BODE PLOT TO TIME DOMAIN

Bode plot to Time-domain relation

In this post, I'll try to draw a relation between the bode plot of a transfer function to its unit step response. We shall restrict this to first order only, but this is equally applicable for second-order too. You will appreciate it for real at the end of this post. 

  

Without being much formal about it, let's start our little discussions.

For a first-order system, the transfer function can be written as 

\[\frac{s+a}{s+b}\]. 

Now for a passive circuit with zero initial conditions (which is considered an informal definition of transfer function also) the zero of the system must be dominant over the pole (for non-zero pole). That is |a| < |b|.

without loss of generality let's take some transfer function 

I)  \(\frac{1}{s}\)

II) \(\frac{1}{s + 10}\)

III)\(\frac{s}{s+10}\)

IV)\(\frac{s+1}{s+10}\)

I)  \(\frac{1}{s}\)

Bode:

Laplace transform of output: 

\(\frac{1}{s} \times \frac{1}{s}=\frac{1}{s^{2}}\)

Time domain plot

Can you relate the above two plots somehow? Well if not let's check for the next example.

II) \(\frac{1}{s + 10}\)

Bode:

Laplace transform of output: 

\(\frac{1}{s} \times \frac{1}{s+10}=\frac{1}{s \times(s+1)}\)

Time domain plot of step response

Still, it doesn't ring a bell? Let's give you a hint. 

Qualitatively compare initial and final values between the Bode plot of the transfer function and unit step response.

Well with this let's check for the last two transfer functions.

III)\(\frac{s}{s+10}\)

Bode:

Laplace transform of output: 

\(\frac{1}{s} \times \frac{s}{s+10}=\frac{1}{(s+1)}\)

Time domain plot of step response

IV)\(\frac{s+1}{s+10}\)

Laplace transform of output: 

\(\frac{1}{s} \times \frac{s+1}{s+10}=\frac{s+1}{s \times (s+1)}\)

Time domain plot of step response

Trick: Let me conclude this with a 2 pole 2 zero system.

TF : \(\frac {(s+2) \times (s+4)}{(s+1) \times (s+10)}\)

From the initial and final value theorem, one can conclude 

1) magnitude y(t=0) = |G( j*high freq)|

2) magnitude y(t= steady state)= |G(0)|

3) We need to traverse w = infinity to w = 0 to get unit-step response. Just it will have mixed exponent nature.