Saikat Mukherjee

Now I ask my name….which I found at first in my way… –A sweet smell of desire…… it has things to say.

Category: Electrical Engineering

With Varying Damping Factor ζ Pole Distribution of a Second order Linear Time Invariant system when subjected to an Unit Step Input

From the time response analysis of a second order system the characteristic equation of a system can be written as-
s^{2} + 2\zeta\omega_{n}s + \omega_{n}^2 = 0
where \omega_n is undamped natural frequency

For this second degree equation we can have 2 values of s , those are-
s_{1} = - \zeta\omega_{n} + j\omega_{n}\sqrt{1-\zeta^2}
and s_{2} = - \zeta\omega_{n} - j\omega_{n}\sqrt{1-\zeta^2}

Now, \mid{s_1} \mid = \mid{s_2} \mid = \sqrt {{(\zeta \omega_n)^2} + (\omega_n \sqrt{1-\zeta^2})^2} = \pm\omega_n

It is evident that for any value of \zeta , the values of the poles ; i.e the value of s remains constant ; i.e \omega_n .
Take a look at the locus of the poles with change in the value of \zeta below-

locus of poles with varying ζ

  • Here  \theta = cos^{-1}\zeta
  • \omega_d is the damped natural frequency.
  • It’s evident from the figure that if \zeta =0 ; i.e for a limitedly stable system,the mod value of the pole will be \omega_n and there will be two poles at \pm\omega_n .
  • With the increase in value of \zeta pole will be shifted along with the perimeter of the half-circle with radius \omega_n .The locus of the pole ; i.e the shown red half circle will be on the left side of the j \omega axis in the value of s_1 s_2 the real part is always negative ;i.e - \zeta\omega_n .
  •  When in case of a critically damped system damping factor \zeta =1 ,both of the poles are at - \omega_n .
  •  In the range 0< \zeta <1 the poles traverse along the perimeter of the half circle(remember the radius is \omega_n ),one pole starting from +j \omega_n and the other one from -j \omega_n , towards the left side of the j \omega axis, creating that half circular locus and they meet at - \omega_n when \zeta becomes 1 .
  • Now when \zeta continues to increase beyond 1 ; i.e for a over damped system, there stays no imaginary part in the expressions of the poles, they break away from the point - \omega_n and start travelling in opposite direction along the real ( \sigma ) axis.

One can say that for \zeta = 0 , the poles are on the imaginary axis. And for \zeta = 0 the response of a second order system to a unit step input keep oscillating like sine wave
Say the response is c(t) .
Then the response for \zeta = 0 becomes-
c(t) = 1- cos {\omega_n}{t} [This will be proved in some other post].

So for basic understanding it can be concluded that for a second order Linear Time Invariant (LTI) system, damping factor \zeta is the decider for the stability of that system.

[Details on LTI system stability will be discussed in some other post].


Fourier Analysis for a Continuous Time Non-periodic Signal

Fourier Series is defined in trigonometric terms but cosine and sine can be defined with complex terms also-

cos{\theta} + isin{\theta} = e^{i\theta}
cos{\theta} - isin{\theta} = e^{-i\theta}

First of all before we go further one must know that along with defining any periodic function by the sum of various harmonics of sine and cosine, these can be defined by the following as well-

f(x) = \sum_{n=-\infty}^{\infty} C_{n}e^{\frac{in\pi\omega t}{l}} ;[Let the period of any periodic function is 2l ]
where angular velocity {\omega} = \frac{2\pi}{T} ; [T is the time period; i.e time required to pass 2\pi angle once]

{\omega} = {2\pi f} [f is the frequency]
Here {\omega} is considered as the frequency.

Here one should know that C_{n}=\frac{1}{2l}\int_{-l}^{l}f(x)e^{\frac{-in\pi x}{l}}\,\mathrm{d}x
Now defining f(x) in terms of \omega is called frequency domain expression and defining in terms of time (where laplace transformation is used to express any system) is called time domain expression.
A case can arise when f(x) is not periodic,this means f(x) has no period and time period T is tending to infinity (T \rightarrow \infty ) and so \omega\rightarrow 0 . As it’s non-periodic function, so obviously its frequency can’t be defined.




This is a non-periodic function where -a\leq\omega t\leq a ,means this function doesn’t repeat after a fixed period of time.
Now f(t) =\sum_{n=-\infty}^{\infty}C_{n}e^{jn\omega t} ; [ l term only comes when period is not 2\pi ,say here time period is 2\pi ]

\therefore C_{n}=\frac{1}{T}\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)e^{-jn\omega t}\,\mathrm{d}t
As here f(t) is non-periodic function, t \rightarrow \infty and \omega\rightarrow 0 ,so 'n' becomes meaningless in the above expression.
Let n\omega \rightarrow \omega
\omega \rightarrow \triangle \omega

\therefore T \rightarrow \frac{2\pi}{\triangle \omega}

\therefore f(t) becomes-

f(t)=\sum_{\omega =-\infty}^{\infty}C_{\omega}e^{j\omega t} [\forall \omega = 0,\pm \triangle \omega , \pm 2\triangle \omega ... ] \longrightarrow (1)

\therefore C_{\omega} = \frac{\triangle \omega}{2\pi}\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)e^{-j\omega t}\,\mathrm{d}t

\therefore Substituting C_{\omega} in equation (1)

f(t) = \frac{1}{2\pi}[\sum_{\omega =-\infty}^{\infty}\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)e^{-j\omega t}\,\mathrm{d}t]e^{j\omega t}\triangle \omega

\because T \rightarrow \infty ,\triangle \omega \rightarrow \,\mathrm{d}\omega and \sum \rightarrow \int

\therefore f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}[\int_{-\infty}^{\infty}f(t)e^{-j\omega t}\,\mathrm{d}t]e^{j\omega t}\,\mathrm{d}\omega

\therefore f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{j\omega t}\,\mathrm{d}\omega \Longrightarrow [Inverse Fourier Transform]

F({\omega}) = \mathscr{F}[f(t)]= \int_{-\infty}^{\infty}f(t)e^{-j\omega t}\,\mathrm{d}t \Longrightarrow [Fourier Transform]
[\mathscr{F} denotes fourier transform ]

So it has been shown that for a non-periodic function we have to first get the fourier trasform of that function (Of course firstly by verifying Dirichlet conditions) to get the fourier series of that function.