Frequency Response

The transfer function \(\mathbf{H}(\omega)\) is the ratio of a circuit's output phasor to its input phasor, as a function of frequency.

Since input and output may each be a voltage or a current, there are four kinds:

Being a ratio of complex quantities, \(\mathbf{H}\) is itself complex and is written

where \(H\) is the magnitude response and \(\phi\) the phase response. Plotting both against frequency gives the frequency response, and that pair of plots is the complete description of what a linear circuit does to any sinusoid.

Written as a ratio of polynomials in \(j\omega\),

the roots of \(\mathbf{N} = 0\) are the zeros and the roots of \(\mathbf{D} = 0\) the poles. A pole makes the response large; a zero makes it small. Locating them is the quickest way to see what a circuit will do.

Frequency responses span enormous ranges — a filter may pass one frequency at unity gain and reject another by a factor of a million. A logarithmic scale is essential, and the one universally used comes from telephone engineering.

The bel measures a power ratio as its logarithm:

The bel proved too large for convenience, so the decibel — one tenth of a bel — is used instead:

The voltage form. If the two powers are developed in equal resistances, then \(P = V^{2}/R\) and

The factor is 10 for power ratios and 20 for voltage or current ratios. Confusing them is the commonest error with decibels, and it produces an answer exactly double or half the correct one. Strictly the voltage form requires equal resistances; in practice it is applied to any voltage ratio, and the convention is universal.

The row worth memorising is \(1/\sqrt{2}\). A voltage ratio of 0.7071 is \(-3\) dB, and since power goes as the square, the power has fallen to exactly one half. This is why the −3 dB point and the half-power point are the same thing, and why cutoff frequencies are quoted at \(-3\) dB.

The great convenience of decibels is that cascaded stages add instead of multiplying, since the logarithm of a product is the sum of the logarithms. This is what makes Bode plots possible.

Bode plots are semilogarithmic plots of magnitude in decibels and phase in degrees against frequency on a logarithmic axis. Their power comes from a single observation: on a logarithmic magnitude scale, the factors of a transfer function add rather than multiply, so each may be plotted separately and the results summed graphically.

To exploit this, the transfer function is first written in standard form, in which every factor is arranged as \(\left(1+j\omega/\text{something}\right)\):

Only four kinds of factor can appear, and each has a simple asymptotic plot:

• a constant gain \(K\);

• a pole or zero at the origin, \(\left(j\omega\right)^{\pm1}\);

• a simple pole or zero, \(\left(1+j\omega/p\right)^{-1}\) or \(\left(1+j\omega/z\right)\);

• a quadratic pole or zero.

Resonance is the condition in which a circuit's impedance is purely real — the inductive and capacitive reactances cancel exactly, leaving voltage and current in phase.

For a series RLC circuit,

Setting the imaginary part to zero:

This is the same \(\omega_{0}\) as in Chapter 8, where it was the undamped natural frequency of the transient. The circuit's tendency to oscillate at \(1/\sqrt{LC}\) when disturbed, and its tendency to respond most strongly to excitation at \(1/\sqrt{LC}\), are the same fact viewed from two domains.

At resonance:

• \(\mathbf{Z} = R\) — the impedance is minimum and purely resistive, so the current is maximum.

• The \(LC\) pair behaves as a short circuit, and the entire source voltage appears across \(R\).

• Voltage and current are in phase, so the power factor is unity.

• The average power dissipated is at its maximum, \(P(\omega_{0}) = \tfrac{1}{2}V_{m}^{2}/R\).

A striking consequence. Although the reactances cancel, they are individually far from zero. The voltage across each is

where \(Q\) is the quality factor defined in the next section. The voltage across the inductor and across the capacitor can be many times the source voltage — they are equal in magnitude and opposite in phase, so they cancel in the KVL sum while each individually is enormous. A series resonant circuit is therefore also called a voltage resonant circuit, and this voltage magnification is a real hazard: components must be rated for \(QV_{m}\), not \(V_{m}\).

The current magnitude in a series RLC circuit is

peaking at \(V_{m}/R\) when \(\omega = \omega_{0}\). At two other frequencies the dissipated power falls to half its maximum, and these are the half-power frequencies \(\omega_{1}\) and \(\omega_{2}\). Half the power means the current has fallen by \(1/\sqrt{2}\), so

Solving this quadratic gives

Multiplying the two roots reveals a neat relationship:

The resonant frequency is the geometric mean of the half-power frequencies — not the arithmetic mean, because the response is not symmetrical about \(\omega_{0}\) on a linear axis. It is symmetrical on a logarithmic axis, which is one more reason Bode plots use one.

The bandwidth is the width of the half-power band:

The parallel RLC circuit is the dual, and by the duality of Section 8.9 every result carries over with current and voltage exchanged.

Its admittance is

and setting the imaginary part to zero gives the same resonant frequency:

But the behaviour at resonance is inverted. Where the series circuit had minimum impedance and maximum current, the parallel circuit has maximum impedance and minimum current. And where the series circuit magnified voltage, the parallel circuit magnifies current: the currents circulating in \(L\) and \(C\) can be \(Q\) times the supply current, which is why it is called a current resonant or tank circuit.

Note the crucial reversal in \(Q\). For the series circuit \(Q = \omega_{0}L/R\) falls as \(R\) rises; for the parallel circuit \(Q = R/\omega_{0}L\) rises with \(R\). A large resistance sharpens a parallel resonance and blunts a series one — exactly the pattern seen with the damping factor in Chapter 8, and for the same reason.

A filter is a circuit designed to pass signals in one band of frequencies and reject those outside it. A passive filter uses only \(R\), \(L\) and \(C\). There are four types.

The transfer functions follow immediately from voltage division. For the low-pass and high-pass pair:

Their squared magnitudes sum to one at every frequency, which is why a single RC network can serve as both — take the output across the capacitor for low-pass, across the resistor for high-pass. Example 14.1 was the low-pass case.

The band-pass filter is a series resonant circuit, so everything from Sections 14.5 and 14.6 applies: its centre frequency is \(\omega_{0} = 1/\sqrt{LC}\), its bandwidth is \(B = R/L\), and its selectivity is \(Q = \omega_{0}/B\). The band-stop filter is the same circuit with the output taken from the other element, so it rejects precisely what the band-pass accepts.

Two practical limitations of passive filters. They cannot have a gain greater than one, since they contain no source of energy. And a filter's response depends on what is connected to it, so cascading two passive stages does not simply multiply their transfer functions — the second loads the first. Both problems are solved in the next section.

An active filter combines resistors and capacitors with an operational amplifier. This solves both limitations of the passive filter at once: the op amp supplies gain, and its low output impedance and high input impedance mean that stages can be cascaded without loading one another.

A third advantage matters more than either at low frequencies. Passive filters need inductors, which at audio frequencies are large, heavy, lossy and expensive. Active filters achieve the same responses with no inductors at all, which is why every audio equaliser and every anti-aliasing filter is active.

The design rests on the Chapter 5 result, extended to impedances in Section 10.11:

First-order active low-pass. Take \(\mathbf{Z}_{1} = R_{1}\) and \(\mathbf{Z}_{f} = R_{f}\parallel C_{f}\):

A low-pass response with a DC gain of \(-R_{f}/R_{1}\) — and note that the gain and the cutoff frequency are set independently, by \(R_{1}\) and \(C_{f}\) respectively. No passive filter offers that.

First-order active high-pass. Move the capacitor to the input, \(\mathbf{Z}_{1} = R_{1}+1/j\omega C_{1}\) with \(\mathbf{Z}_{f} = R_{f}\):

Band-pass by cascading. A low-pass stage followed by a high-pass stage, with the high-pass cutoff below the low-pass cutoff, passes only the band between them. Because op-amp stages do not load one another, the overall transfer function really is the product of the two — and on a Bode plot, the sum.

Higher-order responses come from cascading more stages, and the second-order Sallen–Key topology mentioned in Section 8.8 gives a quadratic pole whose damping is set by a resistor ratio rather than by a physical inductance.

Filter designs are usually tabulated for a normalised prototype with \(R = 1~\Omega\) and \(\omega_{0} = 1~\mathrm{rad/s}\), because the algebra is then trivial. Scaling converts such a prototype into a realisable circuit at any impedance level and any frequency.

Magnitude scaling multiplies every impedance by \(K_{m}\) while leaving the frequency response unchanged. Since \(\mathbf{Z}_{R} = R\), \(\mathbf{Z}_{L} = j\omega L\) and \(\mathbf{Z}_{C} = 1/j\omega C\), this requires

Note that the capacitor scales inversely, because its impedance is inversely proportional to \(C\).

Frequency scaling shifts the response to a higher frequency by \(K_{f}\) without changing impedance levels. Each reactance must reach the same value at \(K_{f}\) times the frequency, so

Combining both gives the general result:

The resistance depends on \(K_{m}\) alone, the inductance on the ratio, and the capacitance on the product. Quality factor, damping ratio and every other dimensionless parameter are unaffected — scaling changes the numbers but never the shape of the response.

Tuning a radio is resonance made audible. A parallel LC circuit in the receiver's front end presents a high impedance only at its resonant frequency, so only that station's signal develops appreciable voltage. All the others — arriving at the antenna simultaneously and at comparable strength — are shunted away.

Since \(\omega_{0} = 1/\sqrt{LC}\), tuning is achieved by varying \(C\) with a fixed \(L\): historically a ganged variable capacitor, today a voltage-controlled varactor diode.

The design constraint is selectivity. Adjacent AM stations are spaced 10 kHz apart, so at 1 MHz the receiver needs \(Q \gtrsim 100\) to separate them — which is exactly the specification of Example 14.6. Too low a \(Q\) and stations interfere; too high, and the sidebands carrying the audio are themselves attenuated.

• The transfer function \(\mathbf{H}(\omega)\) is the output-to-input phasor ratio; its magnitude and phase plots constitute the frequency response.

• Decibels use 10\(\log\) for power ratios and 20\(\log\) for voltage ratios. A voltage ratio of \(1/\sqrt{2}\) is \(-3\) dB and corresponds to half power.

• Bode plots exploit the fact that logarithms add: each factor is plotted asymptotically and the results summed.

• Slopes are \(\pm20\) dB/decade per simple pole or zero and \(\pm40\) for quadratics; the straight-line error at a corner is 3 dB.

• Resonance occurs when the reactances cancel, at \(\omega_{0} = 1/\sqrt{LC}\) for both series and parallel circuits.

• Series resonance gives minimum impedance and magnifies voltage by \(Q\); parallel resonance gives maximum impedance and magnifies current by \(Q\).

• \(Q = \omega_{0}/B\), and \(\omega_{0} = \sqrt{\omega_{1}\omega_{2}}\) — the geometric mean of the half-power frequencies.

• The four passive filter types are low-pass, high-pass, band-pass and band-stop; band-pass and band-stop are resonant circuits.

• Active filters add gain, remove the need for inductors, and allow cascading without loading.

• Scaling converts a normalised prototype to any impedance and frequency without altering the response shape.

• Using \(10\log\) for a voltage ratio. Power ratios take 10, voltage and current ratios take 20. The error is a factor of two in decibels.

• Forgetting to convert to standard form before drawing a Bode plot. Every factor must read \(\left(1+j\omega/\text{corner}\right)\), so \(\left(j\omega+2\right)\) must be written \(2\left(1+j\omega/2\right)\) — and that 2 belongs in the gain term.

• Confusing \(f\) and \(\omega\) in resonance formulas. \(\omega_{0} = 1/\sqrt{LC}\) is in rad/s; \(f_{0}\) is smaller by \(2\pi\). Ratios such as \(Q = \omega_{0}/B\) are safe in either unit provided both are the same.

• Using the series \(Q\) formula for a parallel circuit. They are reciprocal in \(R\): series \(Q = \omega_{0}L/R\), parallel \(Q = R/\omega_{0}L\).

• Taking \(\omega_{0}\) as the arithmetic mean of the half-power frequencies. It is the geometric mean. The arithmetic mean is a good approximation only for \(Q \ge 10\).

• Overlooking voltage magnification at series resonance. A 20 V source across a \(Q = 25\) circuit puts 500 V across the inductor. Components must be rated accordingly.

• Assuming cascaded passive filters multiply their transfer functions. The second stage loads the first. Only buffered or active stages cascade cleanly.

• Applying the straight-line Bode approximation near a lightly damped quadratic pole. The true curve can peak far above the asymptotes when \(\zeta\) is small.

• Scaling the capacitor the same way as the inductor. Under magnitude scaling \(L\) multiplies by \(K_{m}\) while \(C\) divides by it.

• Believing a passive filter can have gain above one. It contains no energy source. Gain requires an active device.

Work these before opening the answers. State clearly whether each frequency is in hertz or radians per second.

1. P14.1 Express as decibels: (a) a power gain of 50, (b) a voltage gain of 50, (c) a voltage gain of 0.25.

   Show answer

   \[\text{(a) } 10\log_{10}50 = 16.99~\mathrm{dB}, \qquad \text{(b) } 20\log_{10}50 = 33.98~\mathrm{dB}\]

   \[\text{(c) } 20\log_{10}0.25 = -12.04~\mathrm{dB}\]

2. P14.2 A series RLC circuit has \(R = 10~\Omega\), \(L = 2~\mathrm{mH}\) and \(C = 20~\mu\mathrm{F}\). Find \(\omega_{0}\), \(Q\) and \(B\).

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   \[\omega_{0} = \frac{1}{\sqrt{\left(2\times10^{-3}\right)\left(20\times10^{-6}\right)}} = \frac{1}{2\times10^{-4}} = 5000~\mathrm{rad/s}\]

   \[Q = \frac{\omega_{0}L}{R} = \frac{\left(5000\right)\left(0.002\right)}{10} = 1.000, \qquad B = \frac{R}{L} = 5000~\mathrm{rad/s}\]

   With \(Q = 1\) the bandwidth equals the resonant frequency — a very broad, unselective resonance.

3. P14.3 A parallel RLC circuit has \(R = 10~\mathrm{k}\Omega\), \(L = 5~\mathrm{mH}\), \(C = 2~\mu\mathrm{F}\). Find \(\omega_{0}\), \(Q\) and \(B\).

   Show answer

   \[\omega_{0} = \frac{1}{\sqrt{\left(5\times10^{-3}\right)\left(2\times10^{-6}\right)}} = \frac{1}{10^{-4}} = 10^{4}~\mathrm{rad/s}\]

   \[Q = \frac{R}{\omega_{0}L} = \frac{10^{4}}{\left(10^{4}\right)\left(5\times10^{-3}\right)} = 200, \qquad B = \frac{\omega_{0}}{Q} = 50~\mathrm{rad/s}\]

4. P14.4 A resonant circuit has \(\omega_{1} = 90\) and \(\omega_{2} = 110~\mathrm{rad/s}\). Find \(\omega_{0}\), \(B\) and \(Q\), and compare the exact \(\omega_{0}\) with the arithmetic mean.

   Show answer

   \[\omega_{0} = \sqrt{\left(90\right)\left(110\right)} = \sqrt{9900} = 99.50~\mathrm{rad/s}\]

   \[B = 20~\mathrm{rad/s}, \qquad Q = \frac{99.50}{20} = 4.975\]

   The arithmetic mean is 100, differing by 0.5 %. Since \(Q < 10\) the approximation is beginning to show, though it is still adequate for most purposes.

5. P14.5 An RC low-pass filter must have a cutoff of 5 kHz using \(C = 0.01~\mu\mathrm{F}\). Find \(R\), and the gain in dB at 50 kHz.

   Show answer

   \[\omega_{c} = 2\pi\left(5000\right) = 3.142\times10^{4}~\mathrm{rad/s}, \qquad R = \frac{1}{\omega_{c}C} = 3183~\Omega\]

   At 50 kHz, \(\omega/\omega_{c} = 10\):

   \[H = \frac{1}{\sqrt{1+100}} = 0.09950 \;\Longrightarrow\; 20\log_{10}0.0995 = -20.04~\mathrm{dB}\]

   One decade above cutoff gives very nearly \(-20\) dB, confirming the asymptotic slope.

6. P14.6 Put \(\mathbf{H}(\omega) = \dfrac{50}{j\omega\left(j\omega+5\right)}\) in standard form and state the low-frequency slope.

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   \[\mathbf{H}(\omega) = \frac{50}{5\,j\omega\left(1+j\omega/5\right)} = \frac{10}{j\omega\left(1+j\omega/5\right)}\]

   A gain of 10, a pole at the origin, and a simple pole at \(\omega = 5\). Below \(\omega = 5\) only the origin pole is active, so the slope is \(-20\) dB/decade; above it, \(-40\) dB/decade.

7. P14.7 Design an active low-pass filter with a gain of 5 and a cutoff of 1 kHz, using \(C_{f} = 10~\mathrm{nF}\).

   Show answer

   \[\omega_{c} = 2\pi\left(1000\right) = 6283~\mathrm{rad/s}, \qquad R_{f} = \frac{1}{\omega_{c}C_{f}} = \frac{1}{\left(6283\right)\left(10^{-8}\right)} = 15.92~\mathrm{k}\Omega\]

   \[R_{1} = \frac{R_{f}}{5} = 3.183~\mathrm{k}\Omega\]

8. P14.8 Scale a prototype with \(R = 1~\Omega\), \(L = 2~\mathrm{H}\), \(C = 0.5~\mathrm{F}\) using \(K_{m} = 500\) and \(K_{f} = 10^{5}\).

   Show answer

   \[R' = 500~\Omega, \qquad L' = \frac{500}{10^{5}}\left(2\right) = 10~\mathrm{mH}, \qquad C' = \frac{0.5}{\left(500\right)\left(10^{5}\right)} = 10~\mathrm{nF}\]

   Check: \(\omega_{0}' = 1/\sqrt{\left(0.01\right)\left(10^{-8}\right)} = 10^{5}~\mathrm{rad/s}\), which is \(K_{f}\) times the prototype's \(\omega_{0} = 1/\sqrt{1} = 1\) \(\checkmark\)

9. P14.9 A series resonant circuit with \(Q = 80\) is driven by a 12 V source at resonance. Find the voltage across the capacitor.

   Show answer

   \[\left|\mathbf{V}_{C}\right| = QV_{m} = \left(80\right)\left(12\right) = 960~\mathrm{V}\]

   Eighty times the source voltage. The capacitor must be rated well above 960 V despite a 12 V supply.

10. P14.10 An FM receiver tunes 88 to 108 MHz with a fixed inductor. What capacitance ratio is required?

    Show answer

    \[\frac{C_{\max}}{C_{\min}} = \left(\frac{f_{\max}}{f_{\min}}\right)^{2} = \left(\frac{108}{88}\right)^{2} = \left(1.227\right)^{2} = 1.506\]

    Only about 1.5:1, compared with 8.78:1 for the AM band of Example 14.9 — FM's narrow fractional bandwidth makes tuning far easier.