Notes for course EE1.1 Circuit Analysis TOPIC 7 FREQUENCY RESPONSE AND FILTERING 1 THE FREQUENCY RESPONSE FUNCTION

Transcription

1 Objectives Notes for course EE. Circuit Analysis TOPIC 7 FREQUENCY RESPONSE AND FILTERING The definition of frequency response function, amplitude response and phase response Frequency response function and filtering Frequency response function and Fourier analysis The series and parallel tuned circuit nd-order passive filters Bode plots. General THE FREQUENCY RESPONSE FUNCTION In the time domain, circuits may be characterised by their transient response The oscilloscope enables us to observe the transient behaviour of a system's input and output signals In the frequency domain, circuits are characterised by their frequency response function The laboratory instruments which measure frequency content of signals and systems are the network analyser, the gain-phase test set and the spectrum analyzer We begin with a definition of frequency response function and explore its use through examples later. Definitions Consider a general linear circuit having no internal independent sources, with a sinusoidal input (or excitation) waveform x(t) and an output (or response) waveform y(t): The phasor equivalent circuit is as follows: The input signal is represented by its phasor, having amplitude X m and angle θ The steady state sinusoidal response phasor will have amplitude Y m and angle θ + φ The system function is defined by: ( ) Y X Y m θ + φ H ω X m θ Y m X m φ We have included the frequency ω as an argument for the system function because, in the case of circuits containing inductors and/or capacitors, the ratio of the output phasor to the input phasor is frequency dependent

2 It is common practice to write H(jω) for H(ω) because ω can arise only from the impedance of an inductor (jωl) or from the impedance of a capacitor /(jωc), where it is always associated with j. H(ω) contains all the information we need to know about the circuit, provided we know its value for each value of ω.3 Amplitude and phase response functions The frequency-domain description of a signal consists of a phasor with amplitude and phase at a particular frequency We adopt the same idea for the system function H(ω) We define the amplitude response function: The phase response function is defines as: A( ω ) H ( ω ) Y m X m φ( ω ) H ( ω ) Notice that since H(ω) is a function of ω, both A and φ are functions of ω We can now state that the amplitude of the circuit output signal is given by: and its phase by Y m A( ω ) X m Y ( ω ) φ( ω) + θ Thus, we multiply the input signal amplitude by the system gain and add its phase angle to the system phase shift to obtain the output amplitude and phase angle, respectively.4 Frequency Response measurement: Gain and Phase Consider the following set-up: We show a laboratory signal generator applying a cosine waveform with unit amplitude and general frequency ω to our system under test The AC steady-state response is measured at a frequency ω, then the input signal is changed to a new frequency ω and the response is measured again By measuring the steady-state output signal, we can determine the value of the gain A(ω) and the phase shift φ(ω) at each frequency We can then plot the frequency response, that is, the gain and the phase versus frequency, as follows:

3 Gain Phase Shift Laboratory instruments can automatically vary the frequency, determine the gain and phase shift at each frequency; they are called gain and phase test sets or network analysers.5 Example Let's look at a simple RC circuit example: Suppose that we run a test by applying a sinusoidal source at the circuit input and adjusting it so that its amplitude is one volt and its frequency is variable We can use AC steady-state analysis to compute the response The AC steady-state equivalent circuit is as follows: where we have treated the frequency ω as a variable Using the voltage divider rule, the response phasor is: Y jωc R + jωc 0 + jωcr 0 + ( ωcr) tan ( ωcr) Since 0 0 is the input phasor X, we may write: Y From the definition of H(ω), we have: Thus: + jωcr 0 + jωcr X H ( ω ) Y X + jωcr A( ω) + ( ωcr) φ( ω) tan ( ωcr) The plot of the amplitude response function A(ω) of the circuit is as follows: 3

4 The amplitude response of this circuit passes low frequencies and suppresses high frequencies Higher-frequency components in any input signal will be attenuated much more than those at low frequencies; a filter with such a response is called a lowpass filter Furthermore, if we make the time constant RC larger, we reduce the high-frequency gain, and if we make the time constant smaller, we increase the high-frequency gain Thus a circuit of resistors and one or more reactive elements not only has a transient response but also has a frequency response function which can effectively filter signals; we consider an example. Illustrative example THE IDEA OF FILTERING We explore the idea of filtering of signals in order to remove noise or interference Suppose a sinusoidal signal is transmitted over a noisy channel that adds an interfering signal that we wish to remove by a filter circuit Let s assume the "noisy" signal is described by: ( ) cos( πt) + 0.5cos( 00πt) x t The term cos( πt )is the desired signal and the term cos( 00πt )represents the additive noise A plot of one period of the signal with noise is shown: If we could come up with a "filter" that would pass the sinusoid at the frequency of π rad/s and block the one at 00 π rad/s, we would have achieved our objective We will assume that the noisy signal is available as a voltage source having the prescribed x(t) as its waveform Let us use the simple RC lowpass filter circuit for which we derived the amplitude response function and plotted it: : 4

5 A( ω ) + ( ωcr) The noise in our signal is a sinusoid with a frequency 00 times that of the signal Consider making the RC time constant such that: Thus, let's see what happens if we adjust it so that π rad/s << << 00π rad/s RC 0π rad/s RC This is 0 higher than our signal frequency and 0 lower than the noise frequency We can calculate the amplitude response of the circuit at the signal frequency and at the nose frequency: A( ω ) A( π ) A( 00π ) + ( ωcr) ω + 0π π 0π π 0π + ( 0) We can use superposition to compute the circuit output signal resulting from our noisy signal as the input signal: y( t) A π ( )cos πt + φ( π ) + A ( 00π )cos 00πt + φ ( 00π ) cos πt 5.7 ( ) cos( 00πt 84.3 ) where we have calculated the phase response at the two frequencies, but the phase is irrelevant in this case The signal component has been almost unchanged (amplitude reduced by about one-half of one percent), but the noise waveform has been attenuated by a factor of /0 5

6 A plot of one period of the response is shown: There is now no visible trace of the contaminating noise We have successfully used our frequency domain approach to come up with the circuit that did the job we wanted it to do The circuit we used is called a st order lowpass filter The corresponding st order highpass filter shown: The circuit is the same as the one we have worked with, except that the capacitor and resistor have been interchanged We can use the voltage divider rule to obtain the frequency response function and the amplitude and phase responses: H ( ω ) Y X R R + jωc A( ω ) + ωcr j ωcr φ ω tan ωcr + ωcr ( ) tan ωcr We have: A( ω ) ω 0 0 A( ω ) ω Hence, for this circuit, a high-frequency signal, considered to be the signal, is passed, and a lowfrequency signal, considered to be the noise, is blocked The plot of the amplitude response is as for the lowpass filter but with the frequency scale inverted. Bandpass Filters and Tuning Another useful application of frequency response is "tuning" a radio receiver Radio stations are limited by law to have a certain bandwidth 6

7 The frequency plot shows the permissible frequency limits for two hypothetical radio stations, KOKA and KOLA The continuous curve represents the amplitude spectrum of the very weak signal that is picked up by the antenna of our radio receiver The signal strength of the one we desire (say KOKA) is weaker than the other (KOLA) If we do not select just one station and reject the other, we will hear a mixture of both stations in our speaker Our solution is to pass the composite signal through a bandpass filter Ideally, the filter should have an amplitude response having the rectangular shape shown in the figure.3 Other Types of Filter Consider the spectrum of the audio signal in a power amplifier in an audio sound system: It is unfortunately true that an amplifier with a very high gain also picks up unwanted disturbances, and one of the most common interfering signals comes from the AC power cables This interference is in the form of a sinusoidal signal whose frequency is 50 Hz, or 00π rad/s, which causes a hum in the speaker Assuming that the audio signal components in a frequency range close to the interfering frequency of 00π rad/s are not very important to the intelligibility of the waveform, we can pass the composite waveform through a band-reject (or bandstop) filter having the frequency response shown above This eliminates only the interfering waveform and passes our desired audio signal relatively unaffected, providing the "notch" is very narrow Thus there are a number of standard filter frequency response types that can be applied in a host of practical situations They are called lowpass, highpass, bandpass and band-reject filters Their ideal gain versus frequency templates are shown: lowpass highpass bandpass band-reject 7

8 Actual filters will not have these "brick wall" responses; that is, they will not change abruptly from one value to another as the frequency changes The first-order RC lowpass filter was far from its ideal template With more circuit elements and more sophisticated design procedures, one can approximate the ideal filter frequency response characteristic much more closely There are catalogues containing tables of pre-designed filters based upon the standard types and some standard types of approximating functions, such as Butterworth, Chebyshev and elliptic In some applications, such as audio, the phase response of a filter has very little effect on perceived sound so it is sufficient to consider only the amplitude response In other applications, such as television, phase characteristics are important In general, for distortion-less filtering of a signal, the filter gain should be constant and the phase should be linear over the frequency range of the waveform One standard filter type we have not mentioned is the all-pass filter The gain is constant with frequency but the phase characteristic can be used to compensate for nonlinear phase characteristic in another circuit Amplitude Phase 3 QUALITY FACTOR FOR INDUCTOR AND CAPACITOR 3. Definition Tuning of a radio receiver clearly requires a bandpass filter The simplest example of a bandpass filter is the LC tuned circuit Ideal inductors and capacitors used in such a circuit should only store energy and not dissipate any Practical inductors and capacitors do dissipate some energy We start by defining a factor that measures the quality of an energy storage element Consider the AC steady-state response of the two-terminal element shown: We define the quality factor, Q-factor or Q, by the equation: peak stored energy Q π energy dissipated per cycle π w P In general, we will pick either the voltage or the current to have zero phase angle as a reference For an ideal lossless inductor or capacitor, the energy dissipated per cycle is zero, implying that the Q-factor is infinite 8 w D

9 Note that Q-factor is dimensionless 3. Q of a lossy inductor Inductors are constructed in the form of a coil of wire having finite resistance Thus, a practical model of an inductor consists of an ideal inductor in series with a small resistance r s : We assume that the current i(t) is sinusoidal and of the form: ( ) I m cos( ωt + φ ) i t The energy stored in an inductor is given by: In AC steady-state terms, the stored energy is: The peak value of stored energy is: w( t) Li ( t) w( t) LI m cos ( ωt + φ) w P LI m The only element that dissipates (or absorbs) energy is the resistor which shares the same current as the inductor Energy is the integral of power; the energy dissipated over one full period is: w D where < > denotes average value We have the general result: Hence T 0 P( t)dt T T T 0 P( t)dt T P( t) w D T r s i ( t) T r s I m cos ( ωt + φ) Tr s I m cos ( ωt + φ) cos x Therefore, the Q of the lossy inductor is: ( ) ( + cos ( x )) w D Tr s I m Q L π w P π LI m w D Tr s I m We have used the fact that T f and f ω π πl Tr s ωl r s 9

10 Example A lossy inductor has a series winding resistance of 0 Ω and a nominal value of 0 mh Find the quality factor at a frequency of 00 krad/s We then have: 3.3 Q of a Lossy Capacitor Q L ωl r s A capacitor is constructed in the form of parallel metal plates (perhaps rolled up or folded in the final construction phase) separated by some sort of dielectric Thus the dielectric s finite resistance can be approximated by the equivalent circuit shown: For a sinusoidal terminal voltage: v( t) V m cos( ωt + φ ) The energy stored on the ideal capacitor as a function of time is: The peak energy stored is, therefore, w( t) Cv ( t) CV m cos ( ωt + φ) w P CV m The energy dissipated in one period in the resistor is: w D T P( t) T V m cos ( ωt + φ ) TV m r p r p Hence the capacitor Q-factor is given by: Q C π w P π CV m w D TV m r p πcr p T ωcr p Example A lossy capacitor has a parallel dielectric resistance of 0 MΩ and a nominal value of 0 nf Find the quality factor at a frequency of 00 krad/s Q C ωcr p Capacitors typically have much higher Q values than do inductors 0

11 3.4 Aide-memoire for inductor and capacitor Q-factor We note that the inductor and capacitor Q-factors may be written in the following form: Q L ω L r s X L r s Q C ωcr p r p ( ωc) r p X C where X L and X C are the inductor and capacitor reactance, respectively We note also that the expressions are dimensionless ratios of impedances involving a wanted (X) and an unwanted (r) quantity Note also that the conditions that make the elements ideal, r s 0, r p, both make Q These considerations allow us to predict Q expressions for other combinations For example, inductor L with parallel resistance r p must have Q for r p ; so the Q expression must be: Q L r p X L r p ω L Capacitor C in series with resistance r s must have Q for r s 0; so the Q expression must be: Q C X C r s ωc r s ωcr s 3.5 Series-to-Parallel Transformation for a Lossy Inductor Consider the lossy inductor: It is possible to find a parallel circuit which is equivalent to it at a specified single frequency: Let's compute the admittance of the series sub-circuit: Y ( jω ) For Q L >>, we can write: Y r s + jω L ( jω ) jq L r s Q L r s jω L r s + ω L ( ) j ω L r s r s + Q j L r s Q L r s Q + L r s ω L r s j ω L r s r s r s jq L + Q L Q + L r s jω L At a single frequency ω, this corresponds to the following equivalent sub-circuit:

12 where r p ' Q L r s L' L Note that this equivalent circuit depends on frequency because Q L is a function of frequency The series and parallel circuits can be equivalent only at one frequency 3.6 The Narrow Band Approximation The equivalence that we have just derived is strictly valid only at a single frequency because r p ' depends on Q L which depends on frequency If we are only interested in small percentage frequency changes relative to the centre frequency, we can assume that Q L is a constant whose value is that assumed at the centre frequency This is called the narrow-band approximation Example 3 Find the parallel equivalent circuit for the lossy inductor of Example with r s 0 Ω, L 0 mh at 00 krad/s. Then find the percentage error in r p varies over a frequency range of 99 krad/s to 0 krad/s Solution We have already computed the inductor Q to be 00 at 00 krad/s Assuming that Q L >>, we have the general expression for r p ': At ω 00 krad/s, we have: At ω 99 krad/s, we have: At ω 0 krad/s, we have: r p ' r p ' r p ' r p ' Q L r s ωl r s ( ωl) 0 r s ( ωl) r s r s ( 5 0 ) 0 ( ) 0 ( 5 0 ) ( ωl) 0 0 r s 0 ( ωl ) r s kω 98.0 kω 0.0 kω Thus, we see that a variation of ± % in the frequency results in only a ± % variation in the resulting parallel resistance Hence, the use of a constant r p ' of 00 kω over this frequency range might be acceptable

13 4 THE PARALLEL TUNED CIRCUIT 4. The Lossless Tuned Circuit and Resonance Consider the lossless LC parallel circuit:. We assume that the driving source is a sinusoidal current source We wish to determine the AC steady-state response for the voltage v(t).as a function of frequency ω The phasor form of the circuit is: The frequency ω appears in the element impedances, so we do not have to make a special note of its value on the phasor circuit diagram The impedance of the two-terminal sub-circuit is: Z ( jω ) jω L jωc jω L + jωc ω L j ω LC jx ω This result means that the impedance is always purely imaginary The reactance X(ω) (the imaginary impedance without the j multiplier) can be plotted versus ω: ( ) The reactance goes to infinity at ω ω 0, where ω 0 LC This phenomenon is called resonance and the frequency ω 0 is called the resonant frequency 4. The Lossy Tuned Circuit Now let's consider the practical situation where the inductor and the capacitor both have finite Q 3

14 If we perform the series-to-parallel transformation on the lossy inductor, we can combine the two resistors into one We then obtain the following equivalent sub-circuit: We note that the parallel resistor is a composite of the loss resistance r p of the capacitor, the (transformed) parallel equivalent resistance r p of the inductor, and any source resistance that might be present (Norton equivalent for the driving source) The narrow band approximation is based on the assumption that the resistance R is a constant over the frequency range of interest Our first objective will be to find the Q of the sub-circuit at the resonant frequency ω 0 We apply a test current source to our sub-circuit and adjust it to be a sinusoid at frequency ω 0 LC The phasor equivalent circuit form is: At the resonant frequency ω 0 LC we know that the part of the sub-circuit consisting of the capacitor and the inductor presents an infinite impedance, Z'(jω 0 ) (this part of the circuit is the same as a lossless tuned circuit we analysed earlier) The impedance at the two terminals of the subcircuit is ( ) R Z jω 0 Thus, the phasor terminal voltage: is: V RI RI m 0 In other words: V m RI m Let's compute the peak energy stored by our sub-circuit We know that the energy stored on the capacitor as a function of time is: 4

15 w C ( t) Cv ( t) CV m cos ( ω 0 t) To find the energy stored by the inductor, we first determlne the inductor current in phasor form: In the time domain, we have: I L The energy stored in the inductor is, thus: V jω 0 L V m 0 ω 0 L 90 V m ω 0 L 90 i L ( t) V m ω 0 L cos ( ω 0 t ) 90 V m ω 0 L sin( ω 0 t) w L ( t) Li L ( t) L V m ω 0 L sin( ω 0 t) V m ( ) ω 0 L sin ω 0 t We can substitute ω 0 LC to show that: w L ( t) CV m sin ( ω 0 t) which we can compare with: w C ( t) CV m cos ( ω 0 t) This shows that the peak energy stored in the inductor is the same as the peak energy stored in the capacitor It also shows that the sum of the inductor and capacitor energies is constant This means that when the capacitor is storing its maximum energy, the inductor is storing no energy, and vice versa The energy is being swapped back and forth between the capacitor and the inductor, and none is coming from the source Because the impedance Z'(jω 0 ) the current into the parallel LC combination is zero when ω ω ο Hence, at ω ω ο the source current of our sub-circuit only feeds the resistor The energy absorbed by the sub-circuit in one period is the same as the energy absorbed by the resistor in that period: w D V m R T 0 5

16 where T 0 is the period corresponding to resonant frequency ω 0 The Q of the sub-circuit at ω ω 0, which we will call Q 0, is: Q 0 π w P w D π CV m V m R T 0 ω 0 RC Using LC /ω 0, we can write an alternative expression for Q 0 : Q 0 ω 0 R Lω 0 R ω 0 L Thus it is shown that the Q-factor of the tuned-circuit is the same as the Q-factor of the inductor or the Q-factor of the capacitor where the same resistor R is used in each case 4.3 Fractional Frequency Deviation Now let's return to our parallel tuned-circuit and compute the impedance at its terminals as a function of the general frequency ω: We redraw the circuit in the phasor domain: For this parallel circuit we have: Z( jω ) R + jωc + jωl R + C jω + ω 0 jω R R + jω 0 RC ω ω 0 ω + jq 0 ω 0 ω 0 ω ω 0 ω This is a standard form for the tuned circuit impedance R + jω 0C ω ω 0 ω 0 ω The expression in the denominator brackets is called the fractional frequency deviation The magnitude and phase of Z(jω) are: Z ( jω ) R ω + Q 0 ω 0 ω 0 ω Z(jω) normalized to R can be plotted as follows: ( ) tan Q 0 ω Z jω ω 0 ω 0 ω 6

17 The peak impedance occurs at ω ω 0 the resonant frequency The response is plotted for two different values of Q 0 The larger the value of Q 0, the more selective is the tuned circuit If we were tuning in a radio station, we would need a large Q 0 if there was another station very close in frequency 4.4 Bandwidth The passband may be defined formally as the range of frequencies for which the normalized gain is greater than / There are two frequencies, one above ω 0 and the other below ω 0, at which the response drops to this value We call these the upper cut-off frequency ω U and the lower cut-off frequency ω L, respectively We define the bandwidth of the tuned circuit by: Let's compute this bandwidth; since: Z ( jω ) R B ω U ω L ω + Q 0 ω 0 ω 0 ω at the upper and lower cut-off frequencies, we must have: or Q 0 ω ω 0 ω 0 ω 7

18 Because ω U > ω 0 and ω L < ω 0, we see that: Q 0 ω Q U 0 ω 0 + ω 0 ω U ω ω 0 ± ω 0 ω We will solve first for the upper cut-off frequency by letting ω Q L 0 ω 0 ω 0 ω L or The solution is x x Q 0 x Q 0 x 0 x Q 0 ± Q 0 We take the positive sign in order to get a positive frequency: ω U ω 0 + Q 0 Q 0 Exchanging ω L. for ω U and changing the sign of Q 0 gives: ω L ω 0 + Q 0 The bandwidth is given by the simple expression: + Q and we have: B ω U ω L ω 0 Q 0 Q 0 ω 0 B ω 0 ω U ω L Notice that ω L and ω U are not arithmetically symmetric relative to ω 0 (this would imply ω L + ω U ω 0 ) It may be shown that they are geometrically symmetric, ie ω L ω U ω 0 How might we measure ω 0 and Q 0 in the laboratory? Looking back at our resonance curve, we see that we can search for the peak frequency, which is ω 0 Next, we find the upper cut-off frequency by noting the frequency at which the response has dropped to / times the peak value (/ 0.707) and identifying it as ω U 8

19 Finally, we find ω L in a similar manner and subtract them to find the bandwidth Then we can compute Q Phase Shift Although we have considered the magnitude or gain characteristic, the phase shift is important for some applications, including TV and digital transmission It is of interest to see how the phase shift of the parallel tuned circuit varies with frequency We now plot φ(ω) as given by the equation we derived for two values of Q 0 : ( ) tan Q 0 ω Z jω ω 0 ω 0 ω Observe that the phase shift is zero at resonance and approaches ±π/ rad (±90 o ) for frequencies below and above the resonant frequency We can define resonance as the frequency at which the tuned-circuit impedance is purely resistive (φ(ω) 0) A look at the equation for φ(ω) shows immediately that this frequency is ω 0 5. The Lossless Series Tuned Circuit Consider an ideal series tuned circuit: 5 THE SERIES TUNED CIRCUIT Its behaviour is very much like that of the parallel tuned circuit, except that all the properties of the latter which we considered on an impedance basis hold for this circuit on an admittance basis The admittance of the series tuned circuit is: where Y ( jω ) Z( jω ) jωl + jωc jω L ( jω ) + ω j ω L 0 ω 0 ω jb ( ω ) ω 0 LC and B(ω) is the susceptance 9

20 A sketch of B(ω) versus ω is shown: Notice that the susceptance approaches infinity at ω ω 0 ; this means that it is equivalent to a short circuit at ω 0 The admittance of the series tuned circuit behaves precisely like the impedance of the parallel tuned circuit 5. The Parallel-to-Series Transformation for the Lossy Capacitor Consider the lossy capacitor equivalent circuit: The admittance is: Y ( jω ) jωc + r P Inverting to obtain the impedance: Z ( jω ) Y ( jω ) + jωc r p r P r P r P jr P Q C + jωcr P + jq C + Q C If we assume that Q C >> (a good approximation for a capacitor), then: Z ( jω ) r P jr P Q C Q C This gives the series equivalent shown: r P Q C + r P jωcr P r P Q C + jωc with r S r P Q C C' C Since Q C depends on frequency, this equivalence is strictly valid only at the frequency for which Q C is determined 0

21 5.3 The Q of the Lossy Series Tuned Circuit at Resonance Look, once again, at the series tuned circuit, assuming that both the inductor and the capacitor are lossy: The series resistance R includes any driving source resistance (considered as a Thevenin equivalent) along with the inductor series loss resistance and the transformed capacitor parallel loss resistance Let's apply a sinusoidal test voltage source to our sub-circuit and test the Q of the entire sub-circuit at the frequency ω 0 We see right away that as jω 0 L + (l/jω 0 C) 0, we have Z'(jω 0 ) l/y'(jω 0 ) 0 So Z(jω 0 ) R Thus, the current is in phase with the source voltage; furthermore, the voltage drop across the inductor/capacitor series combination is zero because its series impedance is zero Hence, we can immediately compute Q 0 π w P π L V m R w D V m R T 0 where T 0 is the period of a sinusoid at the resonant frequency ω 0 By making use of the equivalence: we can obtain an equivalent expression for Q 0 : Q 0 ω 0 R ω 0 Cω 0 LC ω 0 CR ω 0L R As a check, we can see from the circuit that the infinite Q condition is R 0 It is easy to show that the peak energies stored by the inductor and the capacitor are equal to one another - and therefore to the peak energy stored by the sub-circuit itself. 5.4 Using fractional frequency deviation Now compute the impedance of the lossy series tuned circuit:

22 So the admittance is: Z( jω ) R + jωl + Y ( jω ) jωc R + ω 0L jω + ω 0 R ω 0 jω R ω + jq 0 ω 0 ω 0 ω If we compare this with the Z(jω) for the parallel tuned circuit developed in the preceding section, we will see that they are identical in form The magnitude and phase are: Y ( jω ) R ω + Q 0 ω 0 ω 0 ω Y(jω) normalized to l/r and Y(jω) can be sketched: Y ( jω ) tan ω Q 0 ω 0 ω 0 ω The admittance of the series tuned-circuit has the same form as the impedance of the parallel tunedcircuit 5.5 Effect of resistor on resonant frequency A two-terminal circuit is said to be resonant at any nonzero frequency for which the AC steadystate impedance is purely real This does not always occur where an inductor and a capacitor produce equal and opposite reactances, as shown in the next example Example 6 Find the resonant frequency of the following circuit:

23 Solution The AC steady-state impedance is: R jωc Z ( jω ) jω L + R + jωc R + ωcr jω L + ( ) + jω L CR + ( ωcr) Setting the imaginary part to zero gives ω 0 and R + jωcr + ( ω 0 CR) CR L ω 0 LC L CR ω 0 LC L CR ( ) ( ) R jωcr jω L + + ωcr If R, we see that the resonant frequency approaches that of an ideal series tuned circuit The second radical represents the change in resonant frequency due to finite R 6. General 6 ND-ORDER PASSIVE FILTERS The tuned circuits which we have investigated have impedance and asmittance functions which exhibit frequency selectivity We now look into filters; a filter may be defined as a -port circuit whose frequency response function H ω ( ) V o V i exhibits frequency selectivity We begin by considering some examples 6. Filtering Examples The following example investigates a second-order bandpass filter: Example 4 Find the voltage transfer function of the circuit shown and plot its gain and phase characteristics 3

24 Solution Notice that there is a series tuned circuit between the voltage source (the filter input) and the output terminal Calling this impedance Z(jω), we have: H ( ω ) V o V s R L Z jω ( ) + R L Now we already know the functional form of Y(jω) for the lossy series tuned circuit: Hence, we obtain Y ( jω ) Rs ω + jq 0 ω 0 ω 0 ω H ( ω ) R L Y ( jω ) + R L R L ω R L + R s + jr s Q 0 ω 0 ω 0 ω R L RL + R s R + j s ω Q R L + R 0 ω 0 s ω 0 ω H 0 ' ω + jq 0 ω 0 ω 0 ω where R L H 0 R s + R L Q 0 ' R s R s + R L Q 0 Thus, our voltage transfer function has the same form as the admittance of the series tuned circuit or the impedance of the parallel tuned circuit There are only two differences: Qo has been decreased to Q 0 by the additional resistance and Ho is a voltage gain, not an admittance or impedance The gain and phase variations with frequency are shown: Note that they are identical in form with those for the parallel and series tuned circuits The next example discusses a second-order bandstop filter Example 5 Find the voltage gain transfer function of the circuit below and plot its gain and phase versus ω 4

25 Solution We start by recognizing that a major component of this circuit is a parallel tuned circuit, whose impedance we have already determined Thus, we can write the voltage gain transfer function as: where H ( ω ) V o V s R L Z ( jω ) + R L R L R P ω + jq 0 ω + R 0 L ω 0 ω ω + jq 0 ω 0 ω R L ω 0 ω + jq 0 ω 0 ω R P + R L R + j L ω Q R P + R 0 ω H 0 ω 0 0 ' ω L ω 0 ω + jq 0 ω 0 ω 0 ω ω R L + jq 0 ω 0 ω 0 ω ω R P + R L + jq 0 ω 0 ω 0 ω H 0 R L R L + R P Q 0 ' R L R L + R P Q 0 Let's investigate this function by converting it to Euler form by taking the magnitude and angle: and + Q 0 A( ω) H( ω) H 0 φ( ω) H ( ω) tan ω Q 0 We show a plot of these functions: ω ω 0 ω 0 ω ' ω + Q 0 ω 0 ω 0 ω ω 0 tan ' ω Q 0 ω 0 ω 0 ω ω 0 ω 5

26 We have used ω 0, H 0 0. and Q 0 0 for Q 0 ; this makes Q Type of frequency response function by inspection For the above example, we can predict the type of response it has before we carry out detailed analysis Consider the behaviour of the inductor and capacitor at the extreme frequencies ω 0 and ω Element Type Impedance Z ω 0 Z ω Inductor Z L jω L Capacitor Z C jωc Z L 0 (short-circuit) Z C (open-circuit) Z L (open-circuit) Z C 0 (short-circuit) Hence the inductor behaves like a short-circuit at zero frequency and an open-circuit as frequency tends to infinity The capacitor behaves like an open-circuit at zero frequency and a short-circuit as frequency tends to infinity Note that the equivalents for zero frequency are the same as the DC steady state equivalents we used in transient analysis (setting frequency to zero implies that the analysis is a DC analysis) Note further that at resonance, the parallel tuned-circuit is equivalent to an open-circuit and the series tuned-circuit is equivalent to a short-circuit We can now apply this to our band-stop circuit example: At zero frequency the inductor is a short-circuit and at infinite frequency the capacitor is a shortcircuit; therefore v o v s at these extreme frequencies; this implies that the amplitude response is unity and the phase is zero At resonance, the parallel tuned-circuit will become an open-circuit and the circuit simplifies to a potential divider consisting of R p and R L The gain depends on the resistor values and is H 0 0. in this case The phase at resonance must be zero Notice that a good notch filter (this one is not terribly good!) would have a large value of R p relative to the value of R L and a high Q 0 Thus, the notch would be narrower and deeper This by-inspection approach can be applied to most passive circuits in order to determine the type of response (low-pass, high-pass, band-pass, band-stop, all-pass) 6

27 6.4 Canonical Forms for nd-order Filters We now present a set of standard, or canonical, forms for the three most basic standard nd-order filter types: lowpass, bandpass, highpass and bandstop: H LP ( ω ) H BP ( ω ) H HP ( ω ) H BS ( ω ) s may be used as a place-holder for jω/ω 0 These functions all have the same denominator jω ω 0 jω ω 0 jω ω 0 jω ω Q 0 jω ω 0 + jω Q 0 ω 0 + Q 0 jω ω 0 + jω ω 0 jω ω 0 + Q 0 jω ω Q 0 jω ω 0 + If we wish, we can relate these expressions to those for tuned circuit impedances and/or admittances by restructuring the denominator polynomial: ( ) jω D ω ω 0 + jω + jω ω + jq 0 ω 0 Q 0 ω 0 ω 0 Q 0 ω 0 ω The second factor is one that we easily recognize to be the denominator of our tuned circuit admittance or impedance The above forms clearly show the following limiting values: ω 0 ω ω 0 ω H LP jq 0 0 H BP 0 0 H HP 0 jq 0 H BS 0 If Q 0 >>, then the lowpass and highpass filters have substantial gain at the frequency ω 0 However, for the lowpass filter, the actual maximum gain occurs slightly below ω 0 and is a little higher than Q 0 ; it may be shown that:

28 H LP ( ω ) max Q 0 Q 0 at ω p ω 0 Q 0 We can plot amplitude and phase response for the lowpass filter for a range of Q values Amplitude response H(jω) Phase response H(jω) It can be seen that for lower values of Q the frequency of peak gain is significantly less than ω 0 Notice also that the higher the value of Q, the steeper the phase curve at ω ω 0 For the highpass filter, the maximum gain is the same but it occurs slightly above ω 0 : ω p ω 0 Q 0 The plots of gain and phase are the same as for the lowpass filter but with the frequency scale inverted about ω 0 7 MORE GENERAL APPROACH TO ND ORDER PASSIVE FILTER DESIGN 7. General circuit architectures Rather than looking at specific examples, in this section, we take a more general approach Consider three general impedances connected in series: We can regard this circuit as having been obtained by de-activating the sources in a number of possible compete circuits; the short-circuit shows a possible positions for a voltage source The impedance looking into the short-circuit is: Z Z + Z + Z 3 We have dropped the (jω) or (ω) because the use of upper case letters shows that we are working in the frequency domain We now form new -port circuits by introducing a source into the above circuit in such a way that de-activation of the source gives the original circuit: 8

29 Circuit H ( jω ) V o V i Z 3 Z + Z + Z 3 V o V i Z + Z 3 Z + Z + Z 3 We have used voltage division to determine the frequency response functions for these two circuits Notice that in both cases the denominator of the expression is the same as the impedance of the deactivated circuit; therefore the denominator of the circuit depends on the de-activated circuit; it governs the natural response of the circuit independently of sources Consider now three general admittances connected in parallel: We can regard this circuit as having been obtained by de-activating a source in a number of possible compete circuits; the open-circuit shows a possible position for a current source The admittance of the circuit looking into the open-circuit is: Y Y + Y + Y 3 We now form new circuits by introducing sources into the above circuit in such a way that deactivation of the new circuits gives the original circuit: Circuit H ( jω ) V o V i Y Y + Y + Y 3 V o V i Y + Y Y + Y + Y 3 We have used voltage division to determine the frequency response functions for these two circuits 9

30 Notice that in both cases the denominator of the expression is the same as the admittance of the deactivated circuit; again the denominator of the circuit depends on the de-activated circuit; it governs the natural response of the circuit independently of sources We are now ready to consider replacing general impedances and admittances by R, L and C elements 7. Generation of nd order RLC circuits The first of the four architectures, based on series impedances, can generate three useful circuits, which we show together with their frequency response functions: Circuit H(jω) Type H ( jω ) ( jω ) LC + jωcr + Lowpass H ( jω ) H ( jω ) ( jω ) LC ( jω ) LC + jωcr + jωcr ( jω ) LC + jωcr + Highpass Bandpass The frequency response functions can be confirmed by use of the voltage division rule The second architecture generates only one useful circuit: Circuit H(jω) Type H ( jω ) ( jω ) LC + ( jω ) LC + jωcr + Bandstop Notice that all of these four circuits have the same denominator expression; this is expected because they all reduce to the same form when the input voltage source is deactivated The third architecture, based on parallel admittances, can generate three useful circuits: Circuit H(jω) Type H ( jω ) ( jω ) LC + jω L R + Lowpass H ( jω ) ( jω ) LC ( jω ) LC + jω L R + Highpass 30

31 H ( jω ) jω L R ( jω ) LC + jω L R + Bandpass The fourth architecture generates only one useful circuit: Circuit H(jω) Type H ( jω ) ( jω ) LC + ( jω ) LC + jω L R + Bandstop The four circuits based on parallel admittances have the same denominator expression as expected The general nd order denominator expression is: ( ) jω D ω ω 0 where ω 0 is the resonant frequency and Q 0 is the Q-factor 3 + Q 0 jω ω 0 + Comparing this with the expressions in the above tables, the first design equation is: LC ω 0 The second design equation for the first set of circuits is: CR R Q 0 ω 0 Q 0 The second design equation for the first set of circuits is: L C L R ω 0 Q 0 R Q 0 L C The nd order allpass response can not be produced by such simple circuits; we either need to use a passive circuit with a lattice structure or we can realise it with an active circuit using an operational amplifier or transistors 8. Introduction LC Q 0 LC Q 0 8 BODE PLOTS The Bode plot method is a way of rapidly sketching the gain and phase response of a circuit from its transfer function

32 Here, we will concentrate on the gain plot only Let's start with an example: Example 7 Find the transfer function H( ω) Vo Vi for the circuit shown: This circuit is a model for a voltage amplifier, such as the "preamp" in a stereo system Note that the circuit includes a voltage-controlled voltage source Solution As we are interested in frequency response, we use the phasor equivalent circuit: If use a unit of impedance of kω. and a unit of current of ma, the unit of voltage will remain V We can analyse the circuit in stages: V x jω V i jω jω jω jω Let the voltage of the voltage source, 00V x, be denoted V y ; then we have: and V o V y Z Z + Z Putting the three gain terms together, we have: H ( ω ) V o V o V y V x V i V y V x V V y V x 00 Z Y jω07 + jω jω jω jω jω We will be using this result later, so we will write it using symbols: 7 ( jω ) ( )( jω ) 3

33 8. Form of the System Function H( ω) K ( jω ) ( jω + p )( jω + p ) By generalizing upon the preceding example, we see that the system function of any response variable of a circuit constructed from our standard supply of elements, namely R, L, C, and dependent sources, has the form H( ω) V o ( )( jω + z )... ( jω + z m ) ( )( jω + p )... ( jω + p n ) K jω + z V i jω + p We call the factors (jω + z l ), (jω + z )... (jω + z m ) the zero factors of H(w) and the factors (jω + p ), (jω + p )... (jω + n) the pole factors of H(ω) K is the scale factor In the previous example, there is only one zero factor jω with z 0 There are two finite pole factors: p and p 0 7 The scale factor is K These zero and pole factors are all simple, or of order one, ie none are repeated If we plot the gain frequency response of a circuit such as the example circuit, we will obtain a graph with the general shape shown: It is zero at DC (ω 0) and also at infinite frequency (ω ) The response is constant over a very wide frequency range; the value of gain over this frequency range is called the midband gain A mb 8.3 Decibels Consider the following problem: We investigate an experimental plot of A(ω) taken in the lab and discover that A mb 00 We notice something strange about the plot in the low frequency range where A(ω) has a value of approximately As we wish to investigate this effect more closely, we adjust the scale of our plot so that a gain of one corresponds to a height of, say, cm If we wish, however, to closely investigate the shape of the plot in the midband range as well-where A(ω) is l00, we find that our plot must be at least 00 cm tall This would be a very unwieldy plot to handle For this reason, we perform a nonlinear transformation on our scale by setting: The logarithm is to the base 0, not to the base e ( ) A db 0log A ω 33

34 There is a historical reason for the factor of 0 The Bel (after Alexander Graham Bell, inventor of the telephone) was first defined as the unit of the logarithm of a power ratio: Power ratio in Bels log P It was then discovered that for most applications this unit was too large; thus, the decibel became the standard This was defined by the equation P Power ratio in decibels 0log P Assuming that powers are developed in resistors with sinusoidal voltage waveforms, we note that: P i V m R Thus, if both powers are measured relative to the same resistance value, then: Thus, we have the usual equation: Power ratio in decibels 0log V 0log V V V A db 0log A( ω ) There are a number of common values that occur frequently: A(ω) A db / 6 / Notice that a gain of 0 db does not mean that the response is zero-it means that it has exactly the same magnitude as the input Observe also the special values, / have db values of ±3, and, / have db values of ±6 A gain of /x has the same magnitude in db as that of x but opposite sign P 34

35 Doubling or halving gain corresponds to adding or subtracting 6 db Multiplying or dividing gain by a factor of 0 corresponds to adding or subtracting 0 db Example 8 Find the values of gain corresponding to 6 db, 00 db, and 34 db Solution We recall that log(xy) log(x) + log(y); thus, adding db values corresponds to multiplying the actual gain values 6 db 0 db + 6 db 0 db corresponds to a gain of 0 and 6 db to a gain of Thus, 6 db corresponds to a gain of 0 00 db 0og(x) x 0 00/ db 40 db 6 db Gain is Logarithms of a few basic positive integers are worth remembering: log() 0.30 log(3) log(5) log(7) Other values can be easily calculated: log(4) log( ) log() 0.60 log(9) log(3 ) log(3) log(30) 0log(0 3) 0log(0) + 0log(3) db 8.4 The Logarithmic Frequency Scale Now look back at the typical amplifier gain curve and consider the frequencies ω U and ω L Suppose we find in our lab experiment that ω U 0 5 rad/s and ω L 0 rad/s Suppose we want to investigate both regions in the frequency plot carefully, so we make 0 rad/s correspond to cm of horizontal length on our plot We find that in order to include ω U 0 5 rad/s we must make the plot 0 4 0,000 cm long For this reason, we perform a logarithmic transformation on the frequency axis: ω' log ω Note that we do not have the factor of 0 present in this case as we do for the gain Consider a normal ω and a log(ω) scale superimposed: ( ) 35

36 We note that multiplying any given frequency value by 0 results in adding one unit of log frequency; we call such a frequency change a decade Dividing a given frequency by a factor of 0 results in subtracting one unit of log frequency Similarly, multiplying or dividing by a factor of results in adding or subtracting 0.3 unit of log frequency We call such a frequency change an octave The name comes from music, where there are eight notes in a frequency increment of a factor of Note that when using a logarithmic frequency scale, the scale may be labelled in ω or log(ω) units as shown above Logarithmic frequency scales are also commonly used when the frequency is in Hz 8.5 Gain and phase for the general frequency response function A Bode plot is a plot of the gain A(ω) in db versus log(ω) and of the phase φ(ω) versus log(ω) We start from the general expression for the system function of a circuit: H( ω) V o ( )( jω + z )... ( jω + z m ) ( )( jω + p )... ( jω + p n ) K jω + z V i jω + p We then take the absolute value and note that the absolute value of a product is the product of the absolute values and that the absolute value of a ratio is the ratio of the absolute values: A( ω) H( ω) K jω + z jω + z... jω + z m jω + p jω + p... jω + p n Finally, we take the base 0 logarithm and multiply by 0 to get the db value: A db ω ( ) 0logK + 0log jω + z + 0log jω + z log jω + z m 0log jω + p 0log jω + p... 0log jω + p n Our interest is to develop quick approximate methods for sketching the Bode plot Let's investigate the different types of factor one at a time We assume that all the zero and pole constants (z k and p k, respectively) are real 8.6 The Scale Factor We look first at the scale factor, 0log K We have sketched this factor: The solid line represents a scale factor with magnitude greater than unity (positive db value) and the dotted line one with a magnitude less than unity 36

37 The horizontal axis is labelled with values of ω, but distances are plotted in terms of log(ω). Thus, the origin is at ω rad/s 8.7 The General Factor Now suppose that q finite zeros or q finite pole constants are equal Then we have z l z... z q a or p p... p q a in the general system function In this case, there is a zero or a pole factor of order q We will consider factors of the form. F ( ω) ( jω + a) q If the F(ω) term is a qth-order zero factor, it will appear in the numerator If it is a pole factor, it will appear in the denominator Consider the first case for which a 0 Let's consider: We start by taking the absolute value, to obtain: F ( ω) ( jω ) q F ( ω ) ( jω) q jω q ω q We have used the fact that the absolute value of a product is the product of the absolute values Finally, we take the logarithm and multiply by 0: F ( ω ) db 0log ω q ( ) 0qlog( ω ) We have plotted this factor assuming a qth-order zero factor at ω 0 As one must subtract the factor if it represents a pole, we have merely changed the sign in order to obtain the plot for a pole factor: A 0-fold increase or decrease in frequency results in a 0q db increase or decrease in the factor 37

38 This is a slope of ±0q db/decade It is also ±6q db/octave Suppose, now, that a 0: We now have: Next, we take the absolute value: F ( ω) ( jω + a) q F ( ω ) ( jω + a) q jω + a q Finally, taking the logarithm, we have Plotting the log term is not so easy! F ( ω ) db 0log ( jω + a) q 0qlog jω + a 0qlog ω + a For this reason, we resort to an approximation, noting that: ( ) db 0qlog ω F ω ( ); ω >> a 0qlog a; ω << a Note that we have used magnitude signs around the parameter a for it could possibly be negative 8.8 The Straight-Line Approximation We now show the approximation involved in constructing a linearised Bode gain plot where we assume that the factor is equal to 0qlog(ω) for ω a and to 0q log a for ω a The resulting plots for a zero factor and for a pole factor are as follows: The frequency ω a where the staight lines intersect is called the break frequency Note that the extrapolation of the sloping lines always pass through the origin at ( rad/sec, 0 db) 8.9 The Approximation Error It can be sown that the maximum error between the straight line approximation and the actual curve occurs at the break frequency w a and is given by: ( ) ±3q db Error ω a The plus sign will hold for a zero factor and the negative sign for a pole factor The true plot is above the approximate one by 3q db at a zero break frequency and is below it by 3q db at a pole break frequency For a single zero or single pole for which q, the maximum error is ±3 db An octave above and below the break frequency, we have: 38

39 ( ) Error ω a Error ω a A decade above and below the break frequency, we have ( ) Error ω Error ω 0 a for moderately small orders of q ±0.97q q db 0 a ±0.04q 0 db We now give an example showing how to sketch a Bode gain plot. Example 9 Suppose that a circuit or system has the system function Sketch the linearized Bode gain plot. Solution The scale factor is K 0 5, or 00 db It is plotted as follows: H( ω) V o 0 5 V i ( jω) ( jω +0) ( jω +000) There is a simple zero factor at ω 0 The corresponding factor is plotted as follows: For the simple pole factor (jω + 0) the break frequency is 0 rad/s; the plot is as follows: Finally, we plot the Bode factor corresponding to the pole factor (jω + 000) of order : In this last plot, we have not preserved the same scale as for the others because this would lead to problems in presentation Now we simply add all four plots order to obtain the overall Bode gain plot: 39

40 Notice that we have sketched in the curve, which falls below the linearised approximation Because the break frequencies are widely separated, the error at each is almost that computed for each in isolation, i.e., 3 db at 0 rad/s and 6 db at 000 rad/s The gain for the flat part of the curve may be fixed by adding the individual figures at any convenient frequency, eg at 0 rad/sec 8.0 The Quick Method ( ) db A 0 There is a much faster way of doing this type of plot Looking at the Bode plot produced in the above example, we could have started at very low values of frequency with a line of positive slope dictated by the order of the zero factor at ω 0 Then we could make the slope break upward by an amount 0q db/dec at each zero break frequency or downward by an amount 0q db/dec at each pole break frequency, where q is the order of each zero or pole When any frequency being considered is below a given break frequency, we use the constant value; when it is above that break frequency, we use ±0qlog(ω) Example 0 Use the quick method to sketch the Bode gain plot for the following transfer function: Solution ( ) ( jω +0) ( jω +00) ( jω + 500) ( jω ) ( jω +000) jω +0 4 H ω ( ) The first step is to mark the zeros and poles on the log(ω) axis; we use for o for zeros and for poles: The bulk of the solution consists merely of steps in sketching the graph, so we merely sketch the final shape, starting from the low end of the frequency scale: We can evaluate the approximate magnitude at ω rad/s to fix the vertical axis scale: ( ) ( 0) ( 00) ( 500) ( ) ( 000) 0 4 H ( ) db 40