ECE 211 · Lecture 19: Transfer Functions & Frequency Response

§1 · The roadmap

Chapter 14 in five items

The chapter opens with a list, and the list is worth keeping in view, because every later item is a different reading of the same object, the transfer function:

1 Transfer function 2 Frequency response 3 Bode plots 4 Resonance, quality, cutoff 5 Filters

The transfer function is the s-domain relationship between the input and the output, taken with zero initial conditions.
The frequency response is that transfer function evaluated along $s = j\omega$ .
Bode plots present $|H(j\omega)|$ and $\angle H(j\omega)$ against frequency.
Resonance, the quality of resonance, and the cutoff frequency are read from those plots.
Filters are specified and designed from them.

Items 1 and 2 are established on this page. Items 3 to 5 are built on top of them in the lectures that follow, so the definitions collected here are the ones that must be secure.

Zero initial conditions are part of the definition

A transfer function is a property of the circuit, not of any particular startup transient. The initial capacitor voltages and inductor currents are therefore set to zero before the ratio is formed. Initial conditions do not disappear from circuit analysis; they simply belong to the complete s-domain solution rather than to $H(s)$ .

§2 · The three domains

One element, three descriptions

Each element law has now been written three times, and the three versions differ only in what replaces the derivative. Chapters 7 and 8 kept the derivative; Chapter 10 replaced it by $j\omega$ for a single steady-state frequency; Chapter 14 replaces it by the complex variable $s$ , which carries no such restriction.

Element	Time domain (Ch. 7 and 8)	Phasor domain (Ch. 10)	s-domain (Ch. 14)
R	v = iR	Ṽ = ĨR	V(s) = I(s) · R
L	v = L di/dt	Ṽ = jωL Ĩ	V(s) = L[sI(s) − i(0⁺)]
C	i = C dv/dt	Ĩ = jωC Ṽ	I(s) = C[sV(s) − v(0⁺)]

The two reactive laws are read as circuit models. The inductor law contains a term proportional to $I(s)$ and a constant term, so it is drawn as an impedance in series with a voltage source. The capacitor law contains a term proportional to $V(s)$ and a constant term, so it is drawn as an admittance in parallel with a current source.

Inductor: the impedance sL in series with the initial-condition source L i(0⁺).

Capacitor: the impedance 1/sC in parallel with the initial-condition source C v(0⁺).

What survives when the initial conditions are zero

With $i(0^+) = 0$ and $v(0^+) = 0$ , both auxiliary sources vanish and only the impedances remain: $sL$ for the inductor and $1/sC$ for the capacitor. Every circuit on this page is drawn in that reduced form, because the transfer function is defined there. The elements then combine in series and in parallel exactly as resistors do, and the whole toolbox of Lectures 2 to 5 applies unchanged.

§3 · The same circuit, twice

Calculus in one domain, algebra in the other

A single series circuit is analyzed twice, first in the time domain and then in the s-domain, to show that the two routes end at the same differential equation. The point of the exercise is the cost of each route, not the answer.

The circuit

A source $v_s$ drives $R$ , $L$ , and $C$ in series, and the capacitor voltage $v_C$ is the output. One loop carries one current $i$ , and that current is the capacitor current, so $i = C\,v_C'$ .

KVL in the time domain

\tm{rl.vs}{-v_s} + \tm{rl.r rl.i}{R\,i} + \tm{rl.l rl.i}{L\frac{di}{dt}} + \tm{rl.c}{v_C} = 0

Substituting $i = C v_C'$ removes $i$ in favor of the output. The inductor term requires a second derivative, since $v_L = L\,di/dt = L\,\frac{d}{dt}(C v_C') = LC\,v_C''$ :

-v_s + RC\,v_C' + LC\,v_C'' + v_C = 0

The standard second-order form

v_C'' + \tm{rl.r rl.l}{\frac{R}{L}}v_C' + \tm{rl.l rl.c}{\frac{1}{LC}}v_C = \tm{rl.l rl.c}{\frac{1}{LC}}v_s

Reaching this line required one substitution and one differentiation of a product. The work grows with the order of the circuit, and it must be redone for every new output variable.

The same circuit in the s-domain

The elements are replaced by $R$ , $sL$ , and $1/sC$ , with the initial conditions taken as zero. The loop current becomes $I(s) = sC\,V_C(s)$ , which is the s-domain statement of $i = C v_C'$ : the derivative has been replaced by a multiplication.

KVL in the s-domain

\tm{rl.vs}{-V_s} + \tm{rl.r rl.i}{R\,sC\,V_C} + \tm{rl.l rl.i}{s^2 LC\,V_C} + \tm{rl.c}{V_C} = 0

The inductor contributes $sL \cdot I(s) = sL \cdot sC V_C = s^2 LC\,V_C$ . No differentiation is performed anywhere; the same physics is expressed as a polynomial in $s$ .

One line of algebra finishes it

V_C\left[s^2 + \frac{R}{L}s + \frac{1}{LC}\right] = \frac{1}{LC}V_s

Reading $s$ as $d/dt$ recovers the time-domain equation exactly: $v_C'' + \frac{R}{L}v_C' + \frac{1}{LC}v_C = \frac{1}{LC}v_s$ . The correspondence $s \leftrightarrow d/dt$ is what makes the s-domain worth the change of variable.

§4 · The transfer function and the frequency response

Definition

H(s) = \frac{Y(s)}{X(s)} = \frac{\text{s-domain OUT}}{\text{s-domain IN}} \qquad \text{(zero initial conditions)}

The transfer function is a ratio of two s-domain quantities, so it is a property of the circuit alone. For the Section 3 circuit, the bracketed line divides immediately:

H(s) = \frac{V_C}{V_s} = \frac{\dfrac{1}{LC}}{s^2 + \dfrac{R}{L}s + \dfrac{1}{LC}}

The frequency response

The frequency response is that same function evaluated on the imaginary axis, $s = j\omega$ . With $(j\omega)^2 = -\omega^2$ ,

H(j\omega) = \frac{\dfrac{1}{LC}} {-\omega^2 + j\dfrac{R}{L}\omega + \dfrac{1}{LC}}

and the polar form of a quotient gives the magnitude and the angle separately:

|H(j\omega)| = \frac{\dfrac{1}{LC}} {\sqrt{\left(\dfrac{1}{LC} - \omega^2\right)^2 + \left(\dfrac{R}{L}\omega\right)^2}} \qquad\quad \angle H(j\omega) = -\tan^{-1}\!\left[\frac{\dfrac{R}{L}\omega} {\dfrac{1}{LC} - \omega^2}\right]

The arctangent needs its quadrant

The angle of the denominator is the angle of the complex number $\left(\frac{1}{LC} - \omega^2\right) + j\frac{R}{L}\omega$ , whose imaginary part is always positive. Below $\omega = 1/\sqrt{LC}$ the real part is positive and the principal arctangent is correct; above it the real part turns negative and $180°$ must be added before the sign is applied, which carries the phase smoothly from $0°$ through $-90°$ toward $-180°$ . A calculator that returns only the principal value will report a positive phase for high frequencies, which is wrong.

What the transfer function does to a sinusoid

Sinusoidal steady state is the reason the substitution $s = j\omega$ is made. A cosine enters, and a cosine of the same frequency leaves; only the amplitude and the phase are changed, and $H(j\omega)$ carries both changes.

The input phasor is multiplied by $H(j\omega)$ . The frequency is untouched, which is why one complex number per frequency is a complete description of the steady-state behavior.

x(t) = A cos(ωt + θ) B = A · |H(jω)| φ = θ + ∠H(jω)

The response, live

The sliders below drive the Section 3 circuit with $L = 1$ H and $C = 0.01$ F fixed, so that $1/\sqrt{LC} = 10$ rad/s. The resistance sets the sharpness of the peak, and the frequency marker rides both curves.

Magnitude. The response starts at 1 at DC and falls away above the peak.

Phase. It passes through exactly −90° at 10 rad/s, for every value of R.

R (Ω) 10

ω (rad/s) 10

|H(jω)| = 1.000 ∠H(jω) = −90.0° at ω = 10.0 rad/s

cos(ωt) ⟶ 1.000 cos(ωt −90.0°)

The amplitude is scaled and the phase is shifted; the frequency is untouched.

A preview of items 3 to 5

Three features of these curves are named in the lectures that follow. The peak is resonance, its sharpness is the quality of that resonance, and the frequency at which the magnitude has fallen to $1/\sqrt{2}$ of its peak is the cutoff. Reducing $R$ raises and narrows the peak without moving the $-90°$ crossing, which is the first hint that the crossing marks the resonant frequency itself.

§5 · Worked example: a ladder network

From five elements to one transfer function

The circuit is a ladder, so no single divider reaches the output. Source transformation collapses it to one loop, and every combination is performed on impedances exactly as it would be on resistances.

The problem

Find $H(s) = V_o/V_s$ , and then the steady-state output when $v_s(t) = 6 + 12\cos 2t$ V. Given: 10 Ω, 1/20 F, 1 H, 1/10 F, and 5 Ω.

Write the impedances

With zero initial conditions, each reactive element becomes a single impedance: $1/20$ F $\to 20/s$ , $1$ H $\to s$ , and $1/10$ F $\to 10/s$ . The resistors are unchanged. From this point onward the circuit is a resistive network whose values happen to be functions of $s$ .

Transform the source and expose the parallel pairs

The $V_s$ source with its 10 Ω becomes a current source $V_s/10$ in parallel with that same 10 Ω. Two parallel pairs are then visible: the 10 Ω with the 20/s at the input, and the 5 Ω with the 10/s at the output.

Combine each pair

\tm{ld.z1}{10 \parallel \frac{20}{s}} = \frac{10 \cdot \frac{20}{s}}{10 + \frac{20}{s}} = \frac{200}{10s + 20} = \tm{ld.z1}{\frac{20}{s+2}} \qquad\quad \tm{ld.z2}{5 \parallel \frac{10}{s}} = \frac{50}{5s + 10} = \tm{ld.z2}{\frac{10}{s+2}}

Both pairs share the pole at $s = -2$ , which is a coincidence of the chosen element values rather than a general rule.

Transform back, and read one loop

The current source and the 20/(s+2) beside it become a voltage source

\tm{ld.vth}{\frac{V_s}{10} \cdot \frac{20}{s+2}} = \tm{ld.vth}{\frac{2V_s}{s+2}}

in series with the same impedance. One loop remains, carrying $I$ .

Solve the loop

I = \frac{\tm{ld.vth}{\dfrac{2V_s}{s+2}}} {\tm{ld.z1}{\dfrac{20}{s+2}} + \tm{ld.l1}{s} + \tm{ld.z2}{\dfrac{10}{s+2}}} = \frac{2V_s}{20 + s(s+2) + 10} = \frac{2V_s}{s^2 + 2s + 30}

V_o = I \cdot \tm{ld.z2}{\frac{10}{s+2}} = \frac{20 V_s}{(s^2 + 2s + 30)(s+2)} = \frac{20 V_s}{s^3 + 4s^2 + 34s + 60}

The transfer function, and the differential equation it encodes

\boxed{H(s) = \frac{V_o}{V_s} = \frac{20}{s^3 + 4s^2 + 34s + 60}}

Cross-multiplying gives $V_o[s^3 + 4s^2 + 34s + 60] = 20 V_s$ , and reading each power of $s$ as a derivative recovers the differential equation that was never written:

v_o''' + 4v_o'' + 34v_o' + 60 v_o = 20 v_s

A third-order equation was obtained without differentiating anything.

The steady-state response to $v_s(t) = 6 + 12\cos 2t$

The input carries two frequencies, and a linear circuit treats them independently. Each is pushed through $H(j\omega)$ separately, and the results are added.

H(j0) = \frac{20}{0 + 0 + 0 + 60} = \frac{1}{3}

A constant is a cosine of zero frequency. Its phase shift is $0°$ , because $H(j0)$ is real and positive here, and the 6 V term is simply scaled:

6 \cdot \tfrac{1}{3} = \boxed{2\ \text{V}}

H(j2) = \frac{20}{(j2)^3 + 4(j2)^2 + 34(j2) + 60} = \frac{20}{-j8 - 16 + j68 + 60} = \frac{20}{44 + j60}

|H(j2)| = \frac{20}{\sqrt{44^2 + 60^2}} = \frac{20}{74.40} = 0.2688 \qquad \angle H(j2) = -\tan^{-1}\frac{60}{44} = -53.75°

The 12 V amplitude is scaled and the phase is shifted:

12 \cdot 0.2688 = \boxed{3.23\ \text{V}} \qquad 0° + (-53.75°) = \boxed{-53.75°}

v_o(t) = 2 + 3.23\cos(2t - 53.75°)\ \text{V}

Check: does the answer behave sensibly?

Two independent sanity checks are available. First, at DC the two capacitors are open and the inductor is a short, leaving a 10 Ω and a 5 Ω divider: $V_o = 6 \cdot 5/15 = 2$ V, which matches the constant term exactly. Second, the magnitude at 2 rad/s must be smaller than the DC value of $1/3$ , because the denominator has grown from 60 to $|44 + j60| = 74.4$ ; indeed $0.2688 < 0.3333$ .

The board notes round $|H(j2)|$ to $0.27$ and report $3.24$ V. Carrying the unrounded value $20/\sqrt{5536} = 0.26881$ gives $3.226$ V, so $3.23$ V is used here.

§6 · Worked example: an op-amp stage

Two node equations and one transfer function

The closing example replaces the passive ladder by an amplifier stage. Nothing about the method changes: impedances are written in the s-domain, KCL is applied at the nodes, and the ratio of output to input is formed.

The problem

Find $H(s) = V_o/V_s$ . Given: 10 Ω and 20 Ω in the input path, a 1/30 F capacitor from the middle node to ground, and a feedback network of 15 Ω in parallel with 1/60 F. The capacitors become 30/s and 60/s.

The two op-amp rules

No current enters either input terminal, and negative feedback drives the two input voltages together. The non-inverting terminal is grounded, so the inverting terminal sits at 0 V: it is a virtual ground. The node is held at 0 V without being connected to ground, which is exactly what makes the following two equations easy.

KCL at the middle node

\tm{op.r10 op.nvs}{\frac{V_x - V_s}{10}} + \tm{op.r20 op.ninv}{\frac{V_x - 0}{20}} + \tm{op.c30}{\frac{V_x - 0}{30/s}} = 0

Multiplying by 60 gives $6V_x - 6V_s + 3V_x + 2sV_x = 0$ , so

(2s + 9)\,V_x = 6 V_s \;\Longrightarrow\; \boxed{V_x = \frac{6V_s}{2s+9}}

KCL at the virtual ground

\tm{op.r20 op.nx}{\frac{0 - V_x}{20}} + \tm{op.r15 op.nvo}{\frac{0 - V_o}{15}} + \tm{op.c60 op.nvo}{\frac{0 - V_o}{60/s}} = 0

No current enters the op-amp input, so only the three external branches appear. Multiplying by 60 gives $-3V_x - 4V_o - sV_o = 0$ , so

\boxed{V_x = \frac{-(s+4)}{3}\,V_o}

Eliminate $V_x$

\frac{6V_s}{2s+9} = \frac{-(s+4)}{3}V_o \;\Longrightarrow\; H(s) = \frac{V_o}{V_s} = \frac{6}{2s+9} \cdot \frac{3}{-(s+4)} = \frac{-18}{2s^2 + 17s + 36}

\boxed{H(s) = \frac{-9}{s^2 + \tfrac{17}{2}s + 18}}

The minus sign is the inverting configuration showing itself: it contributes a flat $180°$ to the phase at every frequency.

Three frequencies at once

The board leaves this evaluation unfinished. It is completed here for

v_s(t) = 5 + 10\cos(t + 30°) + 20\cos(2t + 45°)\ \text{V}

Each term is carried through $H(j\omega) = \dfrac{-9}{(j\omega)^2 + \frac{17}{2}j\omega + 18}$ on its own. Because the numerator is negative and real, its angle is $180°$ , and the angle of $H$ is $180°$ minus the angle of the denominator.

Input term	ω	H(jω)	\|H\|	∠H	Output term
5	0	−9 / 18	0.500	180.00°	−2.50
10 cos(t + 30°)	1	−9 / (17 + j8.5)	0.4735	153.43°	4.74 cos(t + 183.43°)
20 cos(2t + 45°)	2	−9 / (14 + j17)	0.4087	129.47°	8.17 cos(2t + 174.47°)

v_o(t) = -2.50 + 4.74\cos(t + 183.43°) + 8.17\cos(2t + 174.47°)\ \text{V}

The first phase may be reported equivalently as $-176.57°$ , since angles are defined modulo $360°$ .

How each row was computed

At $\omega = 1$ : the denominator is $-1 + j\tfrac{17}{2} + 18 = 17 + j8.5$ , whose magnitude is $\sqrt{289 + 72.25} = 19.007$ and whose angle is $\tan^{-1}(8.5/17) = 26.57°$ . Hence $|H| = 9/19.007 = 0.4735$ and $\angle H = 180° - 26.57° = 153.43°$ . The output amplitude is $10 \cdot 0.4735 = 4.74$ and the output phase is $30° + 153.43° = 183.43°$ .

At $\omega = 2$ : the denominator is $-4 + j17 + 18 = 14 + j17$ , with magnitude $\sqrt{196 + 289} = 22.023$ and angle $\tan^{-1}(17/14) = 50.53°$ . Hence $|H| = 9/22.023 = 0.4087$ and $\angle H = 180° - 50.53° = 129.47°$ , giving $20 \cdot 0.4087 = 8.17$ at $45° + 129.47° = 174.47°$ .

At $\omega = 0$ : $H(j0) = -9/18 = -1/2$ , a real negative number, so the 5 V constant is inverted and halved to $-2.50$ V. Writing the angle as $180°$ and the magnitude as $0.5$ says the same thing.

Why superposition is legal here

The circuit is linear and time-invariant, so the response to a sum of inputs is the sum of the responses. Each frequency is handled by its own complex number $H(j\omega)$ , and no interaction between frequencies is possible. This is the property that makes the frequency response a complete description of a linear circuit.

§7 · Your turn

Practice: a band-pass output

The circuit below is a series RLC driven by $v_s$ , but the output is taken across the resistor rather than across the capacitor, which changes the transfer function completely. The problem should be attempted on paper before the hints are opened, in order.

Find $H(s) = V_o/V_s$ , and then the steady-state $v_o(t)$ when $v_s(t) = 6 + 10\cos 2t$ V. Given: 1 H, 1/5 F, and 2 Ω.

Hint 1: The circuit is a divider

One loop carries one current $I$ , so the three impedances are in series and the output is the share of $V_s$ that lands on the resistor. Voltage division holds in the s-domain exactly as it does with resistors, provided impedances are used throughout.

Hint 2: Write the impedances first

The inductor is $sL = s$ and the capacitor is $1/sC = 5/s$ . Multiplying the numerator and the denominator by $s$ clears the compound fraction and leaves a ratio of polynomials.

Hint 3: Two frequencies, evaluated separately

The input contains $\omega = 0$ and $\omega = 2$ rad/s. Evaluate $H(j\omega)$ at each, and take note of what the numerator does at $\omega = 0$ before any arithmetic is performed.

Solution

Voltage division across the series impedances gives

H(s) = \frac{R}{sL + \dfrac{1}{sC} + R} = \frac{2}{s + \dfrac{5}{s} + 2} = \boxed{\frac{2s}{s^2 + 2s + 5}}

The numerator carries a factor $s$ , so $H(j0) = 0$ : the constant term is removed entirely. A capacitor in series with the source blocks any steady current, so the resistor holds no DC voltage, and the 6 V term contributes nothing.

At $\omega = 2$ :

H(j2) = \frac{2(j2)}{(j2)^2 + 2(j2) + 5} = \frac{j4}{-4 + j4 + 5} = \frac{j4}{1 + j4}

|H(j2)| = \frac{4}{\sqrt{1 + 16}} = \frac{4}{4.1231} = 0.9701 \qquad \angle H(j2) = 90° - \tan^{-1}4 = 90° - 75.96° = 14.04°

\boxed{v_o(t) = 9.70\cos(2t + 14.04°)\ \text{V}}

Independent check by impedances. At $\omega = 2$ the element impedances are $Z_L = j2$ , $Z_C = 5/(j2) = -j2.5$ , and $Z_R = 2$ , so the loop impedance is $2 - j0.5 = 2.0616\,\angle{-14.04°}$ . The loop current is $10\angle 0° / 2.0616\angle{-14.04°} = 4.851\angle 14.04°$ A, and the resistor voltage is $2 \cdot 4.851 = 9.70$ V at $14.04°$ ✓.

The near-unity magnitude is not an accident: the peak of this response sits at $\omega = \sqrt{5} = 2.236$ rad/s, where the reactances cancel and $|H| = 1$ exactly. The drive at 2 rad/s is close to that peak, which is the resonance behavior named as item 4 of the roadmap.

Chapter 14 in five items

One element, three descriptions

Calculus in one domain, algebra in the other

The circuit

KVL in the time domain

The standard second-order form

The same circuit in the s-domain

KVL in the s-domain

One line of algebra finishes it

Definition

The frequency response

What the transfer function does to a sinusoid

The response, live

From five elements to one transfer function

The problem

Write the impedances

Transform the source and expose the parallel pairs

Combine each pair

Transform back, and read one loop

Solve the loop

The transfer function, and the differential equation it encodes

The steady-state response to vs(t)=6+12cos⁡2tv_s(t) = 6 + 12\cos 2tvs​(t)=6+12cos2t

Two node equations and one transfer function

The problem

The two op-amp rules

KCL at the middle node

KCL at the virtual ground

Eliminate VxV_xVx​

Three frequencies at once

Practice: a band-pass output

The steady-state response to $v_s(t) = 6 + 12\cos 2t$

Eliminate $V_x$