# What is DC gain of sine wave multiple?

I am testing about injection sine wave current to r,c impedance and measurement R + jX see fig.1 .

When scope is Va .

When sine wave multiple (Va*Vb).

I put average to find DC of this signal.Is should be closing 4kV but it is loss some DC gain about 2 times.

What is equation to calculate this DC gain.

• Welcome to SE.SP! I'm not sure what you mean by Is should be closing 4kV but it is loss some DC gain about 2 times. Does that mean the 1.9997 kV is not expected? Also, what do you mean by "DC Gain"? Usually, gain talks about an input output relationship. Which is the input and which is the output?
– Peter K.
Jun 30 at 1:40
• Thank Pete.I mean like this .If we inject DC current 1A to only Resister 4k output should be 4kV right ? When use sine wave I would like to collect DC value from it which is 4kV but when use sine*sine to collect it .I dont know equation to calculate from 1.9997 kV to closing 4kV. Jun 30 at 2:02

In addition to Dan's explanation, the other problem is that there are two impedances in parallel: the 4k$$\Omega$$ resistor and the 4k$$\Omega$$ in series with the capacitor. That means the overall impedance is

\begin{align*} Z_{\tt total} &= \frac{1}{\frac{1}{4000} + \frac{1}{4000 + \frac{1}{\jmath 2 \pi \times 100 \times 0.2 \times 10^{-6}}}}\\ &= 2994.7\Omega - \jmath 999.9.0 \Omega \end{align*}

And it is $$Z_{\tt total}$$ that the maximum of 1 amp is flowing through.

Then, using Ohm's Law we get:

$$V = I Z_{\tt total} = 2994.7 - \jmath 999.9.0\ \ \mbox{V}$$

And if I make your circuit in LT spice

what I get for the voltage across $$R_1$$ is

which should have a peak magnitude of $$3157.2$$ V.

The OP has a correct understanding on how to predict the DC output across the 4K resistor as being the DC current times the resistance.

The "DC current" would be the average of the input signal. The input signal is a sinusoid going from 0 to 1A, so it has an average value of 0.5A. This would be referred to as a "DC offset".

The formula for the input would be:

$$I_1 = 0.5 + 0.5\sin(t)$$

This should be enough to clear up any remaining confusion on predicting the DC voltage at the output as $$V_A$$.