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Consider the relation between the output and input voltages: $$V_o=\frac{V_i}{2}(j-1)$$ where j is an imaginary number, and $$V_i=5sin(1000t)$$ Now, to find the amplitude of \$ V_o \$, should we look at the real part, i.e., $$\frac{-V_i}{2}=\frac{-5}{2}sin(1000t)$$ and say that the amplitude is.$$\frac{5}{2}$$

OR should we take the magnitude of $$\frac{V_o}{V_i}$$ and say that the amplitude is $$\mid V_o\mid= \mid V_i \mid\mid \frac{1}{2}(j-1) \mid$$ where $$\mid V_i \mid = 5$$

Dom Tesilbirth Shira
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  • Note that Vi is a voltage but Vo/Vi is a unitless number so they measure different things. Amplitude is measured in volts, so that rules out using Vo/Vi. – ErikR Jun 16 '21 at 12:11
  • Either use polar notation or complex notation, but don't mix the two. – Chu Jun 16 '21 at 13:09
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    @DomTesilbirthShira please formally accept one answer or raise a comment if further clarification is needed. To understand what I say [read this](https://electronics.stackexchange.com/tour) and note that questions that are adequately answered need formal recognition that they are completed. – Andy aka Jul 11 '21 at 08:54

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Now, to find the amplitude of Vo, should we look at the real part

Not if you are aiming to find the magnitude of \$V_o\$. The magnitude of \$1-j\$ is \$\sqrt2\$ and so, the peak magnitude of \$V_o\$ is \$\dfrac{5}{\sqrt2}\$. The RMS value will of course be 2.5 because you divide the peak magnitude by \$\sqrt2\$.

The term amplitude is not really that well defined if you want to be pedantic.

OR should we take the magnitude of \$\dfrac{V_o}{V_i}\$

That would be the peak gain of the circuit and isn't really relevant as far as I can tell. I mean, in terms of what you appear to be asking about, it doesn't appear relevant.

Andy aka
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Complex notation is a useful representation of a real-world signal, which itself does not contain any imaginary part. So, in complex notation:

  • Amplitude is the magnitude of the complex number (btw as Andy says, you lost a square root along the way)

  • Phase is the argument of the complex number

Splitting the real and imaginary parts is useful for other things. For example:

  • If the variable is a power, then the real and imaginary parts correspond to active power and reactive power.

  • If you use IQ demodulation, and multiply your signal with a sine and a cosine, the averaged and lowpass-filtered outputs of these two multipliers will correspond to the real and imaginary parts of the complex variable that represents your signal.

bobflux
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  • Write your numbers preferably ... real part first, then imaginary ... –  Jun 16 '21 at 11:19