 Hello everyone, I am Mrs. Meenakshi Shrigandhi from Valshan Institute of Technology, Shalapur. Welcome to the video lecture on analysis of first order system. Learning outcome, at the end of this session, students will be able to explain the analysis of first order system for impulse and RAM input. First order system, before we see 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 of the circuit. Now the transfer function of any system is given as the Laplace transform of the output to the Laplace transform of the input. So here we are trying to find the transfer function of the first order system. To find the same, we need to find the Laplace transform of the output and the Laplace transform of the input. So from figure, after applying KVL to the input side, we get Vi of s equal to r into I of s plus 1 upon sc into I of s. This equation is marked as the first equation. Now applying KVL to the output side, we get V of s equal to 1 by sc into I of s. This equation is marked as the second equation. Using equation 1 and 2, the transfer function for the first order system is given as V of s by Vi of s equal to 1 upon 1 plus src. That is equal to 1 upon 1 plus ts, where t is equal to rc. Now let's see what is the unit impulse response of first order system is. Now consider a simple first order system. The main task is to find the output voltage when it is excited by the unit impulse input voltage that is Vi of t. Unit impulse mathematically is defined as Vi of t is equal to 1 for t equal to 0 and it is equal to 0 for t not equal to 0. Now the Laplace transform of unit impulse signal is given as Vi of s is equal to 1. As seen earlier, the transfer function of the first order system is given as V of s by Vi of s is equal to 1 upon 1 plus src, which is written as V of s is equal to Vi of s into 1 upon 1 plus src. Now substitute Vi of s in the above equation. In this case, Vi of s is the Laplace transform of the unit impulse response which is equal to 1. After substituting this in the output response, we get V of s is equal to 1 into 1 by 1 plus src. Now taking rc common from the denominator, this equation can be rewritten as V of s is equal to 1 by rc into s plus 1 by rc. Now replacing rc equal to t, the output response can be written as V of s is equal to 1 upon t into s plus 1 by t. Now the same equation can be rewritten as V of s equal to 1 upon t into 1 upon s plus 1 by t. Now taking the Laplace inverse of this, the output response in time domain is given as V of t is equal to 1 by t into e raise to minus t by t. Now here, e raise to minus t by t is a Laplace inverse of 1 upon s plus 1 by t. Now this is the response of the first order system when a unit impulse response is applied to it. Now if we try to plot the signal for different time instant, then we get the response in this way. Where it can be seen that c of t is an exponential decaying signal for positive values of t and it is 0 for negative values of t. Try to think and answer what will be the response of first order system for a unit RAM signal. Pause the video for some time and note down the answer in your book. Let's see the response of first order system for a unit RAM signal. Now again consider the first order system. Now here for the first order system we are giving the input as unit RAM signal. Now unit RAM signal mathematically is defined as r of t is equal to t for t greater than 0 and r of t is equal to 0 for t less than 0. Now if you find the Laplace transform of this signal then the Laplace transform is given as v i of s equal to 1 by s square. Now as seen earlier the transfer function of the first order system is given as v o of s by v i of s is equal to 1 upon 1 plus s r c. Now taking v i of s on the right side this equation is written as v o of s equal to v i of s into 1 upon 1 plus s r c. Now substitute v i of s in this case for RAM signal v i of s is equal to 1 by s square. Now substitute this equation in v o of s we get v o of s is equal to 1 upon s square into 1 plus s r c. Now using partial fraction method this equation can be written as a by s square plus b by s plus c by 1 plus s r c. After finding the fractions we get a equal to 1, b equal to minus t and c equal to t square. Now substituting these values in the output response which is given as a by s square plus b by s plus c by 1 plus s r c then we get v o of s is equal to 1 by s square minus t by s plus t square by 1 plus s r c. Now taking r c common from the denominator this equation can be written as 1 by s square minus t by s plus t square by t into s plus 1 by t. Here t is equal to r c. After this if we cancel t from the numerator and denominator this equation can be rewritten as 1 upon s square minus t by s plus t by s plus 1 by t. Now applying the inverse Laplace transform to the output response we get the output response in time domain which is given as v o of t is equal to t minus t plus t into e raise to minus t by t. Here t is the Laplace inverse of 1 by s square and 1 is the Laplace inverse of 1 by s and e raise to minus t by t is the Laplace inverse of 1 by s plus 1 by t. Now this is the overall response of the first order system when a unit RAM signal is applied. From this equation you can see that t minus t represent the steady state response and t into e raise to minus t by t represent the transient response. Now after separating the steady state response and the transient state response the steady state term in the unit response can be given as c s s of t is equal to t minus t and the transient term in the unit RAM signal is given as c t of t is equal to t into e raise to minus t by t. Now after plotting the overall response for different time instances then we get the response in this way where it can be seen that the unit RAM response c of t follows the unit RAM input signal for all positive values of t but there is a deviation of t units from the input signal. From this response you can see that this is the actual RAM signal and this is the response which we have got for the first order system after applying the unit RAM signal. So there is a deviation of t units from the input signal. Trying to think an answer for which type of input that is whether it is impulse or RAM does the first order system is stable. Pause the video for some time and note down the answer in your book. Response of first order system. The first order system is stable for impulse input as these responses have bounded output. Whereas the first order system is unstable for RAM input as these responses go on increasing even at infinite amount of time. These are my references. Thank you.