# What is the "steady-state frequency response"?

Discussion in 'Homework Help' started by lll, May 6, 2012.

1. ### lll Thread Starter New Member

Mar 7, 2012
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1
I found the equivalent impedance of the circuit to be Z=R1R2+(R1+R1)jwL

I'm not sure what it means by steady state frequency response. I assumed it meant Vout/Vin, but that gave me a wrong answer:

$V_{out}={\frac{V_{in}}{R_1R_2+(R_1+R_1)jwL}}\frac{jwlR_2}{R_2+jwl}$

$\frac{V_{out}}{V_{in}}=({\frac{jwlR_2}{R_1R_2+(R_1+R_1)jwL}})\frac{1}{R_2+jwl}$

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Yes - you need Vout/Vin.

The equivalent input impedance isn't what you have indicated.

Rather ....

$Z_{in}=R_1+R_2||j\omega L = \frac{R_1R_2+j\omega L(R_1+R_2)}{R_2+j\omega L}$

3. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
Whenever you work a problem, you want to get in the habit of always asking two things:

1) Do the units work out?
2) Does the answer make sense?

In this case, I don't even need to see the circuit to know (not think, not suspect, but KNOW) that your answer is wrong. Why? Because the units don't work out.

Vout/Vin has to be dimensionless. Now jwL has units of ohms and R has units of ohms. Thus, your expression for Vout/Vin has units of 1/ohms, which means it is wrong.

Looking back at your "Z=R1R2+(R1+R1)jwL", this HAS to be wrong because the very first term has units of ohms-squared when impedance has units of ohms. No need to go any further until this is resolved.

As for the "answer making sense" part. Once you have an answer, ask if it reduces to what it should in as many limiting cases as you can. Having now looked at your circuit, the two obvious limiting cases are at DC and at really high frequencies. What does the inductor look like in these two cases? What does the frequency response need to reduce to in order to match.

Using these two rules, we can also eliminate some of the potential answers before we even start to work the problem. The units don't work out on (c) and (d), so there isn't any point in even considering them further. Of the two remaining, both limit properly at DC, but only (b) limits properly at high frequency. So you now only have to discern whether (b) is correct and, if it isn't, then the answer is (e).

From a practical test taking standpoint, consider the power of this approach. Without spending hardly any time at all, you've been able to narrow this question down to a 50/50 guess is push comes to shove and you run out of time. In this case, it's probably even a lot better than that because given a choice between an answer that satisfies the limiting cases and "none of the above", the one that works is almost certainly correct (not guaranteed, but a high likelihood). Hence, if I couldn't get to this problem in time to work it, I would put my money on (b) being correct.

So a good test taking strategy would be to run through all of the problems and eliminate as many incorrect answers as you can quickly (in some cases, this will leave you with only one possible answer). Now you can start working the problems in order to pick from among the remaining answers and you might leave the ones for which you have eliminated the most answers until last so that, if you run out of time, you can simply guess with a much higher probability of getting more of them correct.

But note that my basic advice is independent of whether you are taking a test, or even a class. As a practicing engineer it is your obligation and responsibility to use any and all reasonable error detecting and correcting mechanisms at your disposal. Tracking and checking units is, perhaps, the single most powerful and useful such mechanism, with always asking if the answer makes sense coming in a close second.

lll likes this.

Mar 7, 2012
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