# Plot frquency resonse from transfer function in complex frequency domain

1. Jan 6, 2013

### Steve Collins

I am going through a past paper for an upcoming exam and I want to check that I am approaching this question correctly.

H(s) = 1/(s2 + 8485.28s + 36x10-6)

Calculate and plot frequency response for 0 rad/s, 500 rad/s, 1000 rad/s and 10000 rad/s.

I am under the impression that the 's' term can be replaced with jω which gives:

H(s) = 1/(jω2 + 8485.28jω + 36x10-6)

where ω is the frequency.

I then replace jω with the the frequencies given above.

Is this correct?

2. Jan 6, 2013

### rude man

You replace ω, not jω. Then you get a complex number for which you have to compute its magnitude.

3. Jan 6, 2013

### Steve Collins

so for 500 rad/s,

H(s) = 1/(jω2 + 8485.28jω + 36x10-6)

= 1/(j5002 + 8485.28 x j500 + 36x10-6)

= 1/(j250000 + j4242640 + 36x10-6)

= 1/(36x10-6 + j4492640)

So,

r= 1/√[(36x10-6)2 + 44926402]

= 1/4492640

I've obviously done something wrong here as the imaginary number is so small that it is having no effect on the outcome of the magnitude.

4. Jan 6, 2013

### rude man

If your original H(s) was correct thenn what you did is also correct (but H(s) should be H(jw) of course.

Looks like your transfer function has a high gain at zero frequency (dc) of nearly 30,000 but starts to "roll off" almost immediately (around 5e-9 rad/s if I computed correctly!) at a 20dB/decade rate. At ~8500 rad/s it picks up another 20 dB/decade. (Don't feel bad if you don't dig all this lingo, I'm mostly talking to myself here ...).

This looks very much like the gain of a typical operational amplifier BTW.

5. Jan 6, 2013

### Steve Collins

I'm not a million miles away from what you're saying. I'll continue looking at this tomorrow when I'll attempt the plot and probably have some more questions.

cheers for your help

steve

6. Jan 6, 2013

### rude man

OK, and happy to explain anything you're curious about.
Cheers hence also!

7. Jan 7, 2013

### Steve Collins

I've converted my answers to dBs and got

0 rad/s... 88.874dB
500 rad/s... -133.05dB
1000 rads/s... -139.541dB
10000 rad/s... -165.337dB

Does the frequency along the x-axis have to be logarithmic? If so is there a set standard for spacing?

8. Jan 7, 2013

### rude man

Yes, the standard way is a log-log plot, i.e. dB on the y axis and log10(frequency) on the x axis. The resulting plot consists of straight sections called asymptotes.

The reason is that that way the asymptotes are straight lines. Take H(s) = 1/(Ts+1); the gain plot is a straight line until you get to radian frequency 1/T, then the gain is another straight line but angling downward with slope = 20dB/10:1 change in frequency. The actual gain follows this approximately, with the max. gain error at ω = 1/T. At that point the actual gain is 3dB less than the dc gain. At dc it's 100% accurate and very closely so at high frequencies like ω = 100/T. You should be exposed to how the asymptotes approximate the actual gain in the course of your studies. It gets trickier if you have complex-conjugate poles in your H(s).

Not sure what you mean by "spacing". The x axis has frequency in either Hz or rad/s on log-log paper. Every major division is 10:1 change, e.g. 10Hz, 100Hz, 1000Hz, etc. On the y axis every major division is another 20 dB.

9. Jan 7, 2013

### Steve Collins

I was wondering how to draw a graph with log scale, but looking at the front sheet of the past paper I am working through I see that log paper was provided, obviously!

10. Jan 7, 2013

### rude man

OK, I see what you meant. No one has ever asked me to draw my own log paper either!

But if someone does give you linear-linear, just mark the x axis in powers of 10: 100Hz, 101Hz, 102Hz etc. Half-way between 10 and 100 for example would be √(10*100) = 31.6 Hz etc.

And BTW I think I told you wrong: when you mark off dB on the y axis, that scale is linear in dB. It's the dB themselves that make it effectively logarithmic. So your graph paper should be linear on the y axis and logarithmic on the x.

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