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An amplifier has a non-linear response given by Vo= aVin +bVin^2 at all frequencies. When a sine wave of amplitude 1V is input to the amplifier the output of the amplifier contains a harmonic at -80 dBm. The amplifier output goes to a filter with a 3 dB bandwidth of 1MHz. What is the approximate amplitude of the harmonic at the output of the low pass filter if the input to the amplifier is a 5MHz sine wave with an amplitude that is equal to 0.5

I could not think of an approach to solve this question. Any recommendations to get an idea of solving this question would be helpful.

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    \$\begingroup\$ what do you not understand? maybe begin by writing the output waveform expression. \$\endgroup\$
    – tobalt
    Nov 13, 2021 at 7:43
  • \$\begingroup\$ I could not get what "contains a harmonic at -80dBm" mean \$\endgroup\$
    – Surya
    Nov 13, 2021 at 9:23
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    \$\begingroup\$ Most of the information given is not relevant to the actual question. -80 dBm is an absolute power level, so all you need to do is figure out what the frequency of the harmonic is, and how much the filter attenuates at that frequency. There is still some information missing -- for example, we'll have to assume that the filter's gain at low frequencies is unity (which is sort of implied if both resistors have the same value). You should state that assumption in your answer. \$\endgroup\$
    – Dave Tweed
    Nov 13, 2021 at 11:24
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    \$\begingroup\$ We won't do your homework for you. If you don't understand the question then you really should ask the instructor for clarification. \$\endgroup\$ Nov 13, 2021 at 13:25

3 Answers 3

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The output of amplifier, by inspection, will contain DC, 5 MHz and 10 MHz components (where 5 MHz is fundamental and 10 MHz is extra harmonic they are talking about). LPF is providing a 3 dB loss in output at 1 MHz. So, at a roll off of -20 dB/decade, at 10 MHz the signal would approximately be attenuated by another 20 dBs. So total attenuation at 10 MHz will be -23 MHz. But I really don't know what's the use of "-80 dBm" information here.

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This is a sloppy question, but I think you can work with it. Your reference input is 1 volt - is that 1 volt rms? Assume it is. The harmonic is -80 dBm, which is a power measurement, normally assumed to be into 50 ohms. From this, you can calculate the voltage level of the harmonic, and this will tell you the level of the harmonic for a given voltage input (effectively, it will allow you to calculate b).

So, knowing the harmonic level, you can take your 0.5 volt level (again, assuming it is rms) and calculate the harmonic level of the output. This should be trivial from the numbers involved and you should be able to do it in hour head. From the gain equation, you can find the first harmonic frequency from a 5 MHz input and calculate what will happen to it with your defined low-pass filter.

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Well, notice that the transfer function of the ideal opamp-circuit is given by:

$$\mathscr{H}\left(\text{s}\right):=\frac{\text{v}_\text{f}\left(\text{s}\right)}{\text{v}_\text{o}\left(\text{s}\right)}=\frac{\text{R||}\frac{1}{\text{sC}}}{\text{R}}=\frac{\text{R}\cdot\frac{1}{\text{sC}}}{\text{R}+\frac{1}{\text{sC}}}\cdot\frac{1}{\text{R}}=\frac{1}{1+\text{CRs}}\space\Longleftrightarrow\space$$ $$\text{v}_\text{o}\left(\text{s}\right)=\left(1+\text{CRs}\right)\text{v}_\text{f}\left(\text{s}\right)\tag1$$

Using the convolution property of the Laplace transform, we can write:

$$\text{V}_\text{o}\left(t\right)=\int_0^t\mathscr{L}_\text{s}^{-1}\left[\text{v}_\text{f}\left(\text{s}\right)\right]_{\left(\tau\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[1+\text{CRs}\right]_{\left(t-\tau\right)}\space\text{d}\tau=$$ $$\int_0^t\text{V}_\text{f}\left(\tau\right)\cdot\left(\delta\left(t-\tau\right)+\text{CR}\cdot\delta'\left(t-\tau\right)\right)\space\text{d}\tau=$$ $$\text{V}_\text{f}\left(t\right)+\text{CR}\cdot\left(\delta\left(t\right)\cdot\text{V}_\text{f}\left(0\right)+\text{V}_\text{f}'\left(t\right)\right)\tag2$$

Notice that the time relation between \$\text{V}_\text{o}\left(t\right)\$ and \$\text{V}_\text{in}\left(t\right)\$ is given by:

$$\text{V}_\text{o}\left(t\right)=\alpha\text{V}_\text{in}\left(t\right)+\beta\text{V}_\text{in}^2\left(t\right)=\text{V}_\text{in}\left(t\right)\left(\alpha+\beta\text{V}_\text{in}\left(t\right)\right)\tag3$$

So, we end up with the relation between the input and output:

$$\text{V}_\text{in}\left(t\right)\left(\alpha+\beta\text{V}_\text{in}\left(t\right)\right)=\text{V}_\text{f}\left(t\right)+\text{CR}\cdot\left(\delta\left(t\right)\cdot\text{V}_\text{f}\left(0\right)+\text{V}_\text{f}'\left(t\right)\right)\tag4$$

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