Amplify AC signal only

Question

I amplify a signal which including AC and DC.Why is the output waveform changed when C1 is 0.1u?And how to decide the value of C1?

Answer 1

When Capacitor Reactive Impedance, Xc(f) rises to affect impedance ratio with R such that they are equal, you can measure many responses;

1. the voltage amplitude divider ratio (gain) reduces times \$\dfrac{1}{\sqrt(1+1)} = 0.707 = -3~dB\$
2. the current phase shift changes by 45 deg, (=trig. angle with equal sides of impedance) which as already started to shift with 0.1uF that raises the impedance of Xc(f) to 26.5 kohm
3. the -3dB breakpoint is \$ω_1=1/(R_1C_1)=2\pi f\$
4. you can visualize all of this with a log impedance RLC nomograph and measure the impedance ratio of any XL(f) or C(f) vs V as the division ratio becomes a difference on a log scale. like 1000 is a difference of 3 decades

5. R+ C Series Impedance = \$Zc(f)=R+jXc(f)\$

   6. Also with this you can "ballpark" estimate corner frequencies and LC resonant circuits and Q gain factors for LC intersections with R differences (Ratios). e.g. L//C//R is high impedance
      so $$Q_p=R/X_L(f)=R/X_C(f)$$
   7. at LC value intersection (=resonant f)
      then series resonant f (L+C+R) $$Q_s=\frac{X_L(f)}{R}=\frac{X_C(f)}{R}$$ (using absolute values or ignoring phase due to j

   Here is one example of the RLC Impedance nomograph.