How to Calculate the Output of a Transfer Function by Hand

A transfer function describes how a linear time-invariant (LTI) system changes an input in the Laplace domain. To calculate an output by hand, you need both the transfer function and the input signal. For the usual continuous-time convention, with zero initial conditions, multiply the two Laplace-domain expressions and invert the result.

The input-output relation

If the input is $x(t)$, the output is $y(t)$, and the transfer function is $G(s)$, then

$$G(s)=\frac{Y(s)}{X(s)}, \qquad Y(s)=G(s)X(s).$$

Here, $X(s)$ and $Y(s)$ are the Laplace transforms of the input and output. The transfer-function definition assumes zero initial conditions. The output in time is found by taking the inverse Laplace transform of $Y(s)$.

This gives a practical method:

1. Find $X(s)$ from the specified input.
2. Multiply $G(s)$ by $X(s)$.
3. Simplify the product, often with partial fractions.
4. Invert each term using standard Laplace transform pairs.

Worked example: output for a step input

Suppose a system has transfer function

$$G(s)=\frac{5}{s+2}$$

and the input is a unit step, $x(t)=1$ for $t\geq 0$. The Laplace transform of a unit step is $X(s)=1/s$. Therefore,

$$\begin{aligned} Y(s)&=G(s)X(s)\\ &=\frac{5}{s+2}\cdot\frac{1}{s}\\ &=\frac{5}{s(s+2)}. \end{aligned}$$

To invert this expression, use partial fractions:

$$\frac{5}{s(s+2)}=\frac{A}{s}+\frac{B}{s+2}.$$

Multiplying by $s(s+2)$ gives $5=A(s+2)+Bs$. Setting $s=0$ gives $A=5/2$, and setting $s=-2$ gives $B=-5/2$. Thus

$$Y(s)=\frac{5}{2}\left(\frac{1}{s}-\frac{1}{s+2}\right).$$

Since $1/s$ transforms back to a unit step and $1/(s+2)$ transforms back to $e^{-2t}$ for $t\geq 0$, the output is

$$y(t)=\frac{5}{2}\left(1-e^{-2t}\right), \qquad t\geq 0.$$

The result starts at zero and approaches $5/2$ as time increases. That final value also agrees with the system’s steady-state gain: $G(0)=5/2$, so a stable system with a unit-step input settles to $5/2$.

What changes for other inputs?

The transfer function stays the same; only $X(s)$ changes. For example, an impulse input has transform $X(s)=1$, so its output transform is $Y(s)=G(s)$. For the example above, the impulse response is $y(t)=5e^{-2t}$ for $t\geq 0$.

For a sinusoidal, ramp, or exponential input, first find that input’s Laplace transform, multiply by $G(s)$, and then invert. If the product has repeated poles or irreducible quadratic factors, use the corresponding partial-fraction forms and transform pairs.

Nonzero initial conditions

A transfer function by itself gives the zero-state response: the response caused by the input when the system starts with zero stored energy. If the system has nonzero initial conditions, include their contribution separately.

For instance, the differential equation

$$y'(t)+2y(t)=5x(t)$$

has zero-initial-condition transfer function $G(s)=5/(s+2)$. If instead $y(0)=y_0$, its unilateral Laplace transform gives

$$(s+2)Y(s)-y_0=5X(s),$$

so

$$Y(s)=\frac{5}{s+2}X(s)+\frac{y_0}{s+2}.$$

The first term is the input-driven response; the second is the response due to the initial output value. Ignoring the second term would give an incomplete answer when $y_0\ne 0$.

Common mistakes

• Using the transfer function as the output. The output transform is $G(s)X(s)$, not just $G(s)$, unless the input is an impulse.
• Forgetting the input transform. A unit step has transform $1/s$, while an impulse has transform $1$; confusing them changes the result.
• Assuming initial conditions are included. The usual transfer function describes zero initial conditions. Add initial-condition terms when the problem specifies a nonzero starting state.
• Stopping before the inverse transform. $Y(s)$ is a Laplace-domain expression, not yet the time-domain output $y(t)$.
• Dropping the time interval. For causal signals, state that the result applies for $t\geq 0$ (or include a unit-step factor when the convention calls for it).

For discrete-time systems, the same idea applies with the $z$-transform: multiply the system function by the input’s $z$-transform, then take the inverse $z$-transform.