# What is the difference between frequency response and transfer function?

I would like to understand the difference between frequency response and transfer function. I know the former can be obtained by substituting $s = j\omega$.

But what is the difference in information I can get from both representations? What are the respective limitations and where do I apply which method?

I'd also be glad for some literature recommendations.

Could someone explain the calculations of the second answer (by Chu) a little more extensively? I don't quite get how he determines the values of $\phi$ and X, and how he compares it with setting s equal to $j\omega$ in the transfer function.

• A transfer function is a more general concept than frequency response. For instance, you could have a transfer function for a magnetic core with the hysteresis. A frequency response is more specific and we qualify the response with a transfer function using Laplacian expressions. Dec 16, 2016 at 15:11
• A transfer function is a simple representation of the system you are working with without actual values. Frequency response is more precise with frequency values, components values, etc. Dec 16, 2016 at 15:15
– Null
Mar 23, 2017 at 13:51

A circuit's transfer function is a fully mathematical model that can be used to derive the frequency response and phase response (both together are called the bode plot).

However the same isn't true in reverse - you can't always derive the TF from the bode plot. Sometimes you can but not always.

So, the frequency response is a subset of the bode-plot and the bode-plot is a subset of the transfer function.

Hopefully this picture will help: - Along the top are three bode plot views of a typical frequency response for a 2nd order low pass filter. Bottom left is a 3D view of what lies behind the frequency response - in this example there are two poles (only one shown to make it easier on the eye).

Bottom right is the standard pole zero diagram and this 2D diagram alone embodies the transfer function. So, if you look at the 3D picture and imagine viewing from above, you get the pole zero diagram at bottom right.

The frequency response is a special case of the Laplace transfer function where the transients are assumed to be completely dissipated, leaving the steady state sinusoidal response.

Take, as an example, a sinusoid, $\small \sin(\omega t)\rightarrow \dfrac{\omega}{s^2+\omega^2}$, applied to a simple first order lag, $\small G(s)=\dfrac{1}{1+s}$. The response is: $\small R(s)=\dfrac{\omega}{(s^2+\omega^2)(1+s)}$, and this can be expressed in partial fractions:

$$\small \frac{\omega}{(s^2+\omega^2)(1+s)}=\frac{A+Bs}{(s^2+\omega^2)}+\frac{C}{(1+s)}$$

Inverse LT gives:$$\small r(t)=\frac{A}{\omega}\sin(\omega t)+ B\cos(\omega t)+Ce^{-t/\tau}$$

The exponential term decays to zero, leaving the steady-state response as:

$$\small \frac{A}{\omega}\sin(\omega t)+B\cos(\omega t)= X\sin(\omega t+\phi)$$

Solving for $\small X$ and $\small\phi$ gives $\frac{1}{\sqrt{1+\omega^2}}$, and $\small \arctan{(-\omega)}$, respectively, as is obtained using $\small s\rightarrow j\omega$ in the Laplace TF.

• thanks for this nice example! Could you elaborate a little further what you do at the end? How do you solve for X and $\phi$ and where to you substitute $j\omega$ for s.
– luis
Mar 22, 2017 at 11:27

They are very similar concepts.

The transfer function is a relationship between an output and an input of a linear system.

The frequency response is how some characteristic of a linear system varies over frequency. The thing that varies might be the transfer function. But it might be something else, like the input or output impedance. It might be the variation of something on a system that doesn't have a distinct output and input, like a one-port network.