 Hello everyone, I am Mrs. Meenakshi Shrigandhi from Washington Institute of Technology, Sholapur. Welcome to the video lecture on analysis of first-order system. Learning outcome. At the end of this session, student will be able to explain the analysis of first-order system for step input. First-order system. Let's understand what you mean by order of the system. Order of the system is the highest power of S in the denominator of a closed-loop transfer function. Consider a simple first-order system as shown in the figure, which consists of resistor and capacitor on the input side and capacitor on the output side. Now, the transfer function of any system can be calculated by using this equation that is Vo of S by Vi of S. So, from the figure, apply the KVL to the input side. So, we get Vi of S equal to R of I of S plus 1 upon C of S into I of S. This equation is marked as 1. Now, applying the KVL to the output side, so we get Vo of S equal to 1 upon C S into I of S. This equation is marked as 2. Now, from equation 192, if we solve Vo of S by Vi of S, then the transfer function for the first-order system is given as Vo of S by Vi of S is equal to 1 upon 1 plus S of RC. This equation can be rewritten as 1 upon 1 plus TS, where T is equal to RC. Unit step response of first-order system. Now, consider a simple first-order system. The main task is to find the output response when it is excited by the unit step input voltage of Vi of T. Now, mathematically, the unit step is defined as Vi of T is equal to 1 for P greater than or equal to 0 and it is equal to 0 for P less than 0. The Laplace transform of unit step signal is given as Vi of S is equal to 1 by S. As discussed earlier, the transfer function of the first-order system is given as Vo of S by Vi of S equal to 1 by 1 plus SRC. Now, this equation is rewritten as Vo of S equal to Vi of S into 1 by 1 plus SRC. Now, substituting the Vi of S which is 1 by S for the step input in the above equation, we get Vo of S equal to 1 upon S into 1 plus SRC. Now, by using the partial fraction method, this equation is rewritten as A by S plus B by 1 plus SRC. Now, solving the fraction, we get A equal to 1 and B equal to minus RC. After substituting the values of A and B in the output response, then we get Vo of S equal to 1 upon S minus RC by 1 plus RC. Now, taking RC on the denominator, the equation is rewritten as 1 by S minus 1 by S plus 1 by RC. Now, taking the Laplace inverse, the output response in time domain is written as Vo of T equal to 1 minus E raise to minus T by RC. Here, 1 is representing Laplace inverse of 1 by S and E raise to minus T by RC is representing the Laplace inverse of 1 upon S plus 1 by RC. You also know that the output response consists of two main parts, that is the steady state response and the transient state response. So, the output response can be also written as Co of S S plus Co of T of T, where Co of S S is the steady state response and Co of T of T is the transient state response. So, if we compare this two equation, the steady state response is written as equal to 1 and the transient state response is equal to E raise to minus T by RC. On this, you can see that the transient state response is totally dependent on the values of R and C. If the values of R and C are changed, then the location of the pole, that is minus 1 by RC is also changed. Also, it is observed that the change in the values of R and C is not affecting the steady state response. So, the steady state response is independent of the values of R and C. This table shows the output response for the different values of T. As you can see, for the value of T is equal to 0, if it is substituted in the equation, then the output response is equal to 0. And if T is equal to RC, if it is substituted in this equation, then the output we get is 0.632. Similarly, if we substitute an infinity in this equation, then the output we get is 1. Now, if we try to plot this response, then we get this response. As you can see, the response is purely exponential. Transients are vanishing. This is because the exponential term is containing the negative index and this is because the pole is having a negative part. To think and answer, what will be the response of first order system for a step of A units? Pause the video for some time and note down the answer in your book. Let's see the response of the first order system for a step of A units. Now again, consider the first order system which consists of resistor and capacitor on the input side and capacitor on the output side. Now, here the input applied is the step of A units. So, the Laplace transform of the step of A units is given as Vi of S is equal to A by S. Now, as discussed earlier, the transfer function of the first order system is given as Vo of S by Vi of S is equal to 1 upon 1 plus SRC. Now, this equation is rewritten as Vo of S is equal to Vi of S into 1 upon 1 plus SRC. Now, substitute Vi of S in this case which is A by S for a step of A units in the above equation, then we get Vo of S equal to A by S into 1 plus SRC. Now, using the partial fraction method, this equation is rewritten as A1 by S plus B by 1 plus SRC. Now, solving the fractions, we get values of A1 is equal to A and B equal to minus ARC. Now, substitute the values of A1 and B in the output response which is given as A1 by S plus B by 1 plus SRC. If you substitute in this equation, then we get Vo of S equal to A by S minus ARC by 1 plus SRC. Taking RC in the denominator, this equation is written as A by S minus A by S plus 1 by RC. Taking the Laplace inverse, the output response in the time domain is given as Vo of T is equal to A into 1 minus E raise to minus T by RC. From this output response, it is seen that the rate of decay has not changed, only the steady state value has been changed. Corresponding response is as shown in the figure. Here again it is seen that the transient is totally dependent upon the values of RNC. But if the values of RNC are changed, then the location of the pole that is minus 1 by RC is also changed, which will affect the transient term as the rate of decay will also get changed. Also it is seen that the transient is vanishing. This is because the exponential term is containing the negative index which is because of the pole which is having a negative term. Try to think and answer what are the different factors does the transient response of first order system for step input depends on. Pause the video for some time and note down the answer in your book. The transient response of the first order system. In general the transient response of first order system for step input depends on the poles of the closed loop transfer function, the location of the poles of the closed loop transfer function and it is independent of the magnitude of the input applied. Any change in the magnitude of the selected input will not affect the transient state output. These are my references. Thank you.