 Hello everyone, I am Mrs. Meenakshi Sriyandi from Vulture Institute of Technology, Sholapur. Welcome to the video lecture on analysis of type 0, 1 and 2 systems. Learning outcome, at the end of this session, student will be able to explain the analysis of type 0, 1 and 2 systems for different standard test inputs. Study state performance. A popular method to assess study state performance of unity feedback system is to find their error coefficients k p, k v and k a, where k p is position error constant, k v is velocity error constant, k a is acceleration error constant. In order to find k p, k v and k a, the system must be stable because for an unstable system there is no study state and k p, k v and k a are undefined. Thus, the concept of k p, k v and k a is applicable only if the system is represented in its simple form, only if the system is stable. Let's see the analysis of type 0 system. Let's consider the first reference input of step of magnitude a. Consider an open loop transfer function of type 0 system which is given as g of s h of s equal to k into 1 plus T 1 of s into 1 plus T 2 of s divided by 1 plus T a of s into 1 plus T b of s. For a step input, the error coefficient is given as k p equal to limit s tends to 0 g of s h of s. After solving the limit s tends to 0 for g of s h of s, then we get k p equal to k. Now, the study state error for step input is given as e of s is equal to a by 1 plus k p. After substituting k p equal to k from the above equation we get study state error equal to a by 1 plus k. That is, type 0 system follows the step type of input with finite error that is a by 1 plus k which can be reduced by changing a or k or both as per requirement. Now, k can be increased by introducing a variable gain amplifier in the forward path and error can be reduced. But there is limitation on the increase in the value of k from stability point of view. But increase in k is one way to reduce the error. The corresponding response is as shown in the figure showing finite error of a by 1 plus k. Let us see the analysis of type 0 system for the next reference input of ramp of magnitude a. Consider the open loop transfer function of type 0 system which is given as g of s h of s equal to k into 1 plus t 1 of s into 1 plus t 2 of s divided by 1 plus t a of s into 1 plus t b of s. For a ramp input the error coefficient is given as k v equal to limit s tends to 0 s into g of s into h of s. After solving the limit s tends to 0 for g of s h of s then we get k v equal to 0. Now, the steady state error for ramp input is given as e of s is equal to a by k v. After substituting k v equal to 0 from the avoid equation we get that's why the steady state error is equal to infinity. That is, type 0 system will not follow the ramp input of any magnitude and will give large error in the output which may damage the parameters of system or may cause the saturation in parameters. Hence, ramp input should not be applied to type 0 system. The output may take the form as shown in the figure where the steady state error is in finite. Trying to think and answer what will be the response of type 0 system for a parabolic input? Pause the video for some time and note down the answer in your book. The next reference input is parabolic input of magnitude a for the analysis of type 0 system. Consider our open loop transfer function of type 0 system which is given as g of s h of s equal to k into 1 plus t 1 of s into 1 plus t 2 of s divided by 1 plus t of s into 1 plus t b of s. For a parabolic input, the error coefficient is given as k a equal to limit s tends to 0 s square into g of s h of s. After solving the limit s tends to 0 for g of s h of s, then we get k a equal to 0. Now, the steady state error for parabolic input is given as e of s is equal to a by k a After substituting k a equal to 0 from the above equation, we get that is why the steady state error is equal to infinity. That is, type 0 system will not follow the parabolic input of any magnitude and will give large error in the output which may damage the parameters of system or may cause the saturation in the parameters. Hence, parabolic input should not be applied to type 0 system. The output may take the form as shown where the steady state error is in finite. This table gives the summary of analysis of type 0 system for step, ramp and parabolic input. Let's see the analysis of type 1 system. Let's consider the first reference input of step of magnitude a. Consider the open loop transfer function of type 1 system which is given as g of s h of s equal to k into 1 plus t 1 of s divided by s into 1 plus t a of s into 1 plus t b of s. For a step input the error coefficient is given as k p equal to limit s tends to 0 g of s h of s. After solving the limit s tends to 0 for g of s h of s then we get k p equal to infinity. Now, the steady state error for step input is given as e of s is equal to a by 1 plus k p. After substituting k p equal to infinity from the avoid equation we get and this equals to 0. In general for any type of system more than 0 k p will be infinite and error will be 0. Though mathematically the answer is 0 practically some small error will be present but it will be negligibly small. This is as shown in the figure showing negligibly small error. Let's see the analysis of type 1 system for the next reference input of ramp of magnitude a. Consider the open loop transfer function of type 1 system which is given as g of s h of s equal to k into 1 plus t 1 of s into 1 plus t 2 of s divided by s into 1 plus t a of s into 1 plus t b of s. For a ramp input the error coefficient is given as k v equal to limit s tends to s into g of s h of s. After solving limit s tends to 0 for g of s h of s then we get k v equal to k. Now the steady state error for ramp input is given as e of s is equal to a by k v after substituting k v equal to k from the avoid equation we get steady error equal to a by k. That is type 1 system follow the ramp type of input of magnitude a that is equal to a by k which can be reduced as discussed earlier. The output may take the form as shown in the figure where the steady state error is finite. Try to think and answer what will be the response of type 1 system for a parabolic input. Pause the video for some time and note down the answer in your book. The next reference input is parabolic input of magnitude a for the analysis of type 1 system. Consider the open loop transfer function of type 1 system which is given as g of s h of s equal to k into 1 plus t 1 of s into 1 plus t 2 of s divided by s into 1 plus t a of s into 1 plus t b of s. For a parabolic input the error coefficient is given as k a equal to limit s tends to 0 s square into g of s into h of s after solving the limit s tends to 0 for g of s h of s then we get k a equal to 0. Now the steady state error for parabolic input is given as e of s is equal to a by k a. After substituting k a equal to 0 from the above equation we get the steady state error equal to infinity. That is type 1 system will not follow the parabolic input of any magnitude and will give large error in the output which may damage the parameters of the system or may cause saturation in the parameters. Hence parabolic input should not be applied to type 1 system. The output may take the form as shown in the figure where the steady state error is infinite. This table gives the summary of analysis of type 1 system for step, ramp and parabolic input. Let's see the analysis of type 2 system. Let's consider the first reference input of step of magnitude a. Consider the open loop transfer function of type 2 system which is given as g of s h of s equal to k into 1 plus t 1 of s into 1 plus t 2 of s divided by s square into 1 plus t e of s into 1 plus t way of s. For a step input the error coefficient is given as k p equal to limit s tends to 0 g of s h of s. After solving the limit s tends to 0 for s h of s then we get k p equal to infinity. Now the steady state error for step input is given as e of s is equal to a by 1 plus k p. After substituting k p equal to infinity from the above equation we get a equal to infinity and this equals to 0. In general for any type of system more than 0 k p will be infinite and error will be 0. Though mathematically the answer is 0. Practically some error will be present but it will be negligibly small. This is as shown in the figure showing negligibly small error. Let's see the analysis of type 2 system for the next reference input of ramp of magnitude a. Consider the overlook transfer function of type 2 system which is given as g of s h of s equal to k into 1 plus t 1 of s into 1 plus t 2 of s divided by s square into 1 plus t e of s into 1 plus t b of s. For a ramp input the error coefficient is given as k v equal to limit s tends to 0 s into g of s h of s. After solving the limit s tends to 0 for g of s h of s then we get k v equal to infinity. Now the steady state error for ramp input is given as e of s is equal to a by p. After substituting k v equal to infinity from the above equation we get a by infinity which equals to 0. This is true for any system of type more than 1 hence all system of type 2 and more than 2 follow ramp type of input with negligibly small error. The output may take the form as shown in the figure where the steady state error is negligibly small. Try to think and answer what will be the difference of type 2 system for a parabolic input. Pause the video for some time and note down the answer in your book. The next reference input is parabolic input of magnitude a for the analysis of type 2 system. Consider the open loop transfer function of type 2 system which is given as g of s h of s equal to k into 1 plus t 1 of s into 1 plus t 2 of s divided by s square into 1 plus t e of s into t b of s. For a parabolic input the error coefficient is given as k a equal to limit s tends to 0 s square into g of s h of s after solving the limit s tends to 0 for g of s h of s then we get k a equal to k. Now the steady state error for parabolic input is given as e of s is equal to a by k a after substituting k a equal to k from the above equation we get steady state error equal to a by k and type 2 system will follow parabolic input with finite error equal to a by k which can be controlled by change in a or k or both. The output may take the form as shown in the figure where the steady state error is finite. This table gives the summary of analysis of type 2 system for step, ramp and parabolic input. These are my references Thank you.