 Hello everyone, I am Mrs. Meenakshi Sri Gandhi from Vulture Institute of Technology, Sholapur. Welcome to the video lecture on steady state analysis. Learning outcome? At the end of this session, student will be able to explain the steady state response of a system in terms of steady state error. Steady state analysis? Steady state is that part of the output, which remains after the transients completely vanish from the output. The steady state response mainly has two specifications. How much time the system takes to reach its steady state, which is called as settling time? How far away the actual output is reached from its desired value, which is called as steady state error, which is given as E of ss. Out of the two specifications, the steady state error is the most important specification, which is related only to the steady state. So, let's see on which factors it depends on, how to calculate it and how to reduce it. Steady state error? The steady state error is the difference between the actual output and the desired output. The reference input tells us the level of desired output and the actual output is fed back through feedback element to compare it with the reference input. Hence, it can be defined as the difference between the reference input and the feedback signal. Mathematically, it can be defined in Laplace domain as Laplace of E of t equal to E of s, which is equal to R of s minus E of s B of s. This is for non-unity feedback system and Laplace transform of E of t, which is equal to E of s, that is equal to R of s minus E of s for unity feedback system. Let's see the derivation of steady state error. Consider a simple closed loop system using negative feedback as shown in the figure, where E of s is the error signal and from figure it can be written as E of s equal to R of s minus B of s. B of s is the feedback signal and from figure it can be written as B of s equal to C of s H of s. Now, substituting B of s in equation one, now we get error signal as E of s equal to R of s minus C of s H of s. But C of s is given as E of s G of s. Now, substituting C of s in equation two, now the error signal can be written as E of s equal to R of s minus E of s G of s H of s. Shifting E of s G of s H of s on left side, we get E of s plus E of s into G of s into H of s which equals to R of s. The equation can be rewritten as E of s into 1 plus G of s H of s which equals to R of s. For non-unity feedback signal, E of s can be written as E of s equal to R of s by 1 plus G of s H of s and for unity feedback signal E of s is written as E of s equal to R of s by 1 plus G of s where E of s is the error signal in Laplace domain and is expression in s. In time domain corresponding error will be E of t. Now, the steady state of the system is that state which remain as t tends to infinity and this is given as E of s is equal to limit t tends to infinity E of t. We can relate the steady state error in Laplace domain by using the final value theorem which states that limit t tends to infinity f of t which equals to limit s tends to infinity s of f of s where f of s equal to Laplace of f of t. Therefore, steady state error can be written as E of s is equal to limit t tends to infinity E of t which equals to limit s tends to infinity s of E of s where E of s is equal to Laplace of E of t substituting E of s from the expression derived we can write E of s is equal to limit s tends to 0 into R of s by 1 plus G of s H of s. For negative feedback system use positive sign in the denominator while use negative sign in the denominator if the system uses positive feedback. Try to think and answer what are the different factors that the steady state error depends on pause the video for some time and note down the answer in your book. From the expression derived it can be concluded that the steady state error depends on R of s that is the reference input its type and magnitude G of s H of s that is the open loop transfer function. The dominant non-linearity is present if any. Now let's see the effect of change in input on steady state error. Let's consider the first reference input as step of magnitude a. Consider a system having the open loop transfer function G of s H of s. For unit step R of t is given as R of t is equal to 1 for t greater than or equal to 0 otherwise R of t is equal to 0. Its Laplace transform is given as R of s is equal to a by s. The steady state error as derived earlier is given as E of s is equal to limit s tends to 0 s into R of s by 1 plus G of s H of s. Now substituting R of s in this equation we get E of s is equal to limit s tends to 0 s into a by s by 1 plus G of s H of s. From this we get limit s tends to 0 a by 1 plus G of s H of s. Now taking the limit in the denominator we get E of s is equal to a by 1 plus limit s tends to 0 G of s H of s. For a system selected limit s tends to 0 G of s H of s is a constant and this is called as a positional error coefficient of the system and is denoted as k p. So k p is given as k p equal to limit s tends to 0 G of s H of s which is called as the positional error coefficient and the corresponding error is E of s is equal to a by 1 plus k p. This is shown in the diagram where for a step input the difference between the actual output and the desired output is nothing but the steady state error and in this case it is given as E of s is equal to a by 1 plus k p. Here k p will control the error in the system along with the magnitude of the input applied. Let us consider the second reference input as ramp of magnitude a. Consider a system having open loop transfer function as G of s H of s. Now the Laplace transform for ramp input is given as R of s equal to a by s square. Now the steady state error as derived earlier is given as E of s is equal to limit s tends to 0 s into R of s by 1 plus G of s H of s. Now substituting R of s in this equation we have E of s is equal to limit s tends to 0 s into a by s square divided by 1 plus G of s H of s. Solving this we get E of s is equal to limit s tends to 0 a by s into 1 plus G of s H of s. That is given as limit s tends to 0 a by s plus s into G of s H of s. Now substituting the limits in the denominator we get E of s is equal to a by limit s tends to 0 s into G of s H of s. For a system selected limit s tends to 0 s into G of s H of s is a constant and it is called as velocity error coefficient of the system which is denoted as K v. So K v is given as K v equal to limit s tends to 0 s into G of s H of s which is called as velocity error coefficient and the corresponding error is given as E of s is equal to a by K v. K v will control the error in the system along with the magnitude of the input applied. This is shown in the diagram where for a ramp input the difference between the actual output and the desired output is nothing but the steady state error and in this case it is given as E of s is equal to a by K v. The next reference input is the parabolic input. Consider a system having open loop transfer function G of s H of s. The parabolic input mathematically is given as r of t is equal to a by 2 t square for t greater than or equal to 0. Its Laplace transform is given as r of s equal to a by s cube. The steady state error is given as E of s is equal to limit s tends to 0 s into r of s by 1 plus G of s into H of s. Now substitute r of s in this equation we have limit s tends to 0 s into a by s cube by 1 plus G of s H of s. After solving this we get limit s tends to 0 a by s square into 1 plus G of s H of s. Solving the equation we get limit s tends to 0 a by s square plus s square G of s H of s. Solving the limits we get E of s is equal to a by limit s tends to 0 s square into G of s H of s. For a system selected limit s tends to 0 into s square into G of s H of s is a constant and is called as acceleration error coefficient of the system which is denoted as Ka and is given as Ka is equal to limit s tends to 0 s square into G of s H of s which is called as the acceleration error coefficient and the corresponding error is given as E of s is equal to a by Ka. The acceleration error coefficient Ka will control the error in the system along with the magnitude of the input applied. This is shown in the diagram where for a parabolic input the difference between the actual output and the desired output is nothing but the steady state error and in this case it is given as E of s is equal to a by Ka. Here it gives the summary of the static error coefficients and the corresponding steady state error. These are my references. Thank you.