How to Find a New Output When the Input Changes

A transfer function describes how a linear time-invariant system responds to an input. Once you know the transfer function, you can calculate the output for a different input without solving the system’s differential equation from scratch.

The standard relation is $Y(s)=G(s)U(s)$, where $U(s)$ is the input’s Laplace transform, $Y(s)$ is the output’s Laplace transform, and $G(s)$ is the transfer function. This relation gives the zero-state response: the part caused by the input when the system starts with zero initial conditions.

The method

First obtain the transfer function from the system model. For a differential equation with constant coefficients, take the Laplace transform and set the initial conditions to zero. Rearrange the result to isolate the ratio $Y(s)/U(s)$:

$$G(s)=\frac{Y(s)}{U(s)}.$$

Next, find the Laplace transform of the new input and multiply it by the transfer function:

$$Y_{\mathrm{new}}(s)=G(s)U_{\mathrm{new}}(s).$$

Finally, invert the result to get the time-domain output $y_{\mathrm{new}}(t)$. Use partial fractions or a table of Laplace transforms when needed.

If you already know the output for an old input, you can also use the transfer function to find the new output. You generally cannot determine it from the old output alone: you also need the system model or transfer function and the new input.

Worked example

Consider a first-order system described by

$$\frac{dy}{dt}+2y=u(t),$$

with zero initial condition. Taking the Laplace transform gives

$$sY(s)+2Y(s)=U(s),$$

so its transfer function is

$$G(s)=\frac{Y(s)}{U(s)}=\frac{1}{s+2}.$$

Suppose the input is changed to $u(t)=3e^{-t}$ for $t\geq 0$. Its Laplace transform is $U(s)=3/(s+1)$. Therefore,

$$\begin{aligned} Y(s)&=G(s)U(s)\\ &=\frac{1}{s+2}\frac{3}{s+1}\\ &=\frac{3}{s+1}-\frac{3}{s+2}. \end{aligned}$$

Taking the inverse Laplace transform gives

$$y(t)=3e^{-t}-3e^{-2t},\qquad t\geq 0.$$

The output starts at zero, as required by the initial condition, and then decays toward zero. The response has two exponential terms because the input and the system each contribute a pole to the product in the Laplace domain.

For comparison, if the input had remained a unit step, then $U(s)=1/s$ and the output would be

$$Y(s)=\frac{1}{s(s+2)}=\frac{1}{2s}-\frac{1}{2(s+2)},$$

which gives $y(t)=\tfrac12(1-e^{-2t})$. The transfer function is the same in both cases; only the input transform changes.

If you know the old output

For zero initial conditions, if the old input was $U_1(s)$ and its output was $Y_1(s)$, then $Y_1(s)=G(s)U_1(s)$. If you know $G(s)$ and have the new input $U_2(s)$, use $Y_2(s)=G(s)U_2(s)$. When $U_1(s)$ is nonzero and the ratio is well-defined, you may also write $Y_2(s)=Y_1(s)U_2(s)/U_1(s)$. In practice, using the transfer function directly is usually clearer and avoids dividing by a transform that may have zeros.

What changes with nonzero initial conditions?

The transfer-function formula $Y(s)=G(s)U(s)$ assumes zero initial conditions. For example, with the same equation $y'+2y=u(t)$ but initial value $y(0)=y_0$, the transformed equation is

$$(s+2)Y(s)=U(s)+y_0,$$

so

$$Y(s)=\frac{1}{s+2}U(s)+\frac{y_0}{s+2}.$$

The first term is the response to the input; the second is the natural response from the initial stored state. If the initial condition is unchanged when you replace one input with another, this natural-response term remains in the output. For higher-order systems, each initial value contributes terms in the transformed differential equation.

Common mistakes

• Using an old output as if it were the transfer function. The transfer function is the ratio of output to input transforms under zero initial conditions; an output by itself does not describe the system.
• Ignoring initial conditions. Multiplying $G(s)$ by the new input gives the zero-state response, not necessarily the full response when the system starts with nonzero state.
• Changing the model as well as the input. The simple substitution $Y=GU$ applies when the system itself stays the same and is linear and time-invariant. If parameters change, recalculate $G(s)$.
• Forgetting to invert the transform. $Y(s)$ is an output in the Laplace domain; the requested time-domain signal is $y(t)$.
• Assuming every system is linear and time-invariant. For a nonlinear or time-varying system, one fixed transfer function generally does not describe the input-output relation.