Consider the filter circuit shown in Active Figure 33.22 a . (a) Show that the ratio of the outp

step 1 I'm going to begin by drawing active figure 3 .22A. And so what it looks like is, hey now, this,

Consider the filter circuit shown in Active Figure 33.22 a ....

Consider the filter circuit shown in Active Figure 33.22 $\mathrm{a}$ . (a) Show that the ratio of the output voltage to the input voltage is $$ \frac{\Delta v_{\mathrm{out}}}{\Delta v_{\mathrm{in}}}=\frac{R}{\sqrt{R^{2}+\left(\frac{1}{\omega C}\right)^{2}}} $$ (b) What value does this ratio approach as the frequency decreases toward zero? What value does this ratio approach as the frequency increases without limit? (c) At what frequency is the ratio equal to one-half?

Consider the filter circuit shown in Active Figure 33.22 $\mathrm{a}$ . (a) Show that the ratio of the output voltage to the input voltage is
$$
\frac{\Delta v_{\mathrm{out}}}{\Delta v_{\mathrm{in}}}=\frac{R}{\sqrt{R^{2}+\left(\frac{1}{\omega C}\right)^{2}}}
$$
(b) What value does this ratio approach as the frequency decreases toward zero? What value does this ratio approach as the frequency increases without limit? (c) At what frequency is the ratio equal to one-half?

Key Concepts

Transfer Function

The transfer function is a mathematical representation of the relationship between the output and input of a circuit, expressed in the frequency domain. It encapsulates how different circuit elements such as resistors and capacitors affect the amplitude and phase of a signal as a function of frequency.

Frequency Response

Frequency response describes how a circuit's output amplitude varies with the frequency of the input signal. For RC circuits, this involves understanding how the reactive (capacitive) impedance, which depends on frequency, interacts with resistive elements to attenuate or pass signals over the spectrum.

Low-Frequency and High-Frequency Limits

Analyzing the low-frequency and high-frequency limits of a circuit is crucial for understanding its behavior in extreme cases. As the frequency approaches zero, the capacitive impedance becomes very large, significantly reducing the output. Conversely, at very high frequencies, the capacitive impedance becomes negligible, allowing the output to approach the input signal level.

Determination of Specific Frequency Conditions

Calculating specific frequencies, such as when the transfer function value equals a given fraction of the maximum, involves solving the transfer function equation for the frequency variable. This process is key in identifying critical points like the cutoff or half-amplitude frequency, which characterize the performance of the filter.

Transcript

00:01 I'm going to begin by drawing active figure 3 .22a.

00:07 And so what it looks like is, hey now, this, this voltage source, and it goes back here, delta v out.

00:33 There's a resistor here.

00:37 And there's a capacitor that i did not draw yet.

00:46 So this is a capacitor c.

00:52 And this is called delta vn.

00:59 Okay, so that's what the figure looks like.

01:04 And we're supposed to show the ratio of the voltages.

01:11 All right.

01:15 Well, first we should begin by calculating the impedance, which is going to be the square root of r squared plus x sub c squared, where x sub c is 1 over omega c.

01:40 So putting those two together, i get z is the square root of r squared plus r squared plus 1 over omega -c.

01:51 Mega c squared.

01:58 Current, assuming this is rms, is v over r, i mean, delta v in over z.

02:26 Okay, but delta v out is i times r.

02:39 So, putting these three together, one, two, three, i would get delta v out equals i, which is delta vn over z, which is the square root of r squared plus one over omega c squared squared times r.

03:24 So now just rearranging this, delta v out over delta vn is r over the square root of r squared plus 1 over omega c squared, which is what we wanted to show.

03:49 So that was part a.

03:51 So let's look at part b.

03:59 It's asking us about the frequency going to zero.

04:05 Okay...

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