# Why use transfer functions than differential equations?

I have some simple questions regarding DSP and Control Systems, but found no simple answers. I just need simple and easy to understand answers, not complex answers with huge maths.

1. Why we use Transfer Functions, when we can get a system's output by just solving it's differential equation?
2. What we finally get after solving a transfer function (In Laplace Transform)?
• Hey welcome! So, to pull a teeth right at the beginning: some things are complex and need math to explain. They might still be explained in an easy-to-understand matter. But if someone is always only giving you simple answers, you can be almost certain you're not getting the full picture. Anyway, your questions are relatively easy to answer! – Marcus Müller May 22 '18 at 6:41
• aside from that: your 2. question: What does "solving a transfer function" mean? A transfer function is not a problem to be solved! I'm removing the third question, because a) you should focus on one logical question per post here, and b) it is totally underresearched; honest answer would be go and read a signals and systems theory textbook of any kind or at least ask wikipedia first before asking here about six large terms "Unit Step Function, Unit step response, Impulse Response, Impulse Function, Steady State, State Space", which really are the key terms of 1 year of my formal education. – Marcus Müller May 22 '18 at 6:43
• Answers are appropriate. I just drop this comment that the differential equations and transfer functions are not that much different that someone might assume. Consider $$\frac{Y}{X}=\frac1{3+2s}$$ as $$y(t)=\frac1{3+2\frac{d}{dt}} x(t)$$ except for the Laplace one is in the proper math format. – Arash May 23 '18 at 9:04

Why we use Transfer Functions, when we can get a system's output by just solving it's differential equation?

Because differential equations are unwieldy and hard to deal with, and you can't see the behaviour on different frequencies from these, whereas transfer functions just give you the behaviour of an LTI system given an excitation of given property. Much easier to work with!

1. with ordinary linear differential equations, you end up solving it essentially the same way as you do with the Laplace transformed counterparts. you could come up with an operation that breaks a higher-order differential equation into two lower-order diff eqs. in much the same manner as you factor a high-order polynomial of $s$ into two lower-order polynomials.

so my pea-brain can grasp the factoring operations better with polynomials of $s$ than it does with breaking differential equations (w.r.t. $t$) into smaller differential equations and solving.

1. you get the Laplace Transform of the answer you want.
1. Because handling polynomials are much easier than differential operators. Also multiplication of polynomials are again much easier than convolution of signals.

2. The only reason you learn partial fraction expansion is actually to separate the terms of a compound fraction to simpler items and obtain the eigenmodes, e.g., say you have the following for the impulse response, with zero initial conditions

$$\frac{8s+7}{s^2-s+2} = \frac{3}{s+2}+\frac{5}{s-1}$$ so we know the solution is $3e^{-2t} + 5e^t$ which can also be obtained by solving

$$\ddot y(t) - \dot y(t) + 2y(t) = 8\dot u(t) +7(t)$$

but it would be way more tedious.