So the imaginary component is just -ωCVV*.

[jeff1evesque]

For the following, I was wondering if someone could help me determine the imaginary part of the complex number:
$$Im<GVV* - j\omega CVV*>$$, where $$V*$$ is the complex conjugate of $$V$$.

I was thinking the imaginary component would be $$-\omega CVV*.$$ But I am reluctant on the conclusion, because I am not sure if it would rather be $$\omega CVV*.$$

Could someone enlighten me?

[Is there any chance someone could remind me why complex numbers and phasor's are important for circuits (practicality)? And does every circuit element have a corresponding phasor element, $$e^{j\omega t}?$$ ]

thanks,JL

 

[berkeman]

For the following, I was wondering if someone could help me determine the imaginary part of the complex number:
$$Im<GVV* - j\omega CVV*>$$, where $$V*$$ is the complex conjugate of $$V$$.

I was thinking the imaginary component would be $$-\omega CVV*.$$ But I am reluctant on the conclusion, because I am not sure if it would rather be $$\omega CVV*.$$

Could someone enlighten me?

[Is there any chance someone could remind me why complex numbers and phasor's are important for circuits (practicality)? And does every circuit element have a corresponding phasor element, $$e^{j\omega t}?$$ ]

thanks,


JL

What do you get when you multiply VV* ? Use a simple example to help you if necessary:

$$(a+jb)(a-jb) = ??$$

Is the result complex?

So the imaginary part of any expression is the part that is multiplied by j by definition (not -j). What then is the answer to your problem?

As for why complex numbers are handy for circuits and signals, one big reason is that you can represent both the magnitude and phase of a signal over time as a single complex quantity. You can manipulate the magnitudes and phases of multiple signals, and in the end figure out what the real-time output waveform looks like.

 

[jeff1evesque]

Hello, thanks for responding to my questions.

What do you get when you multiply VV* ?

You get $$|V|^{2}$$

(a+jb)(a-jb) = ??

The answer would be $$a^{2} + b^{2}$$

What then is the answer to your problem?

I think my answer would be $$\omega C|V|^{2}$$, but how does this follow from above?

Thanks again,Jeffrey

 

[Redbelly98]

Hello, thanks for responding to my questions.


You get $$|V|^{2}$$


The answer would be $$a^{2} + b^{2}$$

Yes. Note that these are purely real numbers.

Complex numbers are a good way to represent the impedance -- that is, the ratio of voltage to current amplitudes in an AC circuit. Then the overall impedance of a mixed combination of resistors, capacitors, and inductors can be calculated using the same parallel & series combination rules for resistors that you are (hopefully) familiar with.

 

[Redbelly98]

I think my answer would be $$\omega C|V|^{2}$$, but how does this follow from above?

Well, it means GVV* is purely real, and does not contribute to the imaginary part.

It also means ωCVV* is purely real, therefore -jωCVV* is purely imaginary.