We can define the forward transfer function as $$l(s)=g(s)r(s)$$ Thus, the closed-loop transfer function is


with \$g(s)r(s)\$ the multiplication between all the system transfer functions. It is obviously the main difference is the definition itself (the formula) but, what does it tell us about the system and about the poles/zeros?

  • \$\begingroup\$ You should include a picture that makes it clear what the forward transfer is and what the transfer of the feedback is. What are the poles and zeros of any system? It has to do with the numerator and denominator of the h(s) and their zeros. Due to feedback the location of the poles and zeros changes, this can result in poles in or near the right half plane, this results in a system which is.... \$\endgroup\$ Oct 14 '17 at 20:50
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    \$\begingroup\$ L(s) is called the loop gain and defines the strength of the feedback...Higher the L(s) stronger is the negative feedback....The poles of L(s) determine the stability of the loop, if more than one pole is present before the point where |L(s)| = 1 then the system will be unstable. All stability information is present in the loop gain itself and we don't need to bother about H(s) for that. H(s) tells about the closed loop gain. \$\endgroup\$
    – sarthak
    Oct 15 '17 at 14:58

I just wanted to make things a bit more clear here because it seems that the idea of open loop/closed loop/forward transfer function has got a bit mystified and does not seem exact even though it really is.

If you have a dynamical system with input \$u(s)\$, output \$y(s)\$ defined as: $$ \frac{y(s)}{u(s)} = G(s)$$

Dynamical systems described with transfer functions are idealized, generalized and abstracted, many different systems can be described with the same transfer function. From the transfer function you can ideally find out everything you need to know about the system from the point of view of the control engineer, but that often is not a case. Transfer functions can be stable and unstable:

  • Stable - all poles are negative
    • DC motor (shaft velocity, armature current)
    • Room temperature...
  • Unstable - at least one pole is positive or equal to zero
    • Inverted pendulum
    • Ball on plate
    • Segway, Onewheel,..


In the general, the transfer function's behavior, poles and zeros, time constants and characteristic frequencies are different then you want them to be and there for therefore you need a controller. There are two types of control you can apply to the physical system defined as the one above:

  • Open-loop control
  • Closed-loop control

Open-loop control

Open-loop control procedure does not rely on measurements of the controlled variables and assumes that system behavior is well known and deterministic, therefore it can be controlled without any knowledge what happens with the output value \$y(s)\$.

             enter image description here

The complete open-loop transfer function(also known as forward transfer function) is no longer in between input \$u(s)\$ and output \$y(s)\$ but set point (reference) value of the output \$r(s)\$ and \$y(s)\$: $$ \frac{y(s)}{r(s)} = C(s)G(s)$$

With the poles and zeros of the controller \$C(s)\$ you can tune the behavior of your complete system, even stabilize it in theory. In theory the perfect controller of the open loop procedure would be: $$ C(s) = \frac{1}{G(s)} $$

But what happens in theory is that systems have uncertain stochastic disturbances \$d(s)\$, which you cannot anticipate. And more importantly you cannot compensate without measurement. These disturbances can be a simple as measurement noise, but can be much more complicated and harmful.

      enter image description here

To be able to compensate the parts of the stochastic parts of the system you will need to introduce some kind of measurement. And therefore you need to "close the control loop".

Closed-loop control

Closed-loop control is everywhere and it has well described and documented synthesis procedures and analysis frameworks. The following image shows simple general closed loop block diagram.

enter image description here

The complete transfer function of the closed loop is derived like this: $$ d(s) = 0 $$ $$ y(s) = \Big[r(s) - y(s)\Big]C(s)G(s) $$ $$ y(s)\Big[1 + C(s)G(s)\Big] = r(s) C(s)G(s) $$ $$ \frac{y(s)}{r(s)} = \frac{C(s)G(s)}{1 + C(s)G(s)} $$

Usually, when you are designing the controller \$C(s)\$ you are setting the poles and zeros of the open loop transfer function, using Bode plot, Nyquist plot, root locus, compensation algorithms, loop shaping and similar.

The easiest way to understand this is if you look at the closed loop transfer function denominator. $$ 1 + C(s)G(s) = 1 + G_{open\,loop}$$ What you usually do when you have a transfer function is that you evaluate roots of the denominator - the poles. If you want to know what the behavior of your new transfer function is going to be you have to solve the equation: $$ 1 + C(s)G(s) = 0 $$

By placing the poles and zeros of the closed loop transfer function properly you will be able to get away with a lot of uncertain and stochastic influences in the system, such as:

  • Unknown disturbances
  • Unknown parameters
  • Unknown dynamics
  • System nonlinearity

You can try to follow some tutorials to understand better what the procedures are and what do you gain from using closed loop method. Mathworks tutorials are great for these purposes.


We can define the forward transfer function as $$l(s)=g(s)r(s)$$ while the closed-loop transfer function is


This isn't correct. It would be:


You can redraw the Bode plot into a Nichols chart, that will give you a better representation of closed loop characteristics and gain margin.


Generally speaking, for a linear time-invariant (LTI) system, applying feedback will move the poles to the left on the real-imaginary plane. This can often stabilize an otherwise unstable system.

That said, without specific transfer functions for g(s) and r(s), it's hard to make generalizations about the affect of applying feedback.

If you have matlab, it's relatively simple to create transfer functions and plot the poles and zeros to see what happens when you apply feedback. Alternatively, you can try python-control to do the same.


Most open loop systems are not well defined One cannot be sure what input is needed for the wanted output. It's very usual task to force a not well defined system to a desired state. This means that we need a sensor that measures the output and a controller which decides the desired input of the system.

This new system (=the old system+sensor +controller) we call a closed loop system and if we construct it correctly it will be stable and its output will stay in the wanted state accurately enough.

A good example of the open loop system of a DC motor. We need a speed sensor and a controller to control the speed to be wanted in different loading conditions. When we add the sensor and the control circuit the resulting new system will be the closed loop.

We can analyze that with math.

If we assume the motor to be a linear system, zeros and poles of the transfer function would give to us information on how to design a working circuit to control the motor speed.

For more information about the math behind this see DC motor roots.

  • \$\begingroup\$ -1 for this nonsense: Most open loop systems are unstable. An open loop system cannot be unstable. For instability to occur, there must be feedback i.e. a loop. If you disagree then please prove me wrong by showing an open loop system (no feedback) that is instable. \$\endgroup\$ Oct 14 '17 at 20:46
  • \$\begingroup\$ @Bimpelrekkie Now it's not nonsense, but its value as an answer is still zero or less. \$\endgroup\$
    – user287001
    Oct 14 '17 at 22:09
  • \$\begingroup\$ @Bimpelrekkie: OK. Challenge accepted. I will prove you wrong. The instable open loop system you seek is the DC motor when you want to control its position( not speed ). If you try to control its position without a sensor then you will fail. You need to introduce the feedback to make it stable. \$\endgroup\$
    – Tedi
    Oct 16 '17 at 8:44
  • \$\begingroup\$ OK, and how is that instable then? The motor will just spin in one direction, there is no oscillation going on. You need to introduce the feedback to make it stable Correct and if you do that in the wrong way it will oscillate around that position. Going back and forth around the wanted position: that's an oscillation. Spinning in one direction is not. \$\endgroup\$ Oct 16 '17 at 8:46
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    \$\begingroup\$ @Bimpelrekkie When a system is unstable, the output of the system may be infinite even though the input to the system was finite. Infinity to mathematicians is real but to control engineers is just a limit number in the technical requirements. Can we control it? NO we can't! \$\endgroup\$
    – Tedi
    Oct 16 '17 at 8:52

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