}) versus ωω size 12{ω} {}, with ωω size 12{ω} {} varying from ω=0ω=0 size 12{ω=0} {} to ω=∞ω=∞ size 12{ω= infinity } {}.
A transfer function is the frequency-dependent ratio of the forced function to the forcing function (or of an output to an input). The idea of a transfer function was implicit when we used the concepts of impedance and admittance to relate voltage and current. In general, a linear network can be represented by the block diagram shown in Figure 1.
The transfer function H( ωω size 12{ω} {}) of a circuit is the frequency-dependent ratio of a phasor output Y( ωω size 12{ω} {}) (an element voltage or current) to a phasor input X( ωω size 12{ω} {}) (source voltage or current).
Thus
H ( ω ) = Y ( ω ) X ( ω ) H ( ω ) = Y ( ω ) X ( ω ) size 12{H \( ω \) = { {Y \( ω \) } over {X \( ω \) } } } {} (1)
assuming zero initial conditions. Since the input and output can be either voltage or current at any place in circuit, there are four possible transfer functions:
H ( ω ) = voltage H ( ω ) = voltage size 12{H \( ω \) = ital "voltage"} {} (2)
gain = V 0 ( ω ) V i ( ω ) gain = V 0 ( ω ) V i ( ω ) size 12{ ital "gain"= { {V rSub { size 8{0} } \( ω \) } over {V rSub { size 8{i} } \( ω \) } } } {}
H ( ω ) = current H ( ω ) = current size 12{H \( ω \) = ital "current"} {} (3)
gain = I 0 ( ω ) I i ( ω ) gain = I 0 ( ω ) I i ( ω ) size 12{ ital "gain"= { {I rSub { size 8{0} } \( ω \) } over {I rSub { size 8{i} } \( ω \) } } } {}
H ( ω ) = transfer H ( ω ) = transfer size 12{H \( ω \) = ital "transfer"} {} (4)
impedance = V 0 ( ω ) I i ( ω ) impedance = V 0 ( ω ) I i ( ω ) size 12{ ital "impedance"= { {V rSub { size 8{0} } \( ω \) } over {I rSub { size 8{i} } \( ω \) } } } {}
H ( ω ) = transfer H ( ω ) = transfer size 12{H \( ω \) = ital "transfer"} {} (5)
admit tan ce = I 0 ( ω ) V i ( ω ) admit tan ce = I 0 ( ω ) V i ( ω ) size 12{ ital "admit""tan" ital "ce"= { {I rSub { size 8{0} } \( ω \) } over {V rSub { size 8{i} } \( ω \) } } } {}
where subscripts i and o denote input and output values. Being a complex quantity, H( ωω size 12{ω} {}) has magnitude H( ωω size 12{ω} {}) and a phase φφ size 12{φ} {}; that is H(ω)=H(ω)∠φH(ω)=H(ω)∠φ size 12{H \( ω \) =H \( ω \) ∠φ} {}.
Figure 1: A block diagram reprensentation of a linear network.
To obtain the transfer function using Equation 2, we first obtain the frequency-domain equivalent of the circuit by replacing resistors, inductors, and capacitors with their impedances R, j ωω size 12{ω} {}L, and 1/j ωω size 12{ω} {}C. we then use any circuit technique to obtain the appropriate quantity in Equation 2. We can obtain the frequency response of the circuit by plotting the magnitude and phase of the transfer function as the frequency varies. A computer is a real time-saver for plotting the transfer function.
The transfer function H( ωω size 12{ω} {}) can be expressed in terms of its numerator polynomial N( ωω size 12{ω} {}) and denominator polynomial D( ωω size 12{ω} {}) as
H ( ω ) = N ( ω ) D ( ω ) H ( ω ) = N ( ω ) D ( ω ) size 12{H \( ω \) = { {N \( ω \) } over {D \( ω \) } } } {} (6)
Where N( ωω size 12{ω} {}) and D( ωω size 12{ω} {}) are not necessarily the same expressions for the input and output functions, respectively. The representation of H( ωω size 12{ω} {}) in Equation 6 assumes that common numerator and denominator factors in H( ωω size 12{ω} {}) have canceled, reducing the ratio to lowest terms. The roots of N( ωω size 12{ω} {}) = 0 are called the zeros of H( ωω size 12{ω} {}) and are usually represented as jω=z1,z2,...jω=z1,z2,... size 12{jω=z rSub { size 8{1} } ,z rSub { size 8{2} } , "." "." "." } {} Similarly, the roots of D( ωω size 12{ω} {}) = 0 are the poles of H( ωω size 12{ω} {}) and are represented as jω=p1,p2,...jω=p1,p2,... size 12{jω=p rSub { size 8{1} } ,p rSub { size 8{2} } , "." "." "." } {}
A zero as a root of the numerator polynomial, is a value that results in a zero value of the function. A pole, as a root of the denominator polynomial, is a value for which the function is infinite.
To avoid complex algebra, it is expedient to replace j ωω size 12{ω} {} temporarily with s when working with H( ωω size 12{ω} {}) and replace s with j ωω size 12{ω} {} at the end.

THE DECIBEL SCALE

It is not always easy to get a quick plot of the magnitude and phase of the transfer function as we did above. A more systematic way of obtaining the frequency response is to us Bode plots. Before we begin to construct Bode plots, we should take care of two important issues: the use of logarithms and decibels in expressing gain.
Since Bode plots are based on logarithms, it is important that we keep the following properties of logarithms in mind:
  1. log P 1 P 2 = log P 1 + log P 2 log P 1 P 2 = log P 1 + log P 2 size 12{"log"P rSub { size 8{1} } P rSub { size 8{2} } ="log"P rSub { size 8{1} } +"log"P rSub { size 8{2} } } {}
  2. log P 1 / P 2 = log P 1 − log P 2 log P 1 / P 2 = log P 1 − log P 2 size 12{"log" {P rSub { size 8{1} } } slash {P rSub { size 8{2} } ="log"P rSub { size 8{1} } - "log"P rSub { size 8{2} } } } {}
  3. log P n = n log P log P n = n log P size 12{"log"P rSup { size 8{n} } =n"log"P} {}
  4. log 1 = 0 log 1 = 0 size 12{"log"1=0} {}
In communications systems, gain is measured in bels. Historically, the bel is used to measure the ratio of two levels of power or power gain G; that is,
G = Number of bels = log 10 P 2 P 1 G = Number of bels = log 10 P 2 P 1 size 12{G= ital "Number" {} cSup {} ital "of" {} cSup {} ital "bels"="log" rSub { size 8{"10"} } { {P rSub { size 8{2} } } over {P rSub { size 8{1} } } } } {} (7)
The decibel (dB) provides us with a unit of less magnitude. It is 1/10th1/10th size 12{ {1} slash {"10" rSup { size 8{ ital "th"} } } } {} of a bel and is given by
G dB = 10 log 10 P 2 P 1 G dB = 10 log 10 P 2 P 1 size 12{G rSub { size 8{ ital "dB"} } ="10""log" rSub { size 8{"10"} } { {P rSub { size 8{2} } } over {P rSub { size 8{1} } } } } {} (8)
When P1=P2P1=P2 size 12{P rSub { size 8{1} } =P rSub { size 8{2} } } {}, there is no change in power and the gain is 0 dB. If P2=2P1P2=2P1 size 12{P rSub { size 8{2} } =2P rSub { size 8{1} } } {}, the gain is
G dB = 10 log 10 2 = 3 dB G dB = 10 log 10 2 = 3 dB size 12{G rSub { size 8{ ital "dB"} } ="10""log" rSub { size 8{"10"} } 2=3 ital "dB"} {} (9)
And when P2=0.5P1P2=0.5P1 size 12{P rSub { size 8{2} } =0 "." 5P rSub { size 8{1} } } {}, the gain is
G dB = 10 log 10 0 . 5 = − 3 dB G dB = 10 log 10 0 . 5 = − 3 dB size 12{G rSub { size 8{ ital "dB"} } ="10""log" rSub { size 8{"10"} } 0 "." 5= - 3 ital "dB"} {} (10)
Equation 9 and Equation 10 show another reason why logarithms are greatly used: the logarithm of the reciprocal of a quantity is simply negative the logarithm of that quantity.
Alternatively, the gain G can be expressed in terms of voltage or current ratio. To do so, consider the network shown in Figure 2. If P1P1 size 12{P rSub { size 8{1} } } {} is the input power, P2P2 size 12{P rSub { size 8{2} } } {} is the output (load) power, R1R1 size 12{R rSub { size 8{1} } } {} is input resistance and R2R2 size 12{R rSub { size 8{2} } } {} is the load resistance, then P1=0.5V12/R1P1=0.5V12/R1 size 12{P rSub { size 8{1} } =0 "." 5 {V rSub { size 8{1} } rSup { size 8{2} } } slash {R rSub { size 8{1} } } } {} and P2=0.5V22/R2P2=0.5V22/R2 size 12{P rSub { size 8{2} } =0 "." 5 {V rSub { size 8{2} } rSup { size 8{2} } } slash {R rSub { size 8{2} } } } {}, and Equation 8 becomes
G dB = 10 log 10 P 2 P 1 = 10 log 10 V 2 2 / R 2 V 1 2 / R 1 = 10 log 10 ( V 2 V 1 ) 2 + 10 log 10 R 1 R 2 G dB = 10 log 10 P 2 P 1 = 10 log 10 V 2 2 / R 2 V 1 2 / R 1 = 10 log 10 ( V 2 V 1 ) 2 + 10 log 10 R 1 R 2 size 12{G rSub { size 8{ ital "dB"} } ="10""log" rSub { size 8{"10"} } { {P rSub { size 8{2} } } over {P rSub { size 8{1} } } } ="10""log" rSub { size 8{"10"} } { {V rSub { size 8{2} } rSup { size 8{2} } /R rSub { size 8{2} } } over {V rSub { size 8{1} } rSup { size 8{2} } /R rSub { size 8{1} } } } ="10""log" rSub { size 8{"10"} } \( { {V rSub { size 8{2} } } over {V rSub { size 8{1} } } } \) rSup { size 8{2} } +"10""log" rSub { size 8{"10"} } { {R rSub { size 8{1} } } over {R rSub { size 8{2} } } } } {} (11)
G dB = 20 log 10 V 2 V 1 − 10 log 10 R 2 R 1 G dB = 20 log 10 V 2 V 1 − 10 log 10 R 2 R 1 size 12{G rSub { size 8{ ital "dB"} } ="20""log" rSub { size 8{"10"} } { {V rSub { size 8{2} } } over {V rSub { size 8{1} } } } - "10""log" rSub { size 8{"10"} } { {R rSub { size 8{2} } } over {R rSub { size 8{1} } } } } {} (12)
For the case when R2=R1R2=R1 size 12{R rSub { size 8{2} } =R rSub { size 8{1} } } {}, a condition that is often assumed when comparing voltage levels, Equation 12 becomes
G dB = 20 log 10 V 2 V 1 G dB = 20 log 10 V 2 V 1 size 12{G rSub { size 8{ ital "dB"} } ="20""log" rSub { size 8{"10"} } { {V rSub { size 8{2} } } over {V rSub { size 8{1} } } } } {} (13)
Instead, if P1=I12R1P1=I12R1 size 12{P rSub { size 8{1} } =I rSub { size 8{1} } rSup { size 8{2} } R rSub { size 8{1} } } {} and P2=I22R2P2=I22R2 size 12{P rSub { size 8{2} } =I rSub { size 8{2} } rSup { size 8{2} } R rSub { size 8{2} } } {}, for R1=R2R1=R2 size 12{R rSub { size 8{1} } =R rSub { size 8{2} } } {}, we obtain
G dB = 20 log 10 I 2 I 1 G dB = 20 log 10 I 2 I 1 size 12{G rSub { size 8{ ital "dB"} } ="20""log" rSub { size 8{"10"} } { {I rSub { size 8{2} } } over {I rSub { size 8{1} } } } } {} (14)
Figure 2: Voltage-current relationships for a four-terminal network.
Three things are important to note from Equation 8,Equation 13, and Equation 14:
  1. That 10 log is used for power, while 20 log is used for voltage or current, because of the square relationship between them ( P=V2/R=I2RP=V2/R=I2R size 12{P= {V rSup { size 8{2} } } slash {R} =I rSup { size 8{2} } R} {}).
  2. That the dB value is a logarithmic measurement of the ratio of one variable to another of the same type. Therefore, it applies in expressing the transfer function H in Equation 2 and Equation 3, which are dimensionless quantities, but not in expressing H in Equation 4 and Equation 5.
  3. it is important to note that we only use voltage and current magnitude in Equation 13 and Equation 14. Negative signs and angles will be handled independently as we will see in section 4.
With this in mind, we now apply the concepts of logarithms and decibels to construct Bode plots.

BODE PLOTS

Obtaining the frequency response from the transfer function as we did in section 2 is an uphill task. The frequency range required in frequency response is often so wide that it is inconvenient to use a linear scale for the frequency axis. Also, there is a more systematic way of locating the important features of the magnitude and phase plots of the transfer function. For these reasons, it has become standard practice to use a logarithmic scale for the frequency axis and a linear scale in each of the separate plots of magnitude and phase. Such semilogarithmic plots of the transfer function-known as Bode plots have become the industry standard.
Bode plots are semilog plots of the magnitude (in decibels) and phase (in degrees) of a transfer function versus frequency.
Bode plots contain the same information as the nonlogarithmic plots discussed in the previous section, but they are much easier to construct, as we shall see shortly.
The transfer function can be written as
H = H ∠ φ = He jφ H = H ∠ φ = He jφ size 12{H=H∠φ= ital "He" rSup { size 8{jφ} } } {} (15)
Taking the natural logarithm of both sides,
ln H = ln H + ln e jφ = ln H + jφ ln H = ln H + ln e jφ = ln H + jφ size 12{"ln"H="ln"H+"ln"e rSup { size 8{jφ} } ="ln"H+jφ} {} (16)
Thus, the real part of ln H is a function of the magnitude while the imaginary part is the phase. In a Bode magnitude plot, the gain
H dB = 20 log 10