# AC Op-Amp output voltage expression in time domain

1. Apr 8, 2014

### Maylis

1. The problem statement, all variables and given/known data
For vi(t) =V0 cosωt, obtain an expression for vout(t) in the circuit below and then evaluate it for V0 = 4 V, ω = 400 rad/s, R = 5 kΩ, and C = 2.5 μF.

2. Relevant equations

3. The attempt at a solution
For this problem, I am having difficulty taking that second term with the (1/jωRC)*V0 and transforming it back to the time domain

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2. Apr 8, 2014

### Maylis

Here is my attempt

3. Apr 9, 2014

### Staff: Mentor

You have determined the transfer function, now express it as a gain of magnitude A and phase shift phi. Then apply that gain and phase shift to V0.cos(ωt)

4. Apr 9, 2014

### Maylis

That's where my problem is, I don't know how to make it into that expression

5. Apr 9, 2014

### rude man

Two ways:
1. change your final expression to the form a + jb, then take the magnitude of this and determine the phase angle from the arc tan formula. Remember, the input is a cosine function.

2. change your final expression to the form r exp(jθ). r is your magnitude and θ is your phase angle. Again, the phase angle will be part of a cosine function, not a sine.

6. Apr 9, 2014

### Maylis

I seriously doubt I'm doing this right. All the examples in the book use numbers so I can't tell if I'm doing it right symbolically

7. Apr 9, 2014

### Maylis

Ok I looked up an article on complex number magnitude here is my next attempt

8. Apr 9, 2014

### rude man

I think you have the right approach. But it looks like your computation of r, going from the 1st to the 2nd line, is wrong. You forgot to square the ωRC term in the denominator? Hard to read the photo.

One more thing: if a voltage is V0cos(ωt) the phasor is V0/√2, not V0. This time it didn't matter but sometimes it does.

Finally: careful with your arc tan arguments. arc tan (-1/x) is not the same angle as arc tan(1/-x). I think you did OK there.

9. Apr 9, 2014

### Maylis

Thanks, I redid and fixed that calculation error.

We were working on it in office hours and my professor was stumped for a while. He thought that the input was equal to the output because you only take the real part of the phasor for the time domain, which at first glance appears to be just the input.

However, he forgot that its the real part times e^jwt

10. Apr 9, 2014

### Staff: Mentor

As rude man noted, you are not correctly calculating the magnitude of the complex gain. The calculation needed is to determinine the hypotenuse of a right-angled triangle when you know the base (the real component) and the height (the imaginary component), so use Pythagoras's Theorem.

Somewhere along the way you have forgotten to state that Vo= Vin

11. Apr 9, 2014

### Staff: Mentor

That is not correct.

It is never correct; both are always peak values. The sinusoid is merely the projection onto one axis of the rotating vector, so sinusoid peak = magnitude of phasor.

(You can, of course, elect to deal consistently in RMS values, these being linear systems. But that's merely a scaling factor and changes nothing stated earlier.)

Good point.

12. Apr 9, 2014

### rude man

It's not the real part times exp(jwt). In fact, the magnitude of exp(jwt) is always 1.

exp(jwt) is implicit in phasor notation. A phasor voltage can be V(a + jb) = V{r exp[j arc tan(b/a)]}
with r = √(a2 + b2).

13. Apr 9, 2014

### rude man

The reason it's better to define a phasor as its rms voltage is to simplify power calculations. Power in sinusoids always involves rms, not amplitude, numbers.

For example, if V and I are rms phasors, this results in simple expressions for power:
Active power P = VIcosø
Reactive power Q = VIsinø
with ø the angle betw. V and I.

Further, you can then define a complex power as
S = VI* = P + jQ = VI exp(jø)
where V = V exp(jø)

from which we can compute P without recourse to either impedance or phase angle explicitly:

P = 1/2 (S + S*) = 1/2 (VI* + V*I)

Finally, S = P + jQ = I2Z → P = I2R and Q = I2X

where R and X are the real and imaginary parts of impedance Z. Be careful that I here is real, not complex.

* denotes "complex conjugate" and complex quantities are in bold.

If one decides to forgo these advantages one can indeed define the phasor in amplitude terms, so long as one is consistent. Or we can simply call √2V the phasor and V as something else, which may be conventional. Whatever. Semantics ...

14. Apr 10, 2014

### Staff: Mentor

RMS is the convention in power generation and transmssion line engineering, and phasors in that area are usually based on RMS.

But really, when was the last time you were asked to determine the reactive power component in an OP-AMP's output?? Small signal electronics is far more concerned with issues such as peak power not exceeding specifications, or peak voltage not saturating a device, so phasors here are most useful when they are not scaled to RMS.

If ever you do use RMS of a sinusoid in small-signal OP-AMP analysis it may be prudent to attach the "RMS" subscript, as much to alert others as to remind yourself.

15. Apr 10, 2014

### rude man

If I have an op amp circuit that specifies a 1V input, what's the input? PK-p, amplitude or rms?

If you stick your VOM or DVM into a 115V receptacle, what does it read?

In all cases, rms is understood a priori.

Thing is, while a phasor, rigorously defined, is what you said, "using phasors" always implies rms voltages.

16. Apr 12, 2014

### Staff: Mentor

It will be specified along with the amplitude, either p-p or pk. Certainly it won't be RMS. To prescribe input according to an RMS value would be almost never applicable to an OP-AMP circuit. Whether sinewave, rectangular, or sawtooth, it's as a general rule the pk-pk swing that is the limiting factor, nothing to do with RMS.

Find me a function generator whose amplitude reading defaults to the output waveform's RMS value! Indeed, when was the last time you (or anyone) needed to dial up a squarewave or a sawtooth of 1.00v RMS??