Transfer Function Basics

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Transfer Function Basics

Key Ideas

• Think of signals in an LTI model as pure scaled complex exponentials, like u(t) = Ue^{st}, where U and s are complex numbers.

• While pure complex exponentials are very special solutions of model equations, they are usually informative enough to give a complete picture of system behavior (this is due to linearity and superposition).

• Working with pure complex exponentials allows one to represent LTI models by operations of multiplication, which is usually much more intuitive and easier to handle than differential (or integral) equations.

• Thinking in terms of pure complex exponentials leads to the notion of a transfer function (or system function), which every LTI model can be characterized by.

• In 6.3100, expect transfer functions to be ratios of polynomials with real coefficients, i.e., rational functions of s, like

H(s)=\frac{3s}{s^2+1},\qquad H(s)=0.1s+1.2,\qquad H(s)=-2.3\,\frac{s^2-\sqrt{2}s+1}{s^2+\sqrt{2}s+1}.

• Transfer functions yield straightforward insight into a system's zero-input response (in particular, its stability) and its steady-state responses (in particular, the final value of the step response). They also make combining subsystems of a larger model very straightforward.

A Motivating Example: The Pure Integrator

The scaled pure integrator

M\dot{y}(t) = u(t)

is a common building block. For example, it describes the relation between the total torque u(t) applied to a rotating mass and its angular velocity y(t), where M > 0 is the moment of inertia. We treat u = u(t) as the input and y = y(t) as the output:

This model is linear (differentiation is linear) and time-invariant (its coefficients do not depend on time).

To find the transfer function, assume both signals are complex exponentials: u(t) = Ue^{st} and y(t) = Ye^{st}. Since \dot{y}(t) = sYe^{st}, the equation M\dot{y} = u becomes MsY = U, giving Y = \tfrac{M}{s}U. We call

H(s) = \frac{M}{s}

the transfer function (from u to y) of the model. When s \neq 0, H(s) is a finite complex number, and there exists an initial condition for which y(t) = H(s)Ue^{st}. When s = 0, H(s) = \infty: we say s = 0 is a pole of H, and y(t) = Ye^{0 \cdot t} = Y (any constant) is a zero-input response.

Caution! A common mistake is to think that y(t) = H(s)u(t) holds for all input/output pairs. This is far from true—and the value of s would be undefined unless both u(t) = Ue^{st} and y(t) = Ye^{st} are assumed beforehand. For example, if u(t) = \sin(\omega t) with \omega \neq 0, no initial condition results in a response y(t) = C\sin(\omega t). Instead, using \sin(\omega t) = \frac{1}{2j}e^{j\omega t} - \frac{1}{2j}e^{-j\omega t} and superposition, one can show that

y(t) = \frac{H(j\omega)}{2j}e^{j\omega t} - \frac{H(-j\omega)}{2j}e^{-j\omega t} = -\frac{M}{\omega}\cos(\omega t)

is one possible response to u(t) = \sin(\omega t).

Transfer Function of a Generic LTI Model

For an LTI system with input u = u(t) and output y = y(t):

its transfer function H = H(s) characterizes the relation Y = H(s)U between every pure complex exponential input/output pair u(t) = Ue^{st}, y(t) = Ye^{st}. The value H(s) is either a complex number or \infty; when H(s) = \infty, it means e^{st} is a zero-input response.

In 6.3100, transfer functions are rational functions of s with real coefficients, like

H(s)=\frac{s-1}{s(s-2)},\qquad H(s)=0.1s+\frac{0.2}{s}+0.3,\qquad H(s)=-\frac{1.1}{s^3}.

Some physical phenomena require non-rational transfer functions: the “delay by T > 0” system y(t) = u(t-T) has transfer function H(s) = e^{-sT}, since input u(t) = Ue^{st} results in output y(t) = Ue^{s(t-T)} = e^{-sT}Ue^{st}.

Because the coefficients of rational transfer functions are real, we have H(\bar{s}) = \overline{H(s)}. This leads to the following real-signal interpretation: if the input is u(t) = u_0 e^{\sigma t}\cos(\omega t + \phi), and H(\sigma + j\omega) = re^{j\theta}, then for appropriate initial conditions the output is

y(t) = r\,u_0\,e^{\sigma t}\cos(\omega t + \phi + \theta) = |H(\sigma+j\omega)|\,u_0\,e^{\sigma t}\cos\!\bigl(\omega t + \phi + \angle H(\sigma+j\omega)\bigr).

Transfer Functions for Stability and Steady-State Response

Rational transfer functions inform us in two ways.

• Poles and the ZIR. The poles of H(s) (values where H(s)=\infty) dictate the zero-input response. In particular, stability is equivalent to no poles being in the closed right half-plane \{s \in \mathbb{C} : \operatorname{Re}(s) \ge 0\}.

• Steady-state response. When H(s) \neq \infty, the unique pure complex exponential response to input u(t) = Ue^{st} is y(t) = H(s)Ue^{st}. When \operatorname{Re}(s) exceeds the real part of every pole, e^{st} dominates over the ZIR as t \to +\infty, and y(t) = H(s)Ue^{st} is the steady-state response.

The most important case is s = j\omega (\omega \in \mathbb{R}), since e^{j\omega t} neither grows nor decays. The map \omega \mapsto H(j\omega) is the frequency response. For a stable system, it can be measured by applying sinusoidal inputs

u(t)=u_0\cos(\omega t+\phi)

and measuring the output in steady state. Independently of initial conditions, the output converges to

|H(j\omega)|\,u_0\cos\!\bigl(\omega t + \phi + \angle H(j\omega)\bigr) \quad\text{as } t \to +\infty.

Special Case: Final Value of the Unit Step Response

The unit step response is the zero-state response to the input that equals 0 for t < 0 and 1 for t \ge 0. For a stable system (no poles in the closed right half-plane), the step response converges to the steady-state response to a unit constant input. Since a unit constant is a complex exponential with s = 0 and U = 1, the final value is H(0).

Example: Consider H(s) = \dfrac{s+2}{s^2+bs+1}, where b is a real parameter. When b > 0, all poles have negative real parts (the denominator roots of s^2+bs+1 have \operatorname{Re}(s) < 0), so the final value of the step response is H(0) = 2. When b \le 0, the ZIR does not die out and there is no finite final value.

Complications

Repeated poles. When a pole p has multiplicity m > 1, the ZIR includes not only e^{pt} but also te^{pt},\, t^2e^{pt},\,\ldots,\, t^{m-1}e^{pt}. For example, Newton's law M\ddot{y} = u has transfer function H(s) = M/s^2, with a double pole at s=0; its ZIR includes both constants and ramps.

Input at a pole frequency. If the input u(t) = Ue^{pt} is at a complex frequency p which is also a pole of H, there will generally be no response of the form Ye^{pt}. For a simple pole, the response takes the form (Y_0 + tY_1)e^{pt}—the ratio y(t)/u(t) grows linearly with time, a special form of resonance.

Questions

Question 1: Derive a Transfer Function from Differential Equations

The input u = u(t) and output y = y(t) of a system are related by

\dot{y} = u + w, \qquad \dot{w} = y + 2u,

where w is a hidden state. Substituting u(t)=Ue^{st}, y(t)=Ye^{st}, w(t)=We^{st} gives

sY = U + W, \qquad sW = Y + 2U.

Eliminate W: from the first equation W = sY - U; substituting into the second gives (s^2-1)Y = (s+2)U.

Enter H(s) = Y/U as a Python expression in s (e.g. (s+1)/(s**2+2*s+3)):

H(s) =

Question 2: Match the ZIR to a Transfer Function

For some initial conditions, the system with transfer function

H(s) = \frac{1}{s^2 + 2s + 5}

has a zero-input response of the form y(t) = e^{at}\cos(\omega t), where \omega \ge 0. Find a and \omega.

a =

\omega =

Question 3: Steady-State Response from Transfer Function

An LTI system has transfer function H(s) = (s+1)^{-6}. What is its steady-state response to input u(t) = 3\sin(t)?

Since \sin(t) = \operatorname{Im}(e^{jt}), the steady-state response has amplitude 3|H(j)| and is phase-shifted by \angle H(j). Note that 1+j = \sqrt{2}\,e^{j\pi/4}, so

H(j) = (1+j)^{-6} = \left(\sqrt{2}\,e^{j\pi/4}\right)^{-6} = \tfrac{1}{8}\,e^{-j3\pi/2} = \tfrac{1}{8}\,e^{j\pi/2}.

Amplitude of the steady-state response:

The steady-state response y(t) is proportional to:

Question 4: Transfer Function Samples from Steady-State Response

The steady-state response of a stable LTI system with transfer function H(s) to input u(t) = \sin^2(t) is y(t) = \sin(2t) - 3. Find H(0), |H(2j)|, and \angle H(2j).

Hint: \sin^2(t) = 0.5 - 0.5\cos(2t).

H(0) =

|H(2j)| =

\angle H(2j) in radians, from the (-\pi,\pi] range:

Question 5: Final Value of the Unit Step Response

An LTI system has transfer function H(s) = \dfrac{1}{s^2 + s + a}, where a \in (-\infty,\infty) is a real parameter. Let m(t) denote the unit step response. For which values of a does m(t) converge to a finite limit as t \to +\infty? For those values, what is the limit?

The step response converges to a finite limit when a is greater than:

When a > 0, what is \lim_{t\to+\infty} m(t)? Enter as a Python expression in a:

\displaystyle\lim_{t\to+\infty} m(t) =