Difference between natural response and forced response?

Question

Reference

Second post on EdaBoard.com

Time response of a system is the time evolution of the variables. In circuits, this would be the waveforms of voltage and current versus time.

Natural response is the system's response to initial conditions with all external forces set to zero. In circuits, this would be the response of the circuit with initial conditions (initial currents on inductors and initial voltage on capacitors for example) with all the independent voltages set to zero volts (short circuit) and current sources set to zero amps (open circuit). The natural response of the circuit will be dictated by the time constants of the circuit and in general roots of the characteristics equation (poles).

Forced response is the system's response to an external stimulus with zero initial conditions. In circuits, this would just be the response of the circuit to external voltage and current source forcing function... continue reading

Questions

How can there be even a natural response? Something has to be inputted to create an output? The way I see it is like turning of the main water line and then turning on your faucet and expecting water to come out.
How can we v(t) (from the link above) be solved for if we don't know dv(dt) in order to find the natural response?
If you can please expand on the 2 concepts (natural response and forced response) by explaining their differences in Layman's terms, it would be lovely.

@Felipe_Ribas Can you please confirm this and answer some of the questions? (you can just edit this directly if you want)

Given an an equation 10dy/dt + 24y = 48 means rate of change of output + 24 * output = 48. The initial conditions are y(0)=5 and dy/dt=0.
- That would mean that the input is 48/(24*5) Is that a correct assumption? The solution to that is 0.4 which is the constant input?

score 23 · Accepted Answer · edited Dec 23 '21 at 18:55

23

Think about a simple mechanical system like an elastic bar or a block attached to a spring against gravity, in real world. Whenever you give the system a pulse (to the block or to the bar), they will begin an oscillation and soon they will stop moving.

There are ways that you can analyze a system like this. The two most common ways are:

Complete solution = homogeneous solution + particular solution
Complete response = Natural resopnse (zero input) + forced response (zero state)

As the system is the same, both should result the same final equation representing the same behavior. But you can separate them to better understand what each part means physically (specially the second method).

In the first method, you think more from the point of view of a LTI system or a mathematical equation (differential equation) where you can find its homogeneous solution and then its particular solution. The homogeneous solution can be viewed as a transient response of your system to that input (plus its initial conditions) and the particular solution can be viewed as the permanent state of your system after/with that input.

The second method is more intuitive: natural response means what is the system response to its initial condition. And forced response is what is the system response to that given input but with no initial conditions. Thinking in terms of that bar or block example I gave, you can imagine that at some point you pushed the bar with your hands and you are holding it there. This can be your initial state. If you just let it go, it will oscillate and then stop. This is the natural response of your system to that condition.

Also you can let it go but still keeps giving some extra energy to the system by hitting it repeatedly. The system will have its natural response as before but will also show some extra behavior due to your extra hits. When you find your system complete response by the second method, you can see clearly what is the system natural behavior due to those initial conditions and what is the system response if it had only the input (with no initial conditions). They both together will represent all the system's behavior.

And note that the Zero State response (Forced response) also may consist of a "natural" portion and a "particular" portion. That is because even with no initial conditions, if you give an input to the system, it will have a transient response + permanent state response.

Example response: imagine that your equation represent the following circuit:

RL Circuit

Which your output y(t) is the circuit current. And imagine your source is a DC source of +48v. This way, making the summation of element's voltage in this closed path, you get:

\$\epsilon=V_L+V_R\$

We can rewrite the inductor voltage and resistor voltage in terms of current:

\$\epsilon=L\frac{di}{dt} + Ri\$

If we have a power source of +48VDC and L = 10H and R = 24Ohms, then:

\$48=10\frac{di}{dt}+24i\$

which is exactly the equation you used. So, clearly your input to the system (RL circuit) is your power supply of +48v only. So your input = 48.

The initial conditions you have are y(0) = 5 and y'(0) = 0. Physically it represents that at a t=0 moment, my current of the circuit is 5A but it is not varying. You may think that something happened previously in the circuit which left a current in the inductor of 5A. So in that given moment (initial moment) it sill has those 5A (y(0)=5) but it is not increasing or decreasing (y'(0) = 0).

Solving it:

we first assume the natural response in the format: \$Ae^{st}\$

and then we will find the system behavior due to its initial condition, just as if we had no power supply (\$\epsilon=0\$) which is the Zero-Input response:

\$10sAe^{st} + 24Ae^{st} = 0\$

\$Ae^{st}(10s + 24)=0\$

\$s=-2.4\$

So,

\$i_{ZI}(t)=Ae^{-2.4t}\$

Since we know that i(0) = 5:

\$i(0)=5=Ae^{-2.4(0)}\$

\$A=5\$

\$i_{ZI}(t)=5e^{-2.4t}\$

Note that until now everything is consistent. This last equation represents the system response with no input. If I put t=0, I find i=5 which correspond to the initial condition. And if I put \$t=+\infty\$ I will find i=0 which also makes sense if I do not have any source.

Now we may find the particular solution to the equation which will represent the permanent state due to the power supply presence (input):

we assume now that \$i(t)=c\$ where \$c\$ is a constant value which represents the system output in the permanent state since the input is also a constant. For each system, the output format depends on the input format: if the input is a sinusoidal signal, the output also will be. In this case we have only constant values which makes things easier.

So,

\$\frac{di}{dt}=0\$

then,

\$48 = 10\cdot0 + 24c\$ (using the differential equation)

\$c=2\$

\$i(\infty)=2\$

which also makes sense because we have a DC power supply. So after the transient response of turning the DC power supply ON, the inductor will behave as a wire and we will have a resistive circuit with R=24Ohms. Then we should have 2A of current since the power supply has 48V across it.

But note that if I just add both results to find the complete response, we will have:

\$i(t) = 2 + 5e^{-2.4t}\$

Now I messed things up in the transient state because if I put \$t=0\$ we no longer will find \$i=5\$ as before. And we have to find \$i=5\$ when \$t=0\$ because it is a given initial condition. This is because the Zero-State response has a natural term which is not there and also has the same format as we found before. Adding it there:

\$i(t) = 2 + 5e^{-2.4t} + Be^{st}\$

The time constant is the same so it only left us B:

\$i(t) = 2 + 5e^{-2.4t} + Be^{-2.4t}\$

And we know that:

\$i(t) = 2 + 5 + B = 5\$ (t=0)

So,

\$B=-2\$

Then, your complete solution is:

\$i(t) = 2 + 5e^{-2.4t} - 2e^{-2.4t}\$

you may think of this last term we find as a correction term of the forced response to match the initial conditions. Another way to find it is imagining the same system but no with no initial conditions. Then solving all the way again, we would have:

\$i_{ZS}(t) = 2 + Ae^{-2.4t}\$

But as we now are not considering the initial conditions (i(0)=0), then:

\$i_{ZS}(t) = 2 + Ae^{-2.4t} = 0\$

And when t=0:

\$A=-2\$

so the forced (Zero-State) response of your system is:

\$i_{ZS}(t) = 2 - 2e^{-2.4t}\$

It is a bit confusing but now you can view things from different perspectives.

-Homogeneous/Particular solutions:

\$i(t) = i_p(t)+i_n(t) = 2 + 3e^{-2.4t}\$

The first term (2) is the particular solution and represents the permanent state. The rest of the right side is the transient response, also called homogeneous solution of the equation. Some books call this also Natural response and Forced response since the first part is the forced part (due to the power supply) and the second part is the transient or natural part (system's characteristic). This is the fastest way to find the complete response I think, because you only have to find the permanent state and a natural response once. But may not be clear what is representing what.

-Zero input / zero state:

\$i(t) = i_{ZS}(t)+i_{ZI}(t) = 2 - 2e^{-2.4t} + 5e^{-2.4t}\$

note that is the same equation but the second term is splitted in two. Now, the first two terms (\$2 - 2e^{-2.4t}\$) represent the Zero-State response. In other words, what would happen to the system if there was no initial current and you turned ON the +48V power source.

The second part (\$5e^{-2.4t}\$) represent the Zero-Input response. It shows you what would happen to the system if no input was given (power source remained in 0v). It is only an exponential term which would go to zero since it has no input.

Some people also call this Natural/Forced response format. The natural part would be Zero-Input and the Forced part would be the Zero-State, which by the way is composed by a natural term and particular term.

Again, they all will give you the same result which represents the whole situation behavior including the power source and initial conditions. Just note that in some cases it might be useful to use the second method. One good example is when you are using convolutions and you may find the impulse response to your system with Zero-State. So breaking those terms might help you to see things clearly and also using an adequate term to convolve.

edited Dec 23 '21 at 18:55

Edw590

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answered Dec 09 '13 at 00:15

Felipe_Ribas

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I am confused between `initial condition` and `given input but with no initial conditions`? – bluejamesbond Dec 09 '13 at 00:19
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Think about a child in a swing. If I pull the swing and hold it up there and I say this is my initial moment (t=0). If I just let it go and do not touch the system anymore, the system's behavior (the swing with the child) is purely a response to that initial condition (the swing being hold up there). But still I can can let the swing go and also keeps pushing it every cycle (giving an input). In an electric circuit, initial conditions can be viewed as current or voltage values different from zero in the t=0 moment. – Felipe_Ribas Dec 09 '13 at 00:23
So imagine a RC simple circuit. You can imagine that it has its own characteristic response whenever you give it a +5v short pulse. But you can also have your capacitor already charged from something that happened before. So, now, if you give it the same short +5v pulse, the output of the circuit will be different because it has different initial conditions (capacitor is charged). – Felipe_Ribas Dec 09 '13 at 00:25
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So if I have an equation like `dy2/dt2 + 10dy/dt + 24y = 32`? What is the `32`? What is `dy2/dt2`? What is `dy/dt` and `y`? Logically speaking, why do I even need to know the `dy2/dt2`? Say I have `y(0) = 5` and `dy(dt) = 0`. From your swing example, these are the initial conditions. Am I right? But where are the **additional** pushes or where are the inputs? I know this is very specific, but if you can help me understand this, I couldn't thank you enough. – bluejamesbond Dec 09 '13 at 00:27
Yes, your y(0) and dy/dt(0) are your initial conditions. From the differential equations is a bit difficult to visualize those things you want. In this case, the 32 may represent a constant input to your system. The rest is exaclty the system's behavior described mathematically. So you solve the equation to understand what is the system response. – Felipe_Ribas Dec 09 '13 at 00:33
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Note that Y represent the system output. So your equation relates the output with the variation of the output, with the variation of the variation of the output. So it's complicated to imagine all the meaning of the terms. But generally speaking, the independent part that you put to the right side (32 in this case), generally represents the input of the system. And the Y terms represent the system's behavior. – Felipe_Ribas Dec 09 '13 at 00:34
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If you put that zero, you are saying that you have no input now. But still, if you have your y(0) and dy(0) different from zero (non zero initial conditions), then you still will find some response curve which is the natural response of your system to those conditions (zero input response) – Felipe_Ribas Dec 09 '13 at 00:40
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Now, you can also maintain the 32 and now make y(0)=dy(0)=0. So you are saying that you have null initial conditions. Nothing is charged or nothing is moving in your system (gross way to think). Then if you solve you will have the pure response of the system to that 32 input, which has a transient part and a forced part. – Felipe_Ribas Dec 09 '13 at 00:42
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For the last, if you put y(0)=dy(0)=0 (null initial conditions) and make the 32 to 0, now you have no initial conditions and no input. Probably the complete response you will find will be zero. – Felipe_Ribas Dec 09 '13 at 00:42
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Sure. Just remember: initial conditions is HOW is your system in t=0 so it is all the information that you have of the output (y(t)) when t=0. And input is what you are proiding to the system like pulses or signals, which is represented by the independent part of the differential equation. The rest is the system itself (its behavior). Take your time to run some tests... I had also "lost" great time thinking of this too. – Felipe_Ribas Dec 09 '13 at 00:47
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Woah I understand this a lot better. But I am just confused on your `2,4` notation. Again, I will need to practice some and relate it to your explanation which will take me sometime. Thank you so much!!! And also, in terms of output what is the does `y(t)` now represent in terms of circuits? – bluejamesbond Dec 09 '13 at 02:44
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Sorry about that. You may see that as 24/10 or 2,4 or 2.4. But try to understand the basic idea... I know it is confusing but you may practice to understand better. Good Luck – Felipe_Ribas Dec 09 '13 at 02:48
The y(t) is what I've called i(t). So I'm saying that my system output is the circuit current. y(t) = i(t) – Felipe_Ribas Dec 09 '13 at 02:49
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Note that you MUST have a negative value, otherwise your exponential will go to infinity. So it has to be negative. Take a look at "Solution process" at this page. At some point it finds u(t)=e^(0.196)t. But in the end it shows up a c.e^(-0.196)t. – Felipe_Ribas Dec 09 '13 at 06:44
The three has come from 5e - 2e into a single term – Felipe_Ribas Dec 09 '13 at 06:44
Good luck then. But, just remember: generally a LTI system/linear circuits like RLC circuits will be from the same differential equations type. You do not need this complicated solution that takes into consideration variable coeficients. Maybe you will solve faster with method I put here. But this is just my personal opinion! good luck anyway – Felipe_Ribas Dec 09 '13 at 06:49
Its because we learned differential and I am just trying link these two concepts together. Just trying to get the whole picture. – bluejamesbond Dec 09 '13 at 06:50
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Yeah I know how it feels. Differential equations have a lot of particular cases. But when dealing with circuits and LTI systems, most often you won't need anything of that. Just using Laplace/Fourier you will be alright. And those simple cases you don't even need Laplace. Just assume the output format and use in the differential equation. Simple! – Felipe_Ribas Dec 09 '13 at 06:55
Felipe. I went through what you did and I redid it in an effort to understand more, this is what I got though: https://www.dropbox.com/s/bmbklwact0ltrtp/Photo%20Dec%2009%2C%201%2004%2055%20AM.jpg (Slightly different and has all my work). I used diff eq to solve this I guess. – bluejamesbond Dec 09 '13 at 07:06
There are two simple mistakes: First in the ZI response. Note that you've found k=-5. And you also found that y=-k.e^(-2.4t). So y = -(-5)e^(-2.4t) = 5e^(-2.4t). Be careful with signals. – Felipe_Ribas Dec 09 '13 at 07:21
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The second mistake is that whenever you are finding Zero-State response, as the name already says, you do not take the initial conditions into consideration. Thus, y(0) = 0 in the ZS response (second case). This way you will find Yzs = 2-2e^(-2.4t) – Felipe_Ribas Dec 09 '13 at 07:23
Oh good catch! with the -(-k). For the second mistake. I see it makes sense! – bluejamesbond Dec 09 '13 at 08:49
Now let me try to explain it to you to see I undrstand correctly. Natural response is basically how the system will be running normally. I see it as water in a pond and the natural ripples occurring on its surface. The forced response is basically if I throw a stone into it, there is a large splash. But even these "forced" ripples will eventually reduce down to the natural state or the natural response. In our DC circuit example, the forced response never goes away since it is DC so it always Natural (the initial voltage drop on components) + Forced (DC). (lim->inf = V = 0) => I = 0. Accurate? – bluejamesbond Dec 09 '13 at 09:03
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Yeah, basically. But I'm not really sure you could use that pond analogy because you could think the natural ripple is a consequence of a forced response due to the wind. Without any external disturbance the pond would not have any kind of ripple. You may think that whenever you throw a rock, you give the system some energy. When the rock vanishes, even though you are not disturbing the pond anymore (neither the rock), you still can see the waves generated from the hit. This can be viewed as the natural response of the pond. – Felipe_Ribas Dec 09 '13 at 16:23
But note that this is exaclty the interesting point of it: it all depends on whenever is your t=0. That's why the ZS response (or forced) does contain a homogeneous part and a particular part. Even though your pond is stand still and you throw a rock, this may be a system with only a Zero-State response, since it has no ripple before. BUT, for a second observator who did not see you throw the rock, it can be viewed as only a Zero-Input response (he didnt see the rock) which will only respond some time and then stand still again. – Felipe_Ribas Dec 09 '13 at 16:26
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Thank you! I understand this and it makes total sense. Thank you so much! You have put in a lot of time and effort. – bluejamesbond Dec 09 '13 at 18:01
#Felipe_Ribas Can I ask you a related question? What are we doing convert from `10cos(1000t + 20)` to `e^(20j)` with `w = 1000`? – bluejamesbond Dec 09 '13 at 21:35
I'm not sure I really understood your question. Can you reformulate that? – Felipe_Ribas Dec 09 '13 at 22:16
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Yes I can. I have changed the problem from DC to AC source to show you the scenario. And then I solved it out using some techniques I learned in school. I have put some questions along the solution process so that you can try explaining those parts to me. Page 1: https://www.dropbox.com/s/9x8la375kxybcap/Photo%20Dec%2009%2C%204%2050%2043%20PM.jpg Page 2: https://www.dropbox.com/s/16s1b35uavyzccs/Photo%20Dec%2009%2C%204%2050%2049%20PM.jpg – bluejamesbond Dec 09 '13 at 22:53
let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/11884/discussion-between-felipe-ribas-and-mk1) – Felipe_Ribas Dec 09 '13 at 23:17
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Thank you for all your help. +100 That's the best I can do. – bluejamesbond Dec 12 '13 at 07:15
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Never mind. I just hope you've understood everything or almost everything we talked about. And sorry my english. – Felipe_Ribas Dec 12 '13 at 17:07
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I understood a lot. I just didn't understand your last post on the chat. Your English was great, but I was just having a hard time understanding the concept. – bluejamesbond Dec 12 '13 at 17:30
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FWIW, as far as I comprehend with this answer: permanent state response = steady state response, zero state response = system response with zero initial condition, and zero input response = system initial condition response only. The terminologies may be different from place to place. We must understand the pretext and the context of a given text carefully. – Unknown123 May 03 '19 at 10:13

Kenny · Answer 2 · 2018-03-20T02:32:19.823

This is the problem with text books not clearly defining everything so that everybody can understand the definitions. Natural response is really talking about a system that had (at some point) been 'charged up' such that energy storing elements contain some amount of initial energy, which might translate to an initial voltage in a capacitor or an initial current in an inductor. These result in the initial condition values for capacitors or inductors. Then, at say time t = 0, it is assumed that the magical source that was responsible for energising the circuit, is instantly removed. So, if the magical source had been a voltage source, then 'removing it' could mean physically removing it, or switching it out of the circuit. So, at time t = 0, the natural response will just be the behaviour of maybe a current through an inductor or capacitor, or voltage across a capacitor or inductor. And the circuit is powered only by those initially charged components (because we assume no 'external' source input for time t = 0 onwards).

So, for the natural response, it is really a case where there 'once was' some external input to produce the initial conditions in the inductors and capacitors. Now, if the system were not charged up to begin with, such that all capacitor and inductor voltages and currents were zero to start with, then what would be the natural response of the system? Answer: zero.

Now, the forced response is the response of a circuit (such as a voltage behaviour or current behaviour) for the case where we assume that inductors and capacitors have no initial energy to begin with, which means no initial voltage or initial currents in these components. And then, all of a sudden we apply an external force (source) to the input of the circuit. The behaviour of currents and/or voltages of the circuit for this scenario is just given a name.... called the forced response. Basically, it is a response to a source input based on an assumption that we started off with ZERO energy initial conditions in inductors and capacitors.

Once we've used methods to conveniently obtain the natural response and the forced response, we then just add up both portions to get the complete picture. Kind of like superposition principle.

score 4 · Answer 3 · answered Dec 09 '13 at 00:08

How can there be even a natural response? Something has to be inputted to create an output?

If it helps, think of the natural response as the forced response to an impulse input.

The way I see it is like turning of the main water line and then turning on your faucet and expecting water to come out.

Imagine that the water main is connected to a large holding tank like used in well water systems and you close the valve to the water main.

The tank has been filled with water and is pressurized to the water main pressure before you closed the valve. This is the initial condition.

If you open the tap, water will come out. The holding tank will deliver water for some period of time, as the holding tank empties, and the pressure at the tap will drop. This dwindling flow of water and dropping pressure would be the natural response of the system.

Now, after the holding tank has emptied, you quickly open the water main valve while the tap is still open.

Most of the water flow is initially to "charge" the holding tank and, as the tank fills and pressure builds, water flows at an increasing rate from the tap until the tank is full and the flow and pressure stabilize.

This is the forced response to a step input.

score 1 · Answer 4 · answered Dec 11 '13 at 11:58

I'm not familiar with the term 'forced response' in this context, but here goes. Many systems can be characterised as first order plus dead time (FOPDT). The 'natural response' of such a system to stimulus is an initial delay followed by an exponential approach to a new steady state.

Think of a heater element supplied from a variable voltage source. Initial conditions are power off and heater at ambient temperature. Switch on at say 10 volts. For a short time (the dead time) the heater temperature doesn't change. The temperature then starts to increase, rapidly at first, then gradually settling at a new steady state. If you carefully observed the times involved, you will have three natural characteristics of the system:

Gain - expressed as degrees/volt. If the 10 volts caused a gain of 20 degrees then gain = 2. So for 20 volt input, you should expect a 40 degree increase from ambient.
Dead time - delay to expect in response to a change of input. (inertia)
Time constant or natural frequency - time from start of change to steady state is 5 time constants. (like charging a capacitor)

With this data you can predict how much temperature change to expect for a given voltage change and how long it will take, i.e. natural response.

I presume a 'forced response' would entail over-stimulating the system to get a faster result. So, to increase 30 degrees, we know we need a 15 volt increase in input. By increasing the voltage by 25 volts briefly and then backing off 10 volts, we could reach the desired end temperature faster, i.e. 'forcing' a faster response.

Difference between natural response and forced response?

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