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First Order System Transient Response
To understand the transient response of the first order system, consider the block diagram of a closed loop system with unity negative feedback.
The open loop gain G(s) of the system with unity negative feedback is given by,
$$G(s)=\frac{1}{s\tau}$$
And the closed loop transfer function of the system with unity negative feedback is,
$$\frac{C(s)}{R(s)}=\frac{G(s)}{1+G(s)}=\frac{1}{s\tau+1}\:\:\:...(1)$$
Where,
R(s) = Laplace transform of the input signal r(t),
C(s) = Laplace transform of the output signal c(t),
τ = Time Constant of the system.
As we can see, the power of s is one in the denominator term of the closed loop transfer equation. Hence, the system is said to be first order system.
$$C(s)=\frac{1}{s\tau+1}R(s)\:\:\:...(2)$$
Unit Step Response of the First Order System
Apply the unit step signal as the input to the first order system,
$$r(t)=u(t)$$
Taking Laplace transform on both sides,
$$R(s)=\frac{1}{s}$$
$$C(s)=(\frac{1}{s\tau+1})(\frac{1}{s})=\frac{1}{s(s\tau+1)}\:\:\:....(3)$$
By doing partial fraction of equation (3), we get,
$$\frac{1}{s(s\tau+1)}=\frac{A}{s}+\frac{B}{(s\tau+1)}=\frac{A(s\tau+1)+Bs}{s(s\tau+1)}\:\:\:\:...(4)$$
By comparing LHS and RHS of equation (4), we get,
$$A(s\tau+1)+Bs=1\:\:\:...(5)$$
By equating constant terms on the both side of equation (5), we get, A = 1. By putting A = 1 in the equation (5), we have,
$$B+\tau=0\Rightarrow\:B=-\tau$$
Now, by substituting the value of A and B is the equation (4), we get,
$$C(s)=\frac{1}{s}-\frac{\tau}{s\tau+1}=\frac{1}{s}-\frac{1}{s+\frac{1}{\tau}}\:\:\:...(6)$$
Taking the inverse Laplace transform on both sides of the equation (6),
$$C(t)=(1-e^{\frac{-t}{\tau}})u(t)\:\:\:...(7)$$
The equation (7), represents the response of the first order system for the unit step input, it has steady state term as well as transient term. The value of unit step response c(t) is zero, when t = 0. For all the positive values of t, it will gradually increase from zero to one in steady state.
Unit Ramp Response of the First Order System
Using the Unit Ramp Signal at the input of the first order system.
$$\because\:\:r(t)=t\:u(t)$$
By taking Laplace transform,
$$R(s)=\frac{1}{s^{2}}$$
As the response of the first order system is given by,
$$C(s)=(\frac{1}{s\tau+1})R(s)\:\:\:\:...(8)$$
By putting value of R(s) in the equation (8), we have,
$$C(s)=(\frac{1}{s\tau+1})(\frac{1}{s^{2}})\:\:\:\:...(9)$$
On solving this by doing partial fractions, we get,
$$C(s)=\frac{1}{s^{2}}-\frac{\tau}{s}+\frac{\tau}{s+\frac{1}{\tau}}\:\:\:...(10)$$
Taking the inverse Laplace transform of the equation (10), we have,
$$C(t)=(t-\tau+\tau\:e^{-t/\tau})u(t)\:\:\:\:for\:t\geq0\:\:\:...(11)$$
The equation (11) shows the time response of the first order system for the unit ramp input, the c(t) follows the unit ramp signal for all positive values of t. But, there is a deviation of τ units from the input signal. From the equation (t), it can also be seen that, the c(t) has both steady state terms as well as transient term.
Impulse Response of the First Order System
Applying the Unit Impulse Response at the input of the first order system,
$$r(t)=\delta(t)$$
Taking Laplace transform on both side,
$$R(s)=1$$
As, the response of the first order system is,
$$C(s)=(\frac{1}{s\tau+1})R(s)\:\:\:\:...(12)$$
By substituting, R(s) = 1 in the equation (12), we have
$$C(s)=(\frac{1}{s\tau+1})(1)=(\frac{1}{s\tau+1})$$
$$C(s)=\frac{1}{\tau(s+1/\tau)}=\frac{1}{\tau}(\frac{1}{s+1/\tau})\:\:\:\:...(13)$$
Taking the inverse Laplace on both the sides of the equation (13), we get,
$$C(t)=\frac{1}{\tau}e^{(-t/\tau)}u(t)\:\:\:\:for\:t\geq0\:\:\:\:...(14)$$
The equation (14), shows the unit impulse response of the first order system. The c(t) is an exponential decaying signal for the positive values of t.
Summary of Time Response of the First Order System
| Input | Output |
|---|---|
| Unit Step Signal $r(t)=u(t)\:\:\:For,t\geq0$ | $$c(t)=(1-e^{-t/\tau})u(t)\:\:\:For,t\geq0$$ |
| Unit Ramp Signal $r(t)=t\:u(t)\:\:\:For,t\geq0$ | $$c(t)=(t-\tau+\tau\:e^{-t/\tau})u()t\:\:\:For,t\geq0$$ |
| Unit Impulse Signal $r(t)=\delta(t)\:\:\:For,t\geq0$ | $$c(t)=\frac{1}{\tau}e^{-t/\tau}u(t)\:\:\:For,t\geq0$$ |
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