AC Output Voltage as function of frequency

In summary, the homework statement says that if a sinewave generator of frequency f is connected to a high-pass filter, the output voltage V1(t) will be cosine-wave shaped with a frequency-dependent amplitude. The equation governing the amplitude is found using Kirchoff's voltage and current laws and Ohm's law. If you're familiar with phasors, you can avoid some of the math and derive the equation in the frequency domain. Once you know the amplitude of V1 at any given frequency, you can use Ohm's law to calculate V1/V0.
  • #1
kenef
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Homework Statement


Given a sine-wave generator of frequency f having internal resistance R1 connected to a high-pass filter. If the generator shown in the dotted outlined box produces a voltage V0(t) = V0 cos(2π ft) with no load, derive an expression for the output voltage V1(t) = V1 cos(2π ft + φ) as a function of frequency f.

Homework Equations


V= I R
Xcap = 1 / ωC

The Attempt at a Solution


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Ultimately I am lost at the wording in the question. If one is supposed to take the V1 as the accurate final voltage and simply mathematically manipulate it that is what I have attempted. I know that cos(A+B) = (cos(A)cos(B) - sin(A)cos(B) however from there I do not see much I am allowed to do. Perhaps it is my math that is a little rusty. I do remember a long time ago performing manipulations of expressing one variable in terms of another however involving trigonometric functions seems a bit confusing to me.

I know another method would be to go from this time domain to the frequency domain however this manipulation is unknown to me, as the book we are using only talks about the basis of Time Domain view of RCs and Frequency domain view of RCs... " Learning the Art of Electronics - Hayes"
 

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  • #2
You want to apply Kirchoff's voltage and current laws and Ohm's law to derive a differential equation that relates V0 and V1.

Alternately, you can avoid some of the math if you're familiar with phasors.

https://en.wikipedia.org/wiki/Phasor
 
  • #3
vela said:
You want to apply Kirchoff's voltage and current laws and Ohm's law to derive a differential equation that relates V0 and V1.

Alternately, you can avoid some of the math if you're familiar with phasors.

https://en.wikipedia.org/wiki/Phasor
I was going to do this approach, however once I relate V0 and V1, I am still left with equations in which are of the time domain and not frequency domain
 
  • #4
Why is that a problem?
 
  • #5
Because I need the output voltage in terms of the frequency and not time as it states, at least that's what it means to me. Doing this I am left with the a function in time not frequency...

Maybe I am just over thinking this as well
 
  • #6
The coefficient of the cosine in V1(t) will depend on frequency. That's what the problem means by as a function of frequency. You can still solve the problem in the time domain.
 
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  • #7
Ok so far, I have that V0 cos(2π ft) + R1C dV/dt = V1 cos(2π ft + φ) . Is it correct to assume that R2 should not be considered ?

With this in mind I do know that my final relation should show me what V1/V0 is equal to.

Update:

I have that V0 cos(2π ft) + (R1 + R2) C dV/dt = V1 cos(2π ft + φ), with manipulation I have arrived to V1cosφ - V0 = C( R1 + R2) ∫dV(t) . I do not feel too confident with my approach in this manner, because I have no initial and final values in order to integrate
 
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  • #8
Your differential equation isn't quite right. It's probably simpler if you write a differential equation for the voltage across the capacitor (what you called V) and then relate it to V1. The current in the circuit is ##i = C \frac{dv}{dt}## where ##v## is the current across the capacitor. Applying Ohm's Law to R2, you get ##V_1 = iR_2 = R_2 C\frac{dv}{dt}##. So once you find ##v##, you'll essentially know ##V_1##.

To get the differential equation that governs ##v##, you want to apply KVL to the circuit. You kind of did it, but you made errors. For one thing, source elements and sink elements (like resistors and capacitors) should have different signs. You're also missing the voltage drop across the capacitor.
 
  • #9
You are right, I have not done any circuits in a while and completely missed the fact that I have to subtract my voltage drops across the sink elements! As far as the voltage drop across the capacitor I attempted to perform something that I guess is incorrect, I thought of the capacitor as not charged. Ok so far I have the V0 cos(2π ft) - C {dv}/{dt} (R_1 - R_2 - 1) = 0, is this along the correct path? Doing this I ended up with a rather weird derivation.

I attempted it in the method you described, using only V1 and R2, I came to a solution of V1 = e^ ( (t+ const) / R_2 * C))
 
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  • #10
The plates of the capacitor may not be initially charged, but they acquire a charge as current flows in one end and out the other.

You still have some sign errors, and the constant 1 you threw in there isn't correct. (It doesn't work unit-wise for one thing.) The voltage across the capacitor is simply v; it's not C dv/dt. Keep trying. You're really close.
 

1. What is AC output voltage?

AC output voltage refers to the electrical voltage that is generated by an alternating current (AC) power source. This voltage is characterized by continuously changing polarity and magnitude.

2. How is AC output voltage measured?

AC output voltage is typically measured using a voltmeter or multimeter. These instruments measure the voltage between two points in a circuit, and can display the value in units of volts (V).

3. What is the relationship between AC output voltage and frequency?

The relationship between AC output voltage and frequency is inversely proportional. This means that as the frequency of the AC power source increases, the output voltage decreases, and vice versa.

4. Why does AC output voltage change with frequency?

The change in AC output voltage with frequency is due to the properties of inductors and capacitors in the circuit. These components have reactances that vary with frequency, which affects the overall impedance of the circuit and therefore the output voltage.

5. How does AC output voltage affect electronic devices?

The AC output voltage can have a significant impact on electronic devices, as it determines the amount of power and energy that is supplied to the device. If the voltage is too high, it can cause damage or malfunction, while a low voltage may result in the device not functioning properly.

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